Kerala Syllabus Class 10 Biology Solutions Chapter 2 Windows of Knowledge

Kerala Syllabus 10th Standard Biology Guide Chapter 2 Windows of Knowledge

Question 1.
Why are dogs more capable than human beings in tracking the injured. in such circumstances?
Answer:
The number of receptors influence the efficiency of sense organs. The number of receptors in the sense organs is different in different organisms. The surface of a postage stamp is enough to arrange all the olfactory cells in the nose of a human being. But a large scarf is required to arrange the olfactory cells of a dog.

Question 2.
List various sense organs and the stimuli they receive.
Answer:

Sense organ Receptors Stimuli
Eye Photoreceptors Light
Ear Auditory receptors Sound
Tongue Taste receptors Taste
Nose Olfactory receptors Smell
Skin Various receptors Touch, pain temperature, cold, pressure

Question 3.
Make a table to explain how eyes are protected.
Answer:

Eye socket (orbit) Protection of eyeball.
External eye musc1e Fix the eyeball in the orbit.
Eyebrow Protects the eyes from dust, sweat.
Eyelashes Protect the eyes from dust.
Eyelids Protect the eyes from dust and external shocks.
Conjunctiva Secretes mucus which protects the anterior portion of the eyeball from being dry.
Tears Clean and lubricate the anterior part of the eyeball.

Question 4.
Make a table containing important paris of eye, its peculiarities and functions.
Kerala Syllabus Class 10 Biology Solutions Chapter 2 Windows of Knowledge 1
Answer:

Parts Peculiarity Functions
Sclera Outermost layer, made up of connective tissues. Gives rigidity to the eyeball.
Choroid Middle layer which contains a large number of blood vessels. Gives nutrients and oxygen to the tissues of eye.
Retina The inner most layer of eye. Photoreceptors are seen. Images are formed.
Conjunctiva A thin membrane covers the front part of the eyeball except cornea. Protects the front part of eyeball.
Cornea Slightly projected transparent anterior part of the sclera. Refracts light rays to focus on retina.
Iris Part of the choroid seen behind the cornea. Presence of melanin prevents harmful light rays.
Pupil The aperture seen at the centre of the iris. Contracts and relaxes according to the intensity of light.
Lens Elastic transparent convex lens. Helps to focus light on retina.
Ciliary muscles Circular muscles seen around the lens. The contraction and relaxation of these muscles alter the curvature of lens.
Aqueous chamber The chamber between the lens and the cornea filled with watery fluid – aqueous humour. Gives nutrients to the lens and cornea.
Vitreous chamber The chamber between the retina and the lens; filled with jelly like fluid – vitreous humour. Gives shape to the eyeball.
Yellow spot The part of retina where plenty of photoreceptors are present. It is the point of maximum visual clarity.
Blind spot Photoreceptors are absent. The part of retina from where the optic nerve begins.
Optic nerve Nerve fibre connecting eyeball and brain. Transmits impulses from photo receptors to the visual centre in the brain.

Question 5.
Explain the changes which occur in the pupil based on the intensity of light.
Answer:
Light which passes through the cornea reaches the lens through an aperture called the pupil. The size of the pupil is regulated with the help of muscles in the iris. The pupil dilates in dim light and constricts in bright light. Thus the amount of light falling on the lens is regulated according to the intensity of light.

Question 6.
What is the role of circular muscles and radial muscles in regulating the size of pupil?
Answer:
The size of the pupil is regulated by the action of circular muscles and radial muscles. When the radial muscles contract in dim light, the size of the pupil increases. When the circular muscles contract in intense light, the size of the pupil decreases. Thus the amount of light falling on the lens is regulated according to the intensity of light.

Question 7.
What are the peculiarities of image formed?
Answer:
Real, small, inverted.

Question 8.
Write down the features of the eye that help to view near and distant objects.
Answer:
Kerala Syllabus Class 10 Biology Solutions Chapter 2 Windows of Knowledge 4
While viewing near objects, the curvature of lens increases. As a result, the focal length decreases. The contraction of ciliary muscles and relaxation of ligaments help to increase the curvature of lens.
Kerala Syllabus Class 10 Biology Solutions Chapter 2 Windows of Knowledge 5
While viewing distant objects, the curvature of lens decreases. As a result, focal length increases. The relaxation of ciliary muscles and stretching of ligaments help to decrease the curvature of lens.

Question 9.
What is the power of accommodation of the eye?
Answer:
The capacity of the eye to change the curvature of lens depending on the distance between the eye and the object by adjusting the focal length is called the power of accommodation of the eye.

Question 10.
Name the photoreceptors in the eye. How do they differ from one another?
Answer:
The two photoreceptors present in the retina of the eye are rod cells and cone cells. Rod cells are more in number than cone cells. Rod cells have the shape of a rod and cone cells are cone shaped. Rod cells contain visual pigment called rhodopsin and in cone cells the pigment is photopsin (iodopsin). Rod cells are highly photosensitive and since they are activated even in very dim light, we are able to see objects in dim light. The cone cells provide us with colour vision.
Kerala Syllabus Class 10 Biology Solutions Chapter 2 Windows of Knowledge 6

Question 11.
Prepare a short note on the diversity of cone cells.
Answer:
Three types of cone cells which help us to detect three primary colour of light-red, green and blue are present in our eyes. This is due to the change in amino acid in the opsin molecule.

Question 12.
Explain the function of cone cells and rod cells.
Answer:
The visual pigment seen in cone cell is photopsin (iodopsin) and rod cell is rhodopsin. Both these pigments are formed from a protein named opsin and retinal which is a derivative of Vitamin A. When light rays fall on pigments present in photoreceptors, the pigments dissociate into retinal and opsin. This chemical change leads to the formation of impulses.

Question 13.
How is vision enabled?
Answer:
The visual pigment seen in cone cell is photopsin (iodopsin) and rod cell is rhodopsin. When light rays fall on pigments present in photoreceptors, the pigments dissociate into retinal and opsin. This chemical change leads to the formation of impulses. These impulses are transmitted to cerebrum through the optic nerves and this enables vision.

Question 14.
Prepare a flow chart showing the process of ‘ vision.
Answer:
Light → Cornea → Aqueous humour → Pupil → Lens→ Vitreous humour → Retina → Impulse → Optic nerve → Cerebrum → Sense of sight

Binocular vision
The activity to feel binocular vision:
Stretch your left hand forward. Close your right eye and focus the forefinger of your left hand. Now close your left eye and without changing the direction of your head, focus the I same finger. You can feel the change in the position of finger.

Question 15.
What is binocular vision?
Answer:
The images from two sides of the same object are formed in the left and right eye. These two images are combined together in the visual area of our brain to form a three dimensional image of the object. This is called binocular vision.

Question 16.
What is the difference between night blindness and xerophthalmia?
Answer:

The deficiency of Vitamin A results in the low production of retinal. This in turn creates the deficiency of rhodopsin in rod cells. The resynthesis of rhodopsin also gets blocked. This causes a condition called night blindness. Persons with night blindness cannot see objects clearly in dim light.

Due to the prolonged deficiency of Vitamin A, the conjunctiva and cornea will become dry and opaque. This condition is called xerophthalmia.

Question 17.
Prepare a note on the eye defects and diseases.
Answer:

Night blindness:
The retinal, a part of the visual pigment, is derived from Vitamin A. The deficiency of Vitamin A results in the low production of retinal. This in turn prevents the resynthesis of rhodopsin. In this condition, objects cannot be seen clearly in dim light and this disease is called night blindness.

Xerophthalmia:
If there is a prolonged deficiency of Vitamin A, the conjunctiva and cornea become dry and opaque. This causes xerophthalmia and leads ultimately to blindness.

Colour blindness:
Retina contains cone cells which can detect red, green, and blue colours. Colour blindness is caused due to the defect of cone cells which detect red and green colours. Persons with this defect cannot distinguish green and red colours.

Glaucoma:
If the reabsorption of aqueous humour does not occur, it causes an increase in the pressure inside the eyes which causes a defect called glaucoma. This causes damage to the retina and the photoreceptor cells and ultimately leads to blindness. This defect can be rectified by laser surgery.

Cataract: It is a condition in which the lens of the eyes become opaque resulting in blindness. This can be rectified by replacing the lens with an artificial one, through surgery.

Conjunctivitis:
This is an infection of the conjunctiva. The causative organisms may be bacteria, virus, etc. This disease is transmitted through contact and can be prevented by maintaining personal hygiene.

Question 18.
Food materials rich in Vitamin A and health of the eye
Answer:
The retinal, a part of the visual pigment, is derived from Vitamin A. The deficiency of Vitamin A results in the low production of retinal. This in turn prevents the resynthesis of rhodopsin. So food containing vitamin A is necessary for the health of eye.

Question 19.
Excessive use of mobile, computer, etc and health of the eye.
Answer:
The problems of the eyes caused by the excessive use of cell phones, computer, tablet, etc is called computer vision syndrome. The continuous use of such devices affects the ability of the eye to focus. Headache is its major symptom. Drying of the eye, high pressure in the eye, etc are other symptoms.
Kerala Syllabus Class 10 Biology Solutions Chapter 2 Windows of Knowledge 7

Question 20.
Write the functions performed by ear.
Answer:
Ear helps us in hearing and maintaining the balance of the body.

Question 21.
What are the parts and their function of the externa1 ear?
Kerala Syllabus Class 10 Biology Solutions Chapter 2 Windows of Knowledge 8
Answer:
Pinna – Carries sound waves to the auditory canal.
Auditory canal –
Carries sound waves to the tympanum. Small hairs and wax present inside the canal help to prevent dust and foreign particles from entering the ear.
Tympanum –
A thin circular membrane that separates the middle ear from . the external ear. It vibrates in resonance with sound waves.

Question 22.
What are the parts of middle ear?
Kerala Syllabus Class 10 Biology Solutions Chapter 2 Windows of Knowledge 9
Answer:
Ear ossicles (Malleus, Incus, Stapes), Eustachian tube are the parts of middle ear.

Question 23.
Explain the function of ear ossicles.
Answer:
Malleus, incus and stapes are the bones seen in ear ossicles. The ear ossicles connect the tympanum to the internal ear through the oval window. Ear ossicles amplify and.transmit the vibrations of tympanum to the internal ear.

Question 24.
What is the peculiarity of Eustachian-tube?
Answer:
Eustachian tube connects middle ear and pharynx. It helps in maintaining balance of pressure on either side of the tympanum.

Question 25.
Write the features of labyrinth seen in internal ear.
Answer:
The internal ear is situated inside a bony case in the skull called the bony labyrinth. In this bony case, there are membraneous labyrinths as well.

Question 26.
What are the fluids seen in internal ear?
Answer:
The space inside the membraneous labyrinth is filled with a fluid named endolymph. The space between the membraneous and bony labyrinth is filled with a fluid called perilymph.

Question 27.
Write the main parts and function of internal ear.
Answer:
Kerala Syllabus Class 10 Biology Solutions Chapter 2 Windows of Knowledge 10
Semicircular canals, vestibule and cochlea are the main parts of the internal ear. Semicircular canals and vestibule help in balancing the body whereas cochlea helps in hearing.

Question 28.
Explain oval window and round window.
Answer:
Kerala Syllabus Class 10 Biology Solutions Chapter 2 Windows of Knowledge 11

  • Oval window is a membrane seen attached to the stapes. Spreads the vibration of ear ossicles to the inner ear.
  • Round window helps in the movement of fluid inside the cochlea

Question 29.
How is hearing possible?
Answer:
Cochlea is a coiled tube like a snail shell. It consists of three chambers. Specialized sensory hair cells which are present in basilar membrane that separate the middle and lower chambers, function as auditory recepttjrs. The sound waves which pass through the external ear vibrate the tympanum.

This vibration of the tympanum is transmitted to the ear ossicles which causes the vibration of the membrane in the oval window. This vibration causes the movement of fluid inside the cochlea. As a result, the sensory hair cells seen in the basilar membrane of the cochlea are stimulated and impulses are generated. These impulses reach the cerebrum through the auditory nerve and hearing is effected.

Question 30.
Prepare a flow chart related to the process of hearing.
Answer:
Pinna → Auditory canalTympanum → Ear ossicles → Oval window → Cochlea → Hair cells → Impulse → Auditory nerve → Cerebrum → Sense of hearing.

Question 31.
What are the parts of the internal ear which help in body balancing?
Answer:
Kerala Syllabus Class 10 Biology Solutions Chapter 2 Windows of Knowledge 12
Vestibule and the three semicircular canals play a major role in body balancing.

Question 32.
Explain the maintenance of body balancing?
Answer:
Body balance is maintained in accordance with the movement of the head. Movements of the head bring about the movement of the endolymph present inside the vestibule and the semicircular canals. This causes movement of the sensoiy hair cells and generates impulses. These impulses are transmitted by the vestibular nerves to the cerebellum, and the equilibrium of tl maintained.

Question 33.
How are the taste buds arranged on the tongue?
Answer:
Chemoreceptors seen inside the mouth and tongue help us to detect taste. These are seen mainly on the surface of the tongue. The projected structures seen on the surface of the tongue are called papillae. The parts seen on the papillae that detect taste are the taste buds. We have taste buds that are stimulated by tastes like sweet, salt, sour, bitter, umami, etc.

Question 34.
How do we experience taste?
Answer:
Substances responsible for taste dissolve in saliva, stimulate the chemoreceptors and generate impulses. These impulses reach the brain through the respective nerves and we experience taste.

Question 35.
How do we detect smell?
Answer:
The smell of various substances diffuse in the air and enter the nostrils along with the inhaled air. These aromatic particles dissolve in the mucus inside the nostrils, stimulate olfactory receptors and generate impulses. These impulses from olfactory receptors reach the cerebrum and smell is detected.

Question 36.
What are the receptors in the skin to receive stimuli?
Answer:
Touch receptor, pressure receptor, cold receptor, temperature receptor and pain receptor.

Question 37.
Are the receptors uniformly distributed all over the skin? Prove.
Answer:
The receptors are not uniformly distributed all over the skin. This can be proved by the following activity. Take two refillers of any ball point pen. Ask your friend to close his/her eyes and stretch his/her hand. Place two pointed tips of the refiller firstly at the fingertip and then at the wrist of your friend. The touch receptors are seen more in finger tip. The touch in finger tip is experienced in two points and the touch in hand is experienced in a single point.

Question 38.
Explain about the receptors seen in planaria, housefly, shark and snake.
Answer:
In planaria, the eye spot helps to detect light. The eyes of a housefly consist of cluster of photoreceptors called ommatidia. In shark, lateral lines are seen. There are receptors in the lateral line on either side of the body which help to detect the change in the balance of body. In shark, highly sensitive olfactory receptors are also present. The aromatic particles that stick on the tongue of snake reach Jacobson’s organ seen on the roof of the mouth cavity. The olfactory receptors seen there get stimulated.

Let Us Assess

Question 1.
While viewing nearby objects
a. Ciliary muscles relax
b. Curvature of lens decreases
c. Ciliary muscles contract
d. Focal length increases
Answer:
c. Ciliary muscles contract

Question 2.
Identify the odd one and write down the common feature of others.
Malleus, Eustachian tube, Stapes, Incus
Answer:
Eustachian tube
Others are bones in ear ossicles.

Question 3.
Redraw the figure. Identify the parts
according to the hints and label them.
Hints:
a. The part where the muscles that regulate the size of the pupil are seen.
b. Jelly like fluid.
c. The layer of eye where photoreceptors are seen.
Kerala Syllabus Class 10 Biology Solutions Chapter 2 Windows of Knowledge 13
Answer:
Kerala Syllabus Class 10 Biology Solutions Chapter 2 Windows of Knowledge 14

Question 4.
Figure of Ear is given. Redraw it and name and label the parts mentioned.
Kerala Syllabus Class 10 Biology Solutions Chapter 2 Windows of Knowledge 15
a. Part that receives the vibrations of the eardrum.
b. The tube that connects the pharynx.
c. Part where the auditory receptors are seen.
Answer:
Kerala Syllabus Class 10 Biology Solutions Chapter 2 Windows of Knowledge 16
a. Ear ossicles
b. Eustachian tube
c. Cochlea

Kerala Syllabus 10th Standard Biology Guide