Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination

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Kerala Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination

Plus One Neural Control and Coordination One Mark Questions and Answers

Question 1.
Resting membrane potential is maintained by
(a) Hormones
(b) Neurotransmitters
(c) Ion pumps
(d) None of the above
Answer:
(c) Ion pumps

Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination

Question 2.
The function of our visceral organs is controlled by
(a) Sympathetic and somatic neural system
(b) Sympathetic and para sympathetic neural system
(c) Central and somatic nervous system
(d) None of the above
Answer:
(b) Sympathetic and para sympathetic neural system

Question 3.
Which of the following is not involved in Knee-jerk reflex?
(a) Muscle spindle
(b) Motor neuron
(c) Brain
(d) Inter neurons
Answer:
(c) Brain

Question 4.
Mark the vitamin present in Rhodopsin
(a) Vit A
(b) Vit B
(c) Vit C
(d) Vit D
Answer:
(a) Vit A

Question 5.
Human eyeball consists of three layers and it encloses
(a) Lens, iris, optic nerve
(b) Lens, aqueous humor and vitreous humor
(c) Cornea, lens, iris
(d) Cornea, lens, optic nerve
Answer:
(b) Lens, aqueous humor and vitreous humor

Question 6.
Name the structure which connects two cerebral hemisphere
Answer:
Corpus callosum

Question 7.
Name the region of keenest vision in our eye.
Answer:
Fovea or yellow spot

Question 8.
After playing on a giant wheel, we lost our balance. Why?
Answer:
Change in position of Otolith in ear gives mal information to the brain.

Question 9.
Arrange the ear ossicles in order from inner ear to the tympanum.
Incus Stapes Malleus
Answer:
Stapes → Incus → Malleus

Question 10.
Name the receptors respond to irritants such as ammonia, vinegar or hot chilly pepper.
Answer:
Pain receptors

Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination

Question 11.
The tissues of eye and ear contain photoreceptors and the auditory receptors. Likewise, some are found as film of liquid coating in the membranes of the receptor cells. Name it.
Answer:
Chemoreceptors

Plus One Neural Control and Coordination Two Mark Questions and Answers

Question 1.
Light ray → Lens → retina → pupil → cornea → brain → Vitreous chamber → Aqueous chamber → Optic nerve.
Correct the sequence.
Answer:
Light ray → Cornea → Aqueous chamber → Pupil → Lens → Vitreous chamber → Optic nerve → brain.

Question 2.
Classify the following into 3 groups and give appropriate headings.
Thalamus, Corporaquadrigerhina, Pons, hypothalamus, Cerebrum, Medulla Oblongata.
Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 1

Question 3.

  1. Identify the above stages of nerve impulse conduction
  2. Name the ions involved in this process

Answer:

  1. Identification of above stages
    • I – Resting membrane potential
    • II – Action potential
  2. Na+, K+

Question 4.
Differentiate Blindspot and Yellow spot.
Answer:
1. Blind spot:
The spot at the back of the eye where the optic nerve originates is known as blind spot. Rods and canes are absent here. So the image falling at this spot cannot be carried to the brain.

2. Yellow spot:
Lateral to the blind spot is a depressed area called yellow spot which contains only canes. It is the area of sharpest vision.

Question 5.
‘Sudden death may occur due to the damage of Medulla oblongata’. Why?
Answer:
The medulla oblongata contains several centres which regulate heartbeat, respiration, gastric, secretion, vomiting etc. It carries the nerve fibres which connect spinal cord and cerebrum.

Question 6.
Observe the figure given below.

  1. Identify A & B
  2. Write the main function of part A.

Answer:

  1. Identification of A & B
    • A – Cochlea
    • B – Semicircular canals
  2. Maintenance of balance of the body and posture.

Question 7.
Where do you find bipolar and multipolar neurons in our body?
Answer:

  • Bipolar neuron – Retina of eye
  • Multipolar neuron – Cerebral cortex

Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination

Question 8.
While playing cricket, the ball hit a boy’s head. He immediately vomited and felt difficulty to breath.

  1. Identify the part of brain affected.
  2. Give the function of the affected part.

Answer:

  1. Medulla oblongata
  2. The medulla oblongata contains several centres which regulate heartbeat, respiration salivation, vomiting, etc. It carries the nerve fibers which connect the spinal cord and cerebrum.

Question 9.
Nocturnal animals like bats and owls have vision during night. Give reason.
Answer:
Nocturnal animals like owls have only rods in the retina. The rods are sensitive to dim light and enable to see dim light and at night.

Question 10.
Observe the flow diagram of the pathway of a light ray entering the eye.
Lightray → lens retina →pupil → aqueous chamber → vitreous chamber → cornea → optic nerve → brain

  1. Correct the sequence.
  2. If the light ray falls on the blindspot, what will happen?

Answer:

  1. Light ray → cornea → pupil → aqueous chamber → lens → vitreous chamber → retina → optic nerve → brain
  2. Rods and cones’are absent in blind spot and so the image falling at this spot cannot be carried to the brain.

Question 11.
Arrange the following in the order of reception and transmission of the sound wave from the external auditory canal.
Answer:
Cochlear nerve, eardrum, stapes, incus, malleus, cochlea
Ear drum → malleus → incus → stapes → cochlea → cochlear nerve

Question 12.
Observe the given diagram.
Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 4

  1. Write any one difference between A and B.
  2. Through which neurone the impulse conduction is faster. Justify.

Answer:

  1. The difference between A and B
    • A – non-myelinated neuron
    • B – Myelinated neuron
  2. Impulse conduction is faster through myelinated neuron.

Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination

Question 13.
The given diagram is a part of myelinated nerve fibre.
Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 5

  1. Identify the part where there is no myelin sheath.
  2. Name the type of conduction going on in that type of nerve fibre.

Answer:

  1. Nodes of Ranvier
  2. Saltatory conduction

Question 14.
Observe the diagram.
Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 6
Identify A, B, C and D.
Answer:

  1. A – Nissl’s granule
  2. B – Myelin sheath
  3. C – NodeofRanvier
  4. D – Synaptic knob

Question 15.

  1. Identify the picture.
  2. Write the peculiarity of the picture.

Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 7

Answer:

  1. Bipolar neuron
  2. Bipolar neurons are the neurons with one axon and one dendrite

Question 16.
A patient approaches a doctor with a problem that he was not able to sleep for the last 6 months. The doctor said that it may be due to the defect in the relay centre of his brain and advised him to take sedative pills.

  1. Which part of brain is described here as ‘relay centre’?
  2. What are the actions of sedative pills in body?

Answer:

  1. Thalamus
  2. Sedative pills work in the Thalamus and prevent the transmission of impulses to cerebrum. Depress brain activity produce feelings of calmness, relaxation, drowsiness and deep sleep.

Question 17.
It is said that the number and pattern of convolutions are associated with the degree of intelligence.

  1. Is it true?
  2. If yes, give the scientific reason for it.

Answer:

  1. True
  2. More intelligent forms like mammals especially primates have more convolutions than lower forms.

Question 18.
Diagram below represents a neuron at resting membrane potential (RMP).
Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 8
(a) How RMP is maintained?
(b) Draw the changes at the time of depolarization and explain how it happens.
Answer:
(a) RMP is maintained by
1. A resting membrane is poorly permeable to Na+ ions and Cl ions. But more permeable to K+ ions. The extra cellular fluid has a high concentration of Na+ ion and low concentration of K+ ions. But the intracellular medium has a reverse condition due to the permeability.

2. Sodium-Potassium pump maintain higher concentration of Na+ ions outside the membrane compared to the concentration of Na+ inside

(b)
Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 9
When a neuron is stimulated, there is a momentary reversal of the resting potential. The reversal of polarity is known as depolarisation. When stimulated a resting membrane, sodium pump suddenly stops and sodium ions begin to enter the cells.

The presence of higher concentration of Na+ inside the cell cause the inside membrane +ve and outside become -ve. This condition is called depolarisation.

Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination

Question 19.
Copy the diagram and label A, B, C, D.
Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 10
Answer:

  1. A – Synaptic vesicles
  2. B – Synaptic cleft
  3. C – Presynaptic neuron
  4. D – Post synaptic neuron

Question 20.
A diagram showing the chemical synaptic transmission is given below. Based on the diagram prepare a flow chart showing the process of synaptic transmission.
Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 11
Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 12

Question 21.
While playing cricket the ball hit a boy’s head. He immediately vomited and felt difficulty to breath.

  1. Identify the part of brain which may be affected by the incident.
  2. Give functions of this particular part of brain.

Answer:

  1. Medulla oblongata
  2. Control heart beat, regulate respiration, control circulation, control digestion, control peristalsis etc.

Question 22.
Observe the schematic representation related to maintaining resting membrane potential of an axon.
Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 13

  1. What are the conditions exhibited in the figure for maintaining resting membrane potential?
  2. What are the other conditions of resting membrane potential (not exhibited in the diagram)

Answer:

  1. High permeability of axon wall for K+, Highly negative charge protein in axoplasm, Na+ – K+ pump.
  2. High K+ concentration inside the axon, High Na+ concentration outside the axon, Low Na+ permeability the axon wall, 3Na+ ions for 2K+ ions.

Question 23.
Observe the portion of Brain and answer the following questions.
Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 14

  1. Name the covering of brain.
  2. Identify A, B and C.

Answer:

  1. Meninges
  2. Identification of A, B, and C
    • A – Dura mater
    • B – Arachnoid mater
    • C – Pia mater

Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination

Question 24.
Identify the following figure and label the parts.
Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 15
Answer:
Reflex arc
Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 16

Question 25.
When we step on a thorn, we withdraw our legs suddenly, if we are taking this thorn out, we will not withdraw our legs. Comment on these two statements.
Answer:
1st statement is a reflex action and 2nd statement is a process controlled by Brain.

Question 26.
Observe the picture.
Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 17

  1. Identify the part labelled as (x).
  2. In our retina there are more rod cells than cones. But our vision in darkness is poor. Give scientific explanation to this fact.

Answer:

  1. Yellow spot or fovea
  2. Image forms normally on yellow spot. In yellow spot rods are less and cones are more. So low dim light vision.

Question 27.
Find out the relationships and write the suitable word in the IVth place.

  1. Cornea: Sclera; Yellow spot: _________
  2. Incus: Middle ear; Cochlea: __________
  3. Scala vestibuli Perilymph; Scala media: __________
  4. Rods: Rhodopsin; Cones: ____________

Answer:

  1. Retina
  2. Inner ear
  3. Endolymph
  4. lodopsin

Question 28.
Arrange the organs according to the mechanism of hearing.
Oval window, Perilymph, Organ of Corti, Ear Ossicles, Pinna, Tectorial membrane, Endolymph, Auditory canal, auditory nerve, brain
Answer:
Pinna → Auditory canal → Ear ossicles → oval window → Perilymph → Endolymph → Organ of Corti → Auditory nerve → Brain

Question 29.
Copy the diagram and mark Tectorial membrane and sensory hair cell.
Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 18
Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 19

Question 30.
Analyse the table and fill in the blanks given in the table with appropriate words.

Rods_______________
No ability to detect colour_______________
______________Cone pigments
_______________Cone shaped
Only one type___________________
___________________Photopic vision

Answer:

RodsCones
No ability to detectAbility to detect
colourcolour
RhodopsinCone pigments
Rod shapedCone shaped
Only one typeThree type
ScotopicPhotopic vision

Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination

Question 31.
In animals like bats and owls photoreceptor cells of the retina have mainly rods.

  1. What can you infer from this?
  2. Write down the function of rods and cons.

Answer:

  1. Rod cells bring about vision in night or dim light vision.
  2. The function of rods and cons:
    • Cones are cone shaped sensory cells of retina that bring about vision in day light and also distinguish colours. There are three types of cone cells for sensing primary colours red, green and blue.
    • Rods are rod shaped sensory cells of the retina that bring about vision in night, but cannot distinguish colours.

Question 32.
Ear converts sound waves into neural impulses which are sensed and processed by the brain that enable to recognise sound. Construct a schematic diagram showing the mechanism of hearing the sound of a bell.
Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 20

Question 33.
Arrange the structures found in the retina from inside to outside.
Cone cells, optic nerve, Ganglion cells, Bipolar neuron.
Answer:
Optic nerve → ganglion cells → Bipolar neuron → Cone cells

Question 34.

  1. Name the receptors of smell found as mucous coated thin, yellowish patch of modified pseudo stratified epithelium.
  2. Where is these receptors located?

Answer:

  1. Olfactory epithelium
  2. It is located at the roof of the nasal cavity on either sides of the nasal septum.

Question 35.
The diagram shows a section through a part of the human ear.

  1. Identify the parts labelled A, B and C.
  2. Which parts are involved in the equilibrium of the body.

Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 21

Answer:

  1. Identification of A, B and C.
    • A – Semicircular canal
    • B – Vestibule
    • C – Cochlea
  2. Semicircular canal and vestibule.

Question 36.
Why does red flower look black in dim light?
Answer:
Stimulation of cone cells require high intensity of light. In dim light the cone cells are not stimulated. That is why the red flower looks black in dim light.

Question 37.
The taste buds of Humans are located in pockets around the papillae on the surface and sides of the tongue, but some on the surface of the pharynx and the larynx.

  1. What are the four basic taste senses?
  2. Find out their location in tongue

Answer:

  1. sweet, sour, salt, and bitter
  2. sweet and salty on the front, bitter on the back, and sour on the sides.

Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination

Question 38.
The touch receptors are either free dendritic endings or encapsulated dendritic endings present in the skin.

  1. What are the main functions of receptors of free dendritic endings?
  2. Name the main receptors of encapsulated dendritic endings. Give its functions.

Answer:

  1. They respond to pain and temperature
  2. Meissner’s corpuscles, Pacinian corpuscles
    • Meissner’s corpuscles:
      These are found just beneath the skin epidermis in dermal papillae and abundant in fingertips and soles of the feet. These are light pressure receptors.
    • Pacinian corpuscles:
      These are scattered deep in the dermis and in the subcutaneous tissue of the skin .These are stimulated by deep pressure.

Plus One Neural Control and Coordination Three Mark Questions and Answers

Question 1.
The following steps are involved during synaptic transmission. Rearrange them in correct order.

  1. Release of neurotransmitter at synaptic cleft.
  2. Generation of a new potential at post synaptic neuron
  3. Arrival of impulse at the axon terminal
  4. Binding of neurotransmitter with specific receptor
  5. Movement of synaptic vesicle towards the membrane.

Answer:

  1. Arrival of impulse at the axon terminal
  2. Movement of synaptic vesicle towards the membrane
  3. Release of neurotransmitter at synaptic cleft.
  4. Binding of neurotransmitter with specific receptor.
  5. Generation of a new potential at post synaptic neuron.

Question 2.
A sharp tap is given at your knee cap with rubber hammer. You suddenly stretch your leg.

  1. Give name of this response.
  2. Which nerve centre is involved in this action?
  3. Construct a flow chart of the pathway of impulses in this action.

Answer:

  1. Reflex action
  2. Spinal cord
  3. Receptors in knee ® Sensory fibres ® Interneuron effector organs response → Motor fibres

Question 3.
Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 22

  1. Identify the organ and label A and B.
  2. Gait of a drunkard is not normal. Why?
  3. The death sentence given by the court is always by “hanging” in our country. Why this is preferred to other ways?

Answer:

  1. Organ-Brain
    • A – Gray matter
    • B- White matter
  2. Alcohol effects the cerebellum which control and co-ordinate voluntary muscular action.
  3. Because death is less painful and fast as atlas pierces the medulla oblongata and it is smashed. Medullar oblongata is the control centre of respiratory and cardiac action. Stoppage of which cause death.

Question 4.
Make necessary correction in the flow chart given.
Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 23
Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 24

Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination

Question 5.
Match the following.

AB
CerebrumRespiration, Salivation, Vomiting
CerebellumRelay centre
Medulla oblongataSeat of mind and intelligence
Limbic systemPosture and equilibrium
HypothalamusMaster gland
PituitaryEmotional experiences and expressions.

Answer:

AB
CerebrumSeat of mind and intelligence
CerebellumPosture and Equilibrium
Medulla oblongataRespiration, Salivation, Vomiting
Limbic systemRelay centre
HypothalamusEmotional experiences and expressions
PituitaryMaster gland

Question 6.
Suppose you dramatically escaped from a motor accident. Your heart beat and rate of respiration was increased at that moment.

  1. Name the hormone involved in this change.
  2. Which part of nervous system control these action?
  3. Enlist other physiological changes that you may feel at that time.

Answer:

  1. Adrenalin and Non-adrenaline.
  2. Medulla oblongata.
  3. High metabolic rate, High BP, High body temperature, High level of glucose in blood, Pupil diabetes.

Question 7.
Observe the figure below show two ion channels.
Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 25

  1. This pump does not work on part of the axon. Why?
  2. Which are the ions participated in the process?
  3. Suggest the use of this process in nerve impulse conduction?

Answer:

  1. Myelin sheath: Myelin sheath act as insulator for axon and it increases the speed of nerve impulse conduction by saltatory conduction.
  2. Na+ and K+
  3. It creates action potential when stimulated by allowing the inward movement of Na+ ions through the Na channel and outward movement of K+ through K+ channel.

Question 8.
Neurons are the structural and functional unit of nervous system.

  1. Based on the number of axon and dendrites, how the neurons are classified.
  2. Give examples for each.
  3. Which neuron receives signal from a sensory organ and transmit the impulse to CNS.

Answer:

  1. Multipolar, bipolar and unipolar
  2. Examples
    • Multipolar – found in cerebral cortex
    • bipolar – found in the retina of eye
    • unipolar-found in embryonic state
  3. afferent neurons or sensory neuron.

Question 9.
Observe the figures a and b.
Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 26

  1. Mention the structural difference between the two neurons.
  2. Name the type of impulse transmission through fig B.
  3. Mention the functional difference between the two neutrons.

Answer:

  1. Fig a is non-myelinated neuron and Fig B is myelinated neuron.
  2. Saltatory conduction.
  3. In myelinated impulse transmission is very fast.

Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination

Question 10.
Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 27

  1. Identify the diagrammatic representation
  2. Name P & Q.
  3. Mention the function of P & Q.

Answer:

  1. Reflex action (knee jerk reflex)
  2. Name P & Q
    • P – Afferent pathway
    • Q – Efferent pathway
  3. The afferent pathway receives signal from a sensory organ and transmits the impulse into CNS. The efficient pathway carries signals from CNS to the effector.

Question 11.
A diagram of brain is given below.
Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination - 28

  1. Identify P Q R S
  2. Which part is responsible forthinking, memory and reasoning
  3. Name the nerve band which connects the two hemispheres of brain.

Answer:

  1. Identification of P Q R S
    • P – Cerebrum
    • Q – Thalamus
    • R – Pons
    • S – Corpus callosum
  2. Cerebrum
  3. Corpus callosum

Plus One Neural Control and Coordination NCERT Questions and Answers

Question 1.
Compare the following:

  1. Central neural system (CNS) and Peripheral neural system (PNS)
  2. Resting potential and action potenial
  3. Choroid and retina

Answer:
1. Central Neural System and Peripheral Neural System:
The CNS includes the brain and the spinal cord and is the site of information processing and control. The PNS comprises of all the nerves of the body associated with the CNS (brain and spinal cord).

2. Resting Potential and Action Potential:
The electrical difference across the resting plasma membrane is called as the resting potential. When a stimulus is applied at a site on the polarised membrane, the membrane at the site becomes freely permeable to Na++.

The electrical potential difference across the plasma membrane at the site of stimulus is called the action potential, which is in fact termed as a nerve impulse.

3. Choroid and Retina.
The middle layer, choroid, contains many blood vessels and looks bluish in colour. The choroid layer is thin over the posterior two-third of the eye ball, but it becomes thick in the anterior part to form the ciliary body.

The inner layer is the retina and it contains three layers of cells – from inside to outside – ganglion cells, bipolar cells and photoreceptor cells.

Question 2.
Answer briefly:

  1. How do you preceivethe colour of an object?
  2. Which part of our body helps us in maintaining the body balance?
  3. How does the eye regulate the amount of light that falls on the retina.

Answer:
1. Cones are responsible for color vision. They require brighter light to function than rods require. There are three types of cones, maximally sensitive to long wavelength, medium-wavelength, and short-wavelength light (often referred to as red, green, and blue, repectively, though the sensitivity peaks are not actually at these colors).

2. The inner ear has three semi-circular canals forming cochlea. Cochlea is responsible for maintaining the body balance.

3. The pupil in the eye functions like an aperture. This dilates in case of low light and constricts in case of intense light thereby regulating the amount of light falling on the retina.

Question 3.
The region of the vertebrate eye, where the nerve passes out of the retina is called the
(a) fovea
(b) iris
(c) blind spot
(d) optic chiasma
Answer:
(c) Blind spot

Plus One Neural Control and Coordination Multiple Choice Questions and Answers

Question 1.
Which of the following is not related to the autonomic nervous system?
(a) Peristalsis
(b) Digestion
(c) Excretion
(d) Memory and learning
Answer:
(d) Memory and learning

Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination

Question 2.
Comprehension of spoken and written words take place in the region of
(a) association area
(b) motor area
(c) Wernicke’s area
(d) Broca’s area
Answer:
(c) Wernicke’s area

Question 3.
How many laminae are present in the grey matter of spinal cord?
(a) Four
(b) Six
(c) Eight
(d) Ten
Answer:
(d) Ten

Question 4.
Animals possess nerve or nervous systems to respond to their environment. But the single celled Amoeba does not possesses any nerve cell. so, how it come to know whether a particle it encounters is a grain of sand and not its dinner by?
(a) thermotaxis
(b) skin
(c) hormones
(d) chemotaxis
Answer:
(d) chemotaxis

Question 5.
Thermoregulatory centre of human body is associate with
(a) cerebrum
(b) cerebellum
(c) hypothalamus
(d) medulla oblongata
Answer:
(c) hypothalamus

Question 6.
Sensation of stomach pain is due to
(a) interoceptors
(b) exteroceptors
(c) proprioceptors
(d) chemotactors
Answer:
(a) interoceptors

Question 7.
Bipolar neurons occur in
(a) vertebrate embryos
(b) retina of eye
(c) brain and spinal cord
(d) skeletal muscles
Answer:
(b) retina of eye

Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination

Question 8.
Which foramen is paired in mammalian brain?
(a) Foramen of Luschka
(b) Foramen of Magendie
(c) Foramen of Monro
(d) Inter-ventricular foramen
Answer:
(a) Foramen of Luschka

Question 9.
Which is thickened to form organ of Corti?
(a) Reissner’s membrane
(b) Basilar membrane
(c) Tectorial membrane
(d) All of these
Answer:
(b) Basilar membrane

Question 10.
Skeletal muscles are controlled by
(a) sympathetic nerves
(b) parasympathetic nerves
(c) somatic nerves
(d) autonomic nerves
Answer:
(c) somatic nerves

Question 11.
Which part of human brain is concerned with the regulation of body temperature?
(a) Medulla oblongata
(b) Cerebellum
(c) Cerebrum
(d) Hypothalamus
Answer:
(d) Hypothalamus

Question 12.
Alzheimer’s disease in humans is associated with the deficiency of
(a) dopamine
(b) glutamic acid
(c) acetyleholine
(d) Gamma Amino Butyric Acid (GABA)
Answer:
(c) acetyleholine

Question 13.
The posterior part of the retina, which is just opposite to the lens is
(a) cornea
(b) yellow spot
(c) fovea centralis
(d) Both (b) and (c)
Answer:
(b) yellow spot

Question 14.
In the central nervous system, myelinated fibres form the ______ while the non-myelinated fibre cells form the _________
(a) grey matter, white matter
(b) white matter, grey matter
(c) ependymal cells, neurosecretory cells
(d) neurosecretory cells, ependymal cells
Answer:
(b) white matter, grey matter

Question 15.
The potential difference across the membrane of nerve fibre when it does not shown any physiological activity is called resting potential. It is about
(a) -60mV
(b) -80mV
(c) +60mV
(d) +90mV
(e) -36mV
Answer:
(b) -80mV

Plus One Zoology Chapter Wise Questions and Answers Chapter 10 Neural Control and Coordination

Question 16.
Vomiting centre is located in the
(a) stomach and sometimes in duodenum
(b) gastro-intestinal tract
(c) hypothalamus
(d) medulla oblongata
Answer:
(d) medulla oblongata

Question 17.
The function of vagus nerve innervating the heart
(a) initiate the heart beat.
(b) reduce the heart beat
(c) accelerate the heart beat
(d) maintain constant heart heat
Answer:
(b) reduce the heart beat

Question 18.
The size of pupil is controlled by the
(a) ciliary muscles
(b) suspensory
(c) cornea
(d) iris muscles
Answer:
(d) iris muscles

Question 19.
An action potential in the nerve fibre is produced when positive and negative charges on the outside and the inside of the axon membrane are reversed because
(a) more potassium ions enter the axon as compared to sodium ions leaving it
(b) more sodium ions enter the axon as compared to potassium ions leaving it
(c) all potassium ions leave the axon
(d) all sodium ions enter the axon
Answer:
(b) more sodium ions enter the axon as compared to potassium ions leaving it

Question 20.
A 22 years student goes to his ophthalmologist. He has problem in reading books because he is not able to contract his
(a) suspensory ligament
(b) pupil
(c) iris
(d) ciliary muscles
Answer:
(d) ciliary muscles

Plus One Maths Chapter Wise Questions and Answers Chapter 7 Permutation and Combinations

Students can Download Chapter 7 Permutation and Combinations Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 7 Permutation and Combinations

Plus One Maths Permutation and Combinations Three Mark Questions and Answers

Question 1.
Find the number of different signals that can be made by arranging at least three flags in order on a vertical pole, if 6 different coloured flags are available.
Answer:

3 flags6 × 5 × 4 = 120 ways
4 flags6 × 5 × 4 × 3 = 360 ways
5 flags6 × 5 × 4 × 3 × 2 = 720 ways
6 flags6 × 5 × 4 × 3 × 2 × 1 = 720 ways

Hence the number of different atleast 3 flag signals = 120 + 360 + 720 + 720 = 1920.

Plus One Maths Chapter Wise Questions and Answers Chapter 7 Permutation and Combinations

Question 2.
2. Find the value of n such that

  1. nP5 – 42 × nP3, n>4
  2. (n -1 )P3: nP4 = 1 : 9 (3 score each)

Answer:
1. Given; nP5 = 42 × nP3
⇒ n(n – 1 )(n – 2)(n – 3)(n – 4) = 42 × n(n – 1)(n – 2)
⇒ (n – 3)(n – 4) = 42
⇒ n2 – 7n + 12 = 42
⇒ n2 – 7n – 30 = 0
⇒ (n – 10)(n + 3) = 0
⇒ n = 10; n = -3
The acceptable value is n = 10.

2. Given; 9 × (n – 1)P3 = nP4
Plus One Maths Permutation and Combinations Three Mark Questions and Answers 2

Plus One Maths Chapter Wise Questions and Answers Chapter 7 Permutation and Combinations

Question 3.
Find the value of r if

  1. 5 × 4Pr = 6 × 5Pr – 1
  2. 5Pr = 6Pr – 1 (3 score each)

Answer:
1. Given; 5 × 4Pr = 6 × 5Pr – 1
Plus One Maths Permutation and Combinations Three Mark Questions and Answers 1
⇒ 30 – 11r + r2 = 6
⇒ r2 – 11r + 24 = 0
⇒ (r – 8)(r – 3) = 0
⇒ r = 8, 3
The acceptable answer is r = 3.

2. Given; 5Pr = 6Pr – 1
Plus One Maths Permutation and Combinations Three Mark Questions and Answers 3
⇒ 42 – 13r + r2 = 6
⇒ r2 – 13r + 36 = 0
⇒ (r – 9)(r – 4) = 0
⇒ r = 9, 4
The acceptable answer is r = 4.

Plus One Maths Chapter Wise Questions and Answers Chapter 7 Permutation and Combinations

Question 4.
The letters of the word TUESDAY are arranged in a line, each arrangement ends in S.

  1. How many different arrangements are possible? (2)
  2. How many of them start with letter D? (1)

Answer:

  1. In the word TUESDAY there are 7 letters. When word end in S, there are only 6 possible arrangements. This can be done in 6! = 720
  2. The word start with D and end in S, this can be done in 5! = 120.

Question 5.
Consider the word ANNAMALAI

  1. How many new words can be formed using the given words? (2)
  2. Among the new words how many of them will begin with A and end with I. (1)

Answer:

  1. In the word ANNAMALAI there are 9 letters, of which A appears 4 times, N appears 2 times and the rest ail are different. Therefore the total number of ways is \(\frac{9 !}{4! \times 2 !}\) = 7560.
  2. The word start with A and end in I, this can be done in \(\frac{7 !}{3! \times 2 !}\) = 420.

Plus One Maths Chapter Wise Questions and Answers Chapter 7 Permutation and Combinations

Question 6.
Find the rank of the word NAAGI, if these words are written as in a dictionary.
Answer:
The order of the letters will be A, A, G, I, N
Plus One Maths Permutation and Combinations Three Mark Questions and Answers 4
Therefore the position of ‘NAAGI’ is 24 + 12 + 12 + 1 = 49.

Question 7.
A committee of 3 persons is to be constituted from a group of 2 men and 3 women.

  1. In how many ways can be done? (1)
  2. How many of these committees would consist of 1 man and 2 women? (2)

Answer:
1. The required number of ways
= 5C3 = 5C2 = \(\frac{5 \times 4}{1 \times 2}\) = 10.

2. One man can be selected in 2C1. 2 women can be selected in 3C2
Therefore required number of ways
= 2C1 × 3C2 = 2 × 3 = 6.

Question 8.
It was found at a certain dinner meeting that after every member had shaken hand with every other members, 45 handshakes were interchanged. How many members were present at the metting?
Answer:
Let n be the member of person present in the meeting. The total number of handshakes is same as the number of ways of selecting 2 persons from among n persons.
The total number of handshakes = nC1 = 45
⇒ \(\frac{n(n-1)}{1 \times 2}\) = 45 ⇒ n2 – n – 90 = 0
⇒ (n – 10)(n + 9) = 0
⇒ n = 10, -9
n = 10 is acceptable.

Plus One Maths Chapter Wise Questions and Answers Chapter 7 Permutation and Combinations

Question 9.
In an exam, Arjun has to select 4 questions from each part. There are 6, 7 and 8 question in Part I, Part II and Part III, respectively. What is the number of possible combinations in which he can choose the question?
Answer:
Selecting 4 questions from Part I = 6C4 = 6C2
Selecting 4 questions from Part II = 7C4 = 7 C3
Selecting 4 questions from Part III = 8C4
The required number of ways
= 6C2 × 7C3 × 8C4 = 15 × 35 × 70 = 36750.

Plus One Maths Permutation and Combinations Four Mark Questions and Answers

Question 1.
Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that

  1. all vowels occur together. (2)
  2. all vowels do not occur together. (2)

Answer:
1. DAUGHTER this word has 8 different letters. A, U, E are the vowels. Treat these 3 as one unit, then there are 6 units and can be permuted in 6! ways. The above vowels can be permuted in 3! ways. Hence the total number of words is 6! × 3! = 4320.

2. Number of words all vowels do not occur together = Total number of different words – number of words in which vowels come together
= 8! – 6! × 3! = 6!(8 × 7 – 6) = 2 × 6!(28 – 3)
= 50 × 6! = 36000.

Plus One Maths Chapter Wise Questions and Answers Chapter 7 Permutation and Combinations

Question 2.
How many permutations are there of the 11 letters in MISSISSIPPI

  1. taken all together? (2)
  2. all the I’s not come together? (2)

Answer:
1. In the word MISSISSIPPI there are 11 letters, of which S appears 4 times, I appears 4 times, P appears 2 times and the rest all are different.
Therefore the total number of ways is \(\frac{11 !}{4 ! \times 4 ! \times 2 !}\) = 34650.

2. 4 I’s are kept together and should be counted as one unit, then there are 8 units. The number of ways is \(\frac{8 !}{4 \times 2 !}\) = 840. Therefore the I’s not come together = Total arrangements – 4 I’s together.
= 34650 – 840 = 33810.

Question 3.
Find the number of arrangements of 6 boys and 5 girls in a row so that

  1. no two girls sit together. (2)
  2. boys and girls occupy alternate positions. (2)

Answer:
1. Since no two girls sit together, we have first arrange the 6 boys among themselves. This can be done in 6! ways.
Plus One Maths Permutation and Combinations Four Mark Questions and Answers 5
Now no two girls sit together if we place the girls in between boys. There are 7 places and it should be occupied by 5 girls, can be done in 7P5 ways. Therefore the total number of ways is 6! × 7P5 = 720 × 7 × 6 × 5 × 4 × 3 = 1814400.

2. Boys and girls occupy alternate position can be done as follows. First place the boys whose number is large.
Plus One Maths Permutation and Combinations Four Mark Questions and Answers 6
Boys can be arranged in 6! ways. The place between boys can be filled by 5 girls, can be done in 5! ways. Therefore the total number of ways is 6! × 5! = 720 × 120 = 86400.

Plus One Maths Chapter Wise Questions and Answers Chapter 7 Permutation and Combinations

Question 4.
If the letters of the word DHRONA be permuted and arranged as in a dictionary, find the rank of the word.
Answer:
The order of the letters will be A, D, H, N, O, R
Plus One Maths Permutation and Combinations Four Mark Questions and Answers 7
Plus One Maths Permutation and Combinations Four Mark Questions and Answers 8
Therefore the position of ‘DHRONA’ is 120 + 24 + 6 + 6 + 6 + 2 + 2 + 1 + 1 = 168.

Question 5.
If the letters of the word MOTHER be permuted and arranged as in a dictionary, find the rank of the word.
Answer:
The order of the letters will be E, H, M, O, R, T
Plus One Maths Permutation and Combinations Four Mark Questions and Answers 9
Therefore the position of ‘MOTHER’ is 120 + 120 + 24 + 24 + 6 + 6 + 6 + 2 + 1 = 309.

Plus One Maths Chapter Wise Questions and Answers Chapter 7 Permutation and Combinations

Question 6.
How many

  1. Straight line (2)
  2. Triangles can be formed by joining 12 points, 4 of which are collinear. (2)

Answer:
1. Number of straight lines that can be formed using 12 points = 12C2 = \(\frac{12 \times 11}{1 \times 2}\) = 66. Number of straight lines that can be formed using 4 collinear points = 4C2 = \(\frac{4 \times 3}{1 \times 2}\) = 6.  Since the 4 points are collinear, the required number of lines = 66 – 6 + 1 = 61.

2. Number of triangles that can be formed using 12 points = 12C3 = \(\frac{12 \times 11 \times 10}{1 \times 2 \times 3}\) = 220.
Number of triangles that can be formed using 4 collinear points = 4C3 = 4. Since the 4 points are collinear, the required number of triangles = 220 – 4 = 216.

Question 7.
From 7 men and 4 ladies a committee of 6 is to be formed. In how many ways can this be done when the committee contains

  1. exactly two ladies. (2)
  2. at least two ladies. (2)

Answer:
1. Exactly two ladies can be selected from 4 in 4C2 = \(\frac{4 \times 3}{1 \times 2}\) = 6. The remaining 4 should be selected from 7 men in 7C4 = \(\frac{7 \times 6 \times 5 \times 4}{1 \times 2 \times 3 \times 4}\) = 35. The required number of ways = 6 × 35 = 210.

2. Atleast 2 ladies can be selected as follows;
Plus One Maths Permutation and Combinations Four Mark Questions and Answers 10
The required number of ways = 210 + 140 + 21 = 371.

Plus One Maths Chapter Wise Questions and Answers Chapter 7 Permutation and Combinations

Question 8.
A box contains 6 apples, 5 oranges, and 8 mangoes.

  1. In how many ways a fruit is selected from the box. (1)
  2. In how many different ways can an apple and an orange be selected. (1)
  3. In how many different ways a person take one apple, one orange, and one mango. (2)

Answer:

  1. The box contains 6 + 5 + 8 = 19 fruits, from this one fruit can be selected in 19C1 = 19 ways.
  2. An apple is to be selected from 6 apples and an orange be selected from 5 oranges. The required number of ways = 6C1 × 5C1 = 30.
  3. An apple is to be selected from 6 apples, an orange is to be selected from 5 oranges and one mango is to be selected from 8 mangoes. The required number of ways = 6C1 × 5C 1 × 8C1 = 240.

Question 9.
A student has to answer 6 out of 10 questions which are divided into two parts containing 5 questions each and he is permitted to attempt not more than 4 from any group. In how many ways can he make up his choice?
Answer:
The different possibilities are mentioned below;
Plus One Maths Permutation and Combinations Four Mark Questions and Answers 11
The required number of ways
= 5C1 × 5C2 + 5C3 × 5C3 + 5C2 × 5C1
= 5 × 10 + 10 × 10 + 10 × 5 = 200.

Plus One Maths Chapter Wise Questions and Answers Chapter 7 Permutation and Combinations

Question 10.

  1. How many chords can be drawn through 15 points on a circle? (2)
  2. A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected? (2)

Answer:

  1. Number of chords = 15C2 = \(\frac{15 \times 14}{1 \times 2}\) = 105
  2. Selection of 2 black and 3 red balls = 5C2 × 6C3 = 10 × 20 = 200.

Question 11.

  1. If nC2 = nC8 then find n
  2. Find n if nP5 = 42 × nP3; n> 4

Answer:
1. n = 2 + 8 = 10.

2. DAUGHTER this word has 8 different letters. A, U, E are the vowels. Treat these 3 as one unit, then there are 6 units and can be permuted in 6! ways. The above vowels can be permuted in 3! ways. Hence the total number of words is 6! × 3! = 4320.

Plus One Maths Permutation and Combinations Six Mark Questions and Answers

Question 1.
Find the arrangements of letters of the word INDEPENDENCE. In how many of these arrangements,

  1. do the words start with P. (2)
  2. do all the vowels always occur together. (2)
  3. do the vowels never occur together. (1)
  4. do the words begin with I and end in P? (1)

Answer:
1. In the word INDEPENDENCE there are 12 letters, of which N appears 3 times, E appears 4 times, D appears 2 times and the rest all are different.

When the words start with P, then there are 11 letters to be filled in 11 spaces. Therefore the total number of ways is \(\frac{11 !}{3! \times 2 ! \times 4 !}\) = 138600.

2. The vowels EEEEI are to be kept together and should be treated as one unit. Then these vowels can be arranged in \(\frac{5 !}{4 !}\) ways. This single unit together with 7 letter will count to units, can be arranged in \(\frac{8 !}{3! \times 2 !}\). Therefore the total number of ways \(\frac{5 !}{4 !} \times \frac{8 !}{3 ! \times 2 !}\) = 16800.

3. Number of ways of arrangement with vowels do not come together = Total arrangement – vowels coming together.
= \(\frac{12 !}{3 ! \times 2 ! \times 4 !}\) – 16800 = 1663200 – 16800 = 1646400.

4. When the words start with I and ends with P, then there are 10 letters to be filled in 10 spaces. Therefore the total number of ways is \(\frac{10 !}{3 ! \times 2 ! \times 4 !}\) = 12600.

Plus One Maths Chapter Wise Questions and Answers Chapter 7 Permutation and Combinations

Question 2.
Consider the word ASSASSINATION.

  1. How many permutations are there of the letters of the given word? (2)
  2. How many different ways can be arranged so that the 4S’s come together? (2)
  3. How many different ways can be arranged so that the 4S’s do not come together? (1)
  4. How many begin with A? (1)

Answer:
1. In the word ASSASSINATION there are 13 letters, of which A appears 3 times, S appears 4 times, N appears 2 times, I appears 2 times and the rest all are different. Therefore the total number of ways is \(\frac{13 !}{3! \times 4 ! \times 2 ! \times 2 !}\) = 10810800.

2. 4 S’s are kept together and should be counted as one unit, then there are 10 units. The number of ways is \(\frac{10 !}{3! \times 2 ! \times 2 !}\) = 151200.

3. Number of words in which 4S’s do not come together = Total number of words – 4S’s together = 10810800 -151200 = 10659600.

4. The word will start with any one of the 4 A’s. Then total letter arrange will be 12. Number of words in which begin with A \(\frac{12 !}{2! \times 4 ! \times 2 ! \times 2 !}\) = 2494800.

Plus One Maths Chapter Wise Questions and Answers Chapter 7 Permutation and Combinations

Question 3.
A team of 11 cricket players is to be chosen from 15 players. In how many ways can this be done so as to:

  1. Include a particular player A. (2)
  2. Exclude a particular player B. (2)
  3. Include A and exclude B. (2)

Answer:
1. A particular player A is to be included, then selection of 10 is to be made from 14 players.
The required number of ways 14C10 = 14C4 = \(\frac{14 \times 13 \times 12 \times 11}{1 \times 2 \times 3 \times 4}\) = 7 × 13 × 11 = 1001.

2. A particular player B is to be excluded, then selection of 11 is to be made from 14 players.
The required number of ways = 14C11 = 14C3 = \(\frac{14 \times 13 \times 12}{1 \times 2 \times 3}\)
= 14 × 13 × 2 = 364.

3. A particular player A is to be included and player B is to be excluded, then selection of 10 is to be made from 13 players. The required number of ways
= 13C10 = 13C3 = \(\frac{13 \times 12 \times 11}{1 \times 2 \times 3}\)
= 13 × 2 × 11 = 286.

Plus One Maths Chapter Wise Questions and Answers Chapter 7 Permutation and Combinations

Question 4.
What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these

  1. four cards are of the same suit, (2)
  2. are face cards, (1)
  3. two are red cards and two are black cards, (2)
  4. cards are of the same colour? (1)

Answer:
Selection of 4 cards from 52 = 52C4
1. There are 4 suits in a pack of 52 playing cards. They are Club, Spade, Diamond, and Heart. Selecting 4 from each can be done in,
= 13C4 + 13C4 + 13C4 + 13C4
= 4 × 13C4 = 4 × \(\frac{13 \times 12 \times 11 \times 10}{1 \times 2 \times 3 \times 4}\) = 2860.

2. There are 12 face cards in a pack of 52 playing cards. Selection 4 cards can be done in 12C4 = 495.

3. There are 26 red cards and 26 black in a pack of 52 playing cards. Selection of 2 cards should be done from each colour, this can be done in 26C2 × 26C2 = (325)2 = 105625.

4. Selection of 4 cards from same colour = 26C4 + 26C4 = 29900

Plus One Maths Permutation and Combinations Practice Problems Questions and Answers

Question 1.
There are 3 routes from place A to place B and 2 routes from place B to place C. Find how many different routes are there from A to C.
Answer:
By fundamental principle of counting, there are 2 × 3 = 6 different ways.

Plus One Maths Chapter Wise Questions and Answers Chapter 7 Permutation and Combinations

Question 2.
How many 3 digit numbers can be formed from the digits 1, 2 and 3, assuming that the repetition of digits is not allowed.
Answer:
Plus One Maths Permutation and Combinations Practice Problems Questions and Answers 12
By fundamental principle of counting, there are 1 × 2 × 3 = 6 different 3 digit numbers.

Question 3.
How many two-digit even numbers with distinct digits can be formed from the digits 1, 2, 3, 4, 5.
Answer:

filled in 4 waysfilled by 2, 4 – 2ways

Hence by fundamental principle of counting, there are 4 × 2 = 8 different 2 digit even numbers.

Plus One Maths Chapter Wise Questions and Answers Chapter 7 Permutation and Combinations

Question 4.
Evaluate the following

  1. \(\frac{7 !}{5 !}\)
  2. \(\frac{12 !}{10 ! \times 2 !}\)
  3. 6P4
  4. 9P4
  5. 10P5

Answer:
1. \(\frac{7 !}{5 !}\) = \(\frac{7 \times 6 \times 5 !}{5 !}\) = 45

2.
Plus One Maths Permutation and Combinations Practice Problems Questions and Answers 13

3. 6P4 = 6 × 5 × 4 × 3 = 360

4. 9P4 = 9 × 8 × 7 × 6 = 3024

5. 10P5 = 10 × 9 × 8 × 7 × 6 = 30240

Question 5.
Evaluate the following.

  1. 10C4
  2. 21C3
  3. 19C15
  4. 31C29 (1 score each)

Answer:

  1. 10C4 = \(\frac{10 \times 9 \times 8 \times 7}{1 \times 2 \times 3 \times 4}\) = 10 × 3 × 7 = 210
  2. 21C3 = \(\frac{21 \times 20 \times 19}{1 \times 2 \times 3}\) = 7 × 10 × 19 = 1330
  3. 19C15 = \(\frac{19 \times 18 \times 17 \times 16}{1 \times 2 \times 3 \times 4}\) = = 19 × 3 × 17 × 4 = 3876
  4. 31C29 = \(\frac{31 \times 30}{1 \times 2}\) = 465.

Plus One Maths Chapter Wise Questions and Answers Chapter 7 Permutation and Combinations

Question 6.
How many five digits telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?
Answer:
Digits 6, 7 already used, so only 8 digits available to fill remaining 3 places.
Plus One Maths Permutation and Combinations Practice Problems Questions and Answers 14
Hence by fundamental principle of counting there are 8 × 7 × 6 = 336 different 5 digit telephone numbers.

Question 7.
In how many ways can 5 persons sit in a car, two including the driver in the front seat and 3 in the back seat. If two particular person out of the 5 are to avoid the driver’s seat?
Answer:
Plus One Maths Permutation and Combinations Practice Problems Questions and Answers 15
Hence by fundamental principle of counting there are 3 × 4 × 3 × 2 × 1 = 72 different ways.

Question 8.
How many numbers can be formed from the digits 1, 2, 3 and 9 if repetition of digits is not allowed?
Answer:

Single digit4 ways
Two digit4 × 3 = 12 ways
Three digit4 × 3 × 2 = 24 ways
Four digit4 × 3 × 2 × 1 = 24 ways

Hence the number of total numbers = 4 + 12 + 24 + 24 = 64.

Plus One Maths Chapter Wise Questions and Answers Chapter 7 Permutation and Combinations

Question 9.
How many 3-digit even numbers can be formed from the digit 1, 2, 3, 4, 5, 6 if the digits can be repeated?
Answer:
Plus One Maths Permutation and Combinations Practice Problems Questions and Answers 16
Hence by fundamental principle of counting there are 6 × 6 × 3 = 108 different 3 digit even numbers.

Question 10.
How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that

  1. repetition of the digits is allowed? (1)
  2. repetition of the digits is not allowed (1)

Answer:
1.
Plus One Maths Permutation and Combinations Practice Problems Questions and Answers 17
Hence by fundamental principle of counting there are 5 × 5 × 5 = 125 different 3 digit numbers.

2.
Plus One Maths Permutation and Combinations Practice Problems Questions and Answers 18
Hence by fundamental principle of counting there are 5 × 4 × 3 = 60 different 3 digit numbers.

Plus One Maths Chapter Wise Questions and Answers Chapter 7 Permutation and Combinations

Question 11.
How many 8 letter words, with or without meaning, can be formed using the word EQUATION, using each letter exactly once?
Answer:
EQUATION this word has 8 different letters. Different words that can be made from these letters is 81 = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 =40320.

Question 12.
Find the number of ways in which the letters of the word ASSISTANT can be arranged among themselves.
Answer:
In the word ASSISTANT there are 9 letters, of which S appears 3 times, A appears 2 times, T appears 2 times and the rest all are different. Therefore the total number of ways is
\(\frac{9 !}{3! \times 2 ! \times 2 !}\) = 15120.

Plus One Maths Chapter Wise Questions and Answers Chapter 6 Linear Inequalities

Students can Download Chapter 6 Linear Inequalities Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 6 Linear Inequalities

Plus One Maths Linear Inequalities Three Mark Questions and Answers

Question 1.
Solve the following inequalities.

  1. \(\frac{1}{2}\left(\frac{3 x}{5}+4\right) \geq \frac{1}{3}(x-6)\)
  2. \(\left(\frac{2 x-1}{3}\right) \geq \frac{(3 x-2)}{4}-\frac{(2-x)}{5}\) (3 score each)

Answer:
1. Given;
Plus One Maths Linear Inequalities Three Mark Questions and Answers 1
⇒ 3(3x + 20) ≥ 10(x – 6)
⇒ 9x + 60 ≥ 10x – 60
⇒ 9x – 10x ≥ -60 – 60
⇒ -x ≥ -120 ⇒ x ≤ 120

2. Given;
Plus One Maths Linear Inequalities Three Mark Questions and Answers 2
⇒ 20(2x -1) ≥ 3[15x – 10 – 8 + 4x]
⇒ 40x – 20 ≥ 45x – 54 + 12x
⇒ 40x – 20 ≥ 57x – 54
⇒ 40x – 57x ≥ -54 + 20
⇒ -17x ≥ -34 ⇒ x ≤ 2

Plus One Maths Chapter Wise Questions and Answers Chapter 6 Linear Inequalities

Question 2.
1. Which of the following sets of inequality represent the second quadrant? (1)
(a) x < 0, y < 0
(b) x > 0, y > 0
(c) x < 0, y > 0
(d) x > 0, y < 0
2. Write the system of inequalities that represents the shaded rectangle in the figure given below: (2)
Plus One Maths Linear Inequalities Three Mark Questions and Answers 3
Answer:
1. (a) x < 0, y < 0

2. The shaded figure is a rectangle. The side parallel to x axis are y = -1 and y = 1. The side perpendicular to x axis are x = 2 and x = -2. Hence the inequality that represent the shaded region are
-2 ≤ x ≤ 2; -1 ≤ y ≤ 1.

Plus One Maths Chapter Wise Questions and Answers Chapter 6 Linear Inequalities

Question 3.
Find all pairs of consecutive even positive integers both of which are smaller than10 such that their sum is less than 23.
Answer:
Consecutive even positive integers be x and x + 2. Then, x + x + 2 < 23; x + 2 < 10
⇒ 2x < 23 – 2; x < 10 – 2
⇒ x < \(\frac{21}{2}\) = 10.5; x < 8
⇒ 8 < x ≤ 10 Therefore x can take values 9, 10. Hence the pairs are (9, 10), (10, 9).

Question 4.
The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.
Answer:
Let the shortest side is x, then;
Longest side = 3x;
Third side = Longest side – 2 = 3x – 2
Perimeter = 3x + 3x – 2 + x ≥ 61 ⇒ 7x – 2 ≥ 61
⇒ 7x ≥ 61 + 2 ⇒ x ≥ \(\frac{63}{7}\) = 9.

Plus One Maths Linear Inequalities Four Mark Questions and Answers

Question 1.
Solve the following system of inequalities graphically.

  1. 2x – y > 1; x – 2y < -1
  2. x + y ≤ 9; y > x; x ≥ 0
  3. x – 2y ≤ 3; 3x + 4y ≥ 12; x ≥ 0, y ≥ 1
  4. 2x + y – 3 ≥ 0; x – 2y + 1 ≥ 0; y ≤ 3 (4 score each)

Answer:
1. 2x – y > 1; x – 2y < -1
Plus One Maths Linear Inequalities Four Mark Questions and Answers 4
Plus One Maths Linear Inequalities Four Mark Questions and Answers 5

Plus One Maths Chapter Wise Questions and Answers Chapter 6 Linear Inequalities

2. x + y ≤ 9; y > x ⇒ x – y < 0
Plus One Maths Linear Inequalities Four Mark Questions and Answers 6
Plus One Maths Linear Inequalities Four Mark Questions and Answers 7

3. x – 2y ≤ 3; 3x + 4y ≥ 12
Plus One Maths Linear Inequalities Four Mark Questions and Answers 8
Plus One Maths Linear Inequalities Four Mark Questions and Answers 9

Plus One Maths Chapter Wise Questions and Answers Chapter 6 Linear Inequalities

4. 2x + y ≥ 3; x – 2y ≤ -1
Plus One Maths Linear Inequalities Four Mark Questions and Answers 10
Plus One Maths Linear Inequalities Four Mark Questions and Answers 11

Plus One Maths Linear Inequalities Practice Problems Questions and Answers

Question 1.
Solve the following inequalities.

  1. 4x + 3 < 5x + 7
  2. 3(x – 1) ≤ 2(x – 3) (1 score each)

Answer:
1. Given; 4x + 3 < 5x + 7
⇒ 4x – 5x < 7 – 3 ⇒ -x < 4 ⇒ x > -4.

2. Given; 3(x – 1) < 2(x – 3)
⇒ 3x – 3 ≤ 2x – 6 ⇒ 3x – 2x ≤ -6 + 3
⇒ x ≤ -3.

Plus One Maths Chapter Wise Questions and Answers Chapter 6 Linear Inequalities

Question 2.
Solve the inequality \(\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}\).
Answer:
Given; \(\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}\)
⇒ 9(x – 2) ≤ 25(2 – x)
⇒ 9x – 18 ≤ 50 – 25x
⇒ 9x + 25x ≤ 50 + 18
⇒ 34x ≤ 68 ⇒ x ≤ 2.

Question 3.
Show the solution of each inequality on a number line.

  1. 4x + 3 < 6x + 7
  2. 5x – 3 ≥ 3x – 5
  3. 3(1 – x) < 2(x + 4)
  4. 2 – 3x < 2(x + 6)
  5. -3 ≤ 3 – 2x < 6 (2 score each)

Answer:
1. Given; 4x + 3 < 6x + 7 ⇒ 4x – 6x < 7 – 3
⇒ -2x < 4 ⇒ x > -2
Plus One Maths Linear Inequalities Practice Problems Questions and Answers 12

2. Given; 5x – 3 > 3x – 5 ⇒ 5x – 3x ≥ -5 + 3
⇒ 2x ≥ -2 ⇒ x ≥ 1.
Plus One Maths Linear Inequalities Practice Problems Questions and Answers 13

3. Given; 3(1 – x) < 2(x + 4) ⇒ 3 – 3x < 2x + 8
⇒ -3x – 2x < 8 – 3 ⇒ -5x < 5 ⇒ x > -1.
Plus One Maths Linear Inequalities Practice Problems Questions and Answers 14

4. Given; 2 – 3x < 2(x + 6) ⇒ 2 – 3x < 2x + 12
⇒ -3x – 2x < 12 – 2 ⇒ -5x < 10 ⇒ x > -2
Plus One Maths Linear Inequalities Practice Problems Questions and Answers 15

5. Given; -3 ≤ 3 – 2x < 6
⇒ -3 ≤ 3 – 2x; 3 – 2x < 6
⇒ -3 – 3 ≤ -2x; -2x < 6 – 3
⇒ -6 ≤ -2x; -2x < 3 ⇒ 3 ≥ x; x > \(-\frac{3}{2}\) ⇒ \(-\frac{3}{2}\) < x ≤ 3
Plus One Maths Linear Inequalities Practice Problems Questions and Answers 16

Plus One Maths Chapter Wise Questions and Answers Chapter 6 Linear Inequalities

Question 4.
The marks obtained by a student of class XI in first and second terminal examination are 62 and 48, respectively. Find the minimum marks he should get in the annual examination to have an average of at least 60 marks.
Answer:
Let x denote the mark obtained by the student in Class XI examination, then;
\(\frac{62+48+x}{3}\) ≥ 60 ⇒ 110 + x ≥ 1800 ⇒ x ≥ 70.

Plus One Maths Chapter Wise Questions and Answers Chapter 4 Principle of Mathematical Induction

Students can Download Chapter 4 Principle of Mathematical Induction Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 4 Principle of Mathematical Induction

Plus One Maths Principle of Mathematical Induction Three Mark Questions and Answers

Question 1.
For all n ≥ 1 , prove that
12 + 22 + 32 +……….+ n2 > \(\frac{n^{3}}{3}\)
Answer:
Let p(n): 12 + 22 + 32 + n2
Put n = 1 ⇒ p(1) = 1 > \(\frac{1}{3}\) which is true.
Assuming that true for p(k)
p(k): 12 + 22 + 32 +……….+ k2 > \(\frac{k^{3}}{3}\)
Let p(k + 1): 12 + 22 + 32 +…….+ k2 + (k + 1)2 > \(\frac{k^{3}}{3}\) + (k + 1)2
Plus One Maths Principle of Mathematical Induction Three Mark Questions and Answers 1
Plus One Maths Principle of Mathematical Induction Three Mark Questions and Answers 2
Hence by using the principle of mathematical induction true for all n ∈ N.

Plus One Maths Chapter Wise Questions and Answers Chapter 4 Principle of Mathematical Induction

Question 2.
For all n ≥ 1 , prove that 1 + 2 + 3 +…….+ n < \(\frac{1}{8}\)(2n + 1)2
Answer:
Let p(n): 1 + 2 + 3 +…….+ n ,
Put n = 1 ⇒ p(1) = 1 < \(\frac{9}{8}\) which is true.
Assuming that true for p(k)
p(k): 1 + 2 + 3 +…….+ k < \(\frac{1}{8}\)(2k + 1)2
Let p(k +1): 1 + 2 + 3 +……..+ k + (k +1) < \(\frac{1}{8}\) (2k + 1)2 + (k + 1)
Plus One Maths Principle of Mathematical Induction Three Mark Questions and Answers 3
Hence by using the principle of mathematical induction true for all n ∈ N.

Question 3.
For all n ≥ 1, prove that p(n): 23n – 1 is divisible by 7.
Answer:
p(1): 23(1) – 1 = 8 – 1 = 7 divisible by 7, hence true. Assuming that for p(k)
p(k) : 23k – 1 is divisible by 7.
23k – 1 = 7M
P(k + 1): 23(k + 1) – 1 = 23k + 3 – 1
= 23k23 – 1 = 23k × 8 – 1
= 23k × 8 – 8 + 7 = 8(23k – 1) + 7
= 8(7M) + 7
Hence divisible by 7. Therefore by using the principle of mathematical induction true for all n ∈ N.

Plus One Maths Chapter Wise Questions and Answers Chapter 4 Principle of Mathematical Induction

Question 4.
For all n ≥ 1, prove that p(n): n3 + (n + 1)3 + (n + 2)3 is divisible by 9.
Answer:
p(1): 1 + 23 + 33 = 1 + 8 + 27 = 36 divisible by 9, hence true. Assuming that true for p(k)
p(k): k3 + (k + 1)3 + (k + 2)3 is divisible by 9.
⇒ k3 + (k + 1)3 + (k + 2)3 = 9M
p(k +1 ): (k + 1)3 + (k + 2 )3 + (k + 3)3
= (k +1)3 + (k + 2)3 + k3 + 9k2 + 27k + 27
= [(k + 1)3 + (k + 2)3 + k3] + 9[k2 + 3k + 3]
= 9M + 9[k2 + 3k + 3]
Hence p(k + 1)divisible by 9. Therefore by using the principle of mathematical induction true for all n ∈ N.

Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers

Question 1.
For all n ≥ 1, prove that
Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers 4
Answer:
Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers 5
Hence by using the principle of mathematical induction true for all n ∈ N.

Plus One Maths Chapter Wise Questions and Answers Chapter 4 Principle of Mathematical Induction

Question 2.
For all n ≥ 1, prove that
Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers 6
Answer:
Let
Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers 7
Assuming that true for p(k)
Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers 8
Let p(k + 1):
Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers 9
Hence by using the principle of mathematical induction true for all n ∈ N.

Plus One Maths Chapter Wise Questions and Answers Chapter 4 Principle of Mathematical Induction

Question 3.
For all n ≥ 1 , prove that 1.2.3 + 2.3.4 +………+ n(n + 1)(n + 2) = \(\frac{n(n+1)(n+2)(n+3)}{4}\).
Answer:
Let p(n): 1.2.3 + 2.3.4 +……..+ n(n + 1)(n + 2),
Put n = 1
p(1) = \(\frac{1(1+1)(1+2)(1+3)}{4}\) = 6 which is true.
Assuming that true for p(k)
p(k): 1.2.3 + 2.3.4 +……..+ k(k + 1)(k + 2) = \(\frac{k(k+1)(k+2)(k+3)}{4}\),
Let p(k + 1)
Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers 10
Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers 11
Hence by using the principle of mathematical induction true for all n ∈ N.

Plus One Maths Chapter Wise Questions and Answers Chapter 4 Principle of Mathematical Induction

Question 4.
For all n ≥ 1, prove that
Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers 12
Answer:
Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers 13
Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers 14
Hence by using the principle of mathematical induction true for all n ∈ N.

Plus One Maths Chapter Wise Questions and Answers Chapter 4 Principle of Mathematical Induction

Question 5.
For all n ≥ 1 , prove that p(n): n(n + 1 )(n + 5) is divisible by 3.
Answer:
p(1): 1(1 + 1)(1 + 5) = 12divisible by 3, hence true. Assuming that true for p(k)
p(k): k(k + 1)(k + 5) is divisible by 3.
k(k + 1)(k + 5) = 3M
p(k + 1): (k + 1)(k + 2)(k + 6)
= (k + 1)(k2 + 8k + 12)
= (k + 1)(k2 + 5k + 3k +12)
= (k + 1)[k(k + 5) + 3(k + 6)]
= [k(k + 1)(k + 5) + 3(k + 1)(k + 6)]
= [3M + 3(k + 1)(k + 6)]
= 3[M + (k + 1)(k + 6)]
Hence divisible by 3. Therefore by using the principle of mathematical induction true for all n ∈ N.

Plus One Maths Chapter Wise Questions and Answers Chapter 4 Principle of Mathematical Induction

Question 6.
For all n ≥ 1 , prove that p(n): 2.7n + 3.5n – 5 is divisible by 24.
Answer:
p(1): 2.71 + 3.51 – 5 = 14 + 15 – 5 = 24 divisible by 24, hence true. Assuming that true for p(k)
p(k): 2.7k + 3.5k – 5 is divisible by 24.
⇒ 2.7k + 3.5k – 5 = 24M
p(k + 1): 2.7k + 1 + 3.5k + 1 – 5
= 2.7k.7 + 3.5k.5 – 5
= 2.7k.(6 + 1) + 3.5k.(4 + 1) – 5
= 12.7k + 2.7k + 12.5k + 3.5k – 5
= 12(7k + 5k) + (2.7k + 3.5k) – 5
= 12(7k + 5k) + 24M
7k And 5k are odd numbers, therefore (7k + 5k) will be an even and will be divisible by 24, Hence p(k + 1)divisible by 24. Therefore by using the principle of mathematical induction true for all n ∈ N.

Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement

Students can Download Chapter 9 Locomotion and Movement Questions and Answers, Plus One zoology Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement

Plus One Locomotion and Movement One Mark Questions and Answers

Question 1.
Ribs are attached to
(a) Scapula
(b) Sternum
(c) Clavicle
(d) llium
Answer:
(d) llium

Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement

Question 2.
What is the type of movable joint present between the atlas and axis?
(a) Pivot
(b) Saddle
(c) Hinge
(d) Gliding
Answer:
(a) Pivot

Question 3.
ATPase of the muscle is located in
(a) Actinin
(b) Troponin
(c) Myosin
(d) Actin
Answer:
(c) Myosin

Question 4.
Macrophages and leucocytes exhibit
(a) Ciliary movement
(b) Flagellar movement
(c) Amoeboid movement
(d) Gliding movement
Answer:
(c) Amoeboid movement

Question 5.
Which one of the following is not a disorder of bone?
(a) Arthritis
(b) Osteoporosis
(c) Rickets
(d) Atherosclerosis
Answer:
(d) Atherosclerosis

Question 6.
Suggest a suitable word for the fourth place

  1. thin filament : Actin :: Thick filament : ___________
  2. Pelvic girdle : humerus :: Pelvic girdle : ___________

Answer:

  1. myosin
  2. Femur

Question 7.
Copy the paragraph below about the structure of a striated muscle. Choose the words from the following list & fill up the gaps.

Under lightmicroscope, the striated muscle shows ________ (a) band & _______ (b) band. The distance between 2 adjacent Z lines is known as __________ (c).
(sarcomere, lightband, A band, H-zone, M line, darkband)
Answer:
(a) Ligthband
(b) Darkband
(c) Sarcomere

Question 8.
A person is suffering from joint pain. His blood test shown increased amount of Uric acid. What will be the diagnosis?
Answer:
Gout

Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement

Question 9.
The infants have 33 vertebrae in the vertebral column. But an adult has only 26 vertebrae. What will happen to the remaining vertebrae.
Answer:
Ulna

Plus One Locomotion and Movement Two Mark Questions and Answers

Question 1.
The axial skeleton contains 80 bones. Make a table according to where it is seen and number of bones present in each section.
Answer:

Skull29
Vertebral column26
Ribs24
Sternum1
Total80

Question 2.
One of the following statements is incorrect. Find and correct it.

  1. The number of cervical vertebrae is seven in all mammals except human beings.
  2. Thoracic vertebrae, ribs and sternum together make ribcage.
  3. Accumulation of uric acid in joints leads to gout.

Answer:
Statement 1 is incorrect. The number of cervical vertebrae is seven in all mammals including human beings.

Question 3.
Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement - 1

  1. Name the process shown here, which theory explain this process ?
  2. Draw and complete the process.

Answer:
1. Muscle contraction, Sliding – filament theory

2.

Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement - 2

Question 4.
Red muscle fibers have greater capacity to do work for a prolonged period where as white muscles suffer from fatigue after a short time. Give reason.
Answer:
Red muscle fibers are red in colour due to the presence of large amount of myoglobin. Myoglobin is an oxygen storing pigment. These muscle also contain plenty of mitochondria which can utilise the large amount of oxygen stored in them for ATP production.

So red muscle fibers have greater capacity to do work for a prolonged period. White muscle fibers possess very less quantity of myoglobin and mitochondria.

Question 5.
Shoulder joints are not very stable. But they are freely movable. Give reason?
Answer:
Shoulder joints are freely movable joint or synovial joint. In this type of joint there is presence of a fluid filled synovial cavity between the articulating surfaces of the two bones. Such an arrangement allows considerable movement.

Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement

Question 6.
Name the types of joint between the following.

  1. Atlas/axis
  2. Between Cranial bones
  3. Carpal/Meta carpel of thumb
  4. Between Humerus and pectoral girdle

Answer:

  1. Pivot joint
  2. Fibrous joint
  3. Saddle joint
  4. Ball and socket joint

Question 7.
Analyse the table & fill in the blanks with appropriate words.

RadiumForearm
(a)Femur
Atlas(b)
Frontal bone(c)
HumerusUpper arm

Answer:

RadiumFore arm
Thigh boneFemur
AtlasFirst vertebrae
Frontal boneCranium
HumerusUpper arm

Question 8.
The last two pairs of ribs are called floating ribs.

  1. What do you meant by the term floating ribs?
  2. What are true ribs?

Answer:

  1. Last 2 pairs of ribs are not connected ventrally to the sternum. There ribs are called floating ribs.
  2. First seven pairs of ribs are called true ribs. Dorsally, they pre attached to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilage.

Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement

Question 9.
Some joints in the body are characterized by pressure of a fluid filled cavity between articulating surfaces of two bones. Mention the peculiarity of these types of joint and give examples?
Answer:
This type of joint in called synovial joint or freely movable joint and allows considerable movement.
eg:

  1. Ball and socket joint (between humerus and pectoral girdle)
  2. Hinge joint (knee joint)

Question 10.
A muscle cell viewed under a microscope shown the following characters.

  • Sarcolemma – Present
  • Shape – Cylindrical
  • Nucleus – Multinuclear
  • Striation – Present
  1. Identify the muscle cell.
  2. Calcium ions are necessary for muscle contraction.

Answer:

If a muscle is placed in a solution containing calcium ions. Does the muscle contracts? Substantiate.
Answer:

  1. Skeletal muscle
  2. No. For muscle contraction to occur, presence of Ca++ ions intracellularly is needed.

Question 11.
Pictorial representation of tissue is given in a lab diary with only one labelling – intercalated discusing your knowledge about tissue.

  1. Identify the tissue.
  2. Draw the structure of tissue and label the parts.

Answer:
1. Cardiac muscle

2.
Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement - 3

Question 12.
You can move your hands at your will. But you can’t move your heart at your will. Comment.
Answer:
Muscles of the hand are skeletal muscles and they are voluntary muscle. Muscles of the heart are cardiac muscles and they are involuntary muscles.

Question 13.
Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement - 4

  1. Identify the muscle cell
  2. Mention its function.

Answer:

  1. Non-striated muscle (smooth muscle)
  2. Smooth muscles are involuntary muscles located in the inner visceral organs like alimentary canal, reproductive tract etc. So they help the transportation of food through digestive tract and gamete through the genital tract.

Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement

Question 14.
An investigation was arrived out to find the effect of the temperature on muscle contraction. The results are shown in the graph.
Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement - 5

  1. Describe and suggest an explanation for one effect of temperature on muscle contraction.
  2. Name the filaments involved in muscle contraction.

Answer:

  1. When temperature increases muscle contraction increases, but when temperature decreases the extend of muscle contraction decreases, but the time remain in the contracted state increases,
  2. Actin and myosin

Question 15.
The red colour of the blood is due to the presence of hemoglobin in it. Certain skeletal muscles are red in colour even though they lack hemoglobin. Now explain how the red muscles have that particular colour.
Answer:
Muscle contains a red coloured oxygen storing pigment called myoglobin. Myoglobin content is high in some skeletal muscles which gives a reddish appearance.

Question 16.
Actin, Troponin, Biceps, Muscles of blood vessels, Muscles of heart, Myosin, Muscles of reproductive tract, Tropomyosin, Muscles of Alimentary canal, Triceps.
Rearrange the terms in four columns on the basis of their similarity and give appropriate headings for each columns.
Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement - 6

Question 17.
Identify the myofibril and label the parts given below.
Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement - 7

Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement - 8

Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement

Question 18.
Length of A band remains unchanged during muscular contraction. Is this statement true or false? Justify.
Answer:
True. A band or anisotropic band is the region where both thick filaments and thin filaments are present. During muscular contraction, the length of thick filament or thin filament does not change.

Question 19.
Match the column I with column II.

Column IColumn II
1. Red muscle1. Pelvic girdle
2. Suture2. Globular protein
3. Ischium3. Immovablejoint
4. Troponin4. Mybglobin
5. Sternum

Answer:

Column IColumn II
1. Red muscle4. Mybglobin
2. Suture3. Immovable joint
3. Ischium1. Pelvic girdle
4. Troponin2. Globular protein

Question 20.
Complete the division of human skeletal system by filling the blanks.
Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement - 9
Answer:
(a) Axial
(b) Vertebral column
(c) 12
(d) One

Question 21.
Arrange the following bones into two columns and give proper heading to each columns.
Scapula, Carpals, Femur, Fibula, Tibia, Tarsals, Acetabulum, Metatarsals, Radius, Humerus, Glenoid cavity, Clavicle, Metal Carpals, Phalanges, Patella, Ilium, Ischium, Pubis.
Answer:

Pectoral girdle and
bones of forelimb
Pelvic girdle and Bones
of Hind Limb
ClavicleIlium
ScapulaIschium
HumerusPelvis
RadiusFemur
UlnaPatella
 CarpalsTibia
MetacarpalsFibula
PhalangesMetatarsal
phalanges

Question 22.
Your lungs and Heart are well protected in a cage. Which bone contributes to it?
Answer:
In an infant, there are 33 vertebrae in the vertebral column. Five of these bones fuse to form the sacrum and four of other join to become the coccyx. As a result, an adult vertebral column has 26 vertebrae.

Question 23.
There are joints at your shoulders and elbows. But their movement is different. Give reason.
Answer:

  • Shoulder joint – Ball and socket joint
  • Elbow joint – Hinge joint

Question 24.
What are the different type of movement shown by human cells?
Answer:
Amoeboid, ciliary and muscular and flagellar movement.

Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement

Question 25.
Fill the gaps.
Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement - 10

Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement - 11

Question 26.
Draw a flow chart showing the flow of stimulus that results in muscle contraction. (Start from Neuromuscular junction)
Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement - 12

Question 27.
Match the following.

AB
a. Fixed jointShoulder joint
b. Ball and socket jointsWrist and ankle
c. Hinge jointsPhalanges of hind limbs
d. Pivot jointsBones of skull
e. Gliding jointsElbow and joints of phalanges
f. Ellipsoid jointsSkull and vertebral column.

Answer:

AB
a. Fixed jointBones of skull
b. Ball and socket jointsShoulder joint
c. Hinge jointElbow and joints of Phalanges
d. Pivot jointSkull and vertebral column
e. Gliding jointWrist and ankle
f. Ellipsoid jointPhalanges of hind limbs

Question 28.
Write the odd one out and give the reason for your answer.

  1. Fibrous joints, Ball and socket joints, hinge joints, pivot joints
  2. Humerus, Femur, Radius, Ulna.

Answer:

  1. Fibrous joints: They are immovable joints of the bones of skull whereas all others are freely movable joints.
  2. Femur: It is the thigh bone whereas all other are the bones of the forelimb

Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement

Question 29.
Bones of the older people turn brittle and break quickly.

  1. Name the bone disorder stated above.
  2. List any two reasons for the above disorder.

Answer:

  1. Osteoporosis
  2. reasons for the above disorder
    • Imbalance of hormones like thyrocalcitonin, parathyroid and sex hormones.
    • Deficiency of calcium and Vitamin D.

Question 30.
When you ride a bicycle two majorjointsofyourleg should perform properly.

  1. Name the joints.
  2. To which type of synovial joints these belong?

Answer:

  1. Knee joint, Hip Joint
  2. Hinge joint, Ball and Socket joint

Question 31.
In human body different type of movement shown by some cells.

  1. Name the type of movement shown by human sperm.
  2. Which is the part of sperm help this movement?

Answer:

  1. Flagellar movement
  2. Whip like movement of the tail and the middle piece of the sperm

Plus One Locomotion and Movement Three Mark Questions and Answers

Question 1.
State one difference in each of the following pairs on the basis of what is indicated in brackets.

  1. Glenoid cavity and acetabulum (location and function)
  2. Osteoarthritis and Gout (cause)

Answer:
1. Glenoid cavity:
It is a cap like depression in the pectoral girdle to which the head of the humerus fits in.

Acetabulum:
It is a cap like depression present in the pelvic girdle to which head of the femur bone fits in.

2. Osteoarthritis:
It is caused by the degeneration of the articular cartilage.

Gout:
It is caused by accumulation of uric acid crystals.

Question 2.
Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement - 13

  1. Observe the diagram and identify the type of joint.
  2. Where can you find this type of joints in your body?
  3. Comment on its ability to move.

Answer:

  1. Ball and socket joint
  2. Shoulder joints and Hip joints
  3. In ball and socket joint, one of the bones forms a ball like head while the other forms a cup like socket into which head fits in. It allows the free movement of the bone with the ball like head.

Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement

Question 3.
X-ray photography of forelimb of a person is given here. One of the bones found to be fractured. Name the bone with fracture.
Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement - 14
Answer:
a. (i) Across:
1. Hyoid
3. Cranium
4. Pelvic girdle

(ii) Down:
2. Coccyx
5. Ulna

b. Hyoid and cranium

Question 4.
According to colour, muscles are of two types.

  1. Name the two muscles.
  2. Why are they called so?
  3. Give examples.

Answer:

  1. Red muscle and white muscle.
  2. Red muscles contain large amount of mitochondria and myoglobin whereas the white muscles contain only less amount of mitochondria and myoglobin.
  3. Examples
    • Red muscles – Extensor muscles on the back.
    • White muscles – muscles of the eye ball.

Question 5.
Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement - 15

  1. Observe the above graph and find out the region of graph where muscle show maximum contraction. Justify your answer.
  2. Name the different types of muscle proteins present in thick and thin filaments.
  3. Ca2+ is necessary for muscle contraction. Why?

Answer:
1. Region 2 During Muscular contraction amount of Ca2+ released from ER is very high.

2. Thick filament – Muscle protein – myosin
Thin filament – Muscle protein – actin, troponin, tropomyosin.

3. Muscle contraction is regulated by a regulator protein troponin. Troponin mask the active site of actin molecule from myosin head, Action potential reaching a muscle cause the ER to release Ca2+.

This Ca2+ bind with the specific site of troponin. It causes a conformational change by which the active site of actin molecule is exposed and it results in cross bridge activity and muscle contraction.

Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement

Question 6.
Will muscle contraction occur in the following conditions. Justify your answer.

  1. Troponin is removed from the thin filament.
  2. No ATP is supplied to the muscle.
  3. All endoplasmic reticulum are removed from muscle cells.

Answer:

  1. Muscle contraction occur.
  2. Muscle contraction does not occur. Attachment and detachment of myosin head to actin molecule requires ATP.
  3. No muscle contraction. Ca2+ necessary for muscular contraction is released from endoplasmic reticulum.

Question 7.
The given statements are about the structure of muscles. Categorise them into a table giving appropriate headings.

  1. These muscles have striation.
  2. These are found inside the wall of hollow internal organs.
  3. These muscles have no striations.
  4. They are voluntary muscles.
  5. They are muscles of the arms and legs.
  6. They are involuntary muscles

Answer:

Skeletal MuscleSmooth muscle
1. These muscles have striation2. These muscles are found inside the wall of hollow internal organs.
4. They are voluntary muscles.3. These muscles have no striation
5. They are muscles of the arms and legs6. They are involuntary muscles

Question 8.
Observe the following diagram.
Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement - 16

  1. Identify A, B, C, D.
  2. What is Sarcomere?
  3. List out the changes takes place during muscle contraction to ABC and D.

Answer:

  1. Identification
    • A – Sarcomere
    • B – Hzone
    • C – Aband
    • D – Iband
  2. A sarcomere is the region between two adjacent Z line. It is formed of a complete A band in the middle and halves of two I bands present on either side.
  3. the changes takes place during muscle contraction to ABC and D:
    • A – Two Z lines of the sarcomere come close together resulting in the shortening of the sarcomere.
    • B – H – Zone completely disappears.
    • C – No change in the length of A band
    • D – I – band get reduced

Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement

Question 9.
Give any example of synovial joint.
Answer:

  1. Ball and socket joint (between humerus and pectoral girdle)
  2. Hinge joint – knee joint

Question 10.
Suppose a person is suffering from calcium deficiency for a prolonged time. How does it affect muscular contraction? (Note: Description necessary)
Answer:
Muscle contraction begins when a nerve impulse reaches at the neuromuscular junction. During this time the sarcoplasmic reticulum releases Ca2+. The Ca2+ binds with the specific site of the troponin of the actin filament.

This causes a conformational change in the troponin molecule and expose the active site on the actin molecule. Myosin filament binds with the active site of actin and there is formation a cross bridge. By the movement of cross bridge contraction occurs.

The contraction is followed by relaxation and it occurs when Ca2+ in pumped back into the sarcoplasmic reticulum. As a result, the troponin molecules became free to mask the active site of the myosin head.

Question 11.
Identify the following diseases noting the symptoms given below

  1. The liver is affected, skin and eyes turn yellow due to the deposit of bile pigments.
  2. Difficulty in breathing causing wheezing due to the inflammation of bronchi and bronchioles.
  3. Chronic disorder in which alveolar walls are damaged mainly due to cigarette smoking.
  4. Auto immune disorder affecting neuro muscular junction leading to fatigue, weakening and paralysis of skeletal muscles.
  5. Inflammation of joints due to deposition of uric and crystals.
  6. Age related disorder due to decreased level of estrogen, characterised by decreased bone mass and increased chances of fractures.

Answer:

  1. Jaundice
  2. Asthma
  3. Emphysema
  4. Myasthenia gravis
  5. Gout
  6. Osteoporosis

Question 12.
Identify the following tissues and write down their location.

  1. Loose connective tissue which stores fat.
  2. Dense connective tissue which connect bone to bone.
  3. A contractile tissue which possess intercalatory disc.

Answer:

  1. Adipose tissue – beneath the skin
  2. Ligament – attach one bone to another
  3. Cardiac muscle – heart

Plus One Locomotion and Movement NCERT Questions and Answers

Question 1.
Define sliding filament theory of muscle contraction.
Answer:
Sliding Filament Theory of Muscle Contraction Mechanism of muscle contraction is best explained by the sliding filament theory which states that contraction of a muscle fibre takes place by the sliding of the thin filaments over the thick filaments.

Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement

Question 2.
Describe the important steps in muscle contraction.
Answer:
Muscle contraction is initiated by a signal sent by the central nervous system (CNS) via a motor neuron. A motor neuron alongwith the muscle fibres connected to it constitute a motor unit. The junction between a motor neuron and the sarcolemma of the muscle fibre is called the neuromuscular junction or motor end plate.

1. A neural signal reaching this junction releases a neurotransmitter (Acetylcholine) which generates an action potential in the sarcolemma. This spreads through the muscle fibre and causes the release of calcium ions into the sarcoplasm.

2. Increase in Ca2+ level leads to the binding of calcium with a subunit of troponin on actin filaments and thereby remove the masking of active sites for myosin.

3. Utilising the energy from ATP hydrolysis, the myosin head now binds to the exposed active sites on actin to form a cross bridge. This pulls the attached actin filaments towards the centre of ‘A’ band.

4. The 7’ line attached to these actions are also pulled inwards thereby causing a shortening of the sarcomere, i.e., contraction.

5. It is clear from the above steps, that during shortening of the muscle, i.e., contraction, the T bands get reduced, whereas the A’ bands retain the length.

6. The myosin, releasing the ADP and PI goes back to its relaxed state. A new ATP binds and the cross-bridge is broken. The ATP is again hydroysed by the myosin head and the cycle of cross bridge formation and breakage is repeated causing further sliding.

7. The process continues till the Ca2+ ions are pumped back to the sarcoplasmic cisternae resulting in the masking of actin filaments. This causes the return of 7 lines back to their original position, i.e., relaxation.

Question 3.
Write true or false. If false change the statement so that it is true.

  1. Actin present in thin filament
  2. H-zone of striated muscle fibre represents both thick and thin filaments
  3. The human skeleton has 206 bones
  4. There are 11 pairs of ribs in man.
  5. Stenum is present on the ventral side of the body.

Answer:

  1. True
  2. False, H-zone represents thick filaments
  3. True
  4. False, There are 12 pairs of ribs in man
  5. True

Question 4.
Match column I with column II

Column IColumn II
(a) Smooth muscle(i) Myogobin
(b) Tropomyosin(ii) Thin filament
(c) Red muscle(iii) Sutures
(d) Skull(iv) Involuntary

Answer:
(a) – (iv)
(b) – (ii)
(c) – (i)
(d) – (iii)

Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement

Question 5.
Name the type of joint between the following

  1. Atlas/axis
  2. Carpal/metacarpal of thumb
  3. Between phalanges
  4. Femur/acetabulum
  5. Between carnival bones
  6. Between public bones in the pelvic girdle

Answer:

  1. Pivot joint
  2. Saddle joint
  3. Gliding joint
  4. Ball and socket joint
  5. Fibrous joint
  6. Cartilaginous

Question 6.
Fill in the blank spaces:

  1. All mamnals (expect a few) have __________ cervical vertebra.
  2. The number of phalanges in each limb of human is __________
  3. Thin filament of myofibril contains 2 ‘F’ actins and two other proteins namely __________ and _________
  4. In a muscle fibre Ca2+ is stored in __________
  5. _______and ________ pairs of ribs are called floating ribs.
  6. The human cranium is made of _________ bones.

Answer:

  1. 7
  2. 14
  3. tropomyosin, troponin
  4. sarcoplasmic reticulum
  5. 11th, 12th
  6. 8

Plus One Locomotion and Movement Multiple Choice Questions and Answers

Question 1.
The muscle band that remains unchanged during contraction and relaxation of the skeletal muscle is
(a) I
(b) H
(c) A
(d) A-line
Answer:
(d) A-line

Question 2.
Intervertebral disc consists of a shock absorber connective tissue known as
(a) hyaline cartilage
(b) elastic cartilage
(c) fibrocartilage
(d) reticulo cartilage
Answer:
(c) fibrocartilage

Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement

Question 3.
Common among all mammals is
(a) ventral nerve cord
(b) seven cervical vertebrae
(c) all are carnivores
(d) all are producers
Answer:
(b) seven cervical vertebrae

Question 4.
The lactic acid generated during muscle contraction is converted to glycogen in
(a) muscles
(b) kidney
(c) pancreas
(d) liver
Answer:
(d) liver

Question 5.
Folding and unfolding of actin and myosin leads to amoeboid movement. This is hypothesised by
(a) Allen
(b) Goldacre and Lasch
(c) Berthold
(d) Jennigs
Answer:
(b) Goldacre and Lasch

Question 6.
Muscle fatigue is due to
(a) lactic acid
(b) citric acid
(c) Na
(d) K
Answer:
(a) lactic acid

Question 7.
Which of the following is not syncytial?
(a) Cardiac muscle
(b) Skeletal muscle
(c) Smooth muscle
(d) Interstitial muscle
Answer:
(c) Smooth muscle

Question 8.
Humerus fits into glenoid cavity is example of
(a) ball and socket joint
(b) pivot joint
(c) peg and socket joint
(d) condyloid join
Answer:
(a) ball and socket joint

Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement

Question 9.
Human vertebral column is formed by
(a) 21 vertebrae
(b) 30 vertebrae
(c) 26 vertebrae
(d) 33 vertebrae
Answer:
(d) 33 vertebrae

Question 10.
When body part moves towards the median axis is called
(a) abductor
(b) adductor
(c) supinator
(d) pronator
Answer:
(b) adductor

Question 11.
The thin filaments of a muscle fibre are made up of
(a) actin, troponin, tropomyosin
(b) actin, troponin
(c) niyosin, troponin
(d) actin, tropomyosin
Answer:
(a) actin, troponin, tropomyosin

Question 12.
Which of the following pairs is correctly matched?
(a) Cartilaginous joint Skull bones
(b) Hinge joint – Between vertebrae
(c) Fibrous joint – Between phalanges
(d) Gliding joint – Between zygapophyses of the successive vertebrae
Answer:
(d) Gliding joint – Between zygapophyses of the successive vertebrae

Question 13.
Nucleus purposes is found in
(a) brain
(b) nucleus
(c) intervertebral disc
(d) liver
Answer:
(c) intervertebral disc

Question 14.
Total number of bones found in right upper limb is
(a) 25
(b) 26
(c) 30
(d) 60
Answer:
(c) 30

Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement

Question 15.
In a vertebrate, which germ layer forms the skeletor muscles?
(a) Ectoderm
(b)Endoderm
(c) Mesoderm
(d) Both (a) and (c)
Answer:
(c) Mesoderm

Question 16.
Where did an epidemic bone softening disease itai- itai occurred first in?
(a) South Korea
(b) Japan
(c) China
(d) Burma
Answer:
(b) Japan

Question 17.
Muscle pump is
(a) beating of heart
(b) squeezing effect of muscles upon veins running through them
(c) peristaltic wave that travel along the alimentar canal
(d) None of the above
Answer:
(b) squeezing effect of muscles upon veins running through them

Question 18.
Slow muscle fibres are found in
(a) eye
(b) leg
(c) stomach
(d) heart
Answer:
(b) leg

Question 19.
The gliding joints are important for gliding movements. One example of such a joint is between the
(a) zygapophyses of adjacent vertebrae
(b) humerous and glenoid cavity
(c) occipital condyle and odontoid process
(d) femur and tibio fibula
Answer:
(a) zygapophyses of adjacent vertebrae

Question 20.
Aqueduct of Sylvius (iter) connects
(a) 1st and 2nd ventricles
(b) 3rd and 4th ventricles
(c) 2nd and 3rd ventricles
(d) 4th and 1st ventricles
Answer:
(b) 3rd and 4th ventricles

Plus One Zoology Chapter Wise Questions and Answers Chapter 9 Locomotion and Movement

Question 21.
Volkmann’s canals occur in
(a) internal ear
(b) liver
(c) cartilage
(d) bone
Answer:
(d) bone

Question 22.
For muscle contraction, in myofibrils the formation, of a protein is essential, such protein discovered by
(a) Jean Hanson
(b) Con and Con
(c) lbert Szent Gyorgyi
(d) Hugh Huxley
Answer:
(c) lbert Szent Gyorgyi

Plus One Botany Chapter Wise Questions and Answers Chapter 11 Plant Growth and Development

Students can Download Chapter 11 Plant Growth and Development Questions and Answers, Plus One Botany Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus One Botany Chapter Wise Questions and Answers Chapter 11 Plant Growth and Development

Plus One Botany Plant Growth and Development One Mark Questions and Answers

Question 1.
Ethylene is used for
(a) Retarding ripening of tomatoes
(b) Hastening of ripening of fruits
(c) Slowing down ripening of apples
(d) Both b and c
Answer:
(b) Hastening of ripening of fruits.

Question 2.
Coconut milk contains
(a) ABA
(b) Auxin
(c) Cytokinin
(d) Gibberellin
Answer:
(c) Cytokinin

Question 3.
The affect of apical dominance can be overcome by which of the following hormone:
(a) IAA
(b) Ethylene
(c) Gibberellin
(d) Cytokinin
Answer:
(d) Cytokinin.

Plus One Botany Chapter Wise Questions and Answers Chapter 11 Plant Growth and Development

Question 4.
Growth can be measured in various ways. Which of these can be used as parameters to measure growth
(a) Increase in cell number
(b) Increase in cell size
(c) Increase in length and weight
(d) All the above
Answer:
(d) All the above

Question 5.
ABA acts antagonistic to
(a) Ethylene
(b) Cytokinin
(c) Gibberlic acid
(d) IAA
Answer:
(c) Gibberlic acid.

Question 6.
Name the growth regulator which was isolated from the endosperm of maize.
Answer:
Cytokinin(Zeatin).

Question 7.
What can induce bolting cabbage plant?
Answer:
Gibberellins

Question 8.
Certain plants flower only when they are exposed to low temperature for a few weeks . What do you call for this requirement?
Answer:
Vernalisation.

Question 9.
Some volatile substance from the ripened oranges that hastens the ripening of stored unripened bananas. Name the hormone which induces ripening.
Answer:
Ethylene

Question 10.
In a wheat field, some broad-leaved weeds were found by a farmer. Which phytohormone can be used to eradicate them.
Answer:
2, 4 – D (Dichlorophenoxy acetic acid).

Plus One Botany Chapter Wise Questions and Answers Chapter 11 Plant Growth and Development

Question 11.
Name the phytohormone that can cause the development of seedless fruits.
Answer:
Gibberellic acid.

Plus One Botany Plant Growth and Development Two Mark Questions and Answers

Question 1.
Match the following. (Column I with Column II)

Column IColumn II
a) Cytokinini) Promote senescence
b) IAAii) Delay senescence
c) ABAiii) Initiate rooting
d) Ethyleneiv) lateral shoot growth
v) Plant growth inhibitor

Answer:

Column IColumn II
a) CytokininDelay senescence
b) IAAInitiate rooting
c) ABAPlant growth inhibitor
d) EthylenePromote senescence

Question 2.
Match the following.

AB
1. EthyleneZeatin
2. Abscisic acidOvercome apical dominance
3. CytokininsFruit ripening
4. GibberellinsStomatal closure
Herbicide

Answer:

AB
1. EthyleneFruit ripening
2. Abscisic acidStomatal closure
3. CytokininsZeatin
4. GibberellinsOvercome apical dominance

Question 3.
In plants, the adverse environmental conditions such as severe drought can be overcome by the production of a hormone.

  1. Name the hormone.
  2. Write the role of the hormone.

Answer:

  1. ABA
  2. Stress hormone – Closure of stomata.

Question 4.
Auxin is a growth-promoting phytohormone. Write any two functions of auxin,
Answer:

  1. Production of Seedless fruit
  2. Used as Weedicide

Plus One Botany Chapter Wise Questions and Answers Chapter 11 Plant Growth and Development

Question 5.
Growth pattern of a plant is displayed in graph. Observe the figure.
Plus One Botany Chapter Wise Questions and Answers Chapter 11 Plant Growth and Development img1

  1. What kind of growth form is this?
  2. Why does the graph show a decline to a near-constant level?

Answer:

  1. Sigmoid curve
  2. Stationary phase

Question 6.
Some important functions of a Phytohormone are given below:
Production of Parthenocarpic Fruits Eradication of weeds

  1. Identify the hormone.
  2. Mention any other two important functions of this hormone.

Answer:

  1. Auxin
  2. Root initiation, Apical dominance

Question 7.
In Plants, adverse environmental conditions such as severe drought can be overcome by the use of a hormone.

  1. Name the hormone
  2. Write the role of this hormone.

Answer:

  1. ABA (Abscisic Acid)
  2. It causes the Closure of stomata. This reduces the rate of transpiration.

Question 8.

  1. A farmer is going to plant Tapioca cuttings. Which hormone you will suggest for early rooting of tapioca cuttings?
  2. A defoliated short day plant is kept under short-day conditions for 15 days and another short-day plant with single leaf is kept under short-day conditions for 1 day. Which plant will possibly flower first and give reasons?

Answer:

  1. Auxin
  2. A short day plant with single leaf will flower first Leaf recives the light stimulus and produce Florigen, It is then transported to the vegetative apex. As a result vegetative apex is converted into reproductive apex and bears floral leaves.

Question 9.
Synthetic auxins are of much importance in agriculture. Name any two synthetic auxins.
Answer:

  • 2, 4 – D (2,4 – Dichlorophenoxy acetic acid)
  • 2, 4, 5 – T. ( 2,4,5 – Trichlorophenoxy acetic acid).

Question 10.
Which hormone will add, if you are asked to help a farmer to

  1. Quickly ripens fruit
  2. Bolt a resette plant
  3. Initiate rooting in a twig
  4. Induce stomatal closure in leaves.

Answer:

  1. a) Ethylene
  2. Gibberelline
  3. Auxin
  4. Abscisic acid (ABA)

Question 11.
A gardener finds some broad-leaved dicot weeds growing in his lawns. What can be done to get rid of the weeds efficiently?
Answer:
The dicotyledonous plant grow by their apical shoot meristems while grasses (which make lawns) possess intercalary meristem. Certain auxins, such as synthetic 2, 4-Dichlorophenoxyacetic acid (2,4- D) when applied in excess can damage the shoot apical meristems but they do not cause any damage to the intercalary meristems. Thus, when 2, 4-D is sprayed on lawns, only the dicots get killed and the lawns become free of weeds.

Question 12.
What is the mechanism underlying the phenomenon by which the terminal/apical bud suppresses the growth of lateral buds? Suggest measures to overcome this phenomenon.
Answer:
The phenomenon by which the terminal apical bud suppresses the growth of lateral buds is referred to as apical dominance. This js because of the hormone auxin synthesised in the apical bud that inhibits lateral bud development.

This can be overcome by removing the apical bud (decapitating) and young leaves which will increase branching. It may also be possible to overcome this phenomenon by application of cytokinin and antiauxins like ethylene chlorohydrin, DCA (trichloroanisole), etc.

Plus One Botany Chapter Wise Questions and Answers Chapter 11 Plant Growth and Development

Question 13.
Ethylene is otherwise known as a fruiting hormone. So this is widely used in agricultural fields. Which compound is used as a source of ethylene and what are its merits?
Answer:
Ehephone-Merits:

  1. It promotes female flowers in cucumbers.
  2. It hastens fruit ripening in tomatoes and apples.
  3. It promotes abscission of flowers and fruits.

Question 14.
Plant hormones are organic compounds influencing growth and development. One of the hormones was first isolated from human urine and other is a gaseous hormone.
a) Name the two hormones.
b) Write any two functions of these hormones.
Answer:
Plus One Botany Chapter Wise Questions and Answers Chapter 11 Plant Growth and Development img2

Question 15.
How will you induce lateral branching in a plant which normally does not produce them ? Give reason in support of your answer.
Answer:
When apical bud is removed, it promotes lateral branching. It is due to the removal the auxin from the tip and growth of apical bud is inhibited.

Question 16.
Describe how auxins are related with the bending of shoot towards the source of light.
Answer:
When unilateral light is given, auxin from the illuminated side shifted towards the shaded side and more growth occurs on that side. This causes the bending of the shoot.

Question 17.
How does abscisic acid act antagonistically to auxins and gibberellins?
Answer:
Gibberellin causes the breaking of bud dormancy but auxin check the fruit fall and leaf fall. These two physiological functions are inhibited by ABA.

Question 18.
What is meant by abscission? Name the phytohormone involved in it.
Answer:
It is the shedding of leaves, flowers, fruits, and bark. Abscisic acid(ABA)

Question 19.
What does a stationary phase of sigmoid growth curve indicate?
Answer:
It is otherwise called a declining phase because growth slows down.

Plus One Botany Chapter Wise Questions and Answers Chapter 11 Plant Growth and Development

Question 20.
What is the full form of I BA? Also, mention its one use in agriculture.
Answer:
Indole – 3 -butyric acid
It initiates root formation on stem cuttings by activating root primordia

Question 21
Fill in the places with appropriate word/words.

  1. A phase of growth which is the maximum and fastest is _____.
  2. Apical dominance as expressed in dicotyledonous plants is due to the presence of more _____ in the apical bud than in the lateral ones.
  3. In addition to auxin, _____ must be supplied to culture medium to obtain a good callus in plant tissue culture.
  4. ______ of vegetative plants are the sites of photoperiodic perception.

Answer:

  1. exponential/log phase of an S – curve.
  2. auxin/IAA
  3. cytokinin/Kinetin/6 BAP/Zeatin/etc.
  4. leaves.

Question 22.
Why is abscisic acid called stress hormone?
Answer:
Abscisic acid is also called stress hormone because the synthesis of abscisic acid is stimulated by drought, waterlogging and other adverse environmental conditions. It causes the closure of stomata.

Question 23.
Root and shoot elongation takes place at constant rate in one type of growth but in others zygote develops into embryo. What are the types of growth in both?
Answer:
In the former growth is arithematic type while in the latter initial geometric then arithematic type.

Question 24.
Give the term for the process given below.

  1. formation of vascular cambium and cork cambium
  2. Formation of secondary tissues in dicot stem

Answer:

  1. Dedifferentiation
  2. Redifferentiation

Question 25.

  1. Name the phenomenon of the influence of day length on the flowering of long day and short-day plants
  2. Which is the part of plant shows response to such phenomenon

Answer:

  1. Photoperiodism
  2. Leaf.

Plus One Botany Chapter Wise Questions and Answers Chapter 11 Plant Growth and Development

Question 26.
Identify the true or false statements from the following.

  1. ABA is known as Anti- gibberellin
  2. Auxin promotes flowering in pineapple
  3. Low-temperature treatment not promote flowering in varieties of wheat
  4. Bakane disease associated with ethylene

Answer:

  1. True
  2. True
  3. False
  4. False

Plus One Botany Plant Growth and Development Three Mark Questions and Answers

Question 1.
Identify the hormone have important role from the statement given below

  1. In Tea plantation and hedge making
  2. Increase stem length of sugar cane
  3. ripening of fruits

Answer:

  1. Auxin
  2. Gibberellin
  3. Ethylene

Question 2.
Which one of the plant growth regulators would you use if you are asked to:

  1. Induce rooting in a twig
  2. Quickly ripen a fruit
  3. Delay leaf senescence
  4. ‘bolt’ a rosette plant
  5. Induce immediate stomatal closure in leaves.

Answer:

  1. auxin
  2. ethylene
  3. cytokinin
  4. gibberellins
  5. ABA

Question 3.
The S-shaped growth curve is shown in the diagram. Label ‘a’ to ‘c’ and also write short notes on ‘a’ and ‘b’.
Plus One Botany Chapter Wise Questions and Answers Chapter 11 Plant Growth and Development img3
Answer:

  • a – lag phase (slow phase)
  • b – exponential phase (rapid phase)
  • c – stationary phase (stagnant phase)

Question 4.
Plant growth substances (PGS) have innumerable practical Applications. Name the PGS you should use to

  1. Increase yield of sugar cane
  2. Promote lateral shoot growth
  3. Inhibit seed germination

Answer:

  1. GA3/gibberellinfgibberellic acid
  2. Cytokinin zeatin/kinetin
  3. ABA/Abscissic acid.

Plus One Botany Plant Growth and Development NCERT Mark Questions and Answers

Question 1.
Why is abscisic acid also known as stress hormone?
Answer:
Abscisic acid increases the tolerance of plants to various kinds of stresses. Therefore, it is also called the stress hormone.

Question 2.
‘Both growth and differentiation in higher plants are open’. Comment.
Answer:
Theoretically, growth and differentiation in higher plants are open. This means that there is no limit to the ‘extent a plant part can grow. But it is more correct to say that development and differentiation is open in higher plants. Once a cell loses its capacity to divide then it differentiates Differentiation is the process by which a particular plant starts doing the job it is meant to do.

For example, the job of a leaf is to make food for plant. Sometimes environment or a particular phase of growth can dictate a particular part to behave differently. This is the phase when redifferentiation occurs and the plant part takes on a new role. Thus, it can be said that development and differentiation are open to change under the given environmental conditions and demands of those conditions.

Plus One Botany Chapter Wise Questions and Answers Chapter 11 Plant Growth and Development

Question 3.
Which one of the plant growth regulators would you use if you are asked to:

  1. induce rooting in a twig
  2. quickly ripen a fruit
  3. delay leaf senescence
  4. induce growth in axillary buds
  5. ‘bolt’ a rosette plant
  6. induce immediate stomatal closure in leaves

Answer:

  1. Auxins
  2. Ethylene
  3. Cytokinins
  4. Auxins
  5. Gibberellins
  6. Abscisic Acid

Question 4.
Would a defoliated plant respond to the photoperiodic cycle? Why?
Answer:
The hormone responsible for photoperiodism is found in leaves. So a defoliated plant will be devoid of such hormones and will not respond to the photoperiodic cycle.

Question 5.
What would be expected to happen if:

  1. GA is applied to rice seedlings
  2. dividing cells stop differentiating
  3. a rotten fruit gets mixed with unripe fruits
  4. you forget to add cytokinin to the culture medium

Answer:
1. GAj increases the length of axis. This property will help increase the length of axis so that yield of rice can be increased.

2. When dividing cells stop differentiating then it is the maturity stage of that part of the plant. Further growth of that particular region will be stopped.

3. The ethylene present in rotten fruit will hasten the ripening process of unripe fruit and may lead to premature ripening.

4. There will be lesser cell division and culture will not grow as per the target.

Plus One Botany Chapter Wise Questions and Answers Chapter 11 Plant Growth and Development

Question 6.
Why is not anyone parameter good enough to demonstrate growth throughout the life of a flowering plant?
Answer:
Like most of the organisms plants also go through various phases of growth. When a seed is germinating then the parameters of growth will be different compared to growth of a mature plant. Moreover, plants can be of huge variety, right from perennial plants to plants living hundreds of years. Because of sheer diversity in size and life span any single parameter to define and measure the growth of a plant can never be sufficient.

Plus One Botany Plant Growth and Development Multiple Choice Questions and Answers

Question 1.
The ripening of fruits can be fastened by treatment with
(a) gibberellins
(b) cytokinins
(c) ethylene gas
(d) auxin
Answer:
(c) ethylene gas

Question 2.
Which of the following is the effect of a plant hormone, which is synthesized more in the absence of light?
(a) Inhibits the development of seedless fruits
(b) Responsible for closing of stomata
(c) Induces the dormancy of seeds
(d) Length of intemodes increases
Answer:
(d) Length of intemodes increases

Question 3.
Leaf abscission is caused by
(a) ABA
(b) cytokinin
(c) auxin
(d) gibberellin
Answer:
(a) ABA

Question 4.
A hormone delaying senescence is
(a) auxin
(b) cytokinins
(c) ethylene
(d) gibberellin
Answer:
(b) cytokinins

Question 5.
Which of the following induces bolting in rosette plants?
(a) Gibberellins
(b) Cytokinin
(c) Auxins
(d) Ethylene
Answer:
(a) Gibberellins

Question 6.
Sprouting of potato under storage condition can be prevented by
(a) auxin
(b) gibberellin
(c) ethylene
(d) cytokinin
Answer:
(a) auxin

Question 7.
The plant hormone produced by Rhizobium for nodulation is
(a) IBA
(b) NAA
(c) 2, 4-D
(d) IAA
Answer:
(d) IAA

Plus One Botany Chapter Wise Questions and Answers Chapter 11 Plant Growth and Development

Question 8.
Cell elongation in internodal regions of the green plants takes place due to
(a) indole acetic acid
(b) cytokinins
(c) gibberellins
(d) ethylene
Answer:
(c) gibberellins

Question 9.
How does pruning help in making the hedge dense?
(a) It induces the differentiation of new shoots from the rootstock
(b) It frees axillary buds from apical dominance
(c) it promotes adventitious root growth
(d) it promotes the growth of apical buds
Answer:
(b) It frees axillary buds from apical dominance

Question 10.
Name of a gaseous plant hormone is
(a) IAA
(b) gibberellin
(c) ethylene
(d) abscisic acid
Answer:
(c) ethylene

Question 11.
The maximum growth rate occurs in
(a) stationary phase
(b) senescent phase
(c) lag phase
(d) exponential phase
Answer:
(d) exponential phase

Question 12.
Growth promoting hormone is
(a) gibberellins
(b) ABA
(c) auxins
(d) both a and c
Answer:
(d) both a and c

Question 13.
The discovery of gibberellins is related with one of the following
(a) blast disease of rice
(b) foolish seedling disese
(c) bakane disease of rice
(d) early blight disease of potato
Answer:
(c) bakane disease of rice

Question 14.
Which of the following movements in plants is due the increased concentration of auxin?
(a) Movement of shoot towards the source of light
(b) Nyctinasty
(c) Movement of sunflower towards sun
(d) root differentiation
Answer:
(a) Movement of shoot towards the source of light

Question 15.
The problem of necrosis and gradual senescence, while performing tissue culture can be overcome by
(a) spraying auxins
(b) spraying cytokinins
(c) suspension culture
(d) subculture
Answer:
(b) spraying cytokinins

Question 16.
The following statements are given about plant growth hormones
I. Cytokinins especially help in delaying senescence
II. Auxins are involved in regulating apical dominance
III. Ethylene is especially useful in enhancing seed germination
IV. Gibberellins are responsible for immature falling of leaves
Which of the above statements are correct
(a) I and II only
(b) I and III only
(c) II and III only
(d) II, III, and IV only
Answer:
(a) I and II only

Question 17.
Which plant hormone promotes seed dormancy, bud dormancy and causes stomatal closure?
(a) auxin
(b) Abscisic acid
(c) GA
(d) CH2=CH2
Answer:
(b) Abscisic acid

Question 18.
Abscisic acid is primarily synthesized in
(a) lysosomes
(b) Golgi complex
(c) chloroplast
(d) ribosomes
Answer:
(c) chloroplast

Plus One Botany Chapter Wise Questions and Answers Chapter 11 Plant Growth and Development

Question 19.
Auxin in plant means for
(a) cell elongation
(b) fruit ripening
(c) cell division
(d) cell differentiation
Answer:
(c) cell division

Question 20.
The hormone present in the liquid endosperm of coconut is
(a) cytokinin
(b) gibberellin
(c) ethylene
(d) auxin
Answer:
(a) cytokinin

Plus One Botany Chapter Wise Questions and Answers Chapter 10 Respiration in Plants

Students can Download Chapter 10 Respiration in Plants Questions and Answers, Plus One Botany Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus One Botany Chapter Wise Questions and Answers Chapter 10 Respiration in Plants

Plus One Botany Respiration in Plants One Mark Questions and Answers

Question 1.
Pyruvic acid, the key product of glycolysis can have many metabolic fates. Under the aerobic condition, it forms
(a) Lactic acid
(b) CO2+H2O
(c) Acetyl CoA + CO2
(d) Ethanol + CO2
Answer:
(b) CO2+ H2O

Question 2.
Electron Transport System (ETS) is located in mitochondrial
(a) Outer membrane
(b) Inter membrane space
(c) Inner membrane
(d) Matrix
Answer:
(c) Inner membrane

Plus One Botany Chapter Wise Questions and Answers Chapter 10 Respiration in Plants

Question 3.
Choose the correct statement:
(a) During the conversion of succinyl Co-A to succinic acid, a molecule of ATP is synthesized.
(b) Oxygen is vital in respiration for removal of hydrogen.
(c) Pyruvate is formed in the mitochondrial matrix.
(d) There is complete breakdown of glucose in fermentation.
Answer:
(a) During the conversion of succinyl Co-A to succinic acid, a molecule of ATP is synthesized.

Question 4.
Mitochondria are called powerhouses of the cell. Which of the following observations support this statement?
(a) Mitochondria have a double membrane
(b) The enzymes of the Krebs cycle and the cytochromes are found in mitochondria.
(c) Mitochondria synthesise ATP
(d) Mitochondria are found in almost all plants and animal cells.
Answer:
(c) Mitochondria synthesise ATP.

Question 5.
The end product of oxidative phosphorylation is
(a) NADH
(b) Oxygen
(c) ADP
(d) ATP+H2O
Answer:
(d) ATP+H2O

Question 6.
Fo-F1 particles participate in the synthesis of
Answer:
ATP molecules

Question 7.
The product of aerobic glycolysis in skeletal muscle and anaerobic fermentation in yeast are respectively ……… and …….
Answer:
Lactic acid and Ethyl alcohol.

Question 8.
How many NADH2 molecules are produced from molecule of acetyl co A in TCA cycle?
Answer:
Three.

Plus One Botany Chapter Wise Questions and Answers Chapter 10 Respiration in Plants

Question 9.
Name the final acceptor of electrons in ETC.
Answer:
Oxygen.

Question 10.
Give the number of ATP molecules produced from 1 glucose molecule in aerobic respiration.
Answer:
38.

Question 11.
The process of oxidation in microorganisms are referred as
(a) Cellular respiration
(b) Aerobic respiration
(c) Fermentation
Answer:
(c) Fermentation

Question 12.
Calculate the RQ of tripalmitin with the help of equation given below.
2(C5H98O6) + 145O2 -»102 CO2 + 98H2O + energy
Answer:
\(\mathrm{RQ}=\frac{102 \mathrm{CO}_{2}}{145 \mathrm{O}_{2}}=0.7\)

Question 13.
In the respiration process both in aerobic and anaerobic, the first phase of reactions is the same. Write the name of reaction Answer:
Glycolysis.

Question 14.
A germinating seed released 120 CO2 by utilizing 180 O2 stage. Calculate the RQ. Name the respiratory substrate.
Answer:
The substrate is fat.

Question 15.
Respiratory pathway is an amphibhotic pathway. Comment.
Answer:
It involves both anabolic & catabolic pathways.

Question 16.
Name the unit of oxidative phosphorylation.
Answer:
Oxysomes or Elementary particles.

Question 17.
How many calories are produced by aerobic oxidation of 1 gm mole of glucose?
Answer:
686000 calories.

Plus One Botany Chapter Wise Questions and Answers Chapter 10 Respiration in Plants

Question 18.
We commonly call ATP as the energy currency of the cell. Can you think of some other energy carriers present in a cell? Name any two.
Answer:
NADH2 and FADH2.

Question 19.
RQ value changes for wheat and castor oil seeds. Justify.
Answer:
R Q = 1 carbohydrate (wheat)
R Q < 1 fat(castor)

Plus One Botany Respiration in Plants Two Mark Questions and Answers

Question 1.
Pyruvate, which is formed by the glycolytic catabolism of carbohydrates undergoes oxidative phosphorylation.

  1. Where does it occur?
  2. Who first elucidate this?

Answer:

  1. Cristae of mitochondria
  2. Hans Kreb

Question 2.
Glycolysis is the first stage in the break down of glucose in all organisms. Do you agree with this statement? Where does it take place? Give the net. gain of ATP molecules in glycolysis?
Answer:
Yes, Cytoplasm, 8ATPs

Question 3
Observe the relation of the first pair and fill up the blanks

  1. Symport: Molecules cross the membrance in the same direction
    …………..: Molecules move in the opposite direction
  2. RQ = 1 : Carbohydrate
    RQ > 1: ………..

Answer:

  1. Antiport
  2. Organic acids

Question 4.
‘‘Photorespiration is called a wasteful process.” Comment on it.
Answer:
ATP is utilized. There is no production of ATP during this process. It does not produce any beneficial product.

Question 5
Respirometer is an apparatus used to measure R.Q.

  1. What is R.Q?
  2. R.Q. of glucose is equal to one. Give the reason.

Answer:

  1. Amount of CO2 released during respiration / Amount of O2 absorbed during respiration,
  2. Amount of CO2 released= Amount O2 absorbed

Question 6.
Plant physiologists observed a relationship between respiration and salt absorption. Is absorption of salt increased due to respiration? Explain.
Answer:
Yes. As a result of respiration, the energy is released in the form of ATP. This energy is used for the active absorption of salt.

Plus One Botany Chapter Wise Questions and Answers Chapter 10 Respiration in Plants

Question 7.
Glycolysis is the first step of both aerobic and anaerobic respiration. During this process glucose undergoes partial oxidation.

  1. What is the final product of glycolysis?
  2. What is the total yield of ATP molecules during glycolysis?

Answer:

  1. Pyruvic acid
  2. 8 ATP

Question 8.
The RQ value of a respiratory substrate is 1. Find the type of substrate and comment on RQ.
Answer:
Carbohydrate RQ:
It is the ratio of volume of CO2 evolved to the volume of O2 consumed. The value of RQ is different for respiratory substrates eg fat(0.7), protein (0.9)etc.

Question 9.
Oxidative phosphorylation is an important process in cellular respiration. Explain it?
Answer:
Oxidative phosphorylation takes place in the inner membrane of mitochondria. In the F1 head the inorganic phosphate combined with ADP to form ATP. It occurs due to the entry of proton pair from intermembrane space to the matrix through proton channel.

Question 10.
Why does anaerobic respiration produce less energy than aerobic respiration?
Answer:

  1. The regeneration of NAD fails to produce ATP as the electrons are not shifted to oxygen
  2. The end product of anaerobic respiration can be further oxidized to release energy.

Plus One Botany Chapter Wise Questions and Answers Chapter 10 Respiration in Plants

Question 11.
What is fermentation Name any two organic compounds produced in this process?
Answer:
Anaerobic respiration also called fermentation involves the production of energy from food nutrients in the absence of oxygen.

Question 12.
“Respiration is an amphibolic pathway”. Justify the statement with reason.
Answer:
Respiration involves both the catabolic and anabolic process. Fatty acid undergoes breackdown to form acetyl. co-A.
Acetyl Co-A is again used for the synthesis of Fat. In respiration cycle the two process are involved.i.e catabolism and anabolism. So it is called a Amphibolic Pathway”.

Question 13.
ADP is converted to ATP as a result of phosphorylation which takes place in photosynthesis and respiration during electron transport system. What is the difference between these two?
Answer:
photophosphorylation, inorganic phosphate combines with ADP to form ATP This process takes place in chloroplast in the presence light. But in oxidative phosphorylation inorganic phosphate combines with ADP to form ATP. It takes place in Mitochondria.

Question 14.
When does anaerobic respiration occur in man and yeast?
Answer:
In man during streneous exercise O2, inedequecy is experience in Skeletal tissues. In such a case the respiration is anaerobic. In yeast, during fermentation the respiration takes place is anaerobic.

Question 15.
Mention the important series of events of aerobic respiration that occur in the matrix of the mitochondrion as well as ones that take place in inner membrane of the mitochondrion.
Answer:
Citric acid cycle,Terminal oxidation and electron transport system.

Question 16.
Name the end products aerobic and anaerobic glycolysis. List the two ways by which molecules of ATP are produced in glycolysis during aerobic respiration in a cell.
Answer:
2 Pyruvic acid, 2ATP and 2 NADH2 Two ways of ATP formation in glycolysis are:

  1. During dephosphorylation of 1, 3, DPGA called substrate-level phosphorylation
  2. Reduced NADH2 releases ATP molecules in electron transport chain

Question 17.
Define RQ What is its significance?
Answer:
It is the ratio of volume of CO2 released to volume of O2 taken during respiration. Value of R Q gives the indication of nature of substrate respired by a particular tissue.

Question 18.
Different substrates get oxidized during respiration. How does Respiratory Quotient (RQ) indicate which type of substrate, i.e., carbohydrate, fat or protein is getting oxidized?
R Q=A/B
What do A and B stand for?
What type of substrates have R.Q. of 1, < 1 or> 1?
Answer:
A-Volume CO2 evolved B- Volume O2 consumed RQ=1 (Carbohydrate) RQ=< 1 Fat/protein RQ=> 1 (Organic acids).

Plus One Botany Chapter Wise Questions and Answers Chapter 10 Respiration in Plants

Question 19.
Which of the following will release more energy on oxidation? Arrange them in ascending order,
a. Moffat
b. 1 gm of protein
c. 1 gm of glucose
d. 0.5 g of protein + 0.5g glucose
Answer:
c → d → b → a

Question 20.
What is the importance of F0-F, particle in ATP production during aerobic respiration?
Answer:
F0 is an integral membrane protein complex and has a proton channel while F1 has enzyme ATP synthase to form ATP. One molecule of NADH produces 3 ATP molecules in ETS.

Question 21.
What is Respiratory Quotient? What is its value for fats?
Answer:
It is the ratio of the volume of CO2 evolved to the volume of O2consumed.
\(So, RQ =\frac{\text { Volume of } \mathrm{CO}_{2} \text { Evolved }}{\text { Volume of } \mathrm{O}_{2} \text { consumed }}\)

Question 22.
Give two example for anaerobic respiration occurs in organisms and products.
Answer:
Alcoholic fermentation and Lactic acid fermentation Ethyl alcohol and lactic acid.

Question 23.
The complete oxidation of pyruvic acid that occurs in mitochondria undergoes both oxidation and decarboxylation.

  1. Which is the first step of oxidative-decarboxylation occurs?
  2. Where does substrate-level phosphorylation occurs?

Answer:

  1. Conversion of pyruvic acid to acetyl CoA
  2. Conversion of succinyl CoA to Succinic acid.

Plus One Botany Chapter Wise Questions and Answers Chapter 10 Respiration in Plants

Question 24.
Give the name of intermediate compounds having carbon atoms 6,5,4 and 2 of mitochondrial oxidation.
Answer:

  • 6C – Citric acid, 5C- Alpha keto glutaric acid,
  • 4C- Succinic acid/succinyl coAl malic acid/ OAA,
  • 2C- Acetyl CoA.

Plus One Botany Respiration in Plants Three Mark Questions and Answers

Question 1.
Incomplete oxidation of food material in the absence of oxygen is anaerobic respiration.
(a) Give an example of a microorganism which carry out anaerobic respiration.
(b) Where does anaerobic respiration take place in human beings?
(c) What is the net gain of ATP during this process? Spa) Yeast
(b) Muscle cells
Answer:
(c) 2 ATP

Question 2.
ETS operated in the inner mitochondrial membrane, it involves terminal oxidation and oxidative phosphorylation

  1. Who discovered chemiosmotic hypothesis in mitochondria
  2. Name the elementary particle promotes ATP synthesis
  3. What is chemiosmotic hypothesis

Answer:

  1. Peter Mitchel
  2. Fg-F, particle
  3. Proton gradient leads to ATP production

Question 3.
Yeast cells can respire in the absence of O2

  1. Name the process
  2. Write the end product of the process
  3. Write the net production of ATP during process.

Answer:

  1. Anaerobic respiration
  2. CO2 + C2H5OH (Carbon dioxide and ethyl alcohol)
  3. 2 ATPs

Question 4.
Respiration is a breakdown process it involves various steps

  1. Where does the common step of aerobic and anaerobic process occurs
  2. Find out the number of carbon atoms of a compound as end product of the above reaction
  3. Name the 6 C intermedite compound splits and forms another 3 C intermediates.

Answer:

  1. Cytoplasm
  2. 3 Carbon (Pyruvic acid)
  3. Fructose 1, 6 biphosphate

Question 5.
ATP and NADPH2 molecules synthesised in light reaction of photosynthesis are used for the synthesis of glucose in dark reaction.

  1. Who proposed the dark reaction?
  2. List out three phases in dark reaction.
  3. Location of dark reaction in the chloroplast?
  4. Expense of ATP and NADPH2 for the synthesis of one molecule of glucose in dark reaction?

Answer:

  1. Melvin Calvin
    • Carboxylation
    • Reduction
    • Regeneration of RUBP
  2. Stroma
  3. 12 NADPH2 and 18 ATP molecules.

Question 6.
In the last step of aerobic respiration oxidation of reduced co-enzymes produced in glycolysis and Krebs cycle occur.

  1. What are the important reduced coenzymes?
  2. Why is this process called as terminal oxidation?
  3. Where does terminal oxidation takes place?

Answer:

  1. FADH2, NADH2.
  2. It is the formation of metabolic water at the end of electron transport chain by combining protons, electrons, and O2.
  3. The inner mitochondrial membrane of mitochondria.

Plus One Botany Chapter Wise Questions and Answers Chapter 10 Respiration in Plants

Question 7.
An athlete felt muscular pain after a race.

  1. Explain this in terms of anaerobic respiration?
  2. Name any two microorganisms in which anaerobic respiration occurs
  3. Glycolysis is common for both aerobic and anaerobic respiration. In glycolysis, there is a net gain of 8 ATP. But in anaerobic repiration the net gain is only 2 ATP. Give reason.

Answer:
1. Athletes felt muscular pain is due to the inadequecy of 02. In the absence of 02 partial oxidation takes place. It involves the formation of pyruvic acid followed by lactic acid.

2. Yeast & Lactobacillus.

3. After the formation of pyruvic acid, 2 NADPH molecules are utilised for the formation of Lactic acid. So the net gain of ATP in Anaerobic respiration is 2 ATPs.

Question 8.
Mitochondria is called the “Power House’’ of the cell.

  1. Is the statement correct?
  2. Write down reasons.

Answer:

  1. Yes
    • Synthesis of ATP takes place in mitochondria,
    • Inner membrance acts as electron transport chain.
    • Presence of ATP – synthetase enzyme.
    • Enzymes for aerobic respiration in mitochondrial matrix

Question 9.
Analyse the diagram and answer the questions.
Plus One Botany Chapter Wise Questions and Answers Chapter 10 Respiration in Plants img1

  1. What does the diagram represent?
  2. Write the role of F0-F., unit in the process?
  3. What is oxidative phosphorylation?
  4. Where does it take place?

Answer:

  1. ATP synthesis
  2. Flow of proton
  3. In the presenceC2, oxidation takes place and ADP combines with inorganic phosphate to form ATP
  4. Crystal

Question 10.
Respirometer is an apparatus used to measure R.Q.

  1. What is R.Q?
  2. R.Q. of glucose is equal to one. Give reason.
  3. Name the respiratory substrate for which R.Q. is more than one?

Answer:

  1. Amount of C02 released/Amount of O2 absorbed
  2. In glucose amount of CO2 released = amount of O2 absorbed
  3. Organic acids

Question 11.
RuBP carboxylase, PEPcase, Pyruvate dehydrogenase, ATPase, cytochrome oxidase, Hexokinase, Lactate dehydrogenase. Select/choose enzymes from the list above which are involved in

  1. Photosynthesis
  2. Respiration
  3. Both in photosynthesis and respiration

Answer:

  1. RuBP carboxylase, PEPcase, ATPase,
  2. Pyruvate dehydrogenase, ATPase, cytochrome oxidase, Hexokinase, Lactate dehydrogenase
  3. ATPase

Question 12.
The following statements are about an important event of cellular respiration. Read them carefully and answer the questions.
It is the last step of cellular respiration It synthesise ATP Water is the end product It accepts oxygen.

  1. Name the process
  2. Where does it occur?
  3. Name any 4 components of ‘it’

Answer:

  1. Electron transport system/oxidative phosphorylation
  2. Inner mitochondrial membrane
  3. NADP, FAD, Ubiquinone (Co-Q)
    Cyt. b, cyt – a, cty a3 etc.

Question 13.
Respiration is viewed most simply as the oxidative production of ATR Justify the statement.
Answer:
The energy-releasing process by oxidation of organic food materials in the living cell is respiration. During this process, the energy contained in the food is released and is trapped in the ATP molecules. NADH+H+ and FADH2 formed during various steps of respiration are oxidised, and protons (H+) and electrons (e ) are released.

These electrons are transported to the oxygen through a series of electron carries in the electron transport system (ETS), and their energy is stored in ATP molecules. So, respiration is process of oxidative production of ATP. 45% of energy released during the oxidation of 1 glucose molecule is stored in 38 ATP molecules.

Question 14.
The second phase of aerobic respiration takes place within mitochondria. This phase is called TCA cycle. The different steps of this reaction were found out by a British Biochemist who was awarded Nobel Prize in 1953.

  1. Identify the scientist and name the first product of the reaction.
  2. Write the first step of this reaction, why it is called TCA cycle?
  3. From where Acetyl Co. A comes into mitochondria?
  4. In which step FADHJs formed.

Answer:

  1. Hans Krebs, citric acid(Tricarboxylic acid)
  2. oaa+acetylcoA citric acid .since it has three -COOH group
  3. Cytoplasm
  4. 5th step (Succinic acid to malic acid)

Plus One Botany Chapter Wise Questions and Answers Chapter 10 Respiration in Plants

Question 15.
Glycolysis is common and first phase of both aerobic and anaerobic respiration.

  1. Who discovered this process?
  2. What are the substrate and end products in this reaction?

Answer:

  1. Embeden, Mayerhof & Parnas
  2. Glucose – Substrate.
    The end product – Pyruvic acid.

Plus One Botany Respiration in Plants NCERT Mark Questions and Answers

Question 1.
Discuss “The respiratory pathway is an amphibolic pathway.”
Answer:
Glucose is the favoured substrate for respiration. All carbohydrates are usually first converted into glucose before they are used for respiration. Other substrates can also be respired but then they do not enter the respiratory pathway at the first step.

Since respiration involves breakdown as well as synthesis of substrates, the respiratory process involves both catabolism. That is why respiratory pathway is considered to be an amphibolic pathway rather than as a catabolic one.

Question 2.
What is the significance of step wise release of energy in respiration?
Answer:
If energy will be released at one go then most of it will be most in the form of heat. Cells should be in a position to utilize all the energy to synthesize something. To facilitate proper usage of energy, it is released in a stepwise manner during respiration.

Question 3.
Define RQ. What is its value for fats?
Answer:
The ratio of the volume of CO2 evolved to the volume of O2 consumed in respiration is called the respiratory quotient (RQ) or respiratory ratio.
Plus One Botany Chapter Wise Questions and Answers Chapter 10 Respiration in Plants img2

Question 4.
Distinguish between the following:

  1. Aerobic respiration
  2. Glycolysis and Fermentation
  3. Glycolysis and Citric acid Cycle

Answer:
1. There is incomplete oxidation of glucose during anaerobic respiration, while there is complete oxi¬dation during aerobic respiration.

2. The pyruvic acid formed during glycolysis is first converted to Acetyl coenzyme A, which undergoes citric acid cycle to produce critic acid. At the end of citric acid cycle NADH+H+ is released.

3. In both (b) and (c) glycolysis is the first step cellular respiration. The product of glycolysis is further utilized by either fermentation or critic acid cycle.

Question 5.
What are the assumptions made during the calculation of net gain of ATP?
Answer:
It is possible to make calculations of the net gain of ATP for every glucose molecule oxidised, but in reality this can remain only a theoretical exercise. These calculations can be made only on certain assumptions that:

1. There is a acquential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle and ETS pathway following one after another.

2. The NADH synthesised in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.

3. None of the intermediates in the pathway are utilised to synthesise any other compound.

4. Only glucose is being respired – no other alternative substrates are entering in the pathway at any of the intermediary stages.

Plus One Botany Respiration in Plants Multiple Choice Questions and Answers

Question 1.
Common phase between aerobic and anaerobic modes of respiration is
(a) Krebs cycle
(b) EMP/glycolysis
(c) oxidative phosphorylation
(d) PPP
Answer:
(b) EMP/glycolysis

Question 2.
Complete oxidation of one gram mol. of Glucose give rise to
(a) 6860,000 cal
(b) 686,000 cal
(c) 68,600 cal
(d) 6860 cal
Answer:
(b) 686,000 cal

Question 3.
When a molecule of pyruvic acid is subjected to anaerobic oxidation and forms Lactic acid there is
(a) loss of 3 ATP molecules
(b) loss of 6ATP molecules
(c) loss of 2 ATP molecules
(d) loss of 4 ATP molecules
Answer:
(b) loss of 6ATP molecules

Plus One Botany Chapter Wise Questions and Answers Chapter 10 Respiration in Plants

Question 4.
When one glucose molecule is completely oxidised, it changes
(a) 36 ADP mole into 36 ATP molecules
(b) 32 ADP mole into 32 ATP molecules
(c) 38 ADP mole into 38 ATP molecules
(d) 30 ADP mole into 30 ATP molecules
Answer:
(c) 38 ADP mole into 38 ATP molecules

Question 5.
Substrate phosphorylation occurs during
(a) Fumaric acid → malic acid
(b) Oxalosuccinic acid → alpha-ketoglutaric acid
(c) Succinic acid – fumaric acid
(d) a Ketoglutaric acid – Succinic acid
Answer:
(d) a Ketoglutaric acid – Succinic acid

Question 6.
Glycolysis is significant for energy production in
(a) RBC
(b) fungi
(c) plants
(d) none of the above
Answer:
(a) RBC

Question 7.
In aerobic respiration
(a) 2 PGAL are formed in Glycolysis and none in kerbs cycle
(b) 6 PGAL in glycolysis, 3 PGAL in kerbs cycle
(c) PGAL formation does not occur in respiration
(d) 8 PGAL in glycolysis, 3 PGAL in krebs cycle
Answer:
(a) 2 PGAL are formed in Glycolysis and none in kerbs cycle

Question 8.
Two enzymes common to EMP pathway and C3 cycle are
(a) Aldolase & triosephosphate isomerase
(b) Aldolase & enolase
(c) Cytochrome oxidase & enolase
(d) phosphoglyceromutase & triosephosphate isomerase
Answer:
(a) Aldolase & triosephosphate isomerase

Question 9.
The RQ value of fat is 0.7, it indicates
(a) volume of CO2 is greater than O2
(b) volume of O2 is greater than CO2
(c) volume of CO2 is greater than volume of O2
(d) none of the above
Answer:
(b) volume of O2 is greater than CO2

Question 10.
The net gain of ATP in aerobic respiration is
(a) 18
(b) 36
(c) 38
(d) 8
Answer:
(b) 36

Question 11.
Three carboxylic group occurs in
(a) pyruvic acid
(b) citric acid
(c) malic acid
(d) Aspartic acid
Answer:
(b) citric acid

Question 12.
Which is the first 5C carbon compound formed in TCA cycle
(a) succinic acid
(b) oxaloacetic acid
(c) alpha-ketoglutaric acid
(d) fumaric acid
Answer:
(c) alpha-ketoglutaric acid

Question 13.
The concentration of alcohol in fermentation influence the
(a) death of cells
(b) growth of cells
(c) production of succinic acid
(d) production of lactic acid
Answer:
(a) death of cells

Question 14.
Acetyl coA is called
(a) 5C compound
(b) 2C compound
(c) 3C compound
(d) 4C compound
Answer:
(b) 2C compound

Plus One Botany Chapter Wise Questions and Answers Chapter 10 Respiration in Plants

Question 15.
Wine and beer are produced directly by fermentation. Brandy and whisky require both fermentation and distillation because
(a) fermendation is inhibited at an alcohol level of 10-18%
(b) distillation prolongs storage
(c) distillation improves quality
(d) distillation purifies the beverage.
Answer:
(a) fermendation is inhibited at an alcohol level of 10-18%

Question 16.
The process by which ATP is produced in the inner membrane of a mitochondrion. The electron transport system transfers protons from the inner compartment to the outer, as the protons flow back to the inner compartment, the energy of their movement is used to add phosphate to ADP, forming ATP.
(a) Chemiosmosis
(b) Phosphorylation
(c) Glycolysis
(d) Fermentation
Answer:
(a) Chemiosmosis

Question 17.
In which of the following reactions of glycolysis, a molecule of water is removed from the substrate?
(a) Fructose – 6 – phosphate → Fructose -1, 6 – bisphosphate
(b) 3 – phosphate – glyceraldehyde → 1, 3 bisphosphoglyceric acid
(c) PEP → Pyruvic acid
(d) 2 – phosphoglycerate → PEP
Answer:
(d) 2 – phosphoglycerate → PEP

Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination

Students can Download Chapter 8 Excretory Products and their Elimination Questions and Answers, Plus One zoology Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination

Plus One Excretory Products and their Elimination One Mark Questions and Answers

Question 1
Which one of the following is also known as an antidiuretic hormone?
(a) Oxytocin
(b) Vasopressin
(c) Adrenaline
(d) Calcitonin
Answer:
(b) Vasopressin

Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination

Question 2.
A large quantity of one of the following is removed from our body by lungs.
(a) CO2 only
(b) H2O only
(c) CO2 and H2O
(d) ammonia
Answer:
(c) CO2 and H2O

Question 3.
The pH of human urine is approximately
(a) 6.5
(b) 7
(c) 6
(d) 7.5
Answer:
(c) 6

Question 4.
Which of the following pairs is wrong?
(a) Uricotelic ________ Birds
(b) Ureotelic _________ Insects
(c) Ammonotelic ________ Tadpole
(d) Ureotelic _________ Elephant
Answer:
(c) Ammonotelic ________ Tadpole

Question 5.
Which one of the following statements is incorrect?
(a) The medullary zone of kidney is divided into a few conical masses called medullary pyramids projecting into the calyces.
(b) Inside the kidney the cortical region extends in between the medullary pyramids as renal pelvis.
(c) Glomerulus along with Bowman’s capsule is called the renal corpuscle.
(d) Renal corpuscle, proximal convoluted tabule (PCT) and distal convoluted tubule (DCT) of the nephron are situated in the cortical region of kidney.
Answer:
(b) Inside the kidney the cortical region extends in between the medullary pyramids as renal pelvis.

Question 6.
Take the odd one justify?
GFR, JGA, ANF, ADH, TSH
Answer:
TSH: all others are the hormones regulate urine formation.

Question 7.
Identify the ammonotelic animal from the following and give reason. (Man, Crow, Lizard, Tadpole)
Answer:
Tadpole

Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination

Question 8.
While studying the excretory system in man, a student noticed some words.
RAAS and JGA. Can you help him to expand these terms.
Answer:
RAAS – Renin – Angiotensin – Aldosterone System
JGA – Juxta glomerular apparatus

Question 9.
In ureotelic animals, urea is produced in
(a) Kidney
(b) Liver
(c) Flame cells
(d) Malpighian tubules
Answer;
(b) Liver

Question 10.
State whether true or false and correct the false statements if any:

  1. Flame cells are the excretory structures in round worms.
  2. Fat is absorbed in the form of chylomicrons through the intestinal villi.
  3. Blood is absent in cockroach.
  4. Joint between atlas and axis vertebrae is a pivot joint.

Answer:

  1. False – Flame cells are the excretory structures in flat worms.
  2. True
  3. False – Blood is present in cockroach. It is colourless.
  4. True

Plus One Excretory Products and their Elimination Two Mark Questions and Answers

Question 1.
Draw the schematic diagram of the urine formation. (Hint: Show all the three process involved)
Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination - 1

Question 2.
Complete the following sentences.

  1. Reabsorption of water from DCT is facilitated by the hormone __________
  2. Kidney failure can cause a bone disorder called _________
  3. Angiotensin II activates the adrenal cortex to release __________
  4. In cases of Kidney failure, urea can be removed by the process called __________

Answer:

  1. ADH
  2. Gout
  3. Aldosterone
  4. Hemodialysis

Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination

Question 3.
Pick out the correct word from list provided (Myasthenia gravis, Uremia, Angina, diuresis)

  1. Autoimmune disorder affecting neuromuscular junction is __________
  2. Reduce loss of water through urine is _________
  3. A system of acute chest pain appears when no enough oxygen is reaching the heart muscle is _________
  4. Accumulation of urea in blood is __________

Answer:

  1. Myasthenia gravis
  2. Diuresis
  3. Angina
  4. Uremia

Question 4.
Classify the following animals according to their type of excretion.
Frog, Tadpole, birds, man, Turtle, shark, insect, camel
Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination - 2

Question 5.
Tadpole is ammonotelic while the frog is ureotelic. Comment on this statement.
Answer:
Tadpole larvae’s habitat is aquatic whereas the frog lives in the terrestrial habitat. So in the aquatic habitat due to the presence of water, ammonia can be excreted easily.

But in frog the water must be conserved. So ammonia is changed into urea which needs only small quantity of water for excretion.

Question 6.
Observe the diagram and answer the following questions.
Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination - 3

  1. Label the parts (1) and (2).
  2. Name the functional unit of the kidney.

Answer:

  1. The parts are:
    • (1) – Renal cortex
    • (2) – Ureter
  2. Nephron

Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination

Question 7.
Rearrange the following parts in their correct way.
Glomerulus → Capsular space → loop of Henle → PCT → DCT → Collecting duct → Collecting tubule → Ureter → Renal pelvis → Urethra → Urinary bladder.
Answer:
Glomerulus → Capsular space → PCT → Loop of Henle → DCT →Collecting tubule → Collecting duct → Renal pelvis → Ureter → Urinary bladder → Urethra.

Question 8.
Fill in the gaps:

  1. Ascending limb of Henle’s loop is ____(1)____ to water. Whereas the descending limb is _____(2)_____ to it.
  2. Reabsorption of water from distal parts of the tubules is facilitated by hormone __________
  3. Dialysis fluid contain all the constituents as in plasma except ________
  4. A healthy adult human excretes (on an average) _______ gm of urea/day.

Answer:

  1. (1) – Impermeable, (2) -Permeable
  2. ADH
  3. Nitrogenous wastes
  4. 25-30 gm of urea/day

Question 9.
Brain controls the kidney action.

  1. Name the hormones involved in this.
  2. How do they act on kidney?

Answer:

  1. Antidiuretic Hormone (ADH)
  2. The pituitary gland release the hypothalamic hormone ADH and its is transported to the kidney. The presence of ADH makes the tubules permeable to water. Thus the water is conserved or excreted according to the presence of fluid level in the body.

Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination

Question 10.
A patient approaches a doctor having symptoms like excretion of large amount of urine, excessive thirst and dehydration.

  1. Identify the disease.
  2. Which hormone deficiency causes this disease?

Answer:

  1. Diabetes insipidus
  2. ADH

Question 11.
A table showing the average quantity of urine in a person in two different seasons are given.
Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination - 4

  1. Compare it. Do you agree with this table? State the reason for producing different Quantity in different season.
  2. Name the hormone responsible for that.

Answer:

  1. Amount of water in the body is low in summer season and increase its quantity in winter season. In summer season water lose is very high due to various reasons like evaporation. So body eliminate less amount of waterthrough urine to conserve water in the body
  2. ADH or vasopressin

Question 12.
Teacher asks one of the student to name the excretory organs in man. He names three organs beside kidney. Teacher appreciates him for the correct answer.

  1. What was his answer?
  2. Give the excretory role of the above organs

Answer:

  1. Liver, skin, lungs
  2. the excretory role of the above organs
    • Liver – Excretes cholesterol, bile pigments, etc.
    • Skin – Sweat gland- Excretes excess sodium chloride, small amount of urea lactic acid etc. Sebaceous gland excretes lipids such as waxes, steroles, fatty acid etc.
    • Lungs – Excretes CO2 and water

Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination

Question 13.
Observe the diagram
Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination - 5

  1. Copy the diagram and fill the gaps.
  2. Write the concentration of glomerular filtrate (Isotonic, Hypotonic, Hypertonic) in the given regions of the nephron.
    • Proximal convoluted tubule
    • Descending limb
    • Ascending limb
    • Distal convoluted tubule

Answer:
1.
Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination - 6

2. Concentration of glomerular filtrate

  • Proximal convoluted tubule – isotonic
  • Descending limb – hypertonic
  • Ascending limb – hypotonic
  • Distal convoluted tubule – Hypertonic

Question 14.
A patient with renal failure in waiting for a kidney donor for a transplantation operation. The urea level is high and the patient is developing symptoms of uremia. What method you can suggest to keep the patient live till a donor can be found. Explain the principle behind it.
Answer:
Haemodialysis. The haemodialyser has a cellophane tube bathed in a fluid with ionic concentration similarto that of plasma. The process of separation of small solutes from macromolecular colloids is done hereby the process of diffusion. So the blood will be free from urea, uric acid and creatinine.

Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination

Question 15.
About 125ml of glomerular filtrate is produced per minute in our kidney. However 1ml of urine is produced each minute. What happens to the other 124ml of the filtrate.
Answer:
In man about 125ml of glomerular filtrate is formed per minute. But most of the substances are reabsorbed into tubular network of capillaries. This process is known as selective reabsorption. High threshold substances reabsorbed by active transport and low threshold substances reabsorbed by diffusion.

Question 16.
Consumption of alcohol tends to frequent urination.

  1. Name the hormone that control it.
  2. Draw a flow chart showing the action of ethanol present in alcohol and its consequences on urinary system.

Answer:
1. Antidiuretic hormone (ADH)

2.
Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination - 7

Question 17.
Prepare a flow chart showing hormonal control by Juxta glomerular apparatus (JGA) in regulation of kidney function.
Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination - 8

Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination

Question 18.
Flow of blood in vasa recta is known as counter current system. Give reason.
Answer:
The vasa recta is thin walled capillaries lying parallel to the loop of Henle. The blood entering the descending limb of each vasarecta come in close contact with the outgoing blood in ascending limb. This system is called counter current system.

The two limbs of the loop of Henle form another counter current system. These two systems are concentrating the urine in mammalian kidney by diffusion and osmosis.

Question 19.
‘Micturition is an involuntary as well as a voluntary process.’ Justify the statement.
Answer:
The elimination of urine from the urinary bladder is called micturition. The gradual filling of the urinary bladder causes stretching. When the bladder is filled to its limit, the stretch receptors send impulses to the brain to excrete the urine.

The initiation and inhibition of micturition is voluntary in adult but involuntary in children. Due to lack of nervous control the micturition in some humans become involuntary causing bed wetting.

Question 20.
In alcoholic drinkers the urine is dilute. Why?
Answer:
The process of reabsorption of water by the distal parts ofthe kidney tubules become low efficient. This occurs due to the deficiency of ADH. lt leads to diabetes insipedus.

Question 21.
Distinguish between diabetis mellitus and diabetis insipedus.
Answer:
Insulin deficiency causes large amounts of blood sugar to be lost in the urine. It leads to diabetis melitus while the deficiency of ADH leads to diabetes insipedus i.e dilute urine disease.

Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination

Question 22.
The failure of removing toxic substances from the blood ultimately leads kidney failure.

  1. How is it temporarily solved.
  2. Name the machine is used.

Answer:

  1. Hemodialysis is an artificial process of removing toxic substances from the blood
  2. Artificial kidney or Hemodialysis machine.

Plus One Excretory Products and their Elimination Three Mark Questions and Answers

Question 1.
State the differences between the following.

  1. Tubular reabsorption and Tubular secretion.
  2. ADH and RAAS.

Answer:
1. Tubular reabsorption:
The process of selective reabsorption of most of the useful substances from the glomerular filtrate in the renal tubule.

Tubular secretion:
It is the process of active secretion of certain substances such as Hydrogen ions, ammonia etc. from the blood into the lumen of the urinary tubules by the tubular epithelium during urine formation.

2. ADH:
ADH is the hormone from posterior pituitary gland, makes the wall of the distal convoluted tube and collecting tubule permeable to water so that more water is reabsorbed.

RAAS:
Renin – Angiotensin – Aldosterone system increases the blood pressure by constricting arterioles and increases the blood volume by stimulating proximal convoluted tubule.

Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination

Question 2.
Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination - 9

  1. What does the diagram represent?
  2. Name the parts A, B, C, D.
  3. What is the physiological function taking place in part ‘B.

Answer:

  1. Nephron
  2. The parts are
    • A – Bowman’s capsule
    • B – Henle’s loop
    • C – Proximal convoluted tubule
    • D – Collecting duct

3. Henle’s loop is the site of osmoregulation in human kidney. The descending limb of loop of Henle is permeable to water but almost impermeable to electrolytes. The ascending limb is impermeable to water but allows transport of electrolyte actively or passively.

Therefore, as the concentrated filtrate pass upward, it gets diluted due to the passage of electrolytes to the medullary fluid.

Question 3.
Observe the diagram and answer the questions.
Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination - 10

  1. Identify the labelled parts A, B, C, D.
  2. What does the diagram represents?

Answer:

  1. The labelled parts are
    • A – Afferent arteriole
    • B – Efferent arteriole
    • C – Glomerulus
    • D – Bowman’s capsule
  2. Glomerular Filtration

Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination

Question 4.
Observe the diagram given below which represents the counter current system in nephron and vasa recta. This mechanism helps in maintaining concentration gradient in the medullary interstitium. Substantiate.
Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination - 11
Answer:
The flow of Glomerular filtrate in the two limbs of Henle’s loop is in opposite directions and thus forms a counter current. The flow of blood through the two limbs of vasa recta is also in a counter current pattern.

The proximity between the Henle’s loop and vasa recta, as well as the counter current in them help in maintaining an increasing osmolarity towards the inner medullary interstitium i.e, from 300 mOsmolL1 in the cortex to about 1200 mOsmolL1 in the inner medulla.

This gradient is mainly caused by NACI and urea. NaCl is transported by the ascending limb of Henle’s loop which is exchanged with the descending limb of vasa recta.

Similarly, small amounts of urea enter the thin segment of the ascending limb of Henle’s loop which in transported back to the interstitium by the collecting tubule. This counter current mechanism helps to maintain a concentration gradient in the medullary interstitium.

Question 5.
The steps involved in the treatment of a uremic patient is given below.
Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination - 12

  1. Complete the missing steps (a) and (b)
  2. Diagonise the treatment
  3. Name the organ which is under failure

Answer:

  1. The missing steps are
    • (a) – Anticoagulant like heparin
    • (b) – Anti – heparin
  2. Hemodialysis
  3. Kidney

Question 6.

  1. Name the mechanism by which JGA plays a complex regulatory role.
  2. Draw a flow chart showing the mechanism.
  3. Mention the feed back which act as a check on the above mechanism,

Answer:
1. Renin-Angiotensin mechanism

2.
Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination - 13
3. An increase in blood flow to the atria of the heart can cause the release of Atrial Natriuretic factor (ANF). ANF can cause vasodilation and thereby decrease the blood pressure. ANF mechanism act as a check on the renin-Angiotensin mechanism.

Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination

Question 7.
Note the relationship between the first two words and suggest a suitable word in the missing place.

  1. Cockroach – Malpighian Tubule : Flatworm – _________
  2. Man – Kidney : Earthworm – _________
  3. Prawn – Greengland : Amphioxus – __________

Answer:

  1. Flame cells
  2. Nephridia
  3. Protonephridia

Plus One Excretory Products and their Elimination NCERT Questions and Answers

Question 1.
Define Glomerular Filtration Rate(GFR)
Answer:
Glomerular Filtration Rate (GFR). The amount of the filtrate formed by the kidneys per minute is called glomercular filtration rate (GFR). GFR in a healthy individual is approximately 125ml/minute, i.e., 180 litres per day. On an average, 1100 -1200 ml of blood pumped out by each ventricle of the heart in a minute.

Question 2.
Explain the autoregulatory mechanism of GFR.
Answer:
Regulation of GFR. The kidney have built-in mechanisms for the regulation of glomerular filtration rate. One such efficient mechanism is carried out by juxta glomerular apparatus(JGA).

JGA is a special sensitive region formed by cellular modifications in the distal convoluted tubule and the afferent arteriole at the location of their contact. A fall in GFR can activate the JG cells to release renin which can stimulate the glomercular blood flow and thereby the GFR back to normal.

Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination

Question 3.
Indicate whether the following statements are true or false:

  1. Micturition is carried out by a reflex
  2. ADH helps in water elimination, making the urine hypotonic.
  3. Protein-free fluid is filtered from blood plasma into the Bowman’s capsule.
  4. Henle’s loop plays an important role in concentrating the urine.
  5. Glucose is actively reabsorbed in the proximal convoluted tublue.

Answer:

  1. True
  2. False
  3. True
  4. True
  5. True

Question 4.
Give a brief account of the counter mechanism.
Answer:
Counter current Mechanism. The Henle’s loop and vasa recta play a significant role in this. The flow of filtrate in the two limbs of Henle’s loop is in opposite directions and thus forms a countercurrent.

The flow of blood through the two limbs of vasa recta is also in a countercurrent pattern. The proximity between the Henle’s loop and vasa recta, as well as the counter current in them help in maintaining an increasing osmolarity towards the inner medullary in terstitium, i.e., from 300 mOsmolL-1 in the cortex to about 1200 mOsmolL-1 in the inner medulla.

This gradient is mainly caused by NaCI and urea. NaCI is transported by the ascending limb of Henle’s loop which is exchanged with the descending limb of vasa recta. NaCI is returned to the intestitium by the ascending portion of vasa recta.

Question 5.
What is meant by the term osmoregulation?
Answer:
Osmoregulation is the regulation of blood volume, body fluid volume and ionic concentration

Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination

Question 6.
Match the items of column I with those of column II.
Column I Column II

Column IColumn II
(a) Ammonotelism(i) Birds
(b) Bowman’s capsule(ii) Water reabsorption
(c) Micturition(iii) Bony fish
(d) Uricotelism(iv) Urinary bladder
(e) ADH(v) Renal tubule

Answer:
(a) – (iii)
(b) – (v)
(c) – (iv)
(d) – (i)
(e) – (ii)

Question 7.
Name the following.

  1. A chordate animal having flame cells as excretory structures.
  2. Cortical portions projecting between the medullary pyramids in the human kidney.
  3. A loop of capillary running parallel to the Henle’s loop.

Answer:

  1. Cephalochordata
  2. Columns of Bertini
  3. Vasa Recta

Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination

Question 8.
Fill in the gaps:

  1. Ascending limb of Henle’s loop is ________ to water whereas the descending limb is __________ to it.
  2. Reabsorption of water from distal parts of the tubules is facilitated by hormone ___________
  3. Dialysis fluid contain all the constituents as in plasma except ___________
  4. A healthy adult human excretes (on an average) _________gm of urea/day.

Answer:

  1. impermeable, permeable
  2. ADH
  3. nitrogenous waste
  4. 25 to 30

Plus One Excretory Products and their Elimination Multiple Choice Questions and Answers

Question 1.
What will happen if the stretch receptors of the urinary bladder wall are totally removed’?
(a) Urine will not collect in the bladder
(b) Micturition will continue
(c) Urine will continue to collect normally in the bladder
(d) There will be no micturition
Answer:
(c) Urine will continue to collect normally in the bladder

Question 2.
Green glands present in some arthropods help in
(a) respiration
(b) excretion
(c) digestion
(d) reproduction
Answer:
(b) excretion

Question 3.
Excretory product of spider is
(a) uric acid
(b) ammonia
(c) guanine
(d) None of these
Answer:
(c) guanine

Question 4.
Which one is the component of ornithine cycle
(a) Ornithine, citrulline and fumaric acid
(b) Ornithine, citrulline and arginine
(c) Ornithine, citrulline and alanine
(d) Amino acids are not used
Answer:
(b) Ornithine, citrulline and arginine

Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination

Question 5.
Urea synthesis occurs in
(a) kidney
(b) liver
(c) brain
(d) muscles
Answer:
(b) liver

Question 6.
Renin is secreted from
(a) juxtaglomerular cells
(b) podocytes
(c) nephridia
(d) stomach
Answer:
(a) juxtaglomerular cells

Question 7.
Loop of Henle is associated with
(a) excretory system
(b) respiratory system
(c) reproductive system
(d) digestive system
Answer:
(a) excretory system

Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination

Question 8.
If one litre of water is introduced in human blood, then
(a) BMR increases
(b) RBC collapses and urine production increases
(c) RBC collapses and urine production decreases
(d) BMR decreases
Answer:
(b) RBC collapses and urine production increases

Question 9.
The urine is
(a) hypotonic to blood and isotonic in medullary fluid
(b) hypotonic to bloodand isotonic to medullary fluid
(c) isotonic to blood and hypotonic to medullary fluid
(d) hypertonic to blood and isotonic to medullary fluid
Answer:
(d) hypertonic to blood and isotonic to medullary fluid

Question 10.
Which substance is in higher concentration in blood than in glomerular filtrate’?
(a) Water
(b) Glucose
(c) Urea
(d) Plasma proteins
Answer:
(d) Plasma proteins

Question 11.
A large quantity of fluid is filtered everyday by the nephrons in the kidneys. Only about 1% of it is excreted as urine. The remaining 99% of the filtrate
(a) gets coiected in the renal pelvis
(b) is lost as sweat
(c) is stored in the urinary bladder
(d) is reabsorbed into the blood
Answer:
(d) is reabsorbed into the blood

Question 12.
The characteristic that is shared by urea, uric acid and ammonia is/are
I. They are nitrogenous wastes.
II. They all need very large amount of water for excretion.
III. They are all equally toxic.
IV. They are produced in the kidneys.
(a) I and III
(b) I and IV
(c) I, III and IV
(d) I only
Answer:
(d) I only

Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination

Question 13.
A bird excretes nitrogenous waste materials in the form of?
(a) uric acid
(b) ammonia
(c) urea
(d) amino acids
Answer:
(a) uric acid

Question 14.
Which of the following is correct with reference to haemodialysis?
(a) Absorbs and resends excess of ions
(b) The dialysis unit has a coiled cellophane tube
(c) Blood is pumped back through a suitable artery after haemodialysis
(d) Anti-heparin is added priorto haemodialysis
Answer:
(b) The dialysis unit has a coiled cellophane tube

Question 15.
Find the incorrect statement regarding mechanism of urine formation in man.
(a) the glomerular filtration rate is about 125 mL/min
(b) the ultra filtration is opposed by the colloidal osmotic pressure of plasma
(c) tubular secretion takes place in the PCT
(d) aldosterone induces greater reabsorption of sodium
Answer:
(c) tubular secretion takes place in the PCT

Question 16.
The excretory material of bony fish is
(a) urea
(b) protein
(c) ammonia
(d) amino acid
Answer:
(c) ammonia

Question 17.
The yellow colour of urine is due to the presence of
(a) urea
(b) uric acid
(c) urochrome
(d) bilirubin
Answer:
(c) urochrome

Question 18.
Malpighian tubules are
(a) excretory organs of insects
(b) excretory organs of frog
(c) respiratory organs of insects
(d) endocrine glands of insects
Answer:
(a) excretory organs of insects

Question 19.
The size of filtration slits of glomerulus
(a) 10wn
(b) 15nm
(c) 20nm
(d) 25nm
Answer:
(d) 25nm

Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination

Question 20.
The conversion of dangerous nitrogenous waste into less toxic excretory matter is carried out in man in the
(a) blood
(b) liver
(c) kidney
(d) skin
Answer:
(b) liver

Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants

Students can Download Chapter 9 Photosynthesis in Higher Plants Questions and Answers, Plus One Botany Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants

Plus One Botany Photosynthesis in Higher Plants One Mark Questions and Answers

Question 1.
Which metal ion is a constituent of chlorophyll?
(a) Iron
(b) Copper
(c) Magnesium
(d) Zinc
Answer:
(c) Magnesium

Question 2.
Which pigment acts directly to convert light energy to chemical energy?
(a) Chlorophyll a
(b) Chlorophyll b
(c) Xanthophyll
(d) Carotenoid
Answer:
(a) Chlorophyll a

Question 3.
Which range of wavelength (in nm) is called photosynthetically active radiation (PAR)?
(a) 100 – 390
(b) 390 – 430
(c) 400 – 700
(d) 760 – 100,00
Answer:
(c) 400-700

Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants

Question 4.
Which light range is most effective in photosynthesis?
(a) Blue
(b) Green
(c) Red
(d) Violet
Answer:
(c) Red

Question 5.
Chemosynthetic bacteria obtain energy from
(a) Sun
(b) Infra red rays
(c) Organic substances
(d) Inorganic chemicals
Answer:
(d) Inorganic chemicals

Question 6.
What is the name given to the process of splitting of water during light reaction?
Answer:
Photolysis

Question 7.
What tissues and cells of a leaf contain the chloroplast
Answer:
Mesophyll

Question 8.
What colour of visible spectrum is least effective in the process of photosynthesis?
Answer:
Green

Question 9.
How many molecules of ATP and how many molecules of NADPH are spent to fix three molecules of CO2 in Calvin cycle?
Answer:
ATP and 6 NADPH

Question 10.
Name the two photosynthetic pigments belonging to carotenoides.
Answer:
Carotene and xanthophylls

Question 11
The most abundant enzyme in the world plays a dual role in rice, tomato etc. Identify the enzyme.
Answer:
RUBISCO

Question 12
Which products formed during the light reaction of photosynthesis are used to drive the dark reaction?
Answer:
ATP and NADPH2

Question 13
By looking at which internal structure of a plant can you fill whether a plant is C3 or C4. Explain.
Answer:
C4 – Plants shows kranz anatomy -chloroplast dimorphism

Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants

Question 14
Why does the rate of photosynthesis decrease at higher temperatures?
Answer:
Enzymes present in stroma undergoes denaturation in the presence of high temperatures.

Question 15
Name the only natural process by which oxygen is liberated for the use of respiration.
Answer:
By photolysis of water during photosynthesis

Plus One Botany Photosynthesis in Higher Plants Two Mark Questions and Answers

Question 1.
If we study vertical sections of leaves of C3 plant and C4 plant, there are some anatomical differences.

  1. What are the anatomical peculiarities of a C4 plant?
  2. Explain C4 pathway by listing the main steps.

Answer:
1. C4 plants show kranz anatomy – Chloroplast dimorphism

2.
Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants img1

Question 2.
Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants img2
The above graph indicates absorption spectrum related to chlorophyll pigments. On the basis of. this graph answer the following.

  1. The peak a & b represents for what?
  2. Which pigments shows a & b peak of absorption?

Answer:

  1. Absorption maxima
  2. Chlorophyll

Question 3.
C4 cycle is called so because of the presence of a C4 acid in this cycle.

  1. Name that C4 acid.
  2. How does it form?

Answer:

  1. Oxalic acetic acid.
  2. It is formed by combining CO2 with PEP in the presence of PEP case enzyme.

Question 4
The response of plant towards periods of day/night is termed as photo peroidism’.

  1. Mention three plant groups based on photoperiodism.
  2. Would a defoliated plant respond to the photoperiodic cycle? Why?

Answer:

  1. Long day plants, short-day plants, day-neutral plants.
  2. No. Photo periodic receptors are located in leaves.

Question 5
Melvin Calvin was awarded the nobel prize in 1961 for the discovery of one of the most important biological process in nature.

  1. Name the process.
  2. Name site for photosynthesis.

Answer:

  1. C3 cycle
  2. Chloroplast

Question 6
Analysis the given and write down the photosynthetic response of the C3 and C4 plant in terms of CO2 concentration.
Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants img3
Answer:
High light intensity induces the opening of stoma. When stoma opens CO2 uptake increases. At high CO2 concentration, the rate of photosynthesis becomes high under high light in both C3 and C4 plants.

Question 7.
C4 plants have a special type of anatomy in leaves to avoid photorespiration.

  1. What is the special anatomy called?
  2. Name any two plants having this anatomy.

Answer:

  1. Kranz
  2. Wheat, Sugarcane

Question 8.
Fill up the blanks:

  1. Oxygen liberated during photosynthesis comes from …………
  2. Inhibition of lateral buds growth by terminal bud is known as …………

Answer:

  1. Water
  2. Apical dominance

Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants

Question 9.
Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants img4
Based on the above diagram answer the following questions.
i) Write the steps of
a) Carboxylation
b) Reduction
c) regeneration
ii) Name the major enzyme involved in C3 pathway.
Answer:
Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants img5

Question 10.
Give the diagram of a chloroplast and label the following parts.

  • thylakoids
  • grana
  • stroma
  • fret

Answer:
Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants img6

Question 11.
Certain distinguishing characters of C3 and C4 plants are given below. Arrange them in the correct table provided.

  1. The primary CO2 acceptor is a PEP, RuBP;
  2. The first stable product is PGA, OAA;
  3. Kranz anatomy is seen not seen;
  4. Carboxylase enzyme in RuBisCo, PEP carboxylase, and RubisCo.

Answer:
Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants img7

Question 12.
Given below is the schematic diagram of calvin cycle, which completes in three stages.
Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants img8

  1. Where does this cycle takes place?
  2. State the three stages of calvin cycle?
  3. Calculate the energy budget for each stage to synthesise one molecule of glucose during the process.

Answer:

  1. Stroma
    • Carboxylation
    • Carbon reduction
    • Regeneration
    • 18 ATP
    • 12NADPH2

Question 13.
ADP is converted into ATP as a result of phosphorylation which takes place in photosynthesis and respiration during electron transport system. What is the difference between these two.
Answer:
Photophosphorylation in the case of photosynthesis and oxidative phosphorylation in respiration.

Question 14.
Rubisco is an enzyme that catalyse two different processes in plants.

  1. What are those reactions?
  2. Name the full form of Rubisco.
  3. What are the site of reaction that takesplace in high oxygen concentration?

Answer:

  1. Carboxylation in C3 cycle. Oxygenation in photorespiration
  2. Ribulose biphosphate carboxylase.
  3. Chloroplast, peroxisomes & mitochondria

Question 15.
Some plants are called C3 and some others are called C4.

  1. Why they are called so?
  2. Explain with examples.

Answer:

  1. C3 plants – First stable product is a 3 carbon compound – PGA
  2. C4 plants – First stable product is a 4 carbon compound – OAA

Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants

Question 16.
Do reactions of photosynthesis called, as ‘Dark Reaction’ need light? Explain.
Answer:
photosynthesis is a two-step process which involves light and dark reaction. CO2 fixation takes place in dark reaction.

Question 17.
Where is NADP reductase enzyme located in the chloroplast? What is the role of this enzyme in proton gradient development?
Answer:

  • The NADP reductase enzyme is located on the stroma side of the membrane.
  • Oxidation of NADPH2 is involved in the accumulation of more proton concentration inside the thylakoid than outside.

Question 18.
What conditions enable RUBISCO to function as an oxygenase? Explain.
Answer:
Under high O2 concentration and low CO2 concentration, RUBISCO to function as an oxygenase and doing photorespiration.

Question 19.
Chlorophyll ‘a’ is the primary pigment for light reaction. What are accessory pigments? What is their role in photosynthesis?
Answer:
Clorophyll b,c ,d and carotenoides. These pigments harvest light energy and hand over to chlorophylla.

Question 20.
2H2O → 2H+ + O2 + 4e –
Based on above equation, answer the following questions:

  1. Where does this reaction take place in plants?
  2. What is the significance of this reaction?

Answer:

  1. Lumen of the thylakoids.
  2. O2 is evolved during this reaction, moreover, electrons are made available to PS-II continuously.

Question 21.
What is the basis for designating C3 and C4 pathways of photosynthesis?
Answer:
The first stable product formed in some plants are 3 carbon compound but in others 4 carbon compound, therefore such plants designated as C3 and C4 respectively.

Question 22.
Chlorophyll b and carotenoids are known as accessory pigments. Give reason?
Answer:
Chlorophyll b and carotenoids harvest light energy and hand over to chlorophyll a (reaction centre)

Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants

Question 23.
C4 plants show high rate of photosynthesis than C3 plants. Are you agree with the statement? Give reason?
Answer:
Yes. In C4 plants photorespiration does not occur. It is a wastage process that occur in C3 plants only, because RUBISCO present in C3 plants can act as carboxylase and oxygenase.

Question 24.
How many ATP molecules are required for the synthesis of one molecule of glucose in

  1. C3 pathway
  2. C4 pathway

Answer:

  1. C3 pathway – 18 ATPs
  2. C4 pathway – 30 ATPs

Question 25.
Cyanobacteria and some other photosynthetic bacteria don’t have chloroplasts. How do they conduct photosynthesis?
Answer:
Cyanobacteria and other photosynthetic bacteria have thylakoids suspended freely in the cytoplasm (i.e., they are not enclosed in membrane), and they have bacteriochlorophyll.

Question 26.
What would happen to the rate of photosynthesis in C3 plants if CO2 concentration level almost doubles from the present level in the atmosphere?
Answer:
The rate of photosynthesis in C3 plants increases with increase in CO2 concentration

Question 27.
Does photorespiration pose threat to plants yet it occurs in angiosperms why?
Answer:
Any angiospermic plants bear only one type of chloroplast and CO2 acceptor is RUBP in these cases. Thus there is no other possibility except to undergo C3 pathway which may ultimately lead to photorespiration.

Question 28.
Secondary CO2 fixation is inevitable for C4 plants. Do you agree? Give justification of your answer.
Answer:
Yes. This type of CO2 fixation takes place in bundle sheath chloroplast and carbohydrate is formed.

Question 29.
Carboxylation of C3 plants are interrupted in changed environmental conditions. Why?
Answer:
Normally carboxylation takes place in C3 plants but in high O2 and low CO2 concentration, the energy wasteful process (photorespiration ) occur.

Question 30.
RuBisco is an enzyme that acts both as a carboxylase and oxygenase. Why do you think RuBisco carries out more carboxylationin in C4 plants.
Answer:
1. The C4 plants have dimorophic chloroplasts. In mesophyll cells chloroplast are granal but in bundle sheath cells the chloroplasts are agranal.

2. So in the these plants when CO2 is fixed in bundle sheath cells, the intracellular CO2 concentration is increased that is why RuBisco enzyme carries out more carboxylation in C4 plants.

Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants

Question 31.
Succulents are known to keep their stomata closed during the day to check transpiration. How do they meet their photosynthetic CO2 requirements?
Answer:
1. Succulent (water-storing) plants such as cacti, euphorbias fix CO2 into organic compound using. PEP carboxylase at night, when the stomata are open.

2. PEP + CO2 → OAA → Malic acid
The organic compound (malic acid) accumulates throughout the night and is decarboxylated during the day to produce CO2.

Question 32
In C4 plants RUBISCO is present but photorespiration does not occur. Why?
Answer:
Because intracellular CO2 concentration is high. Hence RUBISCO do not show oxygenase activity.

Question 33
The statements are given below. Identify true and false

  1. Carbohydrate (glucose) is formed in mesophyll chloroplast of C4 pathway
  2. Calvin cycle involves 3 steps
  3. The absorption peak of PS I is 680nm
  4. Calvin cycle takes place in Temperate plants

Answer:

  1. False
  2. True
  3. False
  4. True

Plus One Botany Photosynthesis in Higher Plants Three Mark Questions and Answers

Question 1.
A plant physiologist while studying the light reaction step of photosynthesis using Chlorella, could detect the following facts.

  1. Name the reaction
  2. What is meant by photolysis?
  3. Draw the electron transport schematically.

Answer:

  1. Noncyclic photophosphorylation
  2. Splitting of water

Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants img9

Question 2.
The enzyme catalysing C4 and C3 pathway is different but they are coming under dark reaction.

  1. Name the enzyme shows CO2 fixation in such plants
  2. Where is the second CO2 fixation occurs in C4 plants
  3. Which is the primary CO2 acceptor and stable product of C3 pathway

Answer:

  1. C4 pathway – PEPCO, C3 pathway-RUBISCO
  2. Bundle sheath chloroplast
  3. CO2acceptor-RUBP, Stable product— 3- phosphoglycerate

Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants

Question 3.
CO2 fixation is associated with wasteful process and efficiency is decreased

  1. Name the wasteful process.
  2. Identify the enzyme catalysing the above process
  3. Which is the chief organelle responsible for this?

Answer:

  1. Photorespiration
  2. Oxygenase activity of RUBISCO
  3. peroxisome

Question 4.
Answer the following questions based on carbon fixation by plants.

  1. which is the key enzyme involved in carbon fixation of photosynthesis in C3 plants?.
  2. Rubisco has a dual role. Comment.
  3. Give the name of two cell organells other than chloroplast involved in the process that catalysed by Rubisco.

Answer:

  1. Ribulose biphosphate carboxylase
  2. Carboxylation & oxygenation
  3. Peroxisome & mitochondria

Question 5.
Type of CO2 fixation in tropical plants is different from temperate plants

  1. Identify the plants in which efficient CO2 fixation occurs.
  2. Tropical plants do not show the wasteful process why?
  3. Give two example for such plants

Answer:

  1. C4 plants (Tropical plants)
  2. Because intracellular CO2 concentration is very high
  3. Sugarcane and Maize.

Question 6.
Cyclic and noncyclic photophosphorylation is associated with light reaction of photosynthesis.

  1. Where does photolysis occurs?
  2. Give two difference between cyclic and noncyclic

Answer:

  1. Inside thylakoid
    • Cyclic – It involves PSI (P700)
    • Only ATP is formed
    • Noncyclic – It involves PSI and PSII (P680)
    • ATP and NADPH2 are produced

Question 7.
Do you think that synthesis of glucose occur during light reaction?

  1. If not what is the function of light reaction?
  2. name the CO2 acceptor in C3 cycle and C4 cycle

Answer:

  1. In light reaction, energy-rich molecules like ATP and NADPH2 are produced. These are used in Dark reaction for Glucose synthesis,
    • RUBP – CO2 acceptor in C3 cycle
    • PEP – CO2 acceptor in C4 cycle

Question 8.
Sugar cane has a special leaf anatomy which make the plant photosynthetically more efficient.

  1. Name the leaf anatomy
  2. Write a note on carbon dioxide fixation in sugarcane.

Answer:
1. Kranz anatomy

2. In sugarcane CO2 fixation tapseplace in mesophyll chloroplast. Initially CO2 is accepted by PEP and by using Pepco it is converted into OAA. It is then transported to bundle sheath chloroplast.

After the transport, decarboxylation occurs and forms pyruvic acid which is transported to mesophyll chloroplast and converted into PEP. The CO2 released during decarboxylation enters into calvin cycle for the synthesis of Glucose.

Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants

Question 9.
C4 plants have a special leaf anatomy in which C3 cycle also take place.

  1. Name the primary acceptor of CO2 and first stable product in C4 cycle.
  2. Why is photorespiration absent in C4 plants?
  3. Name the special type of leaf anatomy and explain.

Answer:
1. CO2 acceptor in C4 is PEP and the stable product is OAA.

2. Because in C4 plants, the primary CO2 acceptor is PEP but in C3 plants the primary CO2 acceptor is RUBP. The enzyme RUBISCO present C3 plants have dual function ie, carboxylase and oxygenase activity. Oxygenase activity promotes photorespiration.

3. Kranz anatomy in C4 plants. In this bundle sheath chloroplast are arranged in the form of a ring or wreath.

Question 10.
Explain how during light reaction of photosynthesis, ATP synthesis is a chemiosmotic phenomenon.
Answer:
The conditions helpful in chemiosmotic ATP synthesis are:

1. The splitting of the water molecule takes place on the inner side of the membrane, it causes the accumulation of protons or hydrogen ions within the lumen of the thylakoids.

2. As electrons move through the photosystems, protons are transported across the membrane, and proton is released into the inner side or the lumen side of the membrane.

3. The NADP reductase enzyme is located on the stroma side of the membrane. Along with electrons that come from the accepter of electrons of PS I, protons are necessary for the reduction of NADP+ to NADPH+ H+. These protons are also removed from the stroma.

Question 11.
Chemiosmotic hypothesis explains the mechanism of ATP synthesis during respiration and photosynthesis.

  1. Write the difference in the accumulation of proton during the above process.
  2. What causes the proton gradience across the thylakoid membrane within the chloroplast?

OR

C3 plants complete the Biosynthetic phase through 3 stages and sometimes they are affected by high concentration of O2.

  1. Name the 3 stages of Biosynthetic phase.
  2. Name the assimilatory powers used during this phase and specify the number.
  3. Write the process carried out by C3 plants during high concentration of O2.
  4. How does the high O2 concentration affect RUBISCO?

Answer:
1.

Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants img10
2. Movement of electrones through electron carrier

OR

  1. Carboxylation, carbon reduction and regeneration
  2. ATP & NAD PH2
  3. Photorespiration
  4. RUBISCO – undergoes oxygenase activity

Question 12.
ATP and NADPH2 molecules synthesized in Light reaction of Photosynthesis are used for the synthesis glucose in Dark reaction.
Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants img11

  1. Who proposed the Dark reaction?
  2. List out three phases in Dark reaction.
  3. Location of dark reaction in the chloroplast.
  4. Expense of ATP and NADPH2 for the synthesis of one molecule of Glucose in Dark reaction.

Answer:

  1. Melvin Calvin
  2. Carboxylation, carbon reduction and regeneration
  3. Stroma
  4. 18 ATP and 12 NADPH2

Question 13.
Observe the labeled and write down the following :
Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants img12

  1. Which gas is passed out into the atmosphere at A?
  2. What raw material required for photosynthesis, enters the leaf at B?
  3. Describe the major steps involved at C in the synthesis of sugar.

Answer:

  1. O2
  2. H2O
  3. CO2 is reduced to sugar (glucose or fructose), by the biochemical reaction of photosynthesis (Calvin cycle). It has 3 major steps.

Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants img13

  • Carboxylation – During which CO2 combines with RuBP.
  • Reduction – During which carbohydrate is formed at the expense of the photochemically made ATP and NADPH.
  • Regeneration – During which is the CO2 acceptor (RuBP) is formed again so that the cycle continues.

Plus One Botany Photosynthesis in Higher Plants NCERT Mark Questions and Answers

Question 1.
By looking at a plant externally can you tell whether a plant is C3 or C4? Why and how?
Answer:
Usually plants growing in dry conditions use ( pathways. It cannot be said conclusively if the plant is a C3 or C4 by looking at fleshy leaf structure of C4 plants.

Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants

Question 2.
By looking at which internal structure of a plant can you tell whether a plant is C3 or C4? Explain.
Answer:
The particularly large cells around the vascular bundles of the C4 pathway plants are called bundle sheath cells, and the leaves which have such anatomy are said to have ‘Kranz’ anatomy. ‘Kranz’ means ‘wreath’ and is a reflection of the arrangement of cells.

The bundle sheath cells may form several layers around the vascular bundles; they are characterised by having a large number of chloroplasts, thick walls impervious to gaseous exchange and no intercellular spaces.

Plus One Botany Photosynthesis in Higher Plants Multiple Choice Questions and Answers

Question 1.
The ATP production in photosynthesis is called
(a) Phototropism
(b) phosphorylation
(c) Photo-oxidation
(d) photophosphorylation
Answer:
(d) photophosphorylation

Question 2.
Conditions helpful in photorespiration are
(a) more oxygen and less carbon dioxide
(b) less oxygen and more carbon dioxide
(c) more temperature and less oxygen
(d) more humidity and less temperature
Answer:
(a) more oxygen and less carbon dioxide

Question 3.
What is common between chloroplasts, chromoplasts, and leucoplasts?
(a) Presence of pigments
(b) Possession of thylakoids and grana
(c) Storage of starch, proteins, and lipids
(d) Ability to multiply by a fission-like process
Answer:
(d) Ability to multiply by a fission-like process

Question 4.
A student set up an experiment on photosynthesis as follow: He takes soda water in a glass tumbler and keeps the tumbler exposed sunlight hoping that he has provided necessary ingredient for photosynthesis to proceed (viz, CO2 H2O chlorophyll and light)
What do you think what will happen after, say few hours of exposure of light?
(a) Photosynthesis will take place and glucose will produced
(b) Photosynthesis will take place and starch will be produced which will turn the mixture turbid
(c) Photosynthesis will not take place because CO2 dissolved in soda water escapes into the atmosphere
(d) Photosynthesis will not take place because intact chloroplasts are needed for the process
Answer:
(a) Photosynthesis will take place and glucose will produced

Question 5.
Photochemical reactions in the chloroplasts are directly involved in
(a) Fixation of carbon dioxide
(b) synthesis of glucose and starch
(c) formation of Phosphoglyceric acid
(d) photolysis of water and phosphorylation of ADP
Answer:
(d) photolysis of water and phosphorylation of ADP

Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants

Question 6.
Match the following with correct combination
Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants img14
Answer:
Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants img15

Question 7.
Choose the correct statement.
(a) The C4 – plants do not have RUBISCO
(b) Carboxylation of RuBP leds to the formation of PGA and phosphoglycolate
(c) Carboxylation of phosphoenolpyruvate results in the formation of C4 – acids
(d) Decarboxylation of C4 – acids occurs in the mesophyll cells
Answer:
(c) Carboxylation of phosphoenolpyruvate results in the formation of C4 – acids

Question 8.
In non-cyclic photophosphorylation, there are photolysis of 12 water molecules, How many H+ are formed?
(a) 24 H+
(b) 36 H+
(c) 12 H+
(d) 32 H+
Answer:
(a) 24 H+

Question 9.
In Hatch and Slack pathway.
(a) chloroplasts are of same type
(b) occurs in Kranz anatomy where mesphyll have small Chloroplasts whereas bundle sheath have granal chloroplasts
(c) occurs in Kranz anatomy where mesphyll have small Chloroplasts whereas bundle sheath have larger agranal chloroplasts
(d) Kranz anatomy where mesphyll cells are diffused
Answer:
(c) occurs in Kranz anatomy where mesphyll have small Chloroplasts whereas bundle sheath have larger agranal chloroplasts

Question 10.
To form one molecule of glucose in Calvin cycle
(a) 9ATP and 36 NADPH are required
(b) 6 ATP and 6 NADPH are required
(c) 3 ATP and 2 NADPH are required
(d) 18ATPand12 NADPH are required
Answer:
(d) 18ATPand12 NADPH are required

Question 11.
Photosynthesis is correctly explained by the equation
(a) 6 CO2 + 12H2 O → C6 H12 O6 + 6O2 + 6H2O
(b) 6 CO2 + 6H2 O → C6 H12 O6 + 6O2 + 6H2
(c) 6CO2 + 6 H2O → C6H12 O6+ 6O2
(d) 2CO2+ 12H2O → C6H12 O6+ 2CO2
Answer:
(a) 6 CO2 + 12H2 O C6 H12 O6 + 6O2 + 6H2O

Question 12.
In sugarcane plant, CO2is fixed into malic acid, in which the enzyme that fixes carbon dioxide is
(a) ribulose phosphate kinase
(b) fructose phosphatase
(c) ribulose bisphosphate carboxylase
(d) phosphoenol pyruvate carboxlase
Answer:
(d) phosphoenol pyruvate carboxlase

Question 13.
Which fractions of the visible spectrum of solar radiations are primarily absorbed by carotenoids of higher plants?
(a) Red and violet
(b) Violet and blue
(c) Blue and green
(d) Green and red
Answer:
(b) Violet and blue

Plus One Botany Chapter Wise Questions and Answers Chapter 9 Photosynthesis in Higher Plants

Question 14.
Which factor is not limiting in normal condition for photosynthesis?
(a) Air
(b) Carbon dioxide
(c) Water
(d) Chlorophyll
Answer:
(a) Air

Question 15.
Which of the following is used during discovery of Calvin cycle?
(a) Spirogyra
(b) Volvox
(c) Chlamydomonas
(d) Chlorella
Answer:
(d) Chlorella

Question 16.
Photosynthesis in C4 plants is relatively less limited by atmospheric carbon dioxide levels because
(a) Four carbon acids are the primary initial carbon dioxide fixation products
(b) the primary Fixation of carbon dioxide is mediated via PEP caroxylase
(c) effective pumping of CO2 into bundle sheath cells.
(d) Rubisco in C4_ plants has higher affinity for CO2
Answer:
(b) the primary Fixation of carbon dioxide is mediated via PEP caroxylase

Question 17.
NADP is converted into NADPH2 in
(a) photosystem I
(b) photosystem II
(c) calvin cycle
(d) non cyclic photophosphorylation
Answer:
(d) noncyclic photophosphorylation

Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation

Students can Download Chapter 7 Body Fluids and Circulation Questions and Answers, Plus One zoology Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation

Plus One Body Fluids and Circulation One Mark Questions and Answers

Question 1.
Name the double walled membrane of human heart. (Cuticle, Pleura, Pericardium, Sarcolemma)
Answer:
Pericardium

Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation

Question 2.
Which metal is responsible for blood clotting?
(a) Iron
(b) Calcium
(c) Copper
(d) Magnesium
Answer:
(b) Calcium

Question 3.
In a case sheet, doctor denoted a patients BP as 140/90. What it shows?
Answer:
Hypertension

Question 4.
Mark the pair of substances among the following which is essential for coagulation of blood.
(a) Heparin and calcium ions
(b) Calcium ions and platelet factors
(c) Oxalates and citrates
(d) Platelet factors and heparin
Answer:
(b) Calcium ions and platelet factors

Question 5.
Which one of the following type of cells lack nucleus?
(a) RBC
(b) Neutrophils
(c) Eosinosphils
(d) Monocytes
Answer:
(a) RBC

Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation

Question 6.
ECG depicts the depolarisation and repolarisation processes during the cardiac cycle. In the ECG of a normal healthy individual one of the following waves is not represented.
(a) Depolarisation of atria
(b) Repolarisation of atria
(c) Depolarisation of ventricles
(d) Repolarisation of ventricles
Answer:
(b) Repolarisation of atria

Question 7.
Which one of the following blood cells is involved in antibody production?
(a) B-Lymphocytes
(b) T-Lymphocytes
(c) RBC
(d) Neutrophils
Answer:
(a) B-Lymphocytes

Question 8.
The cells involved in inflammatory reactions are
(a) Basophils
(b) Neutrophils
(c) Eosinophils
(d) Lymphocytes
Answer:
(a) Basophils

Question 9.
Name the incompatibility observed between the Rh-ve blood of a pregnant mother with Rh+ve blood of the foetus.
Answer:
Erythroblastosis foetalis

Question 10.
Tricuspid valve is present in between _______ and _________?
(a) Right auricle and right ventricle
(b) Left auricle and left ventricle
(c) Right auricle and left auricle
(d) Right ventricle and left ventricle
Answer:
(a) Right auricle and right ventricle

Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation

Question 11.
Rhythmic heartbeat is maintained by a highly specialized excitatory and conductive system. The correct sequence of events will be
(a) AV node → Bundle of His → S A node → Purkinje fibers
(b) Purkinje fibers → AV node → S A node → Bundle of His
(c) A V node → S A node → Bundle of His → Purkinje fibers
(d) S A node → A V node → Bundle of His → Purkinje fibers
Answer:
(d) S A node → A V node → Bundle of His → Purkinje fibers

Question 12.
The normal cardiogram of man is shown below.
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 1
In the diagram, QRS wave has largest amplitude, which represents ventricular depolarization. What condition is obtained if QRS wave is enlarged?
Answer:
Myocardial infarction or cardiac arrest.

Plus One Body Fluids and Circulation Two Mark Questions and Answers

Question 1.
The diagrammatic figure given below represents two types of blood circulations.

  1. Identify the A and B
  2. Write hepatic portal system

Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 2

Answer:

  1. A – Pulmonary circulation B – Systemic circulation
  2. Hepatic portal system is a portal system which carries blood from intestine to liver.

Question 2.
Doctor advised to administer Anti Rh antibodies to Seetha immediately after the first delivery. Justify this statement.
Answer:
If the mother’s blood group is Rh- and foetus blood group is Rh+, there is chance for the condition of erythroblastosis foetalis. This can be avoided by administering Anti-Rh antibodies to the mother immediately afterthe delivery of the first child.

Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation

Question 3.
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 3

  1. Identify A & B
  2. Mention the role of A in the myogenic property of Heart.

Answer:
1. A – SA node B – AV node

2. In myogenic heart cardiac impulse in originated from SA node. So the SA node is called pacemaker. The rate of heart beat in determined by the rate of discharge of cardiac impulse from SA node.

Question 4.
Silicosis, T.B., Pneumonia, asbestosis are the diseases affecting respiratory system. Name four diseases affecting circulatory system.
Answer:

  1. Hypertension
  2. Coronary artery disease (CAD)
  3. Angina
  4. Heart failure

Question 5.
The average life span of RBC is 120 days.

  1. Write the places where RBC is produced and destroyed?
  2. Name respiratory pigment in Human RBC.

Answer:

  1. RBS are produced in red bone marrow and destroyed in spleen
  2. Haemoglobin

Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation

Question 6.
The stroke volume of a normal person is 70ml/min. Find out his cardiac output?
(Hint: Cardiac output = stroke volume × No. of heartbeats/min.)
Answer:
HP No.of heartbeats/min = 72
Stroke volume = 70
Cardiac output = 70 × 72 = 5040 ml

Question 7.
Suppose your friend is suffering from obesity, diabetes and hypertension, etc. What are the type of food materials would you suggest?
Answer:
Fried items and bakery items must not be used. Sugar and sugary materials should be avoided. Fiber-rich vegetarian food is recommended.

Question 8.
Study the table. Correct the mistakes to match column A, B & C.
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 4
Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 5

Question 9.
Copy the diagram and fill (a) and (b).
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 6
Answer:
(a) Hypertension

(b) Coronary Artery Disease.

Question 10.
Complete the table.
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 7
Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 8

Question 11.
Observe the diagram.

Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 9

  1. Which step of cardiac cycle is shown here?
  2. Which sound is produced during the step and how is it produced?

Answer:

  1. Ventricular systole
  2. The ‘lub’ sound. This is produced by the closing of AV valve in orderto prevent backward flow of blood to auricle.

Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation

Question 12.
Some of the features of the body fluids are listed below. Analyse the features and categorise them into two groups.

  1. Transparent fluid found in tissue space.
  2. Extra cellular fluid of the blood.
  3. Contains 91% water, 7% proteins and 1% inorganic and organic substances.
  4. Absorbs the digested fats and lipids and transports in the form of chylomicrons.

Answer:

PlasmaLymph
2. Extra cellular fluid of blood1. Transparent fluid found in tissue space
3. Contains 91% water, 7% proteins and 1% inorganic and organic substances4. Absorbs the digested fats and lipids and transports in the form chylomicrons

Question 13.
‘Bundle of His’ is divided into finer branches known as Purkinje Tissue. Impulses are conducted rapidly along these fibres at 5 m/s and spread from these to all parts of the chambers.

  1. What is the function of Purkinje Tissue?
  2. Which are the chambers mentioned in the paragraph?

Answer:

  1. Conduction of impulse from the Bundle of His and rapidly transfers to the wall of the ventricles,
  2. Right and Left ventricles

Question 14.
“All veins carry deoxygenated blood and all arteries carry oxygenated blood.”

  1. Do you agree?
  2. Justify your answer.

Answer:
1. Statement is incorrect.

2. Arteries are the blood vessels which carry blood away from the heart. Veins are the blood vessels which carry blood towards the heart. All arteries carry oxygenated blood except pulmonary artery.

All veins carry deoxygenated blood except pulmonary vein. Pulmonary artery carry deoxygenated blood towards the lungs and pulmonary vein carry oxygenated blood toward heart.

Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation

Question 15.
Using the terms given, prepare a flow chart incorrect order.
Right auricle, aorta, left ventricle, left auricle, superior vena cava, capillaries, lungs, right ventricle, pulmonary vein, pulmonary artery.
Answer:
Right Auricle → Right ventricle → Pulmonary artery → Lungs → Pulmonary vein → Left Auricle → Left ventricle → Aorta → Capillaries → Superior vena cava → Right Auricle.

Question 16.
Complete the table using ‘open’ or ‘close’.
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 10
Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 11

Question 17.
The human blood vascular system is diagrammatically represented below.
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 12

  1. What is the significance of Pulmonary circulation in the cardiac cycle?
  2. Why circulation in human considering as double circulation?

Answer:
1. The to and fro circulation between the lungs and heart is known as Pulmonary circulation. The purpose of pulmonary circulation is oxygenation of blood.

2. In human circulation is double circulation, it involves Pulmonary circulation and systemic circulation. Right half receives and pumps the deoxygenated blood while the left half the oxygenated blood.

Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation

Question 18.
You know when you cut your finger or hurt yourself, your wound does not continue to bleed for long time. Usually blood stops flowing after sometime. State how?
Answer:
When blood oozes out of a cut, it set into gel within a few minutes. This is known as coagulation. The mechanism of blood coagulation is as follows.
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 13

Question 19.
Copy the table and fill in the gap with appropriate words.
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 14
Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 15

Question 20.
Study the diagrammatic representation :
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 16

  1. ‘X’ marked in the figure is a blood vessel. Name the blood vessel.
  2. Give the importance of circulation between these two organs.

Answer:

  1. hepatic portal vein
  2. hepatic portal vein carries nutrients from intestine to liver.

Question 21.
Heparin is a substance which prevents clotting of blood. Histamine is an anti-inflammatory substance.

  1. Mention the leucocytes that secretes the above said substances.
  2. Draw and label its nucleus.

Answer:
1. Basophil
2.
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 17

Plus One Body Fluids and Circulation Three Mark Questions and Answers

Question 1.
Fill up the blanks with appropriate words.

  1. Thrombocytes are produced from _________
  2. Pace maker of the heart is _________
  3. The nervous band that connects the two cerebral hemispheres is _________

Answer:

  1. Megakaryocytes
  2. Sino – atrial node (SAN)
  3. Corpus callosum

Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation

Question 2.
The given figure shows a Systole.
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 18

  1. Identify the Systole in the figure?
  2. How long does it lasts?
  3. What happens to the bicuspid and tricuspid valves during this systole?

Answer:

  1. Ventricular systole
  2. 0.25 seconds
  3. Closed

Question 3.
Draw a Schematic diagram showing the circulation of blood in man, starting from left ventricle of heart and ending in the right auricle.

  1. Name the type of circulation.
  2. Mention the significance of this circulation.

Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 19

  1. Systemic circulation
  2. The systemic circulation provides nutrients, O2 and other essential substances to the tissues and takes CO2 and other harmful substances away for elimination.

Question 4.
A unique vascular connection exists between the digestive tract and liver.

  1. Name the vascular connection.
  2. Name the vascular connection found in frog between kidney and lower parts of body.
  3. What is CAD?

Answer:

  1. Hepatic portal system
  2. Renal portal system
  3. CAD – Coronary Artery Disease

Question 5.
Complete the table using the terms given in bracket. (Lymphocyte, granulocyte, eosinophil, basophil, plasma, leucocyte)
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 20
Answer:

  • a – Plasma
  • b – leucocyte
  • c – granulocyte
  • d – basophil
  • e – Eosinophil
  • f – lymphocyte

Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation

Question 6.
Observe the diagram given below and answer the questions.

  1. Identify the type of circulation.
  2. Prepare a flow chart showing the above circulation.
  3. Write the purpose of this circulation.

Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 21

  1. Pulmonary circulation
  2. Right, ventricle → Pulmonary artery → lungs → Pulmonary vein → Left auricle
  3. Oxygenation of blood

Question 7.
Leukocytes are the soldiers of the body. Justify?
Answer:
Leukocytes provides defence mechanism to the body against infection. For example Neutrophils and monocytes are phagocytic cells which destroy foreign organisms entering the body.

Basophils secrete histamine, heparin etc. and are involved in inflammatory reactions. Eosinophils resist infections and are also associated with allergic reactions. Lymphocytes are responsible for immune responses of the body.

Question 8.
Listed below are some of the features of open and closed circulatory system. Arrange them into two categories.

  1. Usually low-pressure system
  2. Found in vertebrates
  3. Blood returns to heart rapidly
  4. High pressure system
  5. Blood returns to heart slowly
  6. Found in Arthopods

Answer:

Open circulationClosed circulation
1. Usually low pressure system4. High pressure system
5. Blood returns to heart slowly3. Blood returns to heart rapidly
6. Found in Arthropods2. Found in vertebrates

Question 9.
The heart of fish is said to be a venous heart, whether your heart is a venous one or not? Justify.
Answer:
The heart of fish is a venous heart, because it is a 2 chambered heart and only deoxygenated blood flows through the heart chambers.
Human heart is four chambered heart and carry both oxygenated and deoxygenated blood. So human heart is not a venous heart.

Question 10.
Observe the diagram.
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 22

  1. Identify the blood cell A and B.
  2. How can they differentiate?
  3. What are their function?

Answer:

  1. The blood cells A and B
    • A – Monocyte
    • B – Neutrophil
  2. They can be differentiated in two ways
    • Monocyte – Agranulocyte, Horseshoe shaped nucleus
    • Neutrophil – Granulocyte, multilobed nucleus
  3. Their functions
    • Monocyte – They protect the body by engulfing the pathogen (Phagocytosis)
    • Neutrophil – Phagocytic cells

Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation

Question 11.
During a clinical check-up doctor diagnosed that the patient’s heart murmurs.

  1. What are murmurs?
  2. Why does a heart murmur?
  3. The heart sound ‘dub’ is caused by the closure of AV valve. Do you agree with this statement?

Answer:
1. Due to the abnormalities in chambers or valves of the heart, some abnormal sounds may be audible in addition to normal heart sound. These are called murmurs.

2. Heart may murmur due to valvular damage.

3. The statement is incorrect because ‘dub’ is caused by the sudden closure of semilunar valve.

Question 12.
Observe the diagram and answer the following
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 23

  1. Name the area labelled here.
  2. What is the importance of this area?
  3. Name the area associated with this.
  4. Draw a flow chart showing conducting system of heart.

Answer:

  1. Sino-atrial node (SAN)
  2. Cardiac impulse originates from SAN.
  3. AV node, Bundle of His, Purkinje fibres.
  4. SAN → AVN → Bundle of His→ Purkinje fibres → Ventricle.

Question 13.
Match the following.
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 24
Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 25

Question 14.
A person’s blood pressure is found to be about 140/ 100mm Hg.

  1. What is the condition the person is suffering?
  2. How do you measure BP?
  3. Blood pressure is 100/60mm Hg. How you interpret it?

Answer:

  1. Hypertension
  2. Sphygmomanometer
  3. Hypotension

Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation

Question 15.
The major findings of a health survey conducted in a corporation area are

  • 25% of the population are suffering from hypertension.
  • 15% of male above 40 years are suffering from various cardiac disorders.

1. List any three cardio vascular disorders.
2. Suggest any two possible reason for this disorders.
3. Suggest any two advices to the disease.

Answer:

  1. Hypertension, Atherosclerosis, Cardiac arrest
  2. Two possible reason for cardio vascular disorders:
    • Regular intake of fatty food
    • Smoking
    • Lack of exercise
  3. Advices to the disease:
    • Avoid fatty food
    • Do regular exercise

Question 16.
Match Column I with Column II.

Column IColumn II
(a) EosinophilsCoagulation
(b) RBCUniversal Recipient
(c) AB GroupResist infections
(d) PlateletsContraction of Heart
(e) SystoleGas transport

Answer:

Column IColumn II
(a) EosinophilsResist infections
(b) RBCGas transport
(c) AB GroupUniversal Recipient
(d) PlateletsCoagulation
(e) SystoleContraction of Heart

Question 17.
Select the correct statement from the following:

  • Monocyte is a granulocyte.
  • Frog has 4 chambered heart.
  • Blood is a tissue.
  • The respiratory organs of insects are gills.

Answer:

  • Blood is a tissue. (Monocyte is an agranulocyte Frog has 3 chambered heart Respiratory organs of insects are trachea)

Question 18.
Today we can measure blood pressure with a sphygmomanometer at home. This shows that people are very much concerned about high blood pressure. Find out why?
Answer:
High blood pressure can harm the vital organs like heart, brain and kidneys. High blood pressure compels the heart to work excessively due to which the congestive heart disease may set in at an early age. This is the reason why people are very much concerned with their high B.P.

Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation

Question 19.
Erythroblastosis foetalis is an Rh incompatibility disorder.

  1. Describe the circumstances in which erythroblastosis foetalis might arise.
  2. How is the problem usually averted?

Answer:
1. If the first child of a Rh- woman is Rh+, she develops anti-Rh antibodies in her blood. During subsequent pregnancies if the foetus is Rh+ and Rh antibodies enter the foetal blood from the mother blood causing agglutination resulting erythroblastosis foetalis.

2. This problem can be avoided by administering anti-Rh antibodies to the mother immediately after the delivery of the first child.

Question 20.
Write the odd one out and give the reason for your answer.

  1. P wave, QRS wave, T wave, beta wave.
  2. Renal artery, Pulmonary artery, Hepatic artery, Coronary artery.
  3. SA node, lymph node, AV node, Purkinje fibres

Answer:

  1. Beta wave: These are the waves of normal EEG, whereas all others are the components of a normal ECG.
  2. Pulmonary artery: This blood carries deoxygenated blood whereas all others carry oxygenated blood.
  3. Lymph node: It is found in lymphatic system whereas all other structures are found in the mammalian heart related to auto rhythmicity.

Question 21.
Cardiac output increases when we engage in physical exercises. However, it may fall after the exercise stops. We can see that it is a fully heart controlled event.

  1. What is a cardiac output?
  2. How does heart increase cardiac output during exercise?

Answer:

  1. Cardiac output is the volume of the blood ejected by. each ventricle into the arterial system per minute,
  2. In response to the sensory impulses and the adrenal medullary hormone, the excitability of the pacemaker is increased and thereby heart rate is increased during exercise. This in turn increases the heart rate.

Question 22.
Copy and complete the table on blood group, antigen and antibody.
1.
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 26

2. Name the blood groups commonly known as Universal donor and Universal recipient.
Answer:
1.
Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation - 27

2. The blood groups commonly known as Universal donor and Universal recipient.

  • Universal donor: ‘O’ group.
  • Universal recipient: ‘AB’ group.

Plus One Body Fluids and Circulation NCERT Questions and Answers

Question 1.
What is the importance of plasma proteins?
Answer:
Fibrinogen, globulins and albumins are the major proteins. Fibrinogens are needed for clotting or coagulation of blood. Globulins primarily are involved in defense mechanisms of the body and the albumins help in osmotic balance.

Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation

Question 2.
Match Column I with Column II:

Column IColumn II
(a) Eosinophils(i) Coagulation
(b) RBC(ii) Universal Recipient
(c) AB Group(iii) Resist Infections
(d) Platelets(iv) Contraction of Heart
(e) Systole(v) Gas transport

Answer:

Column IColumn II
(a) Eosinophils(iii) Resist Infections
(b) RBC(v) Gas transport
(c) AB Group(ii) Universal Recipient
(d) Platelets(i) Coagulation
(e) Systole(iv) Contraction of Heart

Question 3.
Why do we consider blood as a connective tissue?
Answer:
Blood is considered a connective tissue for two basic reasons:

  1. embryologically, it has the same origin (meso-dermal) as do the other connective tissue types and
  2. Blood connects the body systems together bringing the needed oxygen, nutrients, hormones and other signaling molecules, and removing the wastes.

Question 4.
Sino-atrial node is called the pacemaker of our heart. Why?
Answer:
Sino-atrial node is a bundle of nerves which initiates the conduction of heartbeat. So, this is also called the pacemaker of our heart.

Question 5.
What is the significance of atrioventricular node and atrioventricular bundle in the functioning of heart?
Answer:
AV Node and atrioventricular bundles are responsible for conduction of cardiac beat cycle to further parts of heart.

Question 6.
Define a cardiac cycle and the cardiac output.
Answer:
Cardiac Cycle. The rhythmic contraction and dilatation of different parts of heart in one beat is called cardiac cycle. Cardiac Output. The heart beats 72 times per minute, i.e., that many cardiac cycles are performed per minute. From this it could be deduced that the duration of a cardiac cycle is 0.8 seconds.

During a cardiac cycle, each ventricle pumps out approximately 70mL of blood which is called the stroke volume. The stroke volume multiplied by the heart rate (no. of beats per min.) gives the cardiac output. Therefore, the cardiac output can be defined as the volume of blood pumped out by each ventricle per minute and averages 5000 mL or 5 litres in a healthy individual.

Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation

Question 7.
What is the difference between lymph and bllod?
Answer:

BloodLymph
RBC PresentRBC Absent
Found in arteries, veins and capillariesFound in interstitial space
Can transport gasesCannot transport gases

Plus One Body Fluids and Circulation Multiple Choice Questions and Answers

Question 1.
In higher vertebrates, SA – node helps in
(a) conduction of blood
(b) initiation of heartbeat
(c) opening of tricuspid valve
(d) opening of bicuspid valve
Answer:
(b) initiation of heartbeat

Question 2.
An oval depression called fossa ovalis, is seen on
(a) inter atrial septum
(b) inter ventricular septum
(c) right auriculo-ventricular septum
(d) left auriculo-ventricular septum
(e) papullary muscles
Answer:
(a) inter atrial septum

Question 3.
The deposition of lipids on the wall lining, the lumen of large and medium-sized arteries is referred to as
(a) deep vein thrombosis
(b) Stokes-Adam’s syndrome
(c) osteoporosis
(d) atherosclerosis
Answer:
(d) atherosclerosis

Question 4.
Which of the following are located in tunica media of human blood vessels?
(a) Collagen fibres and smooth muscle
(b) Squamous epithelium and striated muscle
(c) white fibres and smooth muscle
(d) Yellow fibres and smooth muscle
Answer:
(d) Yellow fibres and smooth muscle

Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation

Question 5.
The important function of lymph is to
(a) transport oxygen to the brain
(b) transport carbon dioxide to the lungs
(c) return RBCstothe lymph nodes
(d) return interstitial fluid to the blood
Answer:
(d) return interstitial fluid to the blood

Question 6.
Cardiac output is determined by
(a) heart rate
(b) stroke volume
(c) blood flow
(d) Both (a) and (b)
Answer:
(d) Both (a) and (b)

Question 7.
Which of the following matches correctly?
(a) Inferior vena cava – Receives deoxygenated blood from the head and body
(b) Superior vena – Receives deoxygenated blood from the lower body and organs
(c) pulmonary artery – carries deoxygenated blood to lungs
(d) hepatic artery – carries deoxygenated blood to the gut
Answer:
(c) pulmonary artery – carries deoxygenated blood to lungs

Question 8.
The cardiac pacemaker in a patient fails to function normally. The doctors find that an artificial pacemaker is to be grafted in him. It is likely that it will be grafted at the site made up of
(a) atrioventricular bundle
(b) Purkinje system
(c) sinuatrial node
(d) atrioventricular node
Answer:
(c) sinuatrial node

Question 9.
Generally artificial pacemaker consists of one battery made up of
(a) nickel
(b) dry cadmium
(c) photo sensitive material
(d) lithium
Answer:
(d) lithium

Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation

Question 10.
which type of white blood cells are concerned with the release of histamine and the natural anticoagulant heparin?
(a) Neutrophils
(b) Basophils
(c) Eosinophils
(d) Monocytes
Answer:
(b) Basophils

Question 11.
In humans, blood passes from the post caval to the diastolic right atrium of heart due to
(a) pushing open of the venous valves
(b) suction pull
(c) stimulation of the sino-auricular node
(d) pressure difference between the caval and atrium
Answer:
(d) pressure difference between the caval and atrium

Question 12.
which one of the following has an open circulator system?
(a) Pheretima
(b) Periplaneta
(c) Hirudinaria
(d) Octopus
Answer:
(b) Periplaneta

Question 13.
in higher vertebrates, SA – node helps in
(a) conduction of blood
(b) initiation of heartbeat
(c) opening of tricuspid valve
(d) opening of bicuspid valve
Answer:
(b) initiation of heartbeat

Question 14.
The opening of auricles and ventricles on the right side is guarded by
(a) tricuspid valve
(b) bicuspid valve
(c) semilunar valve
(d) Eustachian tube
Answer:
(a) tricuspid valve

Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation

Question 15.
Which of the following sequences is truly a systemic circulation pathway?
(a) Right ventricle → Pulmonary aorta → Tissues Pulmonary veins → Left auricle
(b) Right auricle → Left ventricle → Aorta → Tissues → Veins → Right auricle
(c) Left auricle → Left ventricle → Pulmonary aorta + Tissues → Right auricle
(d) Left auricle → Left ventricle → Pulmonary aorta → Arteries → Tissues → Veins → Right atrium
Answer:
(d) Left auricle → Left ventricle → Pulmonary aorta → Arteries → Tissues → Veins → Right atrium

Question 16.
Open circulatory system is present in
I. Arthropods
II. Annelids
III. Chordates
IV. Molluscs
(a) III only
(b) III and II
(c) I and IV
(d) IV only
Answer:
(c) I and IV

Question 17.
The artery, which supplies blood to the pericardium is
(a) brachial artery
(b) pulmonary vein
(c) vertebral artery
(d) coronary artery
Answer:
(d) coronary artery

Plus One Zoology Chapter Wise Questions and Answers Chapter 7 Body Fluids and Circulation

Question 18.
In diastole, heart is filled by
(a) mixed blood
(b) venous blood
(c) oxygenated blood
(d) deoxygenated blood
Answer:
(d) deoxygenated blood

Question 19.
You are required to draw blood from patient and to Keep it in a test tube for analysis of blood corpuscles and plasma. You are also provided with the following four types of test tubes. Which of them will you not use for the purpose?
(a) Test tube containing calcium bicarbonate
(b) Chilled test tube
(c) Test tube containing heparin
(d) Test tube containing sodium oxalate
Answer:
(c) Test tube containing heparin

Plus One Botany Chapter Wise Questions and Answers Chapter 8 Mineral Nutrition

Students can Download Chapter 8 Mineral Nutrition Questions and Answers, Plus One Botany Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus One Botany Chapter Wise Questions and Answers Chapter 8 Mineral Nutrition

Plus One Botany Mineral Nutrition One Mark Questions and Answers

Question 1.
Which one of the following roles is not characteristic of an essential element?
(a) being a component of biomolecules
(b) changing the chemistry of soil
(c) being a structural component of energy-related chemical compounds
(d) activation or inhibition of enzymes
Answer:
(b) changing the chemistry of soil

Question 2.
Which one of the following statements can best explain the term critical concentration of an essential element?
(a) essential element concentration below which plant growth is retarded.
(b) essential element concentration below which plant growth becomes stunted.
(c) essential element concentration below which plant remains in the vegetative phase.
(d) none of the above
Answer:
(b) essential element concentration below which plant growth becomes stunted.

Question 3.
During protein synthesis subunits of ribosomes are combined due to the presence of the particular element. Name it.
Answer:
Magnesium

Plus One Botany Chapter Wise Questions and Answers Chapter 8 Mineral Nutrition

Question 4.
Deficiency symptoms of an element tend to appear first in young leaves.
It indicates that the element is relatively immobile. Which one of the following elemental deficiency would show such symptoms?
(a) sulphur
(b) Magnesium
(c) Nitrogen
(d) Potassium
Answer:
(a) sulphur

Question 5.
With regard to the Biological Nitrogen Fixation by Rhizobium in association with soybean, which one of the following statement/statements does not hold true.
(a) Nitrogenase may require oxygen for its functioning
(b) Nitrogenase is MO- Fe protein
(c) Leg-hemoglobin is a pink coloured pigment.
(d) Nitrogenase helps to convert N2 gas into two molecules of ammonia.
Answer:
(a) Nitrogenase may require oxygen for its functioning

Question 6.
Name the free-living microorganisms which can fix nitrogen.
Answer:
Azotobactor and Rhodospirillum

Question 7.
In what form do plants absorb molebdenum from the soil.
Answer:
Molybdate ions MoO22+

Question 8.
Mineral requirements of plants is explained by a method in the nutrient solution is Called ………. and it was demonstrated by ……….. for the first time.
Answer:
Hydroponics,Julius von sachs

Question 9.
Find the odd one out.
Boron, Copper, Zinc, Phosphorous
Answer:
Phosphorous

Question 10.
A plant cell when kept in a certain solution got plasmolysed. What was the nature of the solution?
Answer:
Hypertonic Solution.

Question 11.
Plants can be cultivated in water. Name the type of cultivation.
Answer:
Hydroponics

Question 12.
Find the odd one among the following Carbon, Manganese, potassium, Nitrogen
Answer:
Manganese

Question 13.
A farmer adds Azotobacter culture to the soil before sowing maize. Which mineral element is being replenished?
Answer:
Nitrogen

Question 14.
In the diagram below, Label the cell Ain Nostoc
Plus One Botany Chapter Wise Questions and Answers Chapter 8 Mineral Nutrition img1
Answer:
Heterocyst

Question 15.
Which one of the following statements can best explain the term critical concentration of an essential element?
(a) essential element concentration below which plant growth is retarded.
(b) essential element concentration below which plant growth becomes stunted.
(c) essential element concentration below which plant remains in the vegetative phase.
(d) none of the above
Answer:
(a) essential element concentration below which plant growth is retarded.

Plus One Botany Chapter Wise Questions and Answers Chapter 8 Mineral Nutrition

Question 16.
Deficiency symptoms of an element tend to appear first in young leaves. It indicates that the element is relatively immobile. Which elemental deficiency would show such symptoms?
Answer:
Sulphur or calcium

Question 17.
From where do plants receive hydrogen?
Answer:
Water absorbed by plants.

Plus One Botany Mineral Nutrition Two Mark Questions and Answers

Question 1.
Give example for

  1. Free-living N2 fixing bacteria.
  2. Symbiotic N2 fixing bacteria.

Answer:

  1. Acetobacter, Nitrococcus
  2. Rhizobium

Question 2.
Crop plants cannot grow well in the nitrogen-deficient soil while plants like Drosera and Nepenthes show vigorous growth.

  1. Justify the statement
  2. Mention any two deficiency shown by such crop plants.

Answer:
1. These plants are insectivorous plants. They capture and feed on insects in order to maintain the N2 content of their body.

2. Two deficiency shown by such crop plants

  • Stunted growth
  • Chlorosis

Question 3.
Study the relation of the given pair and fill up the blanks:

  1. Potassium: Stomatal movement:………… Constituent of chlorophyll
  2. ………………: pollen germination Zinc: biosynthesis of auxin.
  3. Ion exchange: Passive absorption ………….: Active absorption
  4. Protoplasmic streaming theory: De Vries: ………….Active absorption

Answer:

  1. Mg
  2. Boron
  3. Carrier concept
  4. Munch

Question 4.
Plants can be grown in defined nutrient solution in the absence of soil.

  1. Name the technique.
  2. Who demonstrated the technique for the first time?

Answer:

  1. Hydroponics
  2. Julius von sachs

Question 5.
How is nitrogenase enzyme protected?
Answer:
In leguminous plants, the enzyme nitrogenase is protected from Oxygen by the pigment leghaemoglobin.

Question 6.
The functional roles of certain essential elements are given below. Identify the element based on the function.

  1. Essential for the formation of Chlorophyll.
  2. Involved in the pollen germination.

Answer:

  1. Mg
  2. Boron

Question 7.
Explain the scientific reason for growing legume crops prior to cereal crops.
Answer:
Increase the fertility of the soil. Biological N2 fixation.

Plus One Botany Chapter Wise Questions and Answers Chapter 8 Mineral Nutrition

Question 8.
Name the pink coloured pigment present in the root nodules? Write its role in N2 fixation?
Answer:
Leghemoglobin is an oxygen scavenger. It creates anaerobic condition for the action of nitrogenase enzyme to fix N2.

Question 9.
Give example for the following taking part in biological nitrogen fixation.

  1. Free-living N2 fixing bacteria
  2. Free-living N2 fixing cyanobacteria
  3. Symbiotic N2 fixing bacteria.
  4. Symbiotic N2 fixing cyanobacteria

Answer:

  1. Azotobacter, Clostridium
  2. Nostoc, Anabaena, Oscillatonia.
  3. Rhizobium leguminosarum
  4. Nostoc, Anabaena

Question 10.
Prepare a flow chart which shows the conversion of N2 to ammonia in biological N2 fixation.
Answer:
Plus One Botany Chapter Wise Questions and Answers Chapter 8 Mineral Nutrition img2

Question 11.
A pigment similar to hemoglobin is present in leguminous plants. Name the pigment and write its role.

OR

How is nitrogenase enzyme protected?
Answer:
Leghemoglobin which is the scavenger of 02. Nitrogenase enzyme needs oxygen-free condition.

Question 12.
Observe the figure given below and answer the following questions.
Plus One Botany Chapter Wise Questions and Answers Chapter 8 Mineral Nutrition img3

  1. Name this technique.
  2. Write two uses of this technique.

Answer:

  1. Hydroponics
  2. uses
    • Used as kitchen garden,
    • Used to study jdeficiency symptoms.

Question 13.
For the normal growth of plants it requires minerals. Write two examples of micro and macro elements.
Answer:

  • Macro – N, P
  • Micro-Boron, copper

Question 14.
In a survey done in an agricultural area the crops in the area shows various symptoms.

  • Chlorosis
  • Necrosis
  • Delay flowering.

Name the elements whose deficiency causes these symptoms.
Answer:

  • N, P, K, Mg
  • Chlorosis – N, K, Mg, S, Fe, Mn, etc.
  • Necrosis – K, Mg, Ca and Cu Delay flowering – N, S, and Mo

Question 15.
Iron is not a structural component of chlorophyll but its deficiency causes yellowing of leaves. Give reason.
Answer:
Iron activates catalise enzyme in the formation of chlorophyll.

Question 16.
Amides are derivatives of amino acids in which the hydroxyl group is replaced by the amino group (NH2)
a) How are asparagine and glutamine formed.
b) Amides contain more than amino acids.
Answer:
Plus One Botany Chapter Wise Questions and Answers Chapter 8 Mineral Nutrition img4

Question 17.
In a survey done in an agricultural area the crops in the area shows various symptoms.

  • Chlorosis
  • Necrosis
  • Delay flowering

Name the elements whose deficiency causes these symptoms.
Answer:

  • N, P, K, Mg
  • Chlorosis – N, K, Mg, S, Fe, Mn, etc.
  • Necrosis – K, Mg, Ca and Cu
  • Delay flowering – N, S, and Mo

Question 18.
A farmer supplies Nitrogen fertilizer to pea plants.

  1. Is there any necessity to supply Nitrogen fertilizer?
  2. Justify your answer with reason.

Answer:

  1. No
  2. Symbiotic N2 fixation

Plus One Botany Chapter Wise Questions and Answers Chapter 8 Mineral Nutrition

Question 19.
Name the respecive mineral nutrient element that

  1. Forms the core constituent of the ring structure of chlorophyll
  2. Activates carboxylases
  3. Forms the components of nitrogenase
  4. Synthesises middle lamella of plant cell.

Answer:

  1. Magnesium
  2. Manganese
  3. Molybdenum and iron
  4. Calcium

Question 20.
Observe the relationship between the first pair and fill in the blanks.
Posassium: Opening and closing of stomata::
Boron: …………
Answer:
Boron: Pollen germination/ pollen tube growth

Question 21.
State whether True or False.

  1. Potassium plays an important role in opening and closing of stomata.
  2. The movement of ions is usually called flux.
  3. In an LHC (Light-Harvesting complex), Chlorophyll b is the reaction center.
  4. The first stable product in calvin cycle is 3-PGA.

Answer:

  1. True
  2. True
  3. False
  4. True

Question 22.
A few inorganic elements are given. Match them with their specific functions.

ElementsFunction
1. Potassiuma. Absorption and translocation
2. Magnesiumb. Pollen germination
3. Molybdenumc. Opening and closing of Stomata
4. Borond. Maintain ribosome structure

Answer:

ElementsFunction
1. PotassiumOpening and closing of stomata
2. MagnesiumMaintain ribosome structure
3. MolybdenumAbsorption and translocation
4. BoronPollen germination

Question 23.
Pick the suitable minerals from the list, for following processes?

  1. Synthesis of auxin
  2. Pollen germination
  3. Photolysis of water
  4. Nitrogen metabolism

Plus One Botany Chapter Wise Questions and Answers Chapter 8 Mineral Nutrition img5
Answer:

  1. Zn
  2. B
  3. Mn
  4. Mo

Question 24.
Rhizobium leguminosarum cannot do N2 fixation outside root nodules. Give reasons.
Answer:
Rhizobium can fix atmospheric N2 only if the symbiotic relationship with leguminose plant is established.

Question 25.
What type of conditions are created by leghaemoglobin in the root nodule of a legume?
Answer:
It acts as 02 scavenger that means it helps in preventing the combining of 02 with nitrogenase.

Plus One Botany Chapter Wise Questions and Answers Chapter 8 Mineral Nutrition

Question 26.
Leguminous plants can be cultivated in between rice cultivation. Why?
Answer:
Because it increases the fertility of soil (enrich N2 content) for next year rice cultivation.

Question 27.
Match the element with its associated functions/ roles.

A. Boroni. Splitting of H2O to liberate o2 during photosynthesis
B. Manganeseii. needed for the synthesis of auxins
C. Molybdenumiii. component of nitrogenase
D. Zinciv. Pollen germination
E. Ironv component of ferredoxin

Answer:

  1. A – iv
  2. B – i
  3. C – iii
  4. D – ii
  5. E – v

Question 28.
To get maximum yeild a farmer added exess amount of manganese-containing fertiliser. The plants showed some deficiency symptoms. Can you give the reason for this?
Answer:
Excess of the micronutrient -Mn causes Toxicity because it prevent the uptake of Iron, Magnesium, and calcium. This results in deficiency symptom.

Question 29.
Plus One Botany Chapter Wise Questions and Answers Chapter 8 Mineral Nutrition img6

  1. Identify the phenomenon?
  2. A pink coloured protein protects nitrogenase from oxygen. Name that protein.

Answer:

  1. Ammonification
  2. Leghemoglobin

Question 30.

AB
MagnesiumOpening and closing of stomata
CalciumSplitting of water
PotassiumMiddle lamella
ChlorineChlorophyll

Answer:

AB
MagnesiumChlorophyll
CalciumMiddle lamella
PotassiumOpening and closing of stomata
ChlorineSplitting of water

Question 31.

  1. Name the oxygen-binding pigment found in the root nodule of plants like pea and bean.
  2. What is the normal colour of this pigment?

Answer:

  1. Leghemoglobin
  2. Pink-colored

Question 32.
How the presence of magnesium fulfills the requirements of the essentiality of elements from growth and development of plants?
Answer:

  1. It is a constituent of chlorophyll molecule and is essential for photosynthesis.
  2. It cannot be replaced by any other element for the same function.
  3. It is also required as a co-factor by many enzymes involved in cellular respiration and other metabolic pathways.

Plus One Botany Chapter Wise Questions and Answers Chapter 8 Mineral Nutrition

Question 33.
Nitrogen fixation is shown by prokaryotes and not eukaryotes. Comment?
Answer:
In prokaryotes, photosynthetic autotrophs (cyanobacteria) like nostoc and Anabaena have special N2 fixing cell which helps in N2 fixation.

Question 34.
How is sulphur important for plants? Name the amino acids in which it is present.
Answer:
Sulphur, besides being present in some amino acids essential for protein synthesis, is also a constituent of several coenzymes, vitamins, and ferrodoxin which are involved in some biochemical pathways.

Question 35.
How do some bacteria carry out nitrification? What are such bacteria called?
Answer:
Such types of bacteria convert ammonia into nitrates and obtain energy for their activities. They are called chemosynthetic autotrophs.

Question 36.
Name the respective mineral nutrient elements of plants that:

  1. Is needed in the synthesis of auxins
  2. Is a constituent of ferredoxin
  3. Forms the core constituent of the ring structure of chlorophyll
  4. forms the components of nitrogenase and nitrate reductase

Answer:

  1. Zinc
  2. Sulphur
  3. Magnesium
  4. Molybdenum

Question 37.
Why do plants of legume family contain more protein than the other plants?
Answer:
Plants in legume family bears root nodules which bear symbiotic nitrogen-fixing bacteria i.e rhizobium

Question 38.
Nepenthes carries out photosynthesis and still traps insects. Why?
Answer:
Nepenthes grow in nitrogen-deficient soils so these trap, digest and absorb amino acids from the insects supplementing their nitrogen supply.

Question 39.
Mycorrohiza is a mutualistic association. How do the organisms involved in this association gain from each other?
Answer:
It is an association of root and fungi in which surface area of root is increased and more minerals can be absorbed. Fungus get nutrients and sugar from the plant.

Question 40.
Carnivorous plants like Nepenthes and Venus flytrap have nutritional adaptations. Which nutrient do they especially obtain and from where?
Answer:
In these, insect body is digested and plant gets Nitrogen as nutrients.

Question 41.
How are organisms like Pseudomonas and Thiobacillus of significance in nitrogen cycle?
Answer:
Pseudomonas and Thiobacillus carry out denitrification process wherein the nitrate present in the soil is reduced to nitrogen thus contributing to the atmospheric nitrogen.

Question 42.
Think of a plant which lacks chlorophyll. From where will it obtain nutrition? Give an example of such a type of plant.
Answer:
This case is observed in total parasites. In such plants they absorb food, water, and minerals from host. Eg cuscuta.

Question 43.
Nitrogen is essential element for plants and is found in abundance as atmospheric nitrogen. But most plants unable to use it. Why is it so and in what form do plants utilize them?
Answer:
Most plants cannot use it because atmospheric nitrogen is inert in its nature. Nitrogen fixers in soil can convert nitrogen gas into nitrates or nitrite or ammonium forms. These compounds enter the plants as nutrients through the roots and are assimilated as organic nitrogen.

Question 44.
How insectivorous plants fulfill their nitrogen requirements?
Answer:
Insectivorous plants fulfill their nitrogen requirements from insects and small animals trapped by their leaves(a pitcher).

Question 45.
All elements that are present in a plant need not be essential to its survival. Give reasons.
Answer:
The Criteria for essentiality are

  1. The element must be necessary for supporting normal growth and reproduction
  2. In the absence of elements, the plants do not complete their life cycle or set the seeds.
  3. The element must be directly involved in the metabolism of the plant.
  4. The requirement of the element must be specific and not replaceable by another.

Question 46.
Why is that in certain plants deficiency symptoms appear first in younger parts of the plant while in others they do so in mature organs?
Answer:

  • If the elements are actively mobilized, the older part of plant show deficiency symptoms.
  • If they are not mobilized, the deficiency symptoms are first observed in younger regions.

Plus One Botany Chapter Wise Questions and Answers Chapter 8 Mineral Nutrition

Question 47.
Mineral elements are re-exported from one place to other or some are immobile and deficiency symptoms observed

  1. Which is the part of the plant show deficiency symptoms in the former?
  2. Name the mineral element is immobile and which part of the plant shows deficiency symptoms?

Answer:

  1. Older regions because minerals are remobilised from older regions into younger regions.
  2. Ca, Deficiency symptoms are noticed in younger tissues because it forms the structural part.

Plus One Botany Mineral Nutrition Three Mark Questions and Answers

Question 1.
Give the technical term for the following.

  1. Leaf yellowing
  2. death of tissues
  3. soilless cultivation

Answer:

  1. chlorosis
  2. Necrosis
  3. Hydroponics

Question 2.
Plants pick up nitrogen from the soil in the form of ammonia ions or nitrate ions, ammonia being the main product of biological nitrogen fixation. From the above view answer the following.
a) Which soil bacteria convert ammonia to NO2 – and NO3.
b) Draw a schematic diagram of progressive reduction of one molecule of nitrogen in the presentee of enzyme nitrogenase to yield two molecule of ammonia.
c) Name the enzyme for nitrogen assimilation in plants.
Answer:
Plus One Botany Chapter Wise Questions and Answers Chapter 8 Mineral Nutrition img7

Question 3.
Macro and micronutrients are required in particular amount for plants.

  1. What will be the moderate increase of micronutrient affect the plants
  2. Name the micronutrient helps in pollen germination
  3. What is the concentration of macro and micronutrients required for plants?

Answer:

  1. It cause toxicity to plants and affect the uptake and utilisation of macronutrients
  2. Boron
  3. Macronutrients – Excess 10 millimole /kilogram of dry matter
    Micronutrients – Less than 10 millimole /kilogram of dry matter

Plus One Botany Chapter Wise Questions and Answers Chapter 8 Mineral Nutrition

Question 4.
During formation of nitrate in plants, different steps occur.

  1. Name the steps leading to the conversion of atmospheric nitrogen to nitrate
  2. Which is the reductive and oxidative steps?
  3. Name the pigments and Enzymes that are required and the steps they act on.

Answer:

  1. Nitrogen fixation, Nitrification
  2. Nitrogen fixation—reductive process, Nitrification—Oxidative process
  3. pigment—Leghaemoglobin, Enzyme— Nirogenase Pigments and enzymes acts on Nitrogen fixation step

Question 5.
Plus One Botany Chapter Wise Questions and Answers Chapter 8 Mineral Nutrition img8

  1. Name the process
  2. Which enzyme catalyses the process
  3. How the enzyme in the process is protected from oxygen in root nodules?

Answer:

  1. N2 fixation
  2. Nitrogenase
  3. Due to the presence of pigment Leghemoglobin. It acts as the scavenger of oxygen.

Plus One Botany Mineral Nutrition NCERT Mark Questions and Answers

Question 1.
‘All elements that are present in a plant need not be essential to its survival’. Comment.
Answer:
The criteria for essentially of an element are given below:

1. The element must be absolutely necessary for supporting normal growth and reproduction. In the absence of the element, the plants do not complete their life cycle or set the seeds.

2. The requirement of the element must be specific and not replaceable by another element. In other words, deficiency of any one element cannot be met by supplying some other element.

3. The element must be directly involved in the metabolism of the plant. All elements that are present in a plant do not fulfill these criteria hence cannot be essential for plant survival.

Question 2.
Why is purification of water and nutrient salts so important in studies involving mineral nutrition using hydroponics?
Answer:
In 1860, Julius von Sachs, a prominent German botanist, demonstrated, for the first time, that plants could be grown to maturity in a defined nutrient solution in complete absence of soil. The essence of all these methods involves the culture of plants in a soil-free, defined mineral solution. These methods require purified water and mineral nutrient salts.

Purification of water and nurtient salt is important to rule out other influencing factors. The presence of pure nutrients will give clearcut scientific results. This will help in making a sound basis for the right prediction.

Question 3.
What are the steps involved in the formation of a root nodule?
Answer:
Steps in the development of root nodules:
1. Rhizobium bacteria contact a susceptible root hair, divide near it.

2. Upon successful infection of the root hair cause it to curl

3. Infected thread carries the bacteria to the inner cortex. The bacteria get modified into rod-shaped bacteriods and cause inner cortical and pericycle cells to divide. Division and growth of cortical and pericycle cells lead to nodule formation.

4. A mature nodule is complete with vascular tissues continuous with those of the root.

Plus One Botany Mineral Nutrition Multiple Choice Questions and Answers

Question 1.
1. Farmers in a particular region were concerned that premature yellowing of leaves of a pulse crop might cause a decrease in the yield. Which treatment could be most beneficial to obtain maximum seed yield?
(a) Frequent irrigation of the crop
(b) Treatment of the plants with cytokinins along with a small dose of nitrogenous fertilizer
(c) Removal of all yellow leaves and spraying the remaining green leaves with 2, 4, 5- thchlorophenoxy acetic acid
(d) Application of iron and magnesium to promote synthesis of chlorophyll
Answer:
(d) Application of iron and magnesium to promote synthesis of chlorophyll

Question 2.
For nitrogen fixation, useful pigment is
(a) nitrogenase
(b) hemoglobin
(c) myoglobin
(d) leghaemoglobin
Answer:
(d) leghaemoglobin

Plus One Botany Chapter Wise Questions and Answers Chapter 8 Mineral Nutrition

Question 3.
Plants cultivated in nutrient solution without soil is called
(a) somatic hybridization
(b) tissue culture
(c) hydroponics
(d) suspension culture
Answer:
(c) hydroponics

Question 4.
The process of decay of dead organic matter is known as
(a) denitrification
(b) nitrification
(c) nitrogen fixation
(d) ammonification
Answer:
(d) ammonification

Question 5.
Zn, Mo, Fe, Cu are
(a) trace elements
(b) non-essentials
(c) macronutrients
(d) both a and b
Answer:
(a) trace elements

Question 6.
An essential element is that which
(a) improves health of the plant
(b) is irreplaceable and indispensable for the growth of plants
(c) is found in plant ash
(d) is available in the soil
Answer:
(b) is irreplaceable and indispensable for the growth of plants

Question 7.
N2 +8e +8H +16ATP → 2NH3+2H+ + 16ADP + 8e The above equation refers to
(a) amnionification
(b) nitrification
(c) nitrogen fixation
(d) denitrification
Answer:
(c) nitrogen fixation

Question 8.
Micronutrients are
(a) as important as macronutrients but are required in small amount
(b) less important than macronutrients
(c) called micro as they play only a minor role in plant nutrition
(d) required greater than 10m mole/Kg of dry matter
Answer:
(a) as important as macronutrients but are required in small amount

Question 9.
Which element is located at the centre of the porphyrin ring in chlorophyll?
(a) Potassium
(b) Manganese
(c) Calcium
(d) Magnesium
Answer:
(d) Magnesium

Question 10.
Which element is required for the germination of pollen grains?
(a) Boron
(b) Calcium
(c) Chlorine
(d) Potassium
Answer:
(a) Boron

Question 11.
Select the correct statement.
(a) Legumes are incapable of fixing nitrogen
(b) Legumes fix nitrogen through bacteria living fruits
(c) Legumes fix nitrogen only by bacteria present in root nodules
(d) frankia forms symbiotic association with algae
Answer:
(c) Legumes fix nitrogen only by bacteria present in root nodules

Plus One Botany Chapter Wise Questions and Answers Chapter 8 Mineral Nutrition

Question 12.
Chlorosis is caused due to deficiency of
(a) magnesium
(b) calcium
(c) boron
(d) manganese
Answer:
(a) magnesium

Question 13.
Enzyme nitrogenase is responsible for
(a) nitrification
(b) nitrogen fixation
(c) nitrite reduction
(d) nitrate reduction
Answer:
(b) nitrogen fixation

Question 14.
Maximum percentage of which element occurs in plant ash?
(a) Magnesium
(b) Zinc
(c) Potassium
(d) Calcium
Answer:
(d) Calcium

Question 15.
Which of the following metals causes bone cancer
(a) Lead
(b) Cobalt
(c) Uranium
(d) strontium90
Answer:
(d) strontium90

Question 16.
Premature leaf fall is due to deficiency of
(a) phosphorus
(b) nitrogen
(c) calcium
(d) potassium
Answer:
(a) phosphorus

Question 17.
The function of leg haemoglobin during biological nitrogen fixation in root nodules of legumes is to
(a) convert atmospheric nitrogen to ammonia
(b) convert ammonia to nitrite
(c) transport oxygen for activity of nitrogenase
(d) protect nitrogenase from oxygen
Answer:
(d) protect nitrogenase from oxygen

Plus One Botany Chapter Wise Questions and Answers Chapter 8 Mineral Nutrition

Question 18.
Which of the following gene is responsible for biological nitrogen fixation?
(a) Nitrogenase
(b) N if gene
(c) Yeast alanine tRNA synthetase
(d) RNA synthetase
Answer:
(b) N if gene

Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases

Students can Download Chapter 6 Breathing and Exchange of Gases Questions and Answers, Plus One zoology Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases

Plus One Breathing and Exchange of Gases One Mark Questions and Answers

Question 1.
Respiration in insects is called direct because
(a) The tissues exchange O2/ CO2 directly with the air in the tubes.
(b) The tissues exchange O2/ CO2 directly with coelomic fluid.
(c) The tissues exchange O2/ CO2 directly with the air outside through body surface.
(d) Tracheal tubes exchange O2/ CO2 directly with the haemocoel which then exchange with tissues.
Answer:
(d) Tracheal tubes exchange O2/ CO2 directly with the haemocoel which then exchange with tissues.

Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases

Question 2.
Mark the true statement among the following with reference to normal breathing
(a) Inspiration is a passive process where as expiration is active
(b) Inspiration is a active process where as expiration is passive
(c) Inspiration and expiration are active processes
(d) Inspiration and expiration are passive processes
Answer:
(b) Inspiration is a active process where as expiration is passive

Question 3.
Inspiration and expiration are passive processes A person breathes in some volume of air by forced inspiration after having a forced expiration. This quantity of air taken in is
(a) Total lung capacity
(b) Tidal volume
(c) Vital capacity
(d) Inspiratory capacity
Answer:
(c) Vital capacity

Question 4.
Incidence of Emphysema – a respiratory disorder is high in cigarette smokers. In such cases
(a) The alveolar walls are found damaged
(b) The plasma membrane is found damaged
(c) The bronchioles are found damaged
(d) The respiratory muscles are found damaged
Answer:
(b) The plasma membrane is found damaged

Question 5.
Respiratory process is regulated by certain specialized centres in the brain. One of the following listed centres can reduce the inspiratory duration upon stimulation.
(a) Medullary inspiratory centre
(b) Pneumotaxic centre
(c) Apneustic centre
(d) Chemosensitive centre
Answer:
(b) Pneumotaxic centre

Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases

Question 6.
CO2 + H2O → H2CO3 → HCO3 + H+
These all the reactions that takes place in accordance with respiration.

  1. Where these reactions takes place
  2. Which is the enzyme present here?

Answer:

  1. In blood
  2. Carbonic anhydrase

Question 7.
Fill in the table with appropriate heading values.
Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases - 1
Answer:

  1. a – Blood (Deoxygenated)
  2. b – Blood (oxygenated)
  3. c – 40
  4. d – 45

Question 8.
In the following diagram A represents Total lung capacity and B represents Vital capacity and then C represents what?
Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases - 2
Answer:
C – Residual volume

Question 9.
Name the biological principle involved in the exchange of gas between lungs and blood.
(Transmittance, Osmosis, Diffusion, Incorporation)
Answer:
Diffusion

Question 10.
What is the percentage of oxygen transported by Haemoglobin in the form of Oxyhaemoglobin? (67%, 97%, 32%, 50%)
Answer:
97%

Plus One Breathing and Exchange of Gases Two Mark Questions and Answers

Question 1.
Complete the flow chart showing inspiration

Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases - 3
Answer:

  1. A – Nasal opening
  2. B – Nasopharynx
  3. C – Trachea
  4. D – Bronchiole

Question 2.
Oxyhaemoglobin → Oxygen + Hb

  1. Where does this process takes place?
  2. What are the factors that influence O2 dissociation curve?

Answer:

  1. In tissues
  2. pO2, pCO2, H+ concentration, temperature.

Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases

Question 3.
The given diagram shows the exchange of gases between alveolus and alveolar capillary.

  1. Identify X and Y.
  2. Name the Physical Process involved in gas exchange.

Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases - 4
Answer:

  1. X – CO2, Y – O2
  2. Diffusion

Question 4.
Bardiagram showing oxygen transport is shown below.
Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases - 5

  • Name the two methods of oxygen transport.
  • How much of oxygen is transported in method 1 and method 2.

Answer:

  • Methods of oxygen transport
    1. O2 is transported by RBC in the blood (oxyhaemoglobin).
    2. O2 is transported in a dissolved state through plasma.
  • About 97% of O2 is transported as oxyhemoglobin and remaining 3% is carried in a dissolved state through the plasma.

Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases

Question 5.
Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases - 6
From this diagram prepare a table which show the pressure difference in various regions in respiration.
Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases - 7

Question 6.
Match the following.

PrawnSkin
CockroachLungs
ViperGills
EarthwormTrachea

Answer:

PrawnGills
CockroachTrachea
ViperLungs
EarthwormSkin

 

Question 7.
‘Emphysema’ is a term linked with ‘Respiratory system’. What is emphysema?
Answer:
Emphysema is respiratory system disorder in which alveolar walls are damaged due to which respiratory surface is decreased. Cigarette smoking is one of the major cause of emphysema.

Question 8.
Draw a flow chart showing the air passage in the human respiratory system.
(Trachea, nasal cavity, nasopharynx, bronchi, nostrils, alveoli, bronchioles, larynx)
Answer:
Nostrils → nasal cavity → nasopharynx → larynx alveoli ← branchioles ← bronchi ← Trachea.

Question 9.
Study the graph and answer the following questions.
Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases - 8

  1. What is represented by the graph?
  2. Find out the pressure at which Haemoglobin is 50% saturated with O2?

Answer:

  1. Oxygen dissociation curve
  2. 28 mm Hg

Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases

Question 10.

  1. Name the structural and functional unit of lungs.
  2. Why the lungs replaces the skin in Mammals as respiratory organ.

Answer:

  1. Alveoli
  2. Lungs contain millions of alveoli which serves to store oxygen and it is richly supplied with blood capillaries. These alveolar surface area exceeds the skin.

Question 11.
The pO2 in cell cytoplasm is always lower than the capillary blood while pCO2 is greater. How do you account for this?
Answer:
In the cell cytoplasm, as a result of metabolism the CO2 accumulates and oxygen content is reduced. So pO2 in cell cytoplasm is always lower than the capillary blood.

Question 12.
A dog expires CO2. How can you correlate the relationship between the dog and grazing cattle in terms of CO2.
Answer:
Dog expires CO2 to atmosphere and it is utilised by grass during photosynthesis and that grass is utilised by cattle.

Question 13
Observe the pie diagram given below and answer the following questions.
Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases - 9
Figure showing the percentage of carbon dioxide transport.

  1. What are the three methods of CO2 transport mentioned by the letters P, Q, R?
  2. If the amount of Haemoglobin in the blood of a patient is very low. What changes you expect in the processes?

Answer:

  1. Methods of CO2 transport
    • P – Carbamino haemoglobin in combination with amino group of protein.
    • Q – Carbonic acid dissolved in plasma
    • R – as bicarbonates
  2. Reduce the formation of carbaminohaemoglobin and formation of bicarbonates also reduces.

Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases

Question 14.
Find the odd one in each group in Column Awrite in Column B and write the reason in Column C.
Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases - 10
Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases - 11

Question 15.
Analyse the concept map given below and if have mistakes reconstruct it.
Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases - 12
Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases - 13

Question 16.
A person exhaled nearly more than 3000ml of air during a breathing exercise. How do you justify that act?
Answer:
During breathing excercise one can exhale the air nearly of his vital capacity which averages to 3500ml. During normal respiration only tidal volume (500ml) is exhaled.

Question 17.
In orderto give the awareness about healthy practices to reduce respiratory disorders, prepare a pamphlet.
Answer:

  • Bronchitis, Bronchial asthma, Emphysema, Pneumonia, occupational lung disease are some respiratory disorders.
  • Stop cigarette smoking to prevent emphysema, bronchitis and bronchial asthma.
  • Use protective clothes and gears in work places to prevent occupational lung diseases.
  • Regular health check-up, use of antibiotic drugs and bronchodilator drugs by the advice of doctor.

Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases

Question 18.
The venous blood in the lungs has a pCO2 of 46mm Hg. Should the alveolar pCO2 exceeds or be less than 46mm Hg to result in diffusion of CO2 from the blood into alveolus?
Answer:
The alveolar pCO2 must be less than 46mm Hg. Then only there will be difference in the partial pressure and diffusion occurs from the higher concentration to a lower concentration.

Question 19.
Human respiration is controlled by intercostal muscles in the ribs and diaphragm. Justify this statement.
Answer:
The human respiration is controlled by the contraction and relaxation of intercostal muscles in the ribs. Due to the contraction the thoracic space increases and the atmospheric air comes inside to maintain the pressure.

When intercostal muscles relax the thoracic volume reduces and due to high pressure the air expels outward. This is known as expiration.

Question 20.
Observe the figure explaining difference in partial pressure of oxygen in different parts of body.
Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases - 14

  1. Explain the importance of partial pressure of oxygen in the gas exchange.
  2. What may be the partial pressure of oxygen in the tissue for effective gas transport?

Answer:

  1. Difference in the partial pressure of oxygen in the alveolar air and in the blood results in the exchange of gases by simple diffusion.
  2. The partial pressure of oxygen in the tissue is less than 90mm Hg.

Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases

Question 21.
While analysing the blood of normal healthy person, it has been found that the amount of HCO3 ions in his venous blood is much higher than his arterial blood. How do you account for this increase of HCO3 ions.
Answer:
The amount of CO2 in venous blood is much higher than that of arterial blood. Since major portion of CO2 is transported through blood as HCO3 ions. The amount of HO3 remains high in venous blood.

Question 22.
Place the following in the correct order.
Lungs expand, rib rise, air enters lungs, external intercostal muscle contract, thorax expands.
Answer:
External intercostal muscle contract → rib rise Lungs expand ← Thorax expands air enters lungs.

Question 23.
Pneumatoxic centre has an important role in regulating rate of respiration. Prepare a flow chart representing the regulation of respiration by pneumatoxic centre during strenuous exercise.
Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases - 15

Plus One Breathing and Exchange of Gases Three Mark Questions and Answers

Question 1.
Transport of CO2 as Bicarbonates.
Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases - 16

  1. Copy the diagram and fill the gaps.
  2. Name the other ways of CO2 transport.
  3. If blood chlorine level decrease, does it affect gas transport? Justify.

Answer:

1.
Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases - 17

2. Other ways of CO2 transport:

  • Carbamino haemoglobin
  • Carbonic acid

3. Yes. Decrease in chloride level affects CO2 transport. During CO2 transport bicarbonate ions formed within the erythrocytes (RBC) diffuse out into the plasma. It upsets the ionic equilibrium and electrical balance between RBC and plasma.

To regain the lost balance an equal number of chloride (Cl ) ions diffuse into RBC from plasma. This is called chloride shift or Hamburger phenomenon.

Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases

Question 2.
Analyse the flow chart.
Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases - 18

  1. Distinguish the function of external and internal intercostal muscles.
  2. Draw a flow chart explaining inspiration.

Answer:

1. During inspiration external intercostal muscle contracts and internal intercostal muscle relaxes. External intercostal muscle and internal intercostal muscles are antagonistic in action

2.
Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases - 19

Question 3.
Pharynx is a common passage for air and food.

  1. Name the part for the passage of food.
  2. Name the part for the passage of air.
  3. How do these passagework without any problems?

Answer:

  1. Oropharynx
  2. Nasopharynx
  3. Glottis is the opening of nasopharynx into trachea. During swallowing glottis can be covered by a thin elastic cartilagenous flap called epiglottis to prevent the entry of food into the larynx.

Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases

Question 4.
Following figure shows a microscopic section through the alveoli in the lungs of man.
Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases - 20
Give the features visible in the diagram that help to increase the rate of diffusion across the wall of the alveoli into the blood.
Answer:

  1. Surface area is increased by alveoli.
  2. Presence of capillaries
  3. Thin wall helps to increase the rate of diffusion. (The diffusion membrane is made up of three layers namely the thin squamous epithelium of alveoli, the endothelium of alveolar capillaries and the basement substance in between)

Question 5.
Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases - 21

  1. Which of the above graph represents normal condition?
  2. What would be the reason of abnormality of graph?
  3. What are the symptoms shown in such condition?

Answer:

  1. ‘A’ represents normal condition.
  2. The abnormality is due to the low pO2 in blood.
  3. The symptoms include headache, nausea, dizziness and even death.

Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases

Question 6.
A person inhales to his maximum capacity and then he exhales also to his maximum.

  1. What term would you use for the volume of air thus breathed out.
  2. What is the volume of air thus breathed out?
  3. Suppose he hold his breath after that maximum expiration for 5 seconds would there be any exchange of respiratory gases occuring in lungs during that period. How?

Answer:

  1. Vital capacity
  2. 3.5 – 4.5 litres
  3. Yes. Because of residual volume. The volume of air that still remains in the lungs after a forceful expiration is called residual volume. It is about 1500ml in man.

Question 7.
Human beings have a significant ability to maintain and moderate the respiratory rhythm to suit the demand of body tissues-substantiate.
Answer:
This is done by the neural system. A specialised centre present in the medulla region of the brain called respiratory rhythm centre is primarily responsible for this regulation. Another centre present in the pons region of the brain called pneumotaxic centre can moderate the functions of the respiratory rhythm centre.

Neural signal from this centre can reduce the duration of inspiration and thereby alter the respiratory rate. Achemosensitive area is situated adjacent to the rhythm centre which is highly sensitive to CO2 and hydrogen ions.

Increase in these substances can be activate this centre and make necessary adjustments in the respiratory process. Receptors associated with aortic arch also send necessary signals to the rhythm centre to adjust respiratory rate.

Question 8.
Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases - 22

  1. Redraw the graph to correct the mistake
  2. Name the graph
  3. Which are the factors that affect this graph.

Answer:
1.
Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases - 23

2. Oxygen dissociation curve

3. pO2, pCO2, H+ concentration and temperature are the factors that affect oxygen dissociation curve.

Plus One Breathing and Exchange of Gases NCERT Questions and Answers

Question 1.
Define vital capacity. What is its significance?
Answer:
Vital capacity (VC):
The maximum volume of air a person can breathe in after a forced expiration. This includes ERV (Expiratory Reserve Volume), TV (Tidal volume) and IRV (Inspiratory Reserve volume) of the maximum volume of air a person can breathe out after a forced inspiration.

Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases

Question 2.
State the volume of air remaining in the lungs after a normal breathing.
Answer:
Functional Residual Capacity (FRC). Volume of air that will remain in the lungs after a normal expiration.
This includes ERV + RV
ERV = 1000 to 1100ml
RV= 1100 to 1200ml
FRC = 2100 to 2300ml

Question 3.
What are the major transport mechanisms for CO2? Explain.
Answer:
Transport of Carbon dioxide CO2 is carried by haemoglobin as carbamino-haemoglobin (about 20-25 per cent). This binding is related to the partial pressure of CO2, pO2 is a major factor which could affect this binding.

When pCO2 is high and pO2 is low as in the tissues, more binding of carbon dioxide occurs whereas, when the pCO2 is low and pO2 is high as in the alveoli, dissociation of CO2 from carbamino- haemoglobin takes place, i.e., CO2 which is bound to Haemoglobin from the tissues is delivered at the alveoli.

RBCs contain a very high concentration of enzyme, carbonic anhydrase and minute quantities of the same is present in the plasma too. This enzyme facilitates the following reaction in both directions
Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases - 24
At the tissue site where partial pressure of CO2 is high due to catabolism, CO2 diffuses into blood (RBCs and plasma) and forms HCO3 and H+. At the alveolar site where pCO2 is low, the reaction proceeds in the opposite direction leading to the formation of CO2 and H2O.

Thus CO2 trapped as bicarbonate at the tissue level and transported to the alveoli is released out as CO2. Every 100 ml of deoxygenated blood delivers approximately 4ml of CO2 to the alveoli.

Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases

Question 4.
What will be the pO2 and pCO2 in the atmospheric air compared to those in the alveolar air?
(i) pO2 lesser, pCO2 higher
(ii) pO2 higher, pCO2 lesser
(iii) pO2 higher, pCO2 higher
(iv) pO2 lesser, pCO2 lesser
Answer:
(ii) pO2 higher will create the pressure gradient to facilitate the movement of O2 from atmosphere to alveoli and pCO2 lesser will create the movement of CO2 from alveoli to atmosphere.

Question 5.
What is the effect of pCO2 on oxygen transport?
Answer:
Binding of oxygen with haemoglobin is primarily related to partial pressure of O2. Partial pressure of CO2, hydrogen ion concentration and temperature are the other factors which can interfere with this binding. Increased partial pressure of CO2 can increase haemoglobin’s affinity towards oxygen and vice-versa is also true.

Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases

Question 6.
What happens to the respiratory process in a man going up a hill?
Answer:
When a man going uphill or doing some strenuous exercise then there is more consumption of oxygen. This decreases the partial pressure of oxygen in haemoglobin resulting in more demand of haemoglobin. As a result there is an increased breathing rate to fill the gap.

Plus One Breathing and Exchange of Gases Multple Choice Questions and Answers

Question 1.
Which one is the cofactor of carbonic anhydrase?
(a) Iron
(b) Zinc
(c) Copper
(d) Magnesium
Answer:
(b) Zinc

Question 2.
Skin is an accessory organ or respiration in
(a) human
(b) frog
(c) rabbit
(d) lizard
Answer:
(b) frog

Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases

Question 3.
Which of the following respiratory organs are present in spiders and scorpions?
(a) gill slit
(b) Gills
(c) Gill books
(d) Book lungs
Answer:
(d) Book lungs

Question 4.
When the body is rapidly oxidising fats, excess ketone bodies accumulate resulting in
(a) pyruvic acid
(b) lactic acid
(c) ketoacidosis
(d) ATP
Answer:
(c) ketoacidosis

Question 5.
Oxygen dissociation curve is
(a) sigmoid
(b) parabolic
(c) hyperbolic
(d) straight line
Answer:
(a) sigmoid

Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases
Question 6.
Which one of the following statements is incorrect?
(a) The residual air in lungs slightly decreases the efficiency of respiration in mammals
(b) The presence of non-respiratory air sacs, increases the efficiency of respiration in birds
(c) In insects, circulating body fluids serve to distribute oxygen to tissues
(d) The principle of countercurrent flow facilitates
Answer:
(a) The residual air in lungs slightly decreases the efficiency of respiration in mammals

Question 7.
Haemoglobin is having maximum affinity with
(a) carbon dioxide
(b) carbon monoxide
(c) oxygen
(d) ammonia
Answer:
(b) carbon monoxide

Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases

Question 8.
Which is called Hamburger shift?
(a) Hydrogen shift
(b) Bicarbonate shift
(c) Chloride shift
(d) Sodium shift
Answer:
(c) Chloride shift

Question 9.
Left shift of oxyhaenoglobin curve is noticed under
(a) normal temperature and pH
(b) low temperature and high pH
(c) low pH and high temperature
(d) low pH and low temperature
Answer:
(b) low temperature and high pH

Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases

Question 10.
Read the following statements and select the correct one.
(a) The H+ released from carbonic acid combines with haemoglobin to form haemoglobinic acid
(b) Oxyhaemoglobjn of erythrocytes is alkaline
(c) More than 70% of carbon dioxide is transferred from tissues to the lungs in the form of carbamino compounds
(d) In a healthy person, the haemoglobin content is . more than 25gm per 100mL
Answer:
(a) The H+ released from carbonic acid combines with haemoglobin to form haemoglobinic acid

Question 11.
In blood, carbon dioxide is transported majorly as
(a) sodium carbonate
(b) carboxyhaemoglobin
(c) bicarbonate
(d) carbon dioxide as such
Answer:
(c) bicarbonate

Question 12.
Most of the carbon dioxide is transported in blood as a gas
(a) gas
(b) carbaminohaemoglobin
(c) serum carbaminoproteins
(d) HCO3
Answer:
(b) carbaminohaemoglobin

Question 13.
How many molecules of oxygen can bind to a molecule of haemoglobin?
(a) One
(b) Two
(c) Three
(d) Four
Answer:

Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases

Question 14.
When carbon dioxide concentration in blood increases, breathing becomes
(a) shallower and slow
(b) there is no effect on breathing
(c) slow and deep
(d) faster and deeper
Answer:

Question 15.
The total number of lobes and alveoli present in both the lungs of man are
(a) 17 and 30 million, respectively
(b) 5 and 300 million, respectively
(c) 19 and 300 million, respectively
(d) 18 and 300 lakh, respectively
Answer:
(b) 5 and 300 million, respectively

Question 16.
What is vital capacity of our lungs?
(a) Inspiratory reserve volume plus tidal volume
(b) Total lung capacity minus expiratory reserve volume
(c) Inspiratory reserve volume plus expiratory reserve volume
(d) Total lung capacity minus residual volume
Answer:
(c) Inspiratory reserve volume plus expiratory reserve volume

Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases

Question 17.
When the oxygen supply to the tissues is inadequate, the condition is
(a) hypoxia
(b) asphyxia
(c) pleuracy
(d) anoxia
Answer:
(a) hypoxia

Question 18.
Severe Acute Respiratory Syndrome (SARS)
(a) is caused by a variant of Pneumococcuspneumoniae
(b) is caused by a variant of the common cold virus (corona virus)
(c) is an acute form of asthma
(d) affects non-vegetarians much faster than
Answer:
(b) is caused by a variant of the common cold virus (corona virus)

Question 19.
During inspiration, the diaphragm
(a) expands
(b) shows no change
(c) contracts and flattens
(d) relaxes to become dome-shaped
Answer:
(c) contracts and flattens

Plus One Zoology Chapter Wise Questions and Answers Chapter 6 Breathing and Exchange of Gases

Question 20.
The oxygen toxicity is related with
(a) blood poisoning
(b) collapsing of alveolar walls
Answer:
(c) failure of venti

Plus One Botany Chapter Wise Questions and Answers Chapter 7 Transport in Plants

Students can Download Chapter 7 Transport in Plants Questions and Answers, Plus One Botany Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus One Botany Chapter Wise Questions and Answers Chapter 7 Transport in Plants

Plus One Botany Transport in Plants One Mark Questions and Answers

Question 1.
Which of the following statements does not apply to reverse osmosis?
(a) it is used for water purification.
(b) In this technique, a pressure greater than the osmotic pressure is applied to the system
(c) It is a passive process
(d) It is an active process
Answer:
(c) It is a passive process

Question 2.
Which one of the following will not directly affect transpiration?
(a) Light
(b) wind speed
(c) temperature
(d) chlorophyll content of leaves
Answer:
(d) chlorophyll content of leaves

Question 3.
The lower surface of leaf will have a number of stomata in a
(a) isobilateral leaf
(b) dorsiventral leaf
(c) both a and b
(d) none of the above
Answer:
(b) dorsiventral leaf

Plus One Botany Chapter Wise Questions and Answers Chapter 7 Transport in Plants

Question 4.
The form of sugar transported through phloem is
(a) Fructose
(b) Sucrose
(c) glucose
(d) ribose
Answer:
(b) sucrose

Question 5.
The process of guttation takes place
(a) when the root pressure is high and the rate of transpiration is low
(b) when the root pressure is low and the rate of transpiration is high
(c) when the root pressure equals the rate of transpiration
(d) when the root pressure, as well as rate of transpiration, are high.
Answer:
(a) when the root pressure is high and the rate of transpiration is low

Question 6.
Which of the following is an example of imbibition
(a) uptake of water by root hair
(b) exchange of gases in stomata
(c) swelling of seed when putting in the soil
(d) opening of stomata
Answer:
(c) swelling of seed when putting in the soil

Question 7.
What will happen to a plant cell when kept in hypotonic solution?
Answer:
Cell become turgid

Question 8.
How are protoplasm of adjacent cells connected to each other?
Answer:
Plasmodesmata

Question 9.
Name the process in which water drops ooze out from margins of leaves.
Answer:
Guttation

Question 10.
The C4 plants are twice as efficient as C3 plants in terms of fixing C02 but lose only as much water as C3 plants for the same amount of C02 fixed.
Answer:
Half

Question 11.
Plants could be grown in a nutrient solution in the complete absence of soil. Name the technique.
Answer:
Hydroponics

Plus One Botany Chapter Wise Questions and Answers Chapter 7 Transport in Plants

Question 12.
Name the hormone which signals the closure of stomata during severe drought or severe solar radiation.
Answer:
Abscisic acid (ABA)

Question 13.
Name the physical phenomena associated with the first step of seed germination.
Answer:
Imbibition

Question 14.
What causes the leaves of grasses to roll in dry weather?
Answer:
Presence of bulliform cells or motor cells

Question 15.
The diagram below shows the cells placed in hypertonic solution and hypotonic solution. If the diagram ‘B’ is ‘turgid’ what will be diagram ‘A’?
Plus One Botany Chapter Wise Questions and Answers Chapter 7 Transport in Plants img1
Answer:
A – Flaccid

Question 16.
The excess of one element may inhibit the uptake of another element. What is this effect called?
Answer:
Toxicity of micronutrients

Question 17.
Name the type of obligatory association found in pinus seeds that promote germination.
Answer:
Mycorrhiza

Question 18.
A plant cell when kept in a certain solution got plasmolysed. What was the nature of this solution?
Answer:
Hypertonic solution

Question 19.
Why turgidity of the cell essential for plants?
Answer:
It helps to maintain plant organs in normal shape

Question 20.
Absorption of water from the soil by dry seeds increases the, thus helping seedlings to come out of the soil.
Answer:
Pressure

Question 21.
A flowering plant is planted in an earthen pot and irrigated. Urea is added to make the plant grow faster, but after some time the plant dies. Give reason.
Answer:
It is due to exosmosis.

Plus One Botany Chapter Wise Questions and Answers Chapter 7 Transport in Plants

Question. 22
Identify a type of molecular movement which is highly selective and requires special membrane proteins, but does not require energy.
Answer:
Facilitated diffusion

Question 23.
Smaller, lipid-soluble molecules diffuse faster through cell membrane, but the movement of hydrophilic substances are facilitated by certain biomolecules. Name it.
Answer:
Protein

Question. 24
The bending of wooden doors during rainy reasons is due to physiological phenomenon. Name it.
Answer:
Imbibition

Plus One Botany Transport in Plants Two Mark Questions and Answers

Question 1.
Give below is a diagram, where hypotonic solution is separated by a semipermeable membrane from a hypertonic solution and kept for a few hours.
Plus One Botany Chapter Wise Questions and Answers Chapter 7 Transport in Plants img2

 

 

  1. In which direction will the solvent move? From X to Y or Y to X.
  2. Name the process of movement of the solvent.

Answer:

  1. X to Y
  2. Osmosis

Question 2.
Differences between the following.
Diffusion and Osmosis
Answer:
1. Diffusion:
It is the movement of solute from the region of higher concentration to region of lower concentration.

2. Osmosis:
It is the movement of solvent from the region of higher concentration to region of lower concentration through semipermiable membrane

Question 3.
Differences between Guttation and transpiration
Answer:
1. Guttation:
Loss of water in the form of water droplets from the margin of leaves through hydathode.
2. Transpiration:
It is the loss of water in the form of water vapour from the surface of leaves through stomata.

Question 4.
Both the xylem and phloem are tubular structures. One is unidirectional and the other is bi-directional. Comment on it.
Answer:
Phloem is bidirectional in order to transport prepared food from leaves to different parts of the plant and translocates stored food from downward to upward for the growth of buds.

Question 5.
Ramu while observing the root nodules of pea plant, noticed the pink colour. Sita said that it is due to the presence of a substance. Can you name the substance and write its role.
Answer:
Leg hemoglobin. It is an 02 scavenger. It create anaerobic condition for enzyme Nitrogenase during N2 Fixation

Plus One Botany Chapter Wise Questions and Answers Chapter 7 Transport in Plants

Question 6.
Match the following :

Column AColumn B
1) Imbibition

2)  Exosmosis

3) Diffusion

4) Transpiration

a) Leaves

b) Perfumes

c) Dried grapes

d)  Pickles

Answer:

Column AColumn B
1) Imbibition

2)  Exosmosis

3) Diffusion

4) Transpiration

Dried grapes

Pickles

Perfumes

Leaves

Question 7.
Water is absorbed by the root hairs, It can move deeper into root layers by two distinct pathways.

  1. Write the name of the pathways.
  2. Which substance controls water transport in the endodermal region?

Answer:

  1. Apoplast pathway & Symplast pathway
  2. Suberin

Question 8.
“Root pressure is inadequate to lift the water molecules in taller plants like Eucalyptus”.

  1. Which theory is successfully explaining the Ascent of sap in plants like Eucalyptus?
  2. Who proposed this theory?

Answer:

  1. Transpiration pull – cohesion tension theory
  2. Dixon & Jolly

Question 9.
Fresh mangoes can survive fora prolonged period of time when they are placed in highly concentrated salt solution. Some physiological changes take place in the cell. Comment on the changes.
Answer:

  • Exosmosis.
  • Bacteria can not survive in hypertonic solutions.

Question 10.
Pure Honey checks the growth of microorganisms. Justify based on your knowledge in osmosis.
Answer:
Pure honey is hypertonic .bacterial cells shrink and die due to plasmolysis.

Question 11.
A fesh mango pieces is placed in water containing high concentration of sodium chloride.
Exosmosis, Plasmolysis, Deplasmolysis, Imbibition

  1. Select and write down the appropriate terms given above to represent the sequence of events leading to shrinking.
  2. What is plasmolysis?

Answer:

  1. Exosmosis – Plasmolysis
  2. The withdrawal of the protoplast from the cell wall due to exosmosis.

Question 12.
Ψws + Ψp

  1. Expand the equation.
  2. Why solute potential is always negative?

Answer:
1.

  • Ψw – Water potential
  • Ψs – Solute potential,
  • Ψp – pressure potential

2. Since adding solute reduces water potential, it is always negative.

Question 13.
Observe the given figure showing the pathways of water movement in the plant body and answer the questions.
Plus One Botany Chapter Wise Questions and Answers Chapter 7 Transport in Plants img3

  1. Identify A and B pathways.
  2. In which of the above pathways does water move beyond endodermis?

Answer:

  1. A – Apoplast pathway
    B – Symplast pathway
  2. Apoplast pathway

Question 14.
Apoplast and symplast pathways are the two distinct pathways of water to move deeper into the root layers. Which is the only pathway through which water can enter the vascular cylinder and why?
Answer:
Symplast – Because endodermal wall is suberised.

Question 15.
When a freshly collected Spirogyra filament is kept in 10% potassium nitrate solution, it is observed that the protoplasm shrinks in size:

  1. What is this phenomenon called?
  2. What will happen if the filament is replaced in distilled water?

Answer:

1. This phenomenon is called plasmolysis. (The shrinkage of protoplast from the cell wall under the influence of a strong solution/ hypertonic solution is called plasmolysis.

2. If filaments are replaced in water the protoplast starts swelling. It comes in contact with cell wall and cell regains its original size.
The swelling up of plasmolyzed protoplast under the influence of a weak solution or water is called deplasmolysis.

Plus One Botany Chapter Wise Questions and Answers Chapter 7 Transport in Plants

Question 16.
1. In an experiment, Deepa put a piece of raw mango in 0.5M salt solution and another piece in 1M salt solution. Which mango piece will get plasmolyzed fast?
2. Water potential of solution A is -0.3 bar and that of solution B is -0.1 bar. If solution A is separated from solution B by a semi-permeable membrane water molecules will flow from to
Answer:

  1. Mango piece put in 1 M salt solution get plasmolysed rapidly than. 0.5M solution.
  2. B (-1.1) to A (-0.3)

Question 17.
An unavoidable waste mechanism is found in higher plants during high temperature.

  1. Name the process.
  2. Why is it said so?

Answer:

  1. Transpiration
  2. It has merits and demerits. Since water and minerals are absorbed and translocated to distant sites by this process but 99% of water absorbed is transpired out.

Question 18.
Water moves up against gravity and even for a tree of 20 m height, the tip receives water within two hours. Which is the most important physiological phenomenon responsible for the upward movement of water is
Answer:
Transpirational pull.

Question 19.
Osmosis is a special kind of diffusion, in which water diffuses across the cell membrane. What are the factors influence rate and direction of osmosis?
Answer:
Pressure and concentration gradient.

Question 20.
When a tea bag is placed in a cup of water, which phenomenon is observed? Explain the phenomenon.
Answer:
Diffusion. It is the movement of molecules from higher concentration to lower concentration.

Question 21.
Phloem transport is bidirectional while xylem transport is unidirectional. Give a reason?
Answer:
Phloem transport is bidirectional because prepared food materials translocated to storage regions. These storage food materials are again transported to growing regions for the growth of buds. So, through phloem downward and upward transport takes place. Xylem transport is unidirectional because water and minerals are absorbed and conducted upwards.

Question 22.
Write the suitable term in each pair

  1. apoplast: through cell wall; ………: through cytoplasm
  2. endosmosis: turgid; exosmosis: ………..
  3. ………..: unidirectional; phloem transport: bidirectional
  4. Mg : macroelement: …………: microelement

Answer:

  1. Symplast
  2. Flaccid
  3. Xylem
  4. Mn

Question 23.
‘Ψw = Ψs + Ψp Expand the equation
Answer:

  • Ψw = Water potential is the sum of
  • Ψs = Solute potential solute potential
  • Ψp = pressure potential and pressure potential.

Question 24.
Bacteria cannot live in highly salted pickle. Why?
Answer:
The bacterial cell undergoes plasmolysis in saltwater. This results shrinking of protoplast. Hence, Bacteria cannot live in saltwater.

Question 25.
Observe the following diagram showing the movement of water in plant cells. Identify 1 & 2.
Plus One Botany Chapter Wise Questions and Answers Chapter 7 Transport in Plants img4
Answer:

  1. Apoplastic pathway
  2. Sym plastic pathway

Question 26.
A few Pieces of potato were weighed and kept in saltwater overnight. In the morning they are weighed again.

  1. What will be the change in weight?
  2. Name the process responsible for the change.

Answer:

  1. The weight of potato is decreased.
  2. Exosmosis results the loss of water from the potato, it leads to plasmolysis.

Plus One Botany Chapter Wise Questions and Answers Chapter 7 Transport in Plants

Question 27.
Before eating Grapes, it is advised to wash them well by keeping them in water for some time to wash off the pesticides on them. However, it is more advisable to keep the Grapes in saltwater than in fresh water to make them safer for consumption. Suggest an explanation for this.
Answer:
The grapes put in saltwater are safer for consumption because poisonous chemicals present in grapes are come out by the process of exosmosis.

Question 28.
The cytoplasm in a plant cell is surrounded by both cell wall and cell membrane. The specificity of transport of substances are mostly across the cell membrane. Give reason.
Answer:
The cell wall is freely permeable to water and substances in solutions but membrane is selectively permeable.

Question 29.
In a girdled plant, when water is supplied to the leaves above the girdle, leaves may remain green for sometime then wilt and ultimately die. What does it indicate?
Answer:
Girdling removes phloem and hence phloem transport is blocked.

Question 30.
Movement of substances in xylem is unidirectional while in phloem it is bidirectional. Why?
Answer:
Xylem conducts”Water and minerals in upward direction only but phloem conducts food in downward direction and in spring season stored food is again transported to upward for bud growth. Hence movement of substances in xylem is unidirectional while in phloem it is bidirectional.

Question 31.
Plants absorb water from the soil through root hairs by two pathways?

  1. Name the pathways
  2. Explain any one of them

Answer:

  1. Apoplastic pathway and symplastic pathway
  2. In apoplastic pathway, water transport takes place through the nonliving part of the cell (cell wall).

Question 32.
What is the water potential of pure water at atmospheric pressure? What happens to it when solutes are added to it?
Answer:
Zero, Water potential is decreased and comes to more negative values.

Question 33.
Transport of hydrophilic substances through cell membrane is difficult. Do you agree ? Give reason.
Answer:
Yes, because the cell membrane contains lipids.

Question 34.
What is meant loading and unloading of phloem?
Answer:
The entering of sugar into sieve tube is called loading and movement of sugar out of sieve tube into storage tissues is called unloading.

Question 35.
Root pressure theory is not adequate for explaining water movement up a long day plant. Why?
Answer:
It is used for explaining water movement up in herbaceous plants but in long plants, it is explained by the transpiration pull theory.

Question 36.
Why is energy required to develop root pressure?
Answer:
Every activity requires energy. Root pressure develops due to activity of living cells of the root.

Question 37.
What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution?
Answer:
When a pressure greater than atmospheric pressure is applied to pure water or a solution, its water potential is increased.

Question 38.
How is mycoorrhizal association helpful in absorption of water and mineral in plants?
Answer:
Mycorrhizal association help in water and mineral absorption. Fungal hyphae have a very large surface area that absorbs mineral ions and water from a much larger volume of soil.

Question 39.
Give an example for obligate mycorrhizal asociation.
Answer:
Mycorrhizal association is found in pinus seeds i.e Pinus seeds do not germinate and establish without mycorrhiza.

Question 40.
Downhill and uphill movement are used in transport of molecules through membrane.

  1. Which is the above process that requires energy?
  2. Which is the above process not dependent on concentration gradient. Why?

Answer:

  1. uphill transport is active transport requires energy.
  2. Downhill transport, because molecules move from the region of higher concentration to the region of lower concentration.

Plus One Botany Chapter Wise Questions and Answers Chapter 7 Transport in Plants

Question 41.
Two pathways of water movement into the deeper layers of root ie; from epidermis to xylem vessels

  1. Name the two pathways
  2. Which is the substance present in the endodermis block one pathway?

Answer:

  1. Apoplastic pathway and symplastic pathway
  2. suberin in casperian strips

Question 42.
In plants translocation of photosynthates occur in 2 ways but water and mineral transport in one way.

  1. Which is the two way suggested?
  2. Name the conducting channel for both.

Answer:
1. Food translocated from leaves to storage regions called downward transport. During spring season food again transported upward for bud growth. So such type of transport is bidirectional.

2. Food transported through phloem and water and mineral transported through the xylem.

Question 43.
In plants stomata opens during day time and close during night time.
How is decreasing water potential of guard cell affect the opening and closing of stomata?
Answer:
If decreasing the water potential of guard cell, water enters into guard cell from the subsidiary and it becomes turgid, stomata open.

Plus One Botany Transport in Plants Three Mark Questions and Answers

Question 1.
“Water is transient in plants. Less than 1% of water reaching the leaves is used in photosynthesis and plant growth’’. What is the fate of remaining percentage of water? Specify the phenomenon? Give its significance in plants?
Answer:
99% of water absorbed is lost through transpiration.

  1. It provides cooling effect to leaves and other organs.
  2. It promotes mineral absorption.
  3. It helps to maintain the shape and structure of plants by keeping up its turgidity.

Question 2.
What are the factors responsible for ascent of xylem sap in plants?
Answer:

  • Cohesion: The mutual attraction between water molecules
  • Adhesion: The attraction of molecules to polar surface e g., the surface of tracheary elements
  • Surface Tension: The water molecules are attracted to each other in liquid phase more than to water in gas phase.

All these properties provide water high tensile strength (an ability to resist a pulling force) as well as high capillarity (ability to rise in a thin tube).
In plants, capillarity is aided by the small diameter of both tracheids and vessels.

Question 3.
The solution taken in the beaker have lower solute concentration than thistle funel, osmosis starts and continues

  1. What will hapen if more solute is added in the thistle funel ?
  2. Which is the solution mentioned above have higher water potential?
  3. Which is the method possible to change flow of solvent molecules?

Answer:

  1. Diffusion takes place faster
  2. solution was taken in the beaker
  3. apply greater pressure than atmospheric pressure in solution of thistle funel.

Question 4.
Look at the following figure in which two chambers, A and B, containing solutions are separated by a semipermeable membrane. Answer the following questions.
Plus One Botany Chapter Wise Questions and Answers Chapter 7 Transport in Plants img5

  1. Solution of which chamber has a lower water potential
  2. In which direction will osmosis occur?
  3. Which solution has a higher solute potential?
  4. If one chamber has a Ψ of -2000KPa and the other -1000 KPa, which is the chamber that has the higher Ψ?

Answer:

  1. Solution in B chamber has lower water potential.
  2. Osmosis occurs from a to b.
  3. A solution has higher solute potential.
  4. The chamber having -ve 1000K Pa has the higher psi

Plus One Botany Chapter Wise Questions and Answers Chapter 7 Transport in Plants

Question 5.
Identify the following.

  1. Stress hormone
  2. The shrinking of protoplasm due to continuous exosmosis
  3. The positive pressure develops within the xylem of roots due to the absorption of water and minerals.

Answer:

  1. Abscisic acid
  2. Plasmolysis
  3. Root pressure

Plus One Botany Transport in Plants NCERT Mark Questions and Answers

Question 1.
What are the factors affecting the rate of diffusion?
Answer:

  • Factors affecting the rate of diffusion:
  • Gradient of Concentration
  • Permeability of membrane
  • Temperature
  • Pressure

Question 2.
What are porins? What role do they play in diffusion?
Answer:
The porins are proteins that form huge pores in the outer membranes of the plastids, mitochondria and some bacteria allowing molecules up to the size of small proteins to pass through. Thus porins faciliate diffusion.

Question 3.
Decribe the role played by protein pumps during active transport in plants.
Answer:
Active transport uses energy to pump molecules against a concentration gradient. Active transport is carried out by membrane – proteins.

Question 4.
Explain why pure water has maximum water potential?
Answer:
Water molecules possess kinetic energy. In liquid and gaseous form they are in random motion that is both rapid and constant. The greater the concentration of water in a system, the greater is its kinetic energy or ‘water potential’.

Hence, it is obvious that pure water will have the greatest water potential if two systems containing water are in contact, random movement of water molecules will result in net movement of water molecules from the system with higher energy to the one with lower energy. Thus water will move from the system containing water at higher water potential to having low wate potential.

This process of movement of substances down a gradient of free energy is called diffusion. Water. potential is denoted by the Greek symbol Psi and is expressed in pressure units such as pascals(Pa). By convention, the water potential of pure water at standard temperatures, which is not under any pressure, is taken to be zero.

Question 5.
What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution?
Answer:
Under constant temerature the volume of liquid varies inversely to pressure. Moreover, if greater pressure is applied then it will change the state of the matter from liquid to soild at a given temperature. Any liquid will abide by these laws.

Plus One Botany Chapter Wise Questions and Answers Chapter 7 Transport in Plants

Question 6.
Discuss the factors responsible for ascent of xylem sap in plants.
Answer:
The transpiration driven ascent of xylem sap depends mainly on the following physical properties of water:

  • Cohesion – the mutual attraction between water molecules.
  • Adhesion – the attraction of water molecules to polar surfaces (such as the surface of tracheary elements).
  • Surface Tension – water molecules are attracted to each other in the liquid phase more than to water in the gas phase.

These properties give water high tensile strength, i.e., an ability to resist a pulling force, and high capillarity, I ‘.e., the ability to rise in thin tubes. In plants capillarity is aided by the small diameter of the tracheary elements – the tracheids and vessel elements.

Plus One Botany Transport in Plants Multiple Choice Questions and Answers

Question 1.
Closure of stomata due to water stress is brought about by
(A) Auxin
(B) Ethylene
(C) Abscisic acid
(D) Cytokinin
Answer:
(C) Abscisic acid

Question 2.
The most common type of transpiration is
(A) Stomatal
(B) Cuticular
(C) Lenticular
(D) Foliar
Answer:
(A) Stomatal

Question 3.
All the following include osmosis except
(A) Passage of water from root hair to adjacent cells
(B) Passage of water from lower xylem to upper xylem
(C) Passage of water from xylem vessel to mesophyll cell
(D) The entry of water into root hair from soil solution
Answer:
(B) Passage of water from lower xylem to upper xylem

Question 4.
Stomatal opening and closure are due to
(A) Presence of gases inside leaves
(B) Gene action
(C) Pressure offered by guard cells
(D) Pressure offered by subsidiary cells
Answer:
(C) Pressure offered by guard cells

Question 5.
Translocation of sugars is correlated by
(A) Boron
(B) Iron
(C) Molybdenum
(E) Calcium
Answer:
(A) Boron

Plus One Botany Chapter Wise Questions and Answers Chapter 7 Transport in Plants

Question 6.
Identify the wrong statement of the following
(A) Root pressure is a positive pressure that develops in the xylem sap of the root of some plants.
(B) Water molecules remain attached to one another by a strong mutual force called cohesion force
(C) Guttation is frequently seen in herbaceous plants growing under the conditions of low soil moisture and low atmospheric humidity
(D) In monocot leaves, guard cells remain dumbbell-shaped and their cell walls are thickened only in the middle
Answer:
(C) Guttation is frequently seen in herbaceous plants growing under the conditions of low soil moisture and low atmospheric humidity

Question 7.
In which of the following plants, will there be no transpiration?
(A) Plants living in deserts
(B) Plants growing in hilly regions
(C) Aquatic submerged plants
(D) Aquatic plants with floating leaves
Answer:
(C) Aquatic submerged plants

Question 8.
When a cell is placed in hypertonic solution
(A) Exosmois occurs and the cell becomes turgid
(B) Endosmosis occurs and the cell becomes turgid
(C) Exosmosis occurs and the cell becomes plasmolysed
(D) Endosmosis occurs and the cell becomes plasmolysed
Answer:
(C) Exosmosis occurs and the cell becomes plasmolysed

Question 9.
The thin film of water covering soil particles and held strongly by attractive forces is called
(A) Hygroscopic water
(B) Runaway water
(C) Gravitational water
(D) Capillary water
Answer:
(A) Hygroscopic water

Question 10.
Plants die when over-fertilized because of the fertilizer
(A) Damages the wall of delicate root hairs
(B) Blocks absorption of nitrogenous ions
(C) Causes dehydration of plants by exosmosis
(D) Upsets soil environment by poisoning
Answer:
(C) Causes dehydration of plants by exosmosis

Question 11.
Movement of individual molecules of substance from a region of their higher concentration to a region of their lower concentration is called
(A) Osmosis
(B) Turgor pressure
(C) Diffusion
(D) Osmotic potential
Answer:
(A) Osmosis

Question 12.
The proteins that form large pores in the outer membranes of the plastids, mitochondria, etc are
(A) Porins
(B) Aquaporins
(C) Symporins
(D) Both A & B
Answer:
(A) Porins

Question 13.
Transport of molecules across the membrane in the same direction is
(A) Antiport
(B) Symport
(C) Permeability
(D) Uniport
Answer:
(B) Symport

Question 14.
Rate of osmosis depends on
(A) Diffusion
(B) Pressure gradient
(C) Concentration gradient
(D) BothB&C
Answer:
(D) BothB&C

Question 15.
One of the following is selectively permeable
(A) Tonoplast
(B) Suberised cell wall
(C) Copperferrocyanide membrane
(D) Cellulosic cell wall
Answer:
(A) Tonoplast

Question 16.
What will happen if a flaccid cell is placed in hypotonic solution?
(A) Plasmolysis
(B) Exosmosis
(C) Endosmosis
(D) None of these
Answer:
(C) Endosmosis

Plus One Botany Chapter Wise Questions and Answers Chapter 7 Transport in Plants

Question 17.
Pressure that occurs in a cell due to osmotic diffusion of water inside is called
(A) Osmotic pressure
(B) Turgor pressure
(C) Water pressure
(D) Diffusion pressure
Answer:
(A) Osmotic pressure

Question 18.
Movement of water through the cell wall without crossing any membrane is
(A) Apoplastic
(B) Symplastic
(C) Tonoplastic
(D) Cytoplastic
Answer:
(A) Apoplastic

Question 19.
Cytoplasmic strands that connects two neighbouring cells through
(A) Pits
(B) Semipermiable membrane
(C) Plasmodesmata
(D) Root hair
Answer:
(C) Plasmodesmata

Question 20.
Guttation occurs in leaves through
(A) Stomata
(B) Epithelium
(C) Hydathode
(D) Leaf margin
Answer:
(C) Hydathode

Question 21.
Dry seeds swells when they are placed in water due to
(A) Imbibition
(B) Plasmolysis
(C) Diffusion
(D) Osmosis
Answer:
(A) Imbibition

Question 22.
Transpiration in plants helps to
(A) The supply of water for photosynthesis
(B) Maintains cells turgidity
(C) Transport minerals from the soil to all parts of the plant
(D) All the above
Answer:
(D) All the above

Plus One Botany Chapter Wise Questions and Answers Chapter 6 Cell Cycle and Cell Division

Students can Download Chapter 6 Cell Cycle and Cell Division Questions and Answers, Plus One Botany Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus One Botany Chapter Wise Questions and Answers Chapter 6 Cell Cycle and Cell Division

Plus One Botany Cell Cycle and Cell Division One Mark Questions and Answers

Question 1.
Meiosis results in
(a) Production of gametes
(b) Reduction in the number of chromosomes
(c) Introduction of variation
(d) all of the above
Answer:
(d) all of the above

Question 2.
At which stage of meiosis does the genetic constitution of gametes is finally decided
(a) Metaphase I
(b) Anaphase II
(c) Metaphase II
(d) Anaphase I
Answer:
(d) Anaphase I

Question 3.
Meiosis occurs in organisms during
(a) Sexual reproduction
(b) Vegetative reproduction
(c) Both sexual and vegetative reproduction
(d) None of the above
Answer:
(a) Sexual reproduction

Plus One Botany Chapter Wise Questions and Answers Chapter 6 Cell Cycle and Cell Division

Question 4.
During anaphase-l of meiosis
(a) Homologous chromosomes separate
(b) Non-homologous autosomes separate
(c) Sister chromatids separate
(d) non-sister chromatids separate
Answer:
(a) Homologous chromosomes separate

Question 5.
Mitosis is characterised by
(a) Reduction division
(b) Equal division
(c) Both reduction and equal division
(d) None of the above
Answer:
(b) Equal division

Question 6.
A bivalent of meiosis-l consists of
(a) Two chromatids and one centromere
(b) Two chromatids and two centromere
(c) Four chromatids and two centromere
(d) Four chromatids and four centromere
Answer:
(c) Four chromatids and two centromere

Question 7.
Cells which are not dividing are likely to be at
(a) G1
(b) G2
(c) Go
(d) S phase
Answer:
(c) Go

Question 8.
Which type cell divisions occur in meristematic cell of root apex?
Answer:
Mitosis

Question 9.
In which stage the actual reduction of chromosome number occurs in meiosis.
Answer:
Anaphase 1

Question 10.
Give the term for the failure of separation of homologous chromosomes.
Answer:
Non-disjunction

Question 11.
Name the cell divisions which help in growth and recombination of genes.
Answer:
Mitosis and meiosis

Question 12.
It is observed that heart cells do not exhibit cell division. Such cells do not divide further and exit ____ phase to enter an inactive stage called ____ of cell cycle. Fill in the blanks.
Answer:
G1 and G0

Plus One Botany Chapter Wise Questions and Answers Chapter 6 Cell Cycle and Cell Division

Question 13.
In the case of plant cells, the formation of new cell wall begins with a simple precursor. What is this precursor?
Answer:
Middle lamella

Question 14.
It is said that the one cycle of cell division in human cells (eukaryotic cells) takes 24 hours. Which phase of the cycle, do you think occupies the maximum part of cell cycle?
Answer:
Interphase.

Question 15.
At what stage of cell cycle does DNA synthesis take place?
Answer:
substage of Interphase

Question 16.
If the failure of division of cytoplasm occurs after nuclear division, What will happen to the cell?
Answer:
Free nucleii are formed

Question 17.
An anther has 1200 pollen grains. How many pollen mother cells must have been there to produce them?
Answer:
300 pollen mother cells

Question 18.
What is the peculiarity of zygotene?
Answer:
The pairing of the homologous chromosome called synapsis

Plus One Botany Chapter Wise Questions and Answers Chapter 6 Cell Cycle and Cell Division

Question 19.
It is the inactive stage of cell division but cell differentiation occurs. Name it.
Answer:
G0 Phase /Quiscent stage.

Plus One Botany Cell Cycle and Cell Division Two Mark Questions and Answers

Question 1.
Meiosis is the type of cell division which maintain the race. Discuss.
Answer:
Reduces the chromosome number to half, so that chromosome number is maintained in the next generation.

Question 2.
Interphase in cell cycle is sometimes referred to as resting phase. Do you consider this statement true? Substantiate your answer.
Answer:
No.

  1. Nucleus and cytoplasm are metabolically very active.
  2. Amount of DNA becomes doubled.

Question 3.
A diagram of typical cell cycle of a higher plant is shown here. Identify each stage of the cycle and explain what happens during these stages.
Plus One Botany Chapter Wise Questions and Answers Chapter 6 Cell Cycle and Cell Division img1
Answer:

  • G1 – pre mitotic gap -synthesis of RNA &Proteins
  • S – phase of synthesis – DNA replication
  • G2 – Post mitotic Gap phase synthesis of RNA &Proteins continues
  • M – Mitotic phase
  • G0 – Inactive phase

Question 4.
Cytokinesis differs in plant and animal cell. Substantiate this statement.
Answer:

  • Plant cell – formation of cell plate.
  • Animal cell – Invagination of cell membrane.

Question 5.
Analyse the column A and B arrange the matter is an appropriate order.

AB
1)  Pachytene

2)  S phase

3) Dyad

4) M phase

a) Interphase

b) Telophase I

c)  Mitotic phase

d) Prophase I

Answer:

AB
1)  Pachytene

2)  S phase

3)  Dyad

4)  M phase

Prophase I

Interphase

Telophase I

Mitotic phase

Question 6.
Identify the stages of Mitosis in which the following events take place:

  1. Synapsis
  2. Terminalisation of chiasmata.

Answer:

  1. Zygotene
  2. Diakinesis

Plus One Botany Chapter Wise Questions and Answers Chapter 6 Cell Cycle and Cell Division

Question 7.
Different stages of Prophase I of Meiosis are given in column A, arrange them in correct order and match them with the events in column B.

AB
i) Diplotenea) Chromosomes become gradually visible under the light microscope
ii) Pachyteneb) The pairing of Homologous chromosomes
iii) Leptotenec) The appearance of recombination nodules and crossing over takes place
iv) Zygotened) Dissolution of synaptonemal complex and separation of bivalents.
e) Terminalisation of Chiasmata

Answer:

  1. (i) – d
  2. (ii) – c
  3. (iii) – a
  4. (iv) – b

Question 8.
The pairing of homologous chromosomes is called synapsis.

  1. Name each pair of homologous chromosomes.
  2. Name the stage of prophase at which it takes place.

Answer:

  1. Bivalent
  2. Zygotene

Question 9.
Specific chromosome number of each species is conserved across generations in sexually reproducing organisms. What is the reason for this? Write the different steps of this process.
Answer:

  1. Meiosis (meiosis I & II)
  2. Prophase Meta phase Ana phase Telophase

Question 10.
The life cycle of a cell is called cell cycle. It consists of four stages such as Gv S, G2, and M.

  1. Construct a pie diagram showing the different stages indicated above.
  2. State the major events occurring in G, S, and G2 phases.

Answer:
1. Pie diagram of cell cycle.
Plus One Botany Chapter Wise Questions and Answers Chapter 6 Cell Cycle and Cell Division img2
2. G1 Phase -1, Cell grows in size and prepares the machinery needed for the DNA replication. RNA and proteins are synthesized. S phase – DNA replication. G2 phase – Synthesis of RNA and proteins.

Question 11.
Arrange the following stages of cell cycle in correct sequence S, G2, G1, M.
Answer:
S, M, G1, S, G2

Question 12.
X’ shaped structure called ‘chiasmata occurs during a particular stage of cell division.

  1. Name the stage?
  2. What is the significance of this type of cell division?

Answer:

  1. Pachytene
  2. The exchange of genes takes place between homologous chromosomes.

Question 13.
Given below are the five phases of prophase I of Meiosis I. Arrange them in correct order.
Zygotene, diakinesis, diplotene, leptotene, pachytene
Answer:
Leptotene, Zygotene, Pachytene, diplotene and diakinesis

Question 14.
Give the scientific term of the following.

  1. Interchange of genetic material between non-sister chromatids of the homologues chromosomes
  2. The plane of alignment of the chromosomes at metaphase

Answer:

  1. Crossing over
  2. Equatorial

Question 15.
Identify the diagram and label a, b,c and write the events during this.
Plus One Botany Chapter Wise Questions and Answers Chapter 6 Cell Cycle and Cell Division img3
Answer:
(a)G1
(b) S
(c)G2

  • G1 – Interval between mitosis and initiation of DNA replication.
  • S – DNA synthesis or replication of DNA occurs.
  • G2 – In this phase proteins are synthesised for mitosis.

Plus One Botany Chapter Wise Questions and Answers Chapter 6 Cell Cycle and Cell Division

Question 16.
You have supplied I set of glass slides showing gametogenesis or garnets formation in an animal In one of the slide you observed the following features. Four cells with haploid number of chromosomes. Your friend told you that this is a meiotic division. (Hint: The diploid number of chromosome is 16.)
Are you agree with this statement. Justify your answer.
Answer:
Yes. Meiosis takes place in diploid cell or meiocyte to form four haploid cells. These cells contain 8 chromosomes each.

Question 17.
Differentiate reduction division from equational division.
Answer:
1. Reduction division:
It occurs in diploid cells to form 4 haploid cells, ie, the Chromosome number reduced to half in Meiosis I and results 2 daughter cells, which again divides to form 4 daughter cells. All cells formed in meiosis are haploid.

2. Equational division:
It is the mitotic division results 2 daughter cells carrying same set of chromosomes as that of parent cell. No genetic variation occurs.

Question 18.
Can there be mitosis without DNA replication in S phase?
Answer:
There cannot be mitosis without DNA replication in 5 phase of interphase because the trigger for mitosis is disturbance of nucleocytoplasmic ratio caused by DNA replication in S phase. Mitosis restores the quantity of genetic material to the species-specific level.

Question 19.
How does anaphase of mitosis and anaphase I of meiosis differ from each other?
Answer:
In anaphase of mitosis chromatids separate while in anaphase 1 of meiosis homologous chromosomes separate.

Question 20.
How does cytokinesis in plant cells differ from that in animal cells?
Answer:
1. In an animal cell a furrow in the plasma membrane.lt joins in the centre and dividing the cell cytoplasm into two.

2. In-plant cells, wall formation starts in the centre of the cell and grows outward to meet the existing lateral walls. Then Cell division occurs.

Question 21.
How cytokinesis is different in an animal and a plant cell?
Answer:
In-plant cell cytokinesis occurs by cell plate formation while in the animal cell it occurs by cell furrow formation.

Question 22.
Why mitosis is called equational division ? Give the occurrence of mitosis.
Answer:
It keeps the chromosome number constant. It occurs in somatic cells.

Question 23.
What is the feature of a metacentric chromosome?
Answer:
The metacentric chromosome has a centromere in the middle region with two equal arms of the chromosome.

Question 24.
What is kinetochore? Give its function.
Answer:
It is a disc-like area in each chromatid and is site of attachment of spindle microtubule.

Question 25.
Why is meiosis essentially in sexually reproducing organisms?
Answer:
Meiosis reduces the chromosome number to half as it is followed by fertilization which restores diploidy.

Plus One Botany Chapter Wise Questions and Answers Chapter 6 Cell Cycle and Cell Division

Question 26.
Name the stage of cells cycle at which one of the following events occurs.

  1. Chromosomes are moved to spindle equator
  2. Centromere splits and chromatids separate
  3. Pairing between homologous chromosomes. takes place
  4. Crossing over between homologous chromosomes takes place.

Answer:

  1. Metaphase
  2. Anaphase
  3. Zygotene
  4. Pachytene

Question 27.
Downs syndrome and Klinefelter’s syndrome occurs due to mistake in cell division. What does it indicate?
Answer:
It is due to the failure of separation of homologous chromosomes during meiosis.

Question 28.
The events occur in prophase and telophase are one opposite to other,

  1. Name the cell structures shown the above events
  2. How many daughter nuclei are formed at the end of mitotic and meiotic prophase?

Answer:

  1. Nuclear membrane, nucleolus and spindle fibres
  2. Two

Question 29.
Mangolism or Trisomy is due to the failure of one event in cell division.

  1. Which is the improperly functioning stage?
  2. What is the event that not occurs?

Answer:

  1. Anaphase I of meiosis
  2. Separation of homologous chromosomes

Question 30.

  1. What will be the ploidy level of dyad and tetrad of cells in meiosis?
  2. How is it occurs?

Answer:

  1. Dyad – haploid, Tetrad – Haploid
  2. It occurs in meiosis due to separation of homologous chromosomes and chromosome number reduced to half.

Plus One Botany Cell Cycle and Cell Division Three Mark Questions and Answers

Question 1.
In which phase of meiosis is the following formed? Choose the answers from hint points given below.

  1. Synaptonemal complex ______
  2. Recombination nodules ______
  3. Appearance/activation of enzyme recombinase _____
  4. Termination of chiasmata _______
  5. Interkinesis ________
  6. Formation of dyad of cells ________

Answer:

  1. Zygotene
  2. pachytene
  3. pachytene
  4. diakinesis
  5. After Meiosis-I before meiosis II
  6. after first cytokinesis

Plus One Botany Chapter Wise Questions and Answers Chapter 6 Cell Cycle and Cell Division

Question 2.
The interphase stage is significant in mitotic and meiotic cell division

  1. Give one specific event
  2. Name the stage of interphase this event occurs
  3. How will you differentiate interphase from interkinesis?

Answer:

  1. DNA replication
  2. S – phase
  3. Interphase – cell prepares for cell division Interkinesis – short interval between meiosis I and meiosis II

Question 3.
The following events occur during the various phases of the cell cycle, Write the phase against each of the events.

  1. Appearance of nucleolus
  2. Division of centromere
  3. Replication of DNA

Answer:

  1. Telophase
  2. Anaphase
  3. Interphase

Question 4.
Life cycle of a cell is called cell cycle. ‘S’ phase is an important phase of cell cycle.

  1. Justify your answer.
  2. Name the stages of cell cycle at which the following events occur.
    • Crossing over of homologous chromosome.
    • Pairingof homologous chromosomes.
    • Chromosomes are arranged at the equatorial plane.

Answer:

  1. phase of DNA synthesis
  2. stages of the cell cycle
    • Pachytene
    • Zygotene
    • Metaphase

Question 5.
Name the stages of cell division in which the following events occur?

  1. Chromosomes are moved to spindle equator.
  2. Centromere splits and chromatids separate.
  3. Crossing over between homologous chromosomes takes place.

Answer:

  1. Metaphase
  2. Anaphase
  3. Pachytene

Plus One Botany Chapter Wise Questions and Answers Chapter 6 Cell Cycle and Cell Division

Question 6.
Match the words listed in column I with suitable words from column II.

Column IColumn II
a) Diplontic cycle

b) Karyokinesis

c) Haplontic cycle

d) Cytokinesis

e) Meiosis

f)  Cell plate

i)  Meiocytes

ii) Gametic meiosis

iii)  Plant cells

iv) Nuclear division

v) Zygotic meiosis

vi) Cytoplasmic division

Answer:

  1. a) – Gametic meiosis
  2. b) – Nuclear division
  3. c) – Zygotic meiosis
  4. d) – Cytoplasmic division
  5. e) – Meiocytes
  6. f) – Plant cells

Plus One Botany Cell Cycle and Cell Division NCERT Mark Questions and Answers

Question 1.
Distinguish cytokinesis from karyokinesis.
Answer:
The division of cytoplasm is called cytokinesis, while the division of the nucleus is called Karyokinesis.

Question 2.
What is G0 (quiescent phase) of cell cycle?
Answer:
Some cells in the adult animals do not appear to exhibit division (e.g., heart cells and many other cells divide only occasionally, as needed to replace cells that have been lost because of injury or cell death.

These cells do not divide further exit G1 phase to enter an inactive stage called quiescent stage (G0) of the cell cycle. Cells in this stage remain metabolically active but no longer proliferate unless called on to do so depending on the requirement of the organism.

Question 3.
Describe the event taking place during interphase.
Answer:
The interphase is divided into three further phases:
1. G1 phase (Gap 1). G1 phase corresponds to the interval between mitosis initiation of DNA replication. During G, phase the cell is metabolically active and continuously grows but does not replicate its DNA.

2. S phase (Synthesis). S or synthesis phase marks the period during which DNA synthesis or replication takes place.

3. During this time the amount of DNA per cell doubles. If the initial amount of DNA is denoted as 2C then it increases to 4C. However, there is no increase in the chromosome number, if the cell had diploid or 2n number of chromosomes at G1, even after S phase the number of chromosomes remains the same, i.e., 2n.

4. G2 phase (Gap 2). In animal cells, during the S phase, DNA replication begins in the nucleus, and the centriole duplicates in the cytoplasm during the G2 phase, proteins are synthesised in preparation for mitosis White cell growth continues.

Plus One Botany Chapter Wise Questions and Answers Chapter 6 Cell Cycle and Cell Division

Question 4.
Why is mitosis called equational division?
Answer:
Since the number of chromosomes remains same in parent and daughter cells so mitosis is also called a equational division.

Question 5.
Name the stage of cell cycle at which one of the following events occur.

  1. Chromosomes are moved to spindle equator.
  2. Centromere splits and chromatids separate.
  3. Pairing between homologous chromosomes takes place.
  4. Crossing over between homologous chromosomes takes place.

Answer:

  1. Metaphase
  2. Anaphase
  3. Metaphase I of meiosis
  4. Prophase I of meiosis

Question 6.
What is the significance of meiosis?
Answer:
Significance of Meiosis:

  1. Maintaining genetic identity by maintaining number of chromosomes.
  2. Bringing variations to ensure better species.
  3. Facilitates sexual reproduction.

Question 7.
Discuss with your teacher about.

  1. haploid insects and lower plants where cell division occurs, and
  2. Some haploid cells in higher plants where cell division does not occur.

Answer:

  1. Male bees, wasps and ants are haploid organisms because they are produced from unfertilized eggs.
  2. Synergids and antipodal cells in the ovule don’t undergo cell division.

Question 8.
Can there be mitosis without DNA replication in ‘S’ phase?
Answer:
DNA replication is necessary for cell division, and cell division cannot happen without DNA replication.

Question 9.
Can there be DNA replication without cell division?
Answer:
DNA replication takes place in order to prepare for cell division. Cell division is the next logical step after DNA replication.

Question 10.
Analyse the events during every stage of cell cycle and notice how the following two parameters change

  1. number of chromosomes (N) per cell
  2. amount of DNA content (C) per cell

Answer:

  1. Number of chromosomes remains same after mitotic cell division and becomes half after meiotic cell division.
  2. During S phase the DNA content doubles, but number of chromosomes remains the same.

Plus One Botany Cell Cycle and Cell Division Multiple Choice Questions and Answers

Question 1.
Cleavage is a unique form of mitotic cell division in which
(a) there is no growth of cells
(b) the nucleus does not participate
(c) no spindle develops to guide the cells
(d) the plasma membranes of daughter cells do not separate.
Answer:
(a) there is no growth of cells

Plus One Botany Chapter Wise Questions and Answers Chapter 6 Cell Cycle and Cell Division

Question 2.
In animal cells, cytokinesis involves
(a) the separation of sister chromatids
(b) contraction of the contractile ring of microfilament
(c) depolymerisation of kinetochore microtubules
(d) a protein kinase that phosphorylates other enzymes
Answer:
(b) contraction of the contractile ring of microfilament

Question 3.
During mitosis, the number of chromosomes gets
(a) change
(b) no change
(c) maybe change if cell is mature
(d) maybe change if cell is immature
Answer:
(b) no change

Question 4.
A diploid living organism develops from zygote by which type of the following repeated cell divisions?
(a) Meiosis
(b) Amitosis
(c) fragmentation
(d) Mitosis
Answer:
(d) Mitosis

Question 5.
If you are provided with root-tips of onion in your class and are asked to count the chromosomes, which of the following stages can you most conveniently look into?
(a) Metaphase
(b) Telophase
(c) Anaphase
(d) Prophase
Answer:
(a) Metaphase

Question 6.
At which stage of mitosis, chromatids separated and passes to different poles
(a) prophase
(b) Metaphase
(c) anaphase
(d) Telophase
Answer:
(c) anaphase

Question 7.
The two chromatids of a metaphase chromosome represent
(a) replicated chromosomes to be separated at anaphase
(b) homologous chromosomes of a diploid set
(c) non-homologous chromosomes joined at the centromere
(d) maternal and paternal chromosomes joined at the centromere
Answer:
(a) replicated chromosomes to be separated at anaphase

Question 8.
The process of cytokinesis refers to the division of
(a) nucleus
(b) chromosomes
(c) cytoplasm
(d) nucleus and cytoplasm
Answer:
(c) cytoplasm

Question 9.
Which of the following serves as mitotic spindle poison?
(a) Ca2
(b) azide
(c) Tubulin
(d) Colchicine
Answer:
(d) Colchicine

Question 10.
Pairing of homologous chromosomes occurs at which stage?
(a) Zygotene
(b) Leptotene
(c) Metaphase
(d) Pachytene
Answer:
(a) Zygotene

Question 11.
In meiosis, division is
(a) I reductional and II equational
(b) I equational and II reductional
(c) Both reductional
(d) Both equational
Answer:
(a) I reductional and II equational

Plus One Botany Chapter Wise Questions and Answers Chapter 6 Cell Cycle and Cell Division

Question 12.
Which type of chromosomes segregate when a cell undergoes meiosis?
(a) Homologous chromosomes
(b) Non-homologous chromosomes
(c) Both (a) and (b)
(d) centric and acentric chromosomes
Answer:
(a) Homologous chromosomes

Question 13.
Chiasmata are most appropriately observed in meiosis during
(a) diakinesis
(b) diplotene
(c) metaphase-ll
(d) pachytene
Answer:
(b) diplotene

Question 14.
During cell division, sometimes there will be failure of separation of homologous chromosomes. This event is called
(a) interference
(b) complementation
(c) non-disjunction
(d) coincidence
Answer:
(c) non-disjunction

Question 15.
The second meiotic division leads to
(a) separation of sex chromosomes
(b) fresh DNA synthesis
(c) separation of chromatids and centromere
(d) separation of homologous chromosomes.
Answer:
(c) separation of chromatids and centromere

Question 16.
Term meiosis was proposed by
(a) Farmer and Moore
(b) Flemming
(c) Strasburger
(d) Darlington
Answer:
(a) Farmer and Moore

Plus One Botany Chapter Wise Questions and Answers Chapter 6 Cell Cycle and Cell Division

Question 17.
Synapsis occurs in the phase of meiosis.
(a) zygotene
(b) diplotene
(c) pachytene
(d) leptotene
Answer:
(a) zygotene

Question 18.
When the number of chromosomes is already reduced to half in the first reductional division of meiosis, where is the necessity of second meiotic division
(a) The division is required for the formation of four gametes
(b) Division ensures equal distribution of haploid chromosomes
(c) Division ensures equal distribution of genes on chromosomes
(d) Division is required for segregation of replicated chromosomes
Answer:
(d) Division is required for segregation of replicated chromosomes

Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption

Students can Download Chapter 5 Digestion and Absorption Questions and Answers, Plus One zoology Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption

Plus One Digestion and Absorption One Mark Questions and Answers

Question 1.
Hepato-pancreatic duct opens into the duodenum and carries
(a) Bile
(b) Pancreatic juice
(c) Both bile and pancreatic juice
(d) Saliva
Answer:
(c) Both bile and pancreatic juice

Question 2.
One of the following is not a common disorder associated with digestive system
(a) Tetanus
(b) Diarrhea
(c) Jaundice
(d) Dysentery
Answer:
(a) Tetanus

Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption

Question 3.
A gland not associated with the alimentary canal is
(a) Pancreas
(b) Adrenal
(c) Liver
(d) Salivary gland
Answer:
(b) Adrenal

Question 4.
Liver is the largest gland and is associated with various functions, choose one which is not correct
(a) Metabolism of carbohydrate
(b) Digestion of fat
(c) Formation of bile
(d) Secretion of  hormone called gastric
Answer:
(d) Secretion of hormone called gastric

Question 5.
Mark the right statement among the following
(a) Trypsinogen is an inactive enzyme
(b) Trypsinogen is secreted by intestinal mucosa
(c) Enterokinase is secreted by pancrease
(d) Bile contains trypsin
Answer:
(a) Trypsinogen is an inactive enzyme

Question 6.
The major health and nutritional problems mainly arises due to low amount of minerals and vitamins in diet. Name the two PEMs.
Answer:
Kwashiorkor, Marasmus

Question 7.
Deficiency symptoms of a disease are given below: Identify it.
Thin Limbs, retarded growth of body and brain, swelling of legs due to retention of water (oedema), reddish hair, pot belly and diarrhea.
Answer:
Kwashiorkor

Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption

Question 8.
Find the set most suitable with “succus entericus”
(a) bilirubin and biliyerdin
(b) salivary amylase and lysozymes
(c) disaccharides, dipeptidases and lipases
(d) Trypsinogen, chymotrypsinogen procarboxypep-tidases
Answer:
(c) Disaccharides, dipeptidase and lipases

Question 9.
Complete the columns a, b, c, d, e & f

  1. Maltose _______a_______ + ________b_________
  2. Sucrose _______c_______ + ________d_________
  3. Lactose _______e_______ + ________f__________

Answer:

  1. Meltose Glucose + Glucose
  2. Sucrose Glucose + Fructose
  3. Lactose Glucose + Galactose

Question 10.
From this which statement is correct?
(a) Human denititian is heterodont
(b) Fat is soluble in water
(c) Fructose absorbed passively
(d) Small intestine devoid of villi
Answer:
(a) Human dentitian is heterodont

Question 11.
The human intestine is long because
(a) Bacteria in food can be killed gradually
(b) It increases surface are for absorption of food
(c) It provides more space for food storage
(d) None of these
Answer:
(b) It increases surface area for absorption of food.

Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption

Question 12.
Which of the following statement is most suitable for gastrin?
(a) It stimulates the secretion of gastric juice.
(b) It stimulates the secretion of bile.
(c) It inhibits gastric gland to produce gastric secretion.
Answer:
(a) It stimulates the secretion of gastric juice.

Question 13.
Name the end product of digestion of following?

  1. Starch
  2. Protein

Answer:

  1. Glucose
  2. Amino acids

Plus One Digestion and Absorption Two Mark Questions and Answers

Question 1.
Copy the diagram and label the parts.
Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption - 1
Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption - 2

Question 2.
A 15 year old boy was admitted in a hospital with severe case of stomach ulcer. At the end of various tests it is found that Mucus glands are absent. Do you agree with this? Justify your answer.
Answer:
Yes. Highly concentrated HCL is presented in gastric juice. Mucus from the mucus gland protect the epithelium lining of the stomach from the action of HCL. So the absence of mucus gland cause stomach ulcer.

Question 3.
Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption - 3

  1. Name the process (a).
  2. Name the enzyme (b).
  3. Name the product (c).

Answer:

  1. (a) – bile salt – emulsification
  2. (b) – pancreatic lipase
  3. (c) – Glycerol

Question 4.
Distinguish between Renin and Rennin.
Answer:
1. Rennin: A proteolytic enzyme found in gastric juice of infants which helps in the digestion of milk proteins.

2. Renin: It is an enzyme releasing from JGA, when there is a fall in glomerular filtration rate. Renin converts angiotensinogen into antigotensin II and regulate urine formation.

Question 5.
Complete the passage.
Fatty acids and glycerol are first incorporated into small droplets called _______ which move into intestinal mucosa. They are reformed into very small _______ coated fat globules called _________ which are transported to the __________ in the villi.
Answer:
Fatty acids and glycerol are first incorporated into small droplets called micelles which move into intestinal mucosa. They are reformed into very small Protein coated fat globules called Chylomicrons which are transported to the lacteals in the villi.

Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption

Question 6.

  1. It is not necessary to produce amylase in an inactive form in our body, but it is not in the case of Trypsin.
  2. In certain circumstances, peptic ulcer are found on the gut wall.

Answer:

1. Proteins are body building molecules. If proteins digesting enzymes are secreted inactive forms, they would hydrolyse the cellular and extracellular proteins of the organism itself. So the protein-digesting enzyme trypsin is secreted in their inactive form.

2. Mucus and bicarbonates present in the gastric and intestinal secretion protects the gut wall from the action of HCL and protein-digesting enzymes. In the absence of mucus and bicarbonates, peptic ulcer are found on the gut wall.

Question 7.
Construct a flow chart showing the passage of food through the alimentary canal of man.
Answer:
Mouth → Pharynx → Oesophagus → Stomach → Small intestine → Large intestine → Rectum → Anus.

Question 8.
Differentiate between Serosa and Lumen.
Answer:
The outermost layer of the wall of alimentary canal is called serosa. The inner hollow space in the alimentary canal is called lumen.

Question 9.
Find the odd one out and justify your answer.
Salivary gland, milk gland, gastric gland, thyroid gland.
Answer:
Thyroid gland. Thyroid gland is an endocrine gland. All others are exocrine gland.

Question 10.
Give below in bracket are the names of some digestive enzymes group them into A – Gastric, B – Pancreatic and C – Intestinal (Trypsin, Dipeptidase, Rennin, Maltase, Pepsin, Chymotrypsin, Nuclease, Amylase)
Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption - 4

Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption

Question 11.
Ramu ate boiled rice, what are the changes that it undergoes before being absorbed in the small intestine.
Answer:
Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption - 5

Question 12.
The dental formula of human is represented as \(i \frac{2}{2}\), \(\mathrm{C} \frac{1}{1}, \mathrm{Pm} \frac{2}{2}, \mathrm{m} \frac{3}{3}\). What do i, c, pm, m indicate?
Answer:
i – Incisors, c-Canines, pm – Pre molars, m – Molars

Question 13.
Describe the Diphyodont and Thecodont characteristics of human dentition.
Answer:
1. Thecodont: It is a dentition in which the teeth are placed in the sockets of jawbones.

2. Diphyodont: It is a dentition in which the teeth appear twice in the lifetime of man. They are milk teeth and permanent teeth.

Question 14.
Observe the figure given below and answer the following questions.
Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption - 6

  1. ‘x’ is a substance needed for digestion. What is the function of ‘x’?
  2. Name the gastrointestinal hormone responsible for the contraction of gall bladder to release ‘x’.

Answer:

  1. Bile – emulsification
  2. Cholecystokinin

Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption

Question 15.
Given below are the layers of human gut. Write a flow chart from inner region.
Serosa, mucosa, muscularis, submucosa
Answer:
Mucosa → Submucosa → Muscularis → Serosa

Question 16.
After observing the model of the lumen of human intestine your friend said that “the villi projected into the lumen actually block the movement of food through the intestine, if the villi are absent food can easily pass through the intestine.” How will you respond to this comment?
Answer:
Villi actually decrease the movement of chyme. The function of villi is to increase the surface area of the intestine for the absorption of digested nutrients.

Question 17.
Out of the following list, pick out the enzymes that take part in carbohydrate digestion.
Maltase, Lactase, Sucrase, Peptidase, Salivary amylase, Isomaltase.
Answer:
Maltase, Lactase, Sucrase, Salivary amylase, Isomaltase

Question 18.
Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption - 7

  1. Mention the factor ‘x’ involved in the process.
  2. Identify the process and write its importance.

Answer:

  1. Bile salts
  2. Emulsification. The conversion large fat globules into tiny fat droplets is called emulsification. Bile salt helps emulsification. Emulsification increase the surface area for lipase action.

Question 19.
Below are given the names of the enzymes and their source of secretion. Match them.

EnzymesSource of secretion
AmylaseStomach
NucleasePancreas
LipaseSmall intestine
ProteaseSalivary gland

Answer:

EnzymesSource of secretion
AmylaseSalivary gland
NucleaseSmall intestine
LipasePancreas
ProteaseStomach

Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption

Question 20.
Strong acidic food in the stomach stimulate the secretion of secretin from duodenal wall.

  1. On which part of the digestive tract, secretion act?
  2. What is the role of secreting in digestion?

Answer:

  1. Secretin acts on exocrine part of pancreas.
  2. Secretin stimulates secretion of water and bicarbonate ions from pancreas. It protects the intestinal mucosa from acid as well as provide an alkaline medium for enzymatic activity in intestine.

Question 21.
Fatty chyme in duodenum decreases the speed of peristalsis. How the fatty chyme affects the speed of peristalsis?
Answer:
Fatty chyme in duodenum stimulate the release of GIP (Gastric Inhibitory Peptide Hormone). GIP produced by duodenum, inhibit gastric mobility.

Question 22.
Mention whether the following statements true or false. Justify your answer.

  1. Chymotrypsin is a fat digesting enzyme.
  2. Deglutition occurs in the intestine.
  3. Intestinal juice is also known as succus entericus.
  4. Function of enterokinase is to break peptide bond in proteins.

Answer:

  1. False – It is prominent protein digesting enzyme present in pancreaticjuice.
  2. False – It is the process of swallowing of food . from the pharynx to oesophagus.
  3. True – The collective secretion of intestinal gland is known as intestinal juice or succus entericus.
  4. False – Enteokinase present in succus entericus activates the inactive trypsinogen into active trypsin.

Question 23.
The following diagram shows the action of protein digesting enzymes of pancreas.
Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption - 8

(a) Label the number 1, 2 and 3.
(b) Trypsinogen and Chymotrypsinogen are proenzymes (inactive enzymes). Why do they being produced inactive?

Answer:
(a) Label the number 1, 2 and 3.

  1. Enterokinase
  2. Chymotrypsin
  3. Peptides

(b) Trypsin and Chymotrypsin are protein digesting enzyme formed from their respective inactive forms at the time of digestion of the protein food. If they are available all time in active form, they would damage the intestinal wall.

Question 24.
The calorific value of respiratory substrates are different. Do you agree? Give reasons.
Answer:
Yes, The calorific values of carbohydrates, proteins and fats are 4.1 kcal /g, 5.65 kcal /g and 9.45 kcal /g, respectively.

Plus One Digestion and Absorption Three Mark Questions and Answers

Question 1.
Gastric glands are formed of three types of secretory cells. Name them and mention their function.
Answer:

  1. Mucous cell – Secrete mucus
  2. Chief cells – Secrete proteolytic enzyme Pepsin
  3. Parietal cells – Secrete HCL

Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption

Question 2.
Observe the flow chart and answer the questions.
Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption - 9

  1. What is the purpose of micelles in fat absorption?
  2. Identify ‘e’.

Answer:
1. Fatty acids and glycerol being insoluble, cannot be absorbed into the blood. They are first incorporated into small droplets called micelles which move into the intestinal mucosa. Micelles are formed with the help of bile salts and phospholipids,

2. Chylomicrons

Question 3.
Observe and analyse the graph and answer the following questions.
Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption - 10

  1. Which region of the alimentary canal has maximum acidic pH?
  2. Why different regions of alimentary canal has different pH How this is maintained?

Answer:
1. Stomach

2. pH is the buccal cavity – in between 6.5 and 7, which is maintained by HCO3 – in the saliva. pH in the stomach is in between 1.8 and 2, which is maintained by HCL, secreted from gastric gland. The pH in the intestine is in between 8 and 9, which is maintained by HCO3– present in the bile and pancreatic juice.

Question 4.
Observe the given diagram.
Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption - 11

  1. Identify the parts a, b, c.
  2. If part ‘c’ has removed from a person due to a disease. What will be the changes in the digestive process of that person?

Answer:
1. a – Liver, b – Gall bladder, c – Pancreas

2. If the pancreas is totally removed from a person, he will die in a short period. Even if pancreas is only partially removed it will collapse the digestive process. In addition to this, the body cannot regulate the glucose level in blood.

The digestive juice secreted by pancreas contains three enzymes trypsin, amylase and lipase. Thus if pancreas is removed, the production of digestive enzyme will collapse.

Question 5.
Complete the flow chart.
Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption - 12
Answer:
(a) Emulsification
(b) Pancreatic lipase
(c) Fatty acids
(d) Glycerol
(e) Micelles
(f) Chylomicrons

Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption

Question 6.
Observe the diagram. Complete the boxes A, B and C appropriately.
Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption - 13
Answer:

Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption - 14

Question 7.
Observe the diagram.
Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption - 15

  1. Identify the ducts A and B.
  2. B is guarded by a sphincter in its opening to duodenum. Name it.
  3. Write any two functions of liver.

Answer:

  1. Ducts A and B are
    • A – Common bile duct
    • B – Hepato – pancreatic duct
  2. Sphincter of Oddi.
  3. Functions of liver are:
    • Production and secretion of bile
    • Production of albumin, Fibrinogen, Prothrombin, etc.
    • Detoxification.

Question 8.
Given the diagram of intestinal villus.
Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption - 16

  1. Identify the part labelled as ‘a’ and ‘b’ in figure.
  2. Give the function of part ‘a’.
  3. What is the role of ‘b’ in digestion?

Answer:

  1. Identify the part labelled as ‘a’ and ‘b’ in figure
    • ‘a’ Lacteal
    • ‘b’ Crypts of Liberkuhn
  2. Absorption of fat and fat soluble vitamins.
  3. Crypts of Lieberkuhn produce intestinal juice helps in digestion.

Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption

Question 9.
Find the odd one out from each group and justify.

  1. Cardiac, Villi, Fundic, Pyloric
  2. Duodenum, Jejunum, ileum, villi
  3. Parotid, gastric, sublingual, submaxillary

Answer:

1. Villi:
Cardiac, fundic and pyloric are the three major parts of stomach. But villi are small finger like foldings in the intestine.

2. Villi:
Duodenum, jejunum and ileum are three regions of intestine but villi are small finger like foldings inside the intestine.

3. Gastric:
Parotid, Sublingual and submaxillary are three pairs of salivary gland produce saliva. But gastric gland is stomach gland produce gastric juice.

Plus One Digestion and Absorption NCERT Questions and Answers

Question 1.
Choose the correct answer among the following:

  • Gastricjuice contains
    1. pepsin, lipase and rennin
    2. trypsin, lipase and rennin
    3. trypsin, pepsin and lipase
    4. trypsin, pepsin and rennin
  • Succus entericus is the name given to
    1. a junction between ileum and large intestine
    2. intestinal juice
    3. swelling in the gut
    4. appendix

Answer:

  • Gastricjuice contains
    1. Pepsin, Lipase and Rennin
  • Succus entericus is the name given to
    2. Intestinal Juice

Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption

Question 2.
Match column I with and Rennin
Column I Column II

Column IColumn II
(a) Bilirubin and biliverdin(i) Parotid
(b) Hydrolysis of starch(ii) Bile
(c) Digestion of fat(iii) Lipases
(d) Salivary gland(iv) Amylases

Answer:
(a) – (ii), (b) – (iv), (c) – (iii), (d) – (i)

Question 3.
Why are villi present in the intestine and not in the stomach?
Answer:
The major portion of absorption takes palce in small intestines. In stomach simple sugars, water and alcohol absorbed. Villi are specialized projections in small intestine with large absorbent area. This facilities better absorbtion of complex molecules as well.

As food is fit to be absorbed only after complete digestion so villi’s turn comes after all the digestive processes are complete.

Question 4.
State the role pancreatic juice in digestion of proteins.
Answer:
The pancreatic juice contains inactive enzymes – trypsinogen, chymotrypsinogen, procarboxypetidases, amylases, lipases and nucleases. Trypsinogen is activated by an enzyme, enterokinase, secreted by the intestinal mucosa into active trypsin which in turn activities the other enzymes in the pancreatic juice.

Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption - 17

Question 5.
Describe the process of digestion of protein in stomach.
Answer:
The proenzyme pepsinogen, on exposure to hydrochloric acid gets converted into the active enzyme pepsin, the proteolytic enzyme of the stomach. Pepsin converts proteins into proteoses and peptones (peptides).

Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption - 18

Question 6.
Give the dental formula of human beings.
Answer:
Arrangement of teeth in each half of the upper and lower jaw in the order I, C, PM, M is represented by a dental formula which in human is as follows:
\(\frac{2123}{2123}\)

Question 7
Bile juice contains no digestive enzymes, yet it is important for digestion. Why?
Answer:
Bile emulsifies the fat making it easier digest by enzymes. Bile juice also helps inactivation of lipase, the enzyme which digests fat. Additionally, bile converts the acidic chyme to alkaline so that other enzymes can work on food.

Question 8.
Describe the digestive role of chymotrypsin. Which two other digestive enzymes of the same category are secreted by its source gland?
Answer:
Chymotrypsin digests proteins and converts it into dipeptides. Other digestive enzymes of the same category are Trypsin and Carboxypeptidase and they perform similar function.

Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption

Question 9.
Discuss the main steps in the digestion of proteins as the food passes through different parts of the alimentary canal.
Answer:
In the stomach the proenzyme pepsinogen gets converted into pepsin which acts on protein to convert it into peptone.

Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption - 19
In small intestine enzymes Chymotrypsin, trypsin and carboxypeptdase do the rest of the digestion of protein.

Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption - 20

Question 10.
Explain the term thecodont and diphyodont.
Answer:
Thecodont. In human beings each tooth is embedded in a socket of jaw bone. This type of attachment is called the codont.

Diphyodont. Majority of mammals including human beings form two sets of teeth during their life, a set of temporary milk or deciduous teeth replaced by a set of permanent or adult teeth. This type of dentition is called diphyodont.

Question 11.
Name different types of teeth and their number in an adult human.
Answer:
An adult human has 32 permanent teeth which are four different types (Heterodont dentition)

Plus One Digestion and Absorption Multiple Choice Questions and Answers

Question 1.
In intestine, food materials are absorbed through
(a) villi
(b) subtentacular cells
(c) sub-mucosa
(d) gastric glands
Answer:
(a) villi

Question 2.
Diastema refers to
(a) gap between the teeth
(b) gap between tongue and teeth
(c) ciliary cells on alimentary wall
(d) cell lining along pharynx
Answer:
(a) gap between the teeth

Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption

Question 3.
Pylorus is present between
(a) small and large intestine
(b) pancreas and small intestine
(c) oesophagus and stomach
(d) stomach and duodenum
Answer:
(d) stomach and duodenum

Question 4.
Elephant tusks are
(a) molars
(b) canines
(c) incisors
(d) premolars
Answer:
(c) incisors

Question 5.
Which of the following cells produce HCL
(a) beta-cells
(b) alpha-cells
(c) Oxyntic cells
(d) Chief cells
Answer:
(c) Oxyntic cells

Question 6.
Most abundant mineral of animal body is
(a) iron
(b) sodium
(c) potassium
(d) calcium
Answer:
(d) calcium

Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption

Question 7.
Bile salts act as activator of which enzyme?
(a) Pepsinogen
(b) Trypsinogen
(c) Lipase
(d) Pancreatic amylase
Answer:
(c) Lipase

Question 8.
When breast feeding is replaced by less nutritive food low in proteins and calories; the infants below the age of one year are likely to suffer from
(a) marasmus
(b) rickets
(c) kwashiorkor
(d) pellagra
Answer:
(a) marasmus

Question 9.
Which one of the following pairs of food components in humans reaches the stomach totally undigested?
(a) protein and starch
(b) Starch and fat
(c) fat and cellulose
(d) Starch and cellulose
Answer:
(c) fat and cellulose

Question 10.
A young infant may be feeding entirely on mother’s milk, which is white in colour but the stools, which the infant passes out is quite yellowish. This yellow colour is due to
(a) intestinal juice
(b) bile pigments passed through bile juice
(c) undigested milk protein casein
(d) pancreaticjuice poured into duodenum
Answer:
(b) bile pigments passed through bile juice

Question 11.
Which one of the following statements is true regarding digestion and absorption of food in humans?
(a) Oxyntic cells in our stomach secrete the proenzyme pepsinogen
(b) Fructose and amino acids arc abosrbed through intestinal mucosa with the help of carrier ions like Na+
(c) Chylomicrons are small lipoprotein particles that are transported from intestine into blood capillaries
(d) About 60°o of starch is hydrolysed by salivary amylase in our mouth
Answer:
(c) Chylomicrons are small lipoprotein particles that are transported from intestine into blood capillaries

Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption

Question 12.
Secretin and cholecystokinin are digestive hormones. They are secreted in
(a) oesophagus
(b) ileum
(c) pancreas
(d) duodenum
Answer:
(d) duodenum

Question 13.
FAD is a coenzyme derived from
(a) riboflavin
(b) vitamin-B12
(c) thiamine
(d) niacin
Answer:
(b) vitamin-B12

Question 14.
The pH of the digestive juices within the human small intestine is between 7.5 and 8.5. This environment is slightly.
(a) basic
(b) acidic
(c) neutral
(d) None of these
Answer:
(a) basic

Question 15.
Lipids, which can be found in oil based salad dressings and ice cream, during digestion are splitted into
(a) fatty acids and glycerol
(b) glycerol and amino acids
(c) glucose and fatty acids
(d) glucose and amino acids
Answer:
(a) fatty acids and glycerol

Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption

Question 16.
Crypts of Leiberkuhn are involved in
(a) secretion of succus entericus
(b) secretion of rennin
(c) secretion of ptyalin
(d) digestion of food
Answer:
(a) secretion of succus entericus

Question 17.
Which of the following is an organic molecule needed by the body in small amounts?
(a) Protein
(b) Zinc
(c) Vitamin-C
(d) Monosaccharide
Answer:
(c) Vitamin-C

Question 18.
Which one is not true about vitamins?
(a) Vitamins are organic catalysts
(b) Vitamins are indispensable for life
(c) Vitamins act as a source of energy
(d) Tocopherol is anti-sterility vitamin
Answer:
(c) Vitamins act as a source of energy

Question 19.
The malnutrition disease in man is
(a) Cridu chat syndrome
(b) Klinefelter’s syndrome
(c) Potbelly syndrome
(d) Edward’s syndrome
Answer:
(c) Potbelly syndrome

Plus One Zoology Chapter Wise Questions and Answers Chapter 5 Digestion and Absorption

Question 20.
The process of resynthesis of food materials from simpler food molecules is called
(a) biosynthesis
(b) catabolism
(c) absorption
(d) assimilation
Answer:
(a) biosynthesis