Plus Two

Kerala State Board New Syllabus Plus Two Chemistry Previous Year Question Papers and Answers.

Kerala Plus Two Chemistry Previous Year Question Paper Say 2018 with Answers

Time: 2 Hours
Cool off time: 15 Minutes
Maximum: 60 Score

General Instructions to candidates:

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2 hrs.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• Read the instructions carefully.
• Calculations, figures, and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

Answer all questions from 1 to 7. Each question carries 1 score. (7 × 1 = 7)

Question 1.
If N spheres are there in a close packing, what is the total number of tetrahedral and octahedral voids present in it?
Answer:
Tetrahedral 2N, octahedral N

Question 2.
What is the order of a reaction, if its half life is independent of initial concentration?
Answer:
1^st order

Question 3.
What is the magnetic moment of an atom having d configuration.
Answer:
Zero

Question 4.
Gabriel synthesis of used forthe preparation of which type of amines?
i) Primary
ii) Secondary
iii) Tertiary
iv) Quaternary
Answer:
i) Primary

Question 5.
Which vitamin is responsible for blood clotting?
Answer:
Vitamin K

Question 6.
Name the linear polymer formed during the condensation polymerization between phenol and formaldehyde.
Answer:
bakelite

Question 7.
Which is the chemical substance discovered by Paul Ehlrich for the treatment of syphilis?
Answer:
Salvarsan or Arephenamine

II. Questions from 8 to 20 carry 2 score each. Answer any 10 questions. (10 × 2 = 20)

Question 8.
Draw the vapour pressure-mole fraction curve for a non-ideal solution having positive deviation, if A and B are the two volatile components.
Answer:

Question 9.
Calculate the depression in freezing point of a 0.2 molal solution if kg for water is 1.86 K kg mol^-1.
Answer:
ΔT_f = k_jm
= 1.86 × 0.2
= 0.372K

Question 10.
Suppose you are given a sample of NaCl salt. How will you prepare chlorine gas in laboratory using the above sample? (Write balanced chemical equations)
Answer:
By the electrolysis of sodium chloride solution Cl₂ gas can be prepared
2NaCl + 2H₂O → 2Na⁺ + 2OH^– + H₂⁺_(g) + Cl_2(g)

Question 11.
Give one use each of Freon 12, DDT, CCl4and CHl3.
Answer:
Freon – 12-Refrigerant
DDT – insecticide
CCl₄ – For the manufacture of refrigerant
CHl₃ – Antiseptic

Question 12.
Write equations showing Wurtz-Fittig reaction and Fittig reaction.
Answer:
Fitting reaction

Question 13.
Identify A and B in the following equations:

Answer:

Question 14.
How the conversion of carbon dioxide to carboxylic acid can be effected using. Grignard reagent?
Answer:

Question 15.
Complete the following equations:


Answer:

Question 16.
Describe primary and secondary structure of proteins.
Answer:
Structure of proteins:
1) Primary structures – amino acids are arranged in sequence.
2) Secondary structure:
a) α – helix – polypeptide chains are coild to form a helical structure eg. Myosine
b) β-pleated structure: amino acid chains lie side by side and bonded by hydrogen bonds, eg. Keratine.

Question 17.
Explain homopolymers and copolymers with examples.
Answer:
Polymers formed by polymerisation of one type of monomer are called homopolymer, eg. Polythene Polymers formed by polymerisation 0 two or more different monomers are called copolymers eg. SBR, Rubber, Nylon 6, 6.

Question 18.
Briefly explain different types of artificial sweetening agents.
Answer:
Commonly used artificial sweetener is saccharin. It is 550 times sweeter than sucrose. Alitame is 1000 times sweet as cane sugar.
Aspartame, monolellin etc are other sweetening agent.

Question 19.
Write the IUPAC names of the following compounds:
a) [Ni(CO)₄]
b) K₃[Fe(C₂O₄)₃]
Answer:
a) Ni(CO)₄ Tetra carbonyl nickel (0)
b) K₃ [Fe(C₂O₄)₃] Potassium tris oxalate ferrate iii

Question 20.
Distinguish Ferromagnetism and Ferrimagnetism.
Answer:
Ferromagnetic substance – Magnetic moments are in one direction.

They are strongly attracted my magnetic field, eg Fe, Co
Ferrimagneticsubstances: Magnetic moments are unequal and in opposite direction.

eg. Fe₃O₄, MgFe₂O₄

III. Questions from 21 to 29 carry 3 score each. Answer any 7 questions. (7 × 3 = 21)

Question 21.
Silver atoms are arranged in CCP lattice structure. The edge length of its unit cell is 408 pm. Calculate the density of silver. (Atomic mass of silver is 108.4)
Answer:

Question 22.
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
Answer:

Question 23.
Explain any three chemical methods for the preparation of Lyophobic colloids with suitable examples.
Answer:
1) Oxidation methods: Oxidation of aqueous solution of H₂S with SO₂
SO₂ + 2H₂S → 3S + 2H₂O
2) Reduction method : Reduction of AuCl₃ solution using SnCl₂
2AuCl₃ + 3SNCl₂ → 3SnCl₄ + 2Au
3) Hydrolysis. By adding a saturated solution of ferric .chloride dropwise to a large excess of boiling water
FeCl₃ + 3H₂O → Fe(OH)₃ + 3HCl

Question 24.
Explain the following refining processes:
a) Distillation
b) Vapour phase refining
c) Zone refining
Answer:
a) Distillation: The impure metal is heated to form pure metals as distillate it is collected and condensed impurities are left behind, eg Zn and Hg.
Only metals with low boiling point can apply this method.

b) Vapour phase refining
i) Van Arkel method – eg Titanium, Zirconium, Thorium etc.

Mond process: Nickel is strongly heated with carbon monoxide to form Nickel tetra carbonyl this is again heated strongly to get pure nickel.

c) The impure metal bar is heated at one end with moving circular heater. The heater is now slowly moved along the rod. The pure metal recrystallises from the melt while impurities remain in the melt. Finally the end where impurities have collected is cut off. The impure metal bar is heated at one end with moving circular heater. The heater is now slowly moved along the rod. The pure metal recrystallises from the melt while impurities remain in the melt. Finally the end where impurities have collected is cut off.

Question 25.
A solution of CuSO₄ is electrolysed for 20 minutes with a current of 1.5 amperes. What is the mass of copper deposited at cathode?
(Atomic mass of copper – 63)
Answer:
Cu²⁺ + 2e → Cu
Q = It   1.5 × 20 × 60 = 1800 C
Mass of Cu deposited by 1800 C
\(\frac{63.5 \times 1800}{2 \times 96500}\) = 0.5875 g

Question 26.
Briefly explain the manufacture of sulphuric acid by contact process.
Answer:
Contact Process
1) Sulphur is burnt in air to form Sulphur Dioxide
S + O₂ → SO₂
2) Sulphur Dioxide is again oxidised to SO₃ with atmospheric oxygen in the presence of

3) SO₃ is treated with Sulphuric acid to get Oleum
SO₃ + H₂SO₄ → H₂S₂O₇
4) Oleum is diluted to get H₂SO₄
H₂S₂O₇ + H₂O → 2H₂SO₄

Question 27.
Explain with the help of equations, preparation of Xenon fluorides.
Answer:
Xe + F₂ → XeF₂
Xe + 2F₂ → XeF₄
Xe + 3F₂ → XeF₆

Question 28.
Describe lanthanoid contraction. Write any two consequences of it.
Answer:
The steady but slow decrease in the size of atoms or ions of the lanthanoids with increase in atomic number is called Lanthanoid Contraction.

Consequences:

1. As the size of the Lanthanoid ions decreases from La to Lu. The covalent character of hydroxides increases and hence the basic strength decreases.
2. The change in ionic radii of lanthanoids is very small their properties are almost similar. This makes the separation of lanthanoids are very difficult.

Question 29.
How the conversion of an aldehyde to acetal can carried out?
(Write chemical equations)
Answer:

IV. Questions from 30 to 33 carry 4 score each. Answer any 3. (3 × 4 = 12)

Question 30.
Predict the products of electrolysis of the following substances at anode and cathode using suitable chemical equations.
a) Aqueous NaCl
b) H₂SO₄ solution
Answer:
Electrolysis of aqeous NaCl
a) At Cathode H⁺ + 1e^– → \(\frac{1}{2}\)H₂
As the standard reduction potential for H⁺ ions are more it is easily reduced at cathode.
At anode Ch ions are oxidised Cl^– → Cl + e
2Cl → Cl_2(g)

b) Electrolysis of Sulphuric acid (dilute)
At anode
2H₂O → O₂ + 4H⁺ + 4e^–
At cathode
H⁺ + 1e → H
H + H → H₂
Electrolysis of concentrated H₂SO₄
At cathode
H⁺ + 1e → H
H + H → H₂
At anode
2SO\(\mathrm{O}_{4}^{2-}\) → S₂O\(\mathrm{O}_{8}^{2-}\) + 2e

Question 31.
Draw a diagram depicting crystal field splitting in an octahedral environment of d-orbitals. Label the diagram properly. Calculate the crystal field stabilization energy for a d³ configuration.
Answer:
Crystal field splitting in octahedral field.

CFSE for d³ configuration in octa hedral field. CFSE for 3 unpaired electrones
0 – \(\frac{2}{5}\)Δ₀ × 3 = \(\frac{-6}{5}\)Δ₀

Question 32.
a) Predict the products A and B.

b) How methanol is prepared industrially?
Answer:
2CH₃ – CH = CH₂ + (BH₃)₂ → (CH₃ – CH₂ – CH₂)₃ B → CH₃ – CH₂ – CH₂ – OH
By the catalytic hydrogenation of carbon monoxide in presence of a catalyst at 573K and under 200 to 300 atmospheric pressure to form methanol

Question 33.
a) Symbolically represent standard hydrogen electrode, when it acts as an anode and as cathode.
b) Write Nernst equation for a Daniel cell.
(Assume activity of metals is unity)
Answer:

Kerala State Board New Syllabus Plus Two Chemistry Previous Year Question Papers and Answers.

Kerala Plus Two Chemistry Previous Year Question Paper March 2019 with Answers

Time: 2 Hours
Cool off time: 15 Minutes
Maximum: 60 Score

General Instructions to candidates:

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2 hrs.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• Read the instructions carefully.
• Calculations, figures, and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

I. Answer all questions from 1 to 7. Each carries 1 score. (7 × 1 = 7)

Question 1.
The monomeric unit of natural rubber is …………
Answer:
Isoprene or 2-methyl 1, 3 butadiene or

Question 2.
The weakest reducing agent among the hydrides of group 15 elements is …………
Answer:
Ammonia or NH₃.

Question 3.
The reaction in which an amide is converted into a primary amine by the action of Br₂ and alcoholic NaOH is known as ………….
Answer:
Hoffmann bromamide reaction.

Question 4.
\(\mathrm{MnO}_{4}^{-}\) and ……… are formed by the disproportionation of \(\mathrm{MnO}_{4}^{2-}\) in acidic medium.
Answer:
MnO₂ or Mn⁴⁺.

Question 5.
In a solution of components‘A’ and ‘B’, at molecular level, A – B interactions are weaker than those between A – A or B – B interactions. Then the type of deviation shown by this solution is called ……………
Answer:
Positive deviation.

Question 6.
Identify the co-ordination compound which can exhibit linkage isomerism, among the following.
a) [Pt(NH₃)2Cl₂)
b) [Co(NH₃)₅(SO₄)]Br
c) [CO(NH₃)₅(NO₂)]Cl₂
d) [Cr(NH₃)₆][CoF₆]
Answer:
c) [CO(NH₃)₅(NO₂)]Cl₂

Question 7.
For the reaction, 2NO_(g) + O_2(g) → 2NO_2(g), the rate law is given as,
Rate = k[NO]² [O₂], The order of the reaction with respect to O₂ is …………
Answer:
one

II. Answer any ten questions from 8 to 20. Each carries 2 scores. (10 × 2 = 20)

Question 8.
Write the chemical equation representing Reimer-Tiemann reaction.
Answer:

Question 9.
What is reverse osmosis? Write any one of its applications.
Answer:
If the pressure applied is larger than osmotic pressure the direction of osmosis gets reversed. It is called reverse osmosis.
Aplication:

1. Desalination of sea water.
2. Purification of drinking water.

Question 10.
Identify the products and give the name of the following reaction:

Answer:
Cannizzaro reaction:

Question 11.
Explain Haloform reaction.
Answer:
Compounds with CH₃ – CO or CH₃ – CH – OH gp only gives haloforms such compounds when react with sodium hypohalite or a mixture of halogen and NaOH gives haloform.

Question 12.
What is meant by step growth polymerisation? Give an example.
Answer:
Step growth polymers are condensation polymers. The eliminate simple molecules like water or ammonia on addition, eg. nylon 6,6

Question 13.
An element crystallizes in F.C.C. manner. What is the length of a side of the unit cell, if the atomic radius of the element is 0.144 nm?
Answer:
a = 2\(\sqrt{2}\)r
r = 0.144 nm
= 2\(\sqrt{2}\) × 0.144 = 0.407 nm

Question 14.
Draw the structure of H₃PO₂ and account for its reducing character.
Answer:

Due to the presence of two P – H bonds it is a strong reducing agent.

Question 15.
2-Bromobutane is optically active. Explain the stereo¬chemical aspect of SN¹ reaction of 2-Bromobutane with OH- ions.
Answer:

Question 16.
Briefly explain the different types of emulsions and give examples for each.
Answer:
Emulsion: Both dispersed phase & dispersion medium are liquids:
1) Oil in water type, e.g. milk, vanishing cream.
2) Water in oil type, butter.

Question 17.
Give the structural formula and IUPAC name of the product formed by the reaction of propanone with CH₃MgBr in dry ether, followed by hydrolysis.
Answer:

Question 18.
Examine the graph given below. Identify the integrated rate equation and the order of the reaction corresponding to it.

Answer:
Order – zero
Rate equation for zero order K = \(\frac{[\mathrm{Ro}]-[\mathrm{R}]}{\mathrm{t}}\)

Question 19.
How is primary amine distinguished from a secondary amine using a chemical test?
Answer:
A) Carbyl amine test: Only primary amine gives carbyl amine test. Secondary amine does not give this test, (foul smelling gas).
(OR)
B) Hinsberg test: Primary amine react with benzene sulphonyl chloride to form a precipitate which is soluble in alkali. But the precipitate formed by the secondary ammine is insoluble in alkali.
(OR)

Question 20.
Predict the products obtained by the reaction of 2-methoxy-2-methyl propane with HI.
Answer:

III. Answer any seven questions from 21 to 29. Each carries 3 scores. (7 × 3 = 21)

Question 21.
Explain the terms, Zeta potential, electrophoresis and electro-osmosis.
Answer:
Zeta potential: The potential difference between the fixed layer and the diffused layer of an electrical, double layer.
Electrophoresis: The movement of colloidal particles under the influence of an electric field.
Electro osmosis: The movement of particles of dispersion median under the influence of electric field.

Question 22.
The rate constant of a reaction at 293 K is 1.7 × 10⁵ s^-1. When the temperature is increased by 20K, the rate constant is increased to 2.57 × 10⁶ s^-1. Calculate Ea and A of the reaction.
Answer:

Question 23.
Identify A, B and C in the following sequence of reactions:

Answer:
A = CH₃CONH₂
B = CH₃-COOH
C = CH₂Br – COOH

Question 24.
Briefly explain different types of neurologically active drugs and give exmaple for each type.
Answer:
Tranquilizers: Medicines or chemicals used for mental-stress, e.g. Equanil, barbituric acid, veronal, luminal, seconal (anyone).
Analgesics are painkillers and are also neurologically active drug, e.g. paracetamol, morphine, heroine.

Question 25.
Write any three applications of d-block and f-block elements.
Answer:

1. They act as catalyst.
2. They form alloys.
3. Cu, Ag Au are coinage metals.
4. They have different oxidation states.

Question 26.
Give the open chain and ring structures of glucose and account for the existence of glucose in two anomeric forms.
Answer:

Question 27.
A 5% solution (by mass) of cane sugar (C₁₂H₂₂O₁₁) in water has a freezing point of 271 K. Calculate the freezing point of 5% (by mass) solution of glucose (C₆H₁₂O₆) in water. Freezing point of pure water is 273.15 K.
Answer:
ΔTf = \(\frac{1000 \mathrm{~K}_{\mathrm{f}} \cdot \mathrm{W}_{2}}{\mathrm{~W}_{1} \cdot \mathrm{M}_{2}}\)
for 5% suger solution
W₂ = 5
W₁ = 100 – 5 = 95
M₂ = 342
Tf = 271 k
T°f = 273.15
ΔTt= T°f – Tf = 2.15 K

For 5% gulcose solution W₂ = 5, W₁ = 100 – 5 = 95g, M₂ = 180

Question 28.
Explain the steps involved in the vapour phase refining of Ni and Zr.
Answer:
Ni by Monds process.
Ni is strongly heated at a temperature of 330 – 350K to form nickel tetra carbonyl. It is again heated strongly 450 – 470 to decompose it.

Zr by Van Arkel method.
Zirconium is treated with iodine to form zirconium tetra iodide. It is electrically heated to about 1800K with tungsten filament. Zr I₄ is decomposed
Zr + Zl₂ → Zrl₄
Zrl₄ → Zr + 2l₂

Question 29.
What are inter halogen compounds? Which interhalogen compound is used to fluorinate Uranium? How is it prepared?
Answer:
Inter halogen compounds are formed by the direct combination of halogen.

CIF₃ or Br F₃ are used to Fluorinate Uranium.
They are compounds of two or more halogen atoms.

IV. Answer any three questions from 30 to 33. Each carries 4 scores. (3 × 4 = 12)

Question 30.
How can the following conversions be effected?
i) Ethanol-Fluroethane
ii) But-l-ene → But-2-ene
Answer:

Question 31.
Diagrammatically represent H₂ – O₂ fuel cell and write the half cell reactions taking place in this cell.
Answer:

Reaction at cathode
O₂g + 2H_2(l)O + 4\(\bar{e}\) → 4O\(\bar{H}\)_(ag)
Reaction at anode
2H_2(g) + 4O\(\bar{H}\)(aq) → 4H₂O_(I)+ 4\(\bar{e}\)
2H₂ + O₂ → 2H₂O

Question 32.
What are point defects? Explain the non-stoichiometric point defects in ionic crystals.
Answer:
Point defect: The irregularities from ideal arrangement around a point or an atom in crystalline substance.
Non-stoichiometric defect: The defect which do not disturb the stoichiometry of the crystalline substance.
I) Metal excess defect:

• Due to anion vacancies. Electron trapped anion vacancies are called F centre which gives colour.
• Due to extra cation at interstitial site.

II) Metal deficiency defect:

• Due to cation vacancies.

Question 33.
i) With the help of a diagram, give the splitting of d-orbitals of Mn²⁺ ion and octahedral crystal field.
ii) On the basis of crystal field theory, explain why [Mn(H₂O)₆]²⁺ contains five unpaired electrons while [Mn(CN)₆]^4- contains only one unpaired electron.
Answer:
i) Splitting of d orbital in octahedral system.

ii) In [Mn (H₂O)₆]²⁺ Mn²⁺ have five electrons.
As (H₂O) is a weak ligand, i.e., Δ < P no pairing occurs and electronic configuration t₂g³ eg².

But in [Mn(CN)₆]^4- Mn²⁺ has five electrons and are in paired state as CN^– is a strong ligend and Δ > P pairing occurs the electronic configuration is t₂g⁵ eg⁰.

Kerala State Board New Syllabus Plus Two Physics Previous Year Question Papers and Answers.

Kerala Plus Two Physics Previous Year Question Paper Say 2018 with Answers

Time: 2½ Hours
Cool off time : 15 Minutes

General Instructions to candidates

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2½ hrs.
• Your are not allowed to write your answers nor to discuss anything with others during the ‘cool off time’.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• All questions are compulsory and only internal choice is allowed.
• When you select a question, all the sub-questions must be answered from the same question itself.
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non programmable calculators are not allowed in the Examination Hall.

You may use following Physical constants wherever necessary.

Mass of proton 1.66 × 10^-27 kg
Mass of Electron 9.11 × 10^-31 kg
Elementary charge, e = 1.6 × 10^-19 C
Velocity of light in vacuum c = 3 × 10⁸ m/s
Permittivity of free space ε₀ = 8.85 × 10^-12 F/m

Question Nos. 1 to 7 carry 1 score each. Answer 6 questions. (6 × 1 = 6)

Question 1.
Write the physical quantities having the following SI unit.
i) cm
ii) Ωm
Answer:
i) Electric dipole moment
ii) Electrical resistivity

Question 2.
A uniform wire of resistance 40Ω is cut into four equal parts and they are connected in parallel. The effective resistance of the combination is
i) 40Ω
ii) 10Ω
iii) 2.5Ω
iv) 4Ω
Answer:

Question 3.
A particle of charge q is moving through a uniform magnetic field of intensity B with a velocity v. Write an expression for the force acting on the particle in vector form.
Answer:
\(\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})\)

Question 4.
The minimum distance between the object and its real image for a concave mirror of focal length f is
i) f
ii) 2f
iii) 4f
iv) zero
Answer:
iv) zero

Question 5.
Variation of photocurrent with collector plate potential for different intensities I₁, I₂, and I₃ of incident radiation is shown below.

Arrange these intensities in the accending order.
Answer:
I₃ > I₂ > I₁

Question 6.
An electron is revolving around the nucleus of a hydrogen atom in an orbit of radius nine times the radius of the first orbit. Angular momentum of the electron in this orbit is.
i) \(\frac{h}{2 \pi}\)
ii) \(\frac{9 h}{2 \pi}\)
iii) \(\frac{3 h}{\pi}\)
iv) \(\frac{3 h}{2 \pi}\)
Answer:
radius of the nucleus r = n²r₀
In this case
∴ n² = 9, n= 3
Angular momentum L = \(\frac{\mathrm{nh}}{2 \pi}\)
for third orbit L = \(\frac{3 h}{2 \pi}\)

Question 7.
The voltage-current characteristics of an optoelectronic junction device is shown below.

Identify the device.
Answer:
Solar cell

Question Nos. 8 – 15 carry 2 scores each. Answer any 7 questions. (7 × 2 = 14)

Question 8.
What is the radius of the circular path of an electron moving at a speed of 3 × 10⁷ m/s in a magnetic field of 6 × 10^-4 T perpendicular to it?
Answer:
radius r = \(\frac{m v}{q B}\)
v = 3 × 10⁷ m/s, m = 9.1 × 10^-31kg
B = 6 × 10^-4, q = 1.6 × 10^-19
r = \(\frac{9.1 \times 10^{-37} \times 3 \times 10^{7}}{1.6 \times 10^{-19} \times 6 \times 10^{-4}}\) = 0.285 m

Question 9.
a) “Parallel currents attract, and antiparallel currents repel”. State whether this statement is true or false.
b) Define the SI unit of current in terms of force between two current-carrying conductors.
Answer:
a) True
b) Ampere is defined as that current which is maintained in two straight parallel conductors of infinite length placed one meter apart in the vacuum will produce between a force of 2 × 10^-7 Newton per meter length.

Question 10.
A circular metallic loop and a current-carrying conductor are placed as shown in the figure.

If the current through the conductor is increasing from A to B, find the direction of the induced current the loop.
Answer:
Current flows in a clockwise direction

Question 11.
a) The electric field vector of an electromagnetic wave is represented as E_x = E_m Sin(kz – ωt). Write the equation for the magnetic field vector.
b) The ratio of the intensity of the magnetic field to intensity of the electric field has the dimensions of
i) velocity
ii) acceleration
iii) reciprocal of velocity
iv) reciprocal of acceleration
Answer:
a) By = B_m Sin (Kz – ωt)
b) reciprocal of velocity

Question 12.
Write the equations for the following nuclear reactions:
a) β⁺decay of \({ }_{6}^{11} c\) to Boron (B)
b) β ^–decay of \({ }_{15}^{32} c\) to Sulphur (S)
Answer:

Question 13.
Write the truth table of the circuit shown below.

Answer:

Question 14.
A TV transmitting antenna is 100 m tall. How much service area can it cover if the receiving antenna is at the ground level? Radius of earth is 6400 km.
Answer:
Hight of antenna, h = 100 m = 0.1 km
Radius of earth = 6400 km
Transmission range d = \(\sqrt{2 R h}\)
∴ Maximum area covered A = πd²
= π(\(\sqrt{2 R h}\))²
= 3.14 × 2 × 6400 × 0.1
= 4019 km²

Question 15.
Two magnetic dipoles P and Q are placed in a uniform magnetic field \(\vec{B}\) gas shown.

a) Both the dipoles do not experience any torque. Why?
b) Identify the dipole which is in most stable equilibrium.
Answer:
a) Dipole moment is parallel to magnetic field.
b) Dipole Q

Question Nos. 16 to 22 carry 3 scores each. Answer any 6 questions. (6 x 3 = 18)

Question 16.
The experimental setup for the comparison of two resistances is shown below.

a) The working principle of the above device is
i) Ohm’s law
ii) Kirchhoffs second law
iii) Wheatstone’s principle
iv) Point Rule
b) In figure, Let X is the effective resistance of series combination of two 3Ω resistors and R is the effective of a parallel combination of two 3Ω resistors. The balance point is obtained at C. If the length AB is 100 cm. find the length AC of the wire.
Answer:
a) Wheatston’s principle
b) x = 6 Ω
R = 1.5 Ω

Question 17.
A magnetic needle has a magnetic moment 6.7 × 10^-7 Am² and moment of inertia 7.5 × 10^-6 kgm². In a uniform magnetic field, it performs 10 complete oscillations in 6.70s. What is the magnitude of the magnetic field?
Answer:
magnetic moment m = 6.7 × 10^-2 Am²
moment of inertria, I = 7.5 × 10^-6 kgm²

Question 18.
With the help of a ray diagram, show the formation of image of a point object by refraction of light at a spherical surface separating two media of refractive indices n₁ and n₂. Using the diagram derive the relation
\(\frac{n_{2}}{v}-\frac{n_{1}}{u}=\frac{n_{2}-n_{1}}{R}\)
Answer:

Consider a convex surface XY, which separates two media having refractive indices n₁ and n₂. Let C be the centre of curvature and P be the pole. Let an object is placed at ‘O’, at a distance ‘u’ from the pole. I is the real image of the object at a distance ‘V’ from the surface. OA is the incident ray at angle ‘i’ and AI is the refracted ray at an angle ‘r’. OP is the ray incident normally. So it passes without any deviation. From snell’s law,
sin i/sin r = \(\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}\)
If ‘i’ and ‘r’ are small, then sin i » i and sin r » r.

From the Δ OAC, exterior angle = sum of the interior opposite angles
i.e., i = α + θ ………(2)
Similarly, from Δ IAC,
α = r + β
r = α – β ………..(3)
Substituting the values of eq(2) and eq(3)in eqn.(1)
we get,
n₁(α + θ) = n₂(α – β)
n₁α + n₁θ = n₂α – n₂β
n₁θ + n2β = n₂α – n₁α
n₁θ + n₂β = (n₂ – n₁)α ………..(4)
From OAP, we can write,

According to New Cartesian sign convection, we can write,
OP = -u, PI = +v and PC = R
Substituting these values, we get

Question 19.
a) Draw a graph showing the variation of stopping potential with the frequency of incident radiation. Mark the threshold frequency in the figure.
b) Using Einstein’s photoelectric equation, show that photoelectric emission is not possible if the frequency of incident radiation is less than the threshold frequency.
Answer:
a)

b) According to Einsten’s Equation
hν = hν₀ + KE
if = ν < ν₀, kinetic energy of electron becomes negative.

Question 20.
a) The energy levels of an atom are as shown in the figure.

Which transition corresponds to emission of radiation of maximum wavelength?
b) Sketch the energy level diagram for hydrogen atom and mark the transitions corresponding to Balmer series.
Answer:
a) In between 0 to -2ev
b)

Question 21.
The decay rate of a radioactive ample is called its activity.
a) What is the SI unit of activity?
b) The half life of \({ }_{92}^{238} U\) against alpha decay is 1.5 × 10^-17 s. Calculate the activity of a sample of \({ }_{92}^{238} U\) having 25 × 10²⁰ atoms.
Answer:
a) becquerel
b) N = 25 × 10²⁰, half life T_1/2 = 1.5 × 10¹⁷s
activity, R = λN
R = \(\frac{0.693}{\mathrm{~T}_{1 / 2}} \times \mathrm{N}=\frac{0.693}{1.5 \times 10^{17}} \times 25 \times 10^{20}\)
R = 11550 B_q

Question 22.
a) Draw the circuit diagram where a zener diode is used as a direct voltage regulator.
b) In a zener regulated power supply a zener diode with zener voltage 4V is used for regulation. The load current is to be 4 m A and zener current is 20 mA. If the unregulated input is 10V, What should be the value of resistor that is to be connected in series with the diode?
Answer:
a)

b) Zener voltage
V₂ = 4V, I_L = 4 × 10^-3 A, I₂= 20 × 10^-3 A
Unregulated input = 10 v
Voltage across resister = 10 – 4 = 6v
Current through the resister = I_L+ I₂
= (4 + 20) mA
= 24 × 10^-3 A
∴ resistance R = \(\frac{V}{I}\)
R = \(\frac{6}{24 \times 10^{-3}}\) = 250 Ω

Question Nos. 23 to 26 carry 4 scores each. Answer any 3 questions. (3 × 4 = 12)

Question 23.
Coulomb’s Law is a quantitative statement about the force between two point charges.
a) Write the mathematical expression of the above law.
b) Two ions carrying equal charges repel with a force of 1.48 × 10^-8 N when they are separated by a distance of 5 × 10^-10 m. How many electrons have been removed from each iron?
Answer:

Question 24.
a) Potential energy of a system of charges is directly proportional to the product of charges and inversely to the distance between them.
a) Prove the above statement.
b) Two-point charges 3 × 10^-8 C and -2 × 10^-8 C are separated by a distance of 15cm. At what point on the line joining the charges the potential is zero?
Answer:
a) Potential of a system is the work done to assemble the system,
∴ w = qV
= \(\mathrm{q}_{1} \times \frac{1}{4 \pi q_{0}} \frac{\mathrm{q}_{2}}{\mathrm{r}}\)
In the above equation it is dear that energy is di-rectly proportial to charges and inversely proportional to the distance between them.
b) q₁ = 3 × 10^-8

Question 25.
The circuit shown can be analysed using Kirchhoff’s rules.

a) Apply Kirchhoffs first law to the point B.
b) State Kirchhoffs second law.
c) Apply Kirchhoffs second lawtothe mesh ABFGA.
Answer:
a) At the point B,
I₁ = I₂ = I₃
b) In any closed circuit emf is eqaual to sum of potential drops.
c) -I₂ R₃ – I₁ R₂ – I₁R₁ + E₁ = 0

Question 26.
Biot – Savart’s law relates current with the magnetic field produced by it.
a) Write the mathematical expression of the above law in vector form.
b) Using the law derive an expression for intensity of magnetic field at a point at distance x from the centre and on the axis of a circular current loop.
Answer:

Consider a circular loop of radius ‘a’ and carrying current “I’. Let P be a point on the axis of the coil, at distance x from A and r from ‘O’. Consider a small length dl at A.
The magnetic field at ‘p’ due to this small element dI,

The dB can be resolved into dB cos Φ (along with Py) and dBsinΦ (along with Px).
Similarly consider a small element at B, which produces a magnetic field ‘dB’ at P. If we resolve this magnetic field we get.
dB sinΦ (along px) and dB cosΦ (along py¹)
dBcosΦ components cancel each other because they are in opposite direction. So only dB sinΦ components are found at P, so the total filed at P is

Let there be N turns in the loop then,
B = \(\frac{\mu_{0} \text { NIa }^{2}}{2\left(\mathrm{r}^{2}+\mathrm{a}^{2}\right)^{3 / 2}}\)

Question Nos. 27 to 29 carry 5 scores each. Answer any 2 questions. (2 × 5 = 10)

Question 27.
A series LCR circuit shows the phenomenon called resonance.
a) Write the condition for resonance and obtain an equation for resonant frequency.
b) Obtain the Q value of a series LCR circuit with L = 2.0 H, c = 32µ F and R = 10 Ω.
c) Complete the following table using the suitable words from the bracket for two series LCR circuits.
(Current and applied voltage are in the same phase, currently leads the applied voltage, currently lags the applied voltage)

Answer:
a) Condition for resonance X_L = X_C

c) i) Cuurrent lags the applied voltage
ii) Current and applied voltage are in same phase.

Question 28.
Telescope is used to provide angular magnification of distant objects.
a) If f₀ is the focal length of the objective and f_e that of the eyepiece, the length of the telescope tube is ………..
b) Draw the ray diagram of a refracting type telescope when it is in normal adjustment.
c) Write any two advantages of reflecting type telescope over refracting type telescopes.
Answer:
a) f₀ + f_e
b)

c) i) There is no chromatic aberration
ii) There is no spherical aberration

Question 29.
A wavefront is defined as a surface of constant phase.
a) The energy of the wave travels in a direction …….. to the wavefront.
b) Explain the Huygen’s principle.
c) Using Huygen’s Principle, prove that angle of incidence is equal to angle of reflection.
Answer:
a) Perpendicular

b) 1) Every point in a wavefront acts as a source of secondary wavelets.
2) The secondary wavelets travel with the same velocity as the original value.
3) The envelope of all these secondary wavelets gives a new wavefront.

c)

AB is the incident wavefront and CD is the reflected wavefront. T is the angle of incidence and Y is the angle of reflection. Let c, be the velocity of light in the medium. Let PO be the incident ray and OQ be the reflected ray.

The time taken for the ray to travel from P to Q is


O is an arbitrary point. Hence AO is a variable. But the time to travel for a wavefront from AB to CD is a constant. So eq.(2) should be independent of AO. i.e., the term containing AO in eq.(2) should be zero.
∴ \(\frac{A O}{C_{1}}\) (sin i – sin r) = 0
sin i – sin r = 0
sin i = sin r
i = r
This is the law of reflection.

Kerala State Board New Syllabus Plus Two Physics Previous Year Question Papers and Answers.

Kerala Plus Two Physics Previous Year Question Paper March 2019 with Answers

Time: 2½ Hours
Cool off time: 15 Minutes

General Instructions to candidates

• There is a ‘cool off-time’ of 15 minutes in addition to the writing time of 2½ hrs.
• You are not allowed to write your answers nor to discuss anything with others during the ‘cool off time’.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• All questions are compulsory and the only internal choice is allowed.
• When you select a question, all the sub-questions must be answered from the same question itself.
• Calculations, figures, and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

The given value of constants can be used wherever necessary.

Charge of proton = 1.6 x 10^-19C
Mass of proton = 1.67 x 10^-27kg

Answer any three questions from 1 to 4. Each carries 1 score. (3 × 1 = 3)

Question 1.
A charged particle enters a uniform magnetic field at an angle of 40°. It’s path becomes …………
Answer:
Helical

Question 2.
Figure shows the symbolic representation of ……….

i) OR gate
ii) NAND gate
iii) NOR gate
iv) NOT gate
Answer:
iii) NOR gate

Question 3.
Write the unit of mobility.
Answer:

Question 4.
If ‘h’ is Planck’s constant, the momentum of a photon of wavelength 1 A° is
i) h
ii) 10^-10
iii) 10¹⁰ h
iv) 10 h
Answer:

Answer any six questions from 5 to 11. Each carries 2 scores. (6 × 2 = 12)

Question 5.
a) The ratio of electric field on the equatorial point and at the axial point at equal distances from the centre of a short electric dipole is ………..
b) A closed surface encloses an electric dipole. What is the electric flux through the surface?
Answer:

Question 6.
A series LCR circuit connected to an ac source is shown below:

a) Write an expression for impedance offered by its circuit.
b) Under what condition this circuit is used for tuning radio?
Answer:
a) Impedence Z = \(\sqrt{R^{2}+\left(x_{L}-x_{C}\right)^{2}}\)
b) at X_L = X_C this circuit is used as tuner circuit.

Question 7.
Which electromagnetic waves are used for the following purposes?
i) Diagnostic tool in medicine.
ii) Kill germs in water purifiers.
iii) Cellular phones.
iv) In remote switches of household electronic systems.
Answer:
i) X-rays
ii) Uv rays
iii) Radio Waves
iv) IR rays

Question 8.
Calculate the effective capacitance between ’a’ and ‘b’ from the figure given below:
C₁ = C₃ = 100 µF, C₂ = C₄ = 200 µF

Question 9.
Write any two uses of polaroids.
Answer:

1. Polaroids are used in sunglasses
2. Polaroids are used to produce 3D motion films

Question 10.
The temperature dependence of resistivity of a material is shown below:

a) Identify the type of material.
b) Write the relation between resistivity and average collision time for electron.
Answer:
a) Semi conductor
b) ρ = \(\frac{\mathrm{m}}{\mathrm{ne}^{2} \tau}\)

Question 11.
What is meant by the half-life of a radioactive substance? Write its relation with a decay constant.
Answer:
Half-life is the time taken for a radioactive substance to reduce half of its initial value.
\(T_{1 / 2}=\frac{0.693}{\lambda}\)

Answer any six questions from 12 to 18. Each carries 3 scores. (6 × 3 = 18)

Question 12.
A spherical shell of radius R’ is uniformly charged with charge +q. By Gauss’stheorem, find the electric field intensity at a point ‘p’.
a) Outside the spherical shell and
b) Inside the spherical shell.
Answer:
Field Due To A Uniformly Charged Thin Spherical Shell: Consider a uniformly charged hollow spherical conductor of radius R. Let ‘q’ be the total charge on the surface.

To find the electric field at P (at a distance r from the centre), we imagine a Gaussian spherical surface having radius ‘r’.

Then, according to Gauss’s theorem we can write,
\(\int \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{s}}=\frac{1}{\varepsilon_{0}} \mathrm{q}\)
The electric field is constant,at a distance T. So we can write,

b) E = 0.

Question 13.
The equipotential surface through a point is normal to the electric field at that point.
a) What is meant by equipotential surface?
b) What is the work done to move a charge on an equipotential surface?
c) Draw the equipotential surfaces for a uniform electric field.
Answer:
a) The surface over which potential is constant is equipotential surface.
b) Workdone = pd × charge
= 0 × q = 0
c)

Question 14.
The elements of earth’s magnetic field at a place are declination, dip and horizontal intensity.
a) A magnetic needle free to move in horizontal plane is shown below:

Which element of earth’s magnetic field is represented by θ in the figure?
b) The vertical component of earth’s magnetic field at a given place is \(\sqrt{3}\) times its horizontal component. If total intensity of earth’s magnetic field at the place is 0.4 G find the value of horizontal component of earth’s magnetic field.
Answer:
a) Declination
b) tan θ = \(\frac{B_{V}}{B_{H}}=\sqrt{3}\)
∴ θ = 60°
B_H = B cos θ, B = 0.4 G
B_H = 0.4 × cos 60 = 0.2 G
= 0.2 × 10^-4 T

Question 15.
A transformer is used to change the alternating voltage to a high or low value.
a) What is the principle of a transformer?
b) A power transmission line feeds input power of 2300 V to a stepdown transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary in order to get output power at 230 V?
Answer:
a) Mutual Induction
b) V_p = 2300 v, N_p = 4000 turns
V_s = 230 v

Question 16.
Describe Young’s double-slit experiment and derive an expression for the bandwidth of the interference band.
Answer:
Expression for bandwidth

S₁ and S₂ are two coherent sources having wave length λ. Let ‘d’ be the distance between two coherent sources. A screen is placed at a distance D from sources. ‘O’ is a point on the screen equidistant from S₁ and S₂.

Hence the path difference, S₁O – S₂O = 0
So at ‘O’ maximum brightness is obtained.
Let ‘P’ be the position of n^th bright band at a distance x_n from O. Draw S₁A and S₂B as shown in figure.
From the right angle ΔS₁AP

But we know constructive interference takes place at P, So we can take
(S₂P – S₁P) = nλ
Hence eq (1) can be written as

Let x_{n + 1} be the distance of (n + 1)^th bright band from centre O, then we can write

This is the width of the bright band. It is the same for the dark band also.

Question 17.
The schematic diagram of an experimental setup to study the wave nature of electron is shown below:

a) Identify the experiment.
b) Explain how this experiment verified the wave nature of electrons.
Answer:
a) Davisson and Germer Experiment
b)

Experimental setup: The Davisson and Germer Experiment consists of filament ‘F’, which is connected to a low tension battery. The Anode Plate (A) is used to accelerate the beam of electrons. A high voltage is applied in between A and C. ’N’ is a nickel crystal. D is an electron detector. It can be rotated on a circular scale. Detector produces current according to the intensity of incident beam.

Working: The electron beam is produced by passing current through filament F. The electron beam is accelerated by applying a voltage in between A (anode) and C. The accelerated electron beam is made to fall on the nickel crystal. The nickel crystal scatters the electron beam to different angles. The crystal is fixed at an angle of Φ = 50° to the incident beam. The detector current for different values of the accelerating potential ‘V’ is measured. A graph between detector current and voltage (accelerating) is plotted. The shape of the graph is shown in figure.

Analysis of graph:

The graph shows that the detector current increases with accelerating voltage and attains maximum value at 54V and then decreases. The maximum value of current at 54 V is due to the constructive interference of scattered waves from nickel crystal (from different planes of the crystal). Thus wave nature of electrons is established.

The experimental wavelength of electron: The wavelength of the electron can be found from the formula
2d sinθ = nλ …….. (1)
From the figure, we get
θ + Φ + θ = 180°
2θ = 180 – Φ, 2θ = 180 – 50°
θ = 65°
for n = 1
equation (1) becomes
λ = 2d sinθ ……….. (2)
for Ni crystal, d = 0.91 A°
Substituting this in eq. (2), we get
wavelength λ = 1.65 A°

The theoretical wavelength of the electron:
The accelerating voltage is 54 V
Energy of electron E = 54 × 1.6 × 10^-19 J

Discussion: The experimentally measured wave-length is found in agreement with de-Broglie wavelength. Thus wave nature of electron is confirmed.

Question 18.
The energy required to separate all the nucleons inside a nucleus is called binding energy of the nucleus.
a) Write an expression for binding energy in terms of mass defect.
b) Draw the graph showing the variation of binding energy per nucleon as a function of mass number.
c) Which nucleus possess maximum binding energy per nucleon?
Answer:
a) BE = Δ mc² or
BE = (Zm_p + (A – Z) m_n – M)c²
b) Graph

c) Fe (nucleus of iron)

Answer any three questions from 19 to 22. Each carries 4 scores. (3 × 4 = 12)

Question 19.
Niels Bohr made certain modification in Rutherford’s model by adding the ideas of quantum hypothesis.
a) State Bohr’s second postulate of quantisation of angular momentum.
b) Derive an expression for the radius and energy of the electron in the nth orbit of hydrogen atom.
Answer:
a) The orbital angular momentum of electron is an integral multiple of \(\frac{\mathrm{h}}{2 \pi}\)

b) Radius of the hydrogen atom: Consider an electron of charge ‘e’ and mass m revolving round the positively charged nucleus in circular orbit of radius ‘r’.
The force of attraction between the nucleus and the electron is

This force provides the centripetal force for the orbiting electron

According to Bohr’s second postulate, we can write
Angular momentum, mvr = \(\frac{\mathrm{nh}}{2 \pi}\)
i.e., v = \(\frac{\mathrm{nh}}{2 \pi \mathrm{mr}}\)
Substituting this value of V in equation (2), we get

Energy of the hydrogen atom:
The K.E. of revolving electron is

Substituting the value of equation (5) in equation (9)
we get

Question 20.
Two long co-axial solenoids of same length are shown below:

a) Define mutual inductane of the pair of coils.
b) Derive an expression for mutual inductance of two co-axial solenoids.
c) Write the dimension of mutual inductance.
Answer:
a) Φ = MI, when I = 1 A, Φ = M
The mutual inductance of a pair of coils is numerically equal to the magnetic flux linked with one coil when unit current flows through the other.

b) Consider a solenoid (air core) of cross sectional area A and number of turns per unit length n. Another coil of total number of turns N is closely wound over the first coil. Let I be the current flow through the primary.
Flux density of the first coil B = μ₀nI
Flux linked with second coil, Φ = BAN
Φ = μ₀nIAN ………. (1)
But we know Φ = MI ………. (1)
From eq (1) and eq (2) weget
∴ MI = μ₀nIAN
M = μ₀nAN
If the solenoid is covered over core of relative per-meability μ_r
then M = μ_rμ₀nAN
c) ML²T^-2A^-2

Question 21.
A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5 cm. What is the magnifying power of the telescope for viewing distant objects when
a) the telescope is in normal adjustment?
b) the final image is formed at the least distance of distinct vision.
Answer:

Question 22.
In Amplitude Modulation, the amplitude of the carrier wave is varied in accordance with the information signal.
a) What is meant by modulation index?
b) A message signal of frequency 10 kHz and peak value of 10 V used to modulate a carrier of frequency 1 MHz and peak voltage of 20 V. Determine the modulation index.
c) The block diagram of a transmitter is shown below. Identify the elements labelled X and Y.

Answer:
a) Modulation index, μ = \(\frac{A_{m}}{A_{c}}\)
b) Am = 10 V, Ac = 20 V
∴ μ = \(\frac{A_{m}}{A_{c}}=\frac{10}{20}\) = 0.5

Answer any three questions from 23 to 26. Each carries 5 scores. (3 × 5 = 15)

Question 23.
The cyclotron is a device used to accelerate charged particles.
a) With a suitable diagram briefly explain the working of a cyclotron and obtain an expression for cyclotron frequency.
b) Acyclotron oscillator frequency is 10 MHz. What should be the operating magnetic field for accelerating protons?
Answer:
a) Principles: Cyclotron is based on two facts

1. An electric field can accelerate a charged particle.
2. A perpendicular magnetic field gives the ion a circular path.

Constructional Details: Cyclotron consists of two semicircular dees D₁ and D₂, enclosed in a chamber C. This chamber is placed in between two magnets. An alternating voltage is applied in between D₁ and D₂. An ion is kept in a vacuum chamber.

Working: At certain instant, let D₁ be positive and D₂ be negative. Ion (+ve) will be accelerated towards D₂ and describes a semicircular path (inside it). When the particle reaches the gap, D₁ becomes negative and D₂ becomes positive. So ion is accelerated towards D₁ and undergoes a circular motion with larger radius. This process repeats again and again.

Thus ion comes near the edge of the dee with high K.E. This ion can be directed towards the target by a deflecting plate.

Mathematical expression: Let ‘v’ be the velocity of ion, q the charge of the ion and B the magnetic flux density.
If the ion moves along a semicircular path of radius Y, then we can write

Eq. (2) shows that time is independent of radius and velocity.

Resonance frequency (cyclotron frequency): The condition for resonance is half the period of the accelerating potential of the oscillator should be ‘t’. (i.e., T/2 = t or T = 2t). Hence period of AC
T = 2t

Question 24.
The experimental set up to find an unknown resistance using a metre bridge is shown below:

a) What is the principle of a metre bridge?
b) If the balance point is found to be at 39.5 cm from the end ‘A’, the resistor ‘S’ is of 12.5 Ω. Determine the resistance ‘R’. Why are the connections between resistors in a metre bridge made of thick copper strips?
c) If the galvanometer and cell are interchanged at the balance point of the bridge would the galvanometer show any current?
Answer:

The resistors in metre bridge are made of thick copper strips to minimise the resistance of connection.
c) No. The galvanometer will not show any current.

Question 25.
The circuit used to change alternating voltage to direct voltage is called rectifier.
a) With a neat diagram, explain the working of a full wave rectifier having two diodes.
b) What is the output frequency of a full wave rectifier if the input frequency is 50 Hz?
c) Draw the output wave form across the load resistance connected in the full wave rectifier circuit.
Answer:
a) Full wave rectifier:
Circuit details

Full wave rectifier consists of transformer, two diodes and a load resistance R_L. Input a.c signal is applied across the primary of the transformer. Secondary of the transformer is connected to D₁ and D₂. The output is taken across R_L.

Working: During the +ve half cycle of the a.c signal at secondary, the diode D₁ is forward biased and D₂ is reverse biased. So that current flows through D₁ and R_L.

During the negative half cycle of the a.c signal at secondary, the diode D₁ is reverse biased and D₂ is forward biased. So that current flows through D₁ and R_L. Thus during both the half cycles, the current flows through R_L in the same direction. Thus we get a +ve voltage across R_L for +ve and -ve input. This process is called full wave rectifcation.

b) 100 Hz

Question 26.
A ray of light parallel to the principal axis of a spherical mirror falls at a point M as shown in the figure below:

a) Identify the type of mirror used in the diagram.
b) By drawing a suitable ray diagram, obtain the mirror equation.
c) If the mirror is immersed in water, its focal length ………….
Answer:
a) Concave mirror

b)

Let points P, F, C be pole, focus and centre of curvature of a concave mirror. Object AB is placed on the principal axis. A ray from AB incident at E and then reflected through F. Another ray of light from B incident at pole P and then reflected. These two rays meet at M. The ray of light from point B is passed through C. Draw EN perpendicular to the principal axis.
Δ IMF and Δ ENF are similar.

but IF = PI – PF and NF = PF (since aperture is small)
hence eq. (1) can be written as

This is called mirror formula or mirror equation,

c) Remain the same

Kerala State Board New Syllabus Plus Two Zoology Chapter Wise Previous Questions and Answers Chapter 8 Biodiversity and Conservation.

Kerala Plus Two Zoology Chapter Wise Previous Questions Chapter 8 Biodiversity and Conservation

Question 1.
Gopalan cultivated a variety of fruit crops and plants in this field and Raman destroyed the fruit crops and plants and cultivated rubber trees which are double in number. (MARCH-2010)
a) In your opinion which method is suitable for the ecosystem? Give reason.
b) Mention any three factors for the extinction of species.
Answer:
Method by Gopalan
Biodiversity depends on variety of species.  Extinction of species is due to

1. Over exploitation
2. Habitat loss and fragmentation
3. Co-extinction

Question 2.
a) The “Evil Quartet” is the nickname used to describe the causes of biodiversity losses. Explain the reason leading to accelerated rates of extinction of flora and fauna.  (MAY-2010)
b) Philosophically or spiritually, we need to realize that every plant or animal species has an intrinsic value and we have a moral duty to protect them. Justify the statement and write down the protective measures.
Answer:
a) (1) Habitat loss and fragmentation
(2) Over-exploitation
(3) Alien species invasions
(4) Co-extinction
b) Agreed with the statement as we have a moral duty to care for their well-being and pass on our biological legacy in good order to future generations.
Protective measures-
Insitu conservation – Biosphere reserve, National park, Wild Life Sanctuary etc.
Exsitu conservation – Zoo, Botanical gardens, seed bank etc.

Question 3.
The year 2010 has been declared as the International Biodiversity Year by United Nations (UN)  (MARCH-2011)
a) Point out the levels of diversity in nature.(1 Score)
b) Give a brief description of The Evil Quartet.
Answer:
a) Genetic, Species, Ecosystem level diversity.
b) i) Habitat loss and fragmentation
ii) Overexploitation
iii) Alien species Invasion
iv) Co-extinction etc.

Question 4.
The given graph shows the distribution of insects in different latitudes of earth.  (MARCH-2012)

a) What is your observation?
b) List the three reasons for greater biodiversity in tropical regions.
c) Write two causes of biodiversity lossess.
Answer:
a) Species richnes decreases from equator to poles.
b) Climate is constant and predictable Glaciations were absent. Tropical region get more solar energy.
c) Habital loss and fragmentation Invansion of alien species.

Question 5.
Last twenty years alone have witnessed the disappearance of 27 animal species from earth.  (MAY-2012)
a) Name an animal disappeared recently.
b) What may be the causes for this loss?
c) How can we conserve biodiversity?
Answer:
a) Out of Syllabus
b) i) Habitat loss and fragmentation
ii) Over-exploitation
iii) Alien species invasions
iv) Co-extinctions
c) It is through in situ (on site) conservation and ex situ (offsite) conservation

Question 6.
Gopalan cultivated a variety of fruit crops and plants . in this field and Raman destroyed the fruit crops and plants and cultivated rubbertrees which are double in number.  (MARCH-2013)
a) In your opinion which method is suitable for the ecosystem? Give reason.
b) Mention any three factors for the extinction of species.
Answer:
Method by Gopalan
Biodiversity depends on variety of species. Extinction of species is due to
1. Over exploitation
2. Habitat loss and fragmentation
3. Co-extinction

Question 7.
While preparing the species area relationship graph of 4 areas, the following z values are obtained.  (MAY-2013)
Area A = 0.1 Area B = 0.8 Area C = 1.2 Area D = 0.3
a) Which area shows maximum species richness?
b) What are the expected reasons for the loss of biodiversity in areas with low species richness?
Answer:
a) Area a = 0.1
b) (i) Habitat loss and fragmentation
(ii) Over-exploitation
(iii) Alien species invasions
(iv) Co-extinctions

Question 8.
“Nature provides all for the need of man but not for his greed.” (MARCH-2014)
a) Do you agree with this statement? Justify your answer.
b) Distinguish between two types of biodiversity conservations.
Answer:
a) yes, For example forest is used for some basic needs of a man but not for clearing of trees .
b) Exitu conservation – conservation of flora and fauna outside the natural habitat
Eg- Botanical garden
Insitu conservation – conservation of flora and fauna in the natural habitat
Eg-wild life sanctuaries and national parks

Question 9.
a) Variety of species are present around us, what they constitute and comment.  (MAY-2014)
b) Comment on in situ conservation and ex situ conservation.
c) In these aspects explain biodiversity hot spots with example – give importance to recent issues with regard to Western Ghats.
Answer:
a) Biodiversity
b) Exitu conservation – conservation of flora and fauna outside the natural habitat Insitu conservation- conservation.of flora and fauna in the natural habitat
c) Biodiversity hotspots are regions with very high levels of species richness and high degree of endemism Western ghats is the hotspots where accelerated habitat loss occurs.

Question 10.
We have a moral responsibility to take good care of earth’s biodiversity and pass it on in good order to next generation.  (MARCH-2015)
a) Define Biodiversity.
b) Write causes for biodiversity losses.
c) Name two types of biodiversity conservation.
Answer:
a) It is the variety within and between all species of plants, animals and micro-organisms and the ecosystems within which they live and interact.
b) i) Habitat loss and fragmentation
ii) Over-exploitation
iii) Alien species invasions
iv) Coextinctions
c) Insitu conservation, Exitu conservation

Question 11.
Two approaches for the conservation of biodiversity is shown as A and B.  (MAY-2015)

a) Identify the type of biodiversity conservation shown in A and B.
b) Write the difference between the two, types of biodiversity conservation shown in A and B.
c) Which of the above approach is more desirable’ when there is an urgent need to save a species?
Answer:
a) A-Insitu conservation B-Exitu conservation
b) A- It is the conservation of animals in natural habitat
B- It is the conservation of plants outside the natural habitat
c) exitu conservation

Question 12.
Observe the concept diagram of the Evil Quartet of biodiversity loss. (MARCH-2016)

a) Write A and B
b) What is Co-Extinction?
Answer:
a) A- Habitat loss and fragmentation
B -Alien species invasions
b) When a species becomes extinct, the plant and animal species associated with it also become extinct

Question 13.
Read the statement and choose the correct option:   (MARCH-2016)
A: Sacred grooves are examples of in situ conservation
B: Biodiversity hotspots have low degree of endemism.
C: Biodiversity increases when number of organisms in a particular species increases.
a) Statement ‘A’ alone is correct
b) Statement ‘A’ and ‘B’ are correct
c) Statement ‘A’ and ‘C’ are correct.
d) Statement ‘C’ alone is correct
Answer:
Statement ‘A’ alone is correct.

Question 14.
a) “When we conserve and protect the whole ecosystem, its biodiversity at all levels is protected.” Based on this statement explain the strategies of biodiversity conservation.  (MAY-2016)
OR
b) “When need turns to greed, it leads to biodiversity loss.” Substantiate this statement by explaining two causes of biodiversity loss.
Answer:
a) Insitu conservation- It is the conservation of plant and animal sp. in their natural habitat.
Eg-biosphere reserves, national parks and sanctuaries.
Exitu conservation- It is the conservation of plant and animal sp outside the natural habitat.
Eg- Zoological parks, botanical gardens and wildlife safari parks.
OR
b) i) Habitat loss and fragmentation: The degradation of many habitats by pollution affects the survival of many species. It results large habitats are broken up into small fragments.
Amazon rain forest is cleared for cultivating soya beans or for conversion to grasslands for raising beef cattle
ii) Over-exploitation : It leads to the over-exploitation of natural resources. For example the extinction of Steller’s sea cow, passenger pigeon was due to humans.
iii) Alien species invasions : The introduction of foreign species cause the reduction or extinction of indigenous species.
The Nile perch introduced into Lake Victoria in east Africa led to the extinction of more than 200 species of cichlid fish in the lake
iv) Co-extinctions : If two species are in obligatory relationship the extinction of one species affect the other.
Eg – coevolved plant-pollinator mutualism where extinction of one species leads to the extinction of the other.

Question 15.
Z – values of a frugivorous bat species are given below. Which value is not applicable to continents?  (MARCH-2017)
1) 0.6
2) 0.65
3) 0.20
4) 0.68
Answer:
Incorrect option

Question 16.
Distinguish in situ conservation from ex situ conservation with one example each.  (MARCH-2017)
Answer:
Insitu conservation- Ms the conservation of plant and animal sp. in natural habitat.
Eg- biosphere reserves, national parks and sanctuaries.
Exitu conservation- It is the conservation of plant and animal sp outside the natural habitat.
Eg- Zoological parks, botanical gardens and wildlife safari parks.

Question 17.
What are the advantages of biofertilizers over chemical fertilizers? Give an example for biofertilizer.  (MARCH-2017)
Answer:
a) 1) It prevents pollution
2) It improves soil structure and function
b) biofertilizer- Mycorrhiza

Question 18.
Explain the three levels of biodiversity. (MAY-2017)
OR
Explain different types of biodiversity conservation with example.
Answer:
1. Genetic diversity – A single species might show high diversity at the genetic level over its distributional range
2. Species diversity – Diversity at species level
3. Ecological diversity – Diversity at ecosystem level
OR
In situ conservation – the species are protected in their natural habitat.
example: National Park, Wildlife sanctuaries etc.
ex situ conservation – threatened animals and plants are protected outside their natural habitat.
example: Zoological Park, botanical gardens etc.

Kerala State Board New Syllabus Plus Two Zoology Chapter Wise Previous Questions and Answers Chapter 7 Microbes in Human Welfare.

Kerala Plus Two Zoology Chapter Wise Previous Questions Chapter 7 Microbes in Human Welfare

Question 1.
Ramu cultivated pea plants as an intercrop in his paddy field. After harvesting, he allowed the roots of the pea plants remain in the soil for some period. (MARCH-2010)
a) Do you think the action of Ramu is reasonable?
b) Justify your answer.
Answer:
Yes.
Rhizohium found in the root nodules in the pea plant can fix atmospheric nitrogen into nitrates and in-creases fertility of soil.

Question 2.
Match column I with II. (MAY-2010)

Answer:

Question 3.
A bacterial infection was effectively controlled by using a specific anitibiotic for a long time. But nowadays this antibiotic is not found to be so effective to control the said infection. (MARCH-2011)
Give a scientific explanation for this phenomenon based on evolution.
Answer:
Evidence for Natural selection and explanations like origin of antibiotic resistant varieties or elimination of sensitive varieties or Natural selection by Anthropo-genic action.

Question 4.
Rearrange the columns B & C with respect to A. (MARCH-2012)

Answer:
Monascus – Statin – Cholesterol
Streptococcus – Streptokinase – Cholestrol lowering
Pencillium – Penicillin – Antibiotic
Trichoderma – Cyclosporin-A – Immuno suppressant

Question 5.
Match the following (MAY-2012)

Answer:
(A) – (3)
(B) – (2)
(C) – (1)
(D) – (5)

Question 6.
Some bioactive molecules, their source and their medical impotance are given in the table below. Fill up the missing parts. (MARCH-2013)

Answer:
a) streptokinase
b) Trichoderma polysporum
c) Immunosuppressant in organ transplant patients
d) Statins

Question 7.
Complete the illustration appropriately. (MAY-2013)

Answer:
a) Biofertilisers
b) Yeast,ethanol production,
c) (1) Lady bird and dragon flies are useful in the elimination of aphids and mosquitoes.
(2) Bacillus thuringiencis
d) Dough for making dosa and idly is fermented by Bacteria.

Question 8.
The meaning of ‘antibiotics’is ‘against life’, whereas with reference to human beings they are ‘pro life’. Substantiate the statement with suitable example. (MARCH-2014)
Answer:
Penicillin was widely used to kill bacteria during infection.
During second world war,it was widely used to treat soldiers against bacterial infection

Question 9.
In our state waste managenfient is a problem. Government promotes and give subsidy to Biogas plants. Comment the functioning of biogas plants with the help of microbes. (MAY-2014)
Answer:
Methanobacterium is used to produce biogas and can be used as source of energy as it is inflammable. It is an anaerobic bacterium used in sludge digesters.

Question 10.
Microbes can also be used as a source of energy. (MARCH-2015)
Substantiate with suitable examples.
Answer:
Methanobacterium is used to produce biogas and can be used as source of energy as it is inflammable.

Question 11.
BOD of some water samples are given below: (MAY-2015)
A. Sample 1 – 200 mg/L
B. Sample 2 – 80 mg/L
C. Sample 3 – 300 mg/L
D. Sample 4 – 25 mgIL
a) Which of the above water sample is most polluted?
b) What is meant by floes ? What is its role in sewage treatment?
Answer:
a) Sample 4
b) Floes- masses of bacteria associated with fungal filaments to form mesh like structures.
They consume the major part of the organic matter in the effluent. This is significantly reduces the BOD of the effluent.

Question 12.
“BOD is commonly calculated as an index of water pollution”. (MARCH-2016)
a) Do you agree with this statement? Why?
b) Expand BOD.
Answer:
a) Yes, if pollution load increases ,BOD increases
b) biochemical oxygen demand

Question 13.
Choose the correct answer from the bracket. (MAY-2016)
Cyclosporin A is produced by ______
a) Aspergellus
b) Clostridium
c) Trichoderma
d) Acetobacter
Answer:
Trichoderma

Question 14.
Select a bio-control agent from the given microbes: (MAY-2016)
a) Baculo virus
b) Rhino virus
c) Picrona virus
d) Adenovirus
Answer:
a) Baculovirus

Question 15.
Complete the table by filling A, B, C and D using hints from the bracket:  (MAY-2017)
(Gobargas, Biological Control, Anabaena, Saccharomyces cerevisiae, Propionibacterium
sharmanii)

Answer:
A -Gobargas
B – Saccharomyces cerevisiae
C – Anabaena
D-Biological control

Kerala State Board New Syllabus Plus Two Zoology Chapter Wise Previous Questions and Answers Chapter 6 Human Health and Disease.

Kerala Plus Two Zoology Chapter Wise Previous Questions Chapter 6 Human Health and Disease

Question 1.
“More and more children in Metro cities of India suffer from allergies and asthma.” (MARCH-2010)
a) Do you agree with this statement?
b) Justify your answer.
Answer:
a) Yes
b) Absence of previous encounter with the allergies in the over protected environment reduces immunity and children become more prone to allergy/ over exposure to allergens or pollutants in the metro cities can be considered.

Question 2.
Widespread incidence of diseases like H₁H₁, Chikungunya, dengue fever etc. are reported recently. As a science student, prepare an action plan in your school to control those diseases. (MARCH-2010)
Answer:
Measures to control mosquitoes, precaution to prevent the spread of disease. Awareness campaign for students etc.

Question 3.
The flow chart given below represents the life cycle of malarial parasite. Complete the flow chart and answer the following: (MAY-2010)

a) Which species of malarial parasite causes malignant malaria ?
b) Can you suggest two methods to control malaria?
Answer:
2 – Liver, 4 – RBC
a) Plasmodium falci parum
b) 1. Public hygiene and proper waste disposal
2. Periodic cleaning and disinfection of water reservoirs
3. Control of mosquitoes and their breeding grounds (any methods to control insect vectors can be added)

Question 4.
The list given below includes various stages of HIV infection. Arrange them according to correct sequence. (MAY-2010)
Answer:
Viral replication – Macrophage – Helper T cell – Formation of DNA – Progeny Virus – Viral particle – Release of progeny virus

Question 5.
Lakshmi and Sujitha are two Nursery students. Sujitha gets common cold very often. Lakshmi not (MARCH-2011)
a) How do you interpret this in an immunological aspects?
b) What are the common barriers protecting Lakshmi from cold?
Answer:
a) Low level of innate or weak or poor immunity to Sujitha / High immunity to Lakshmi.
b) Two correct points (Physical, Physiological, Cellular and Cytokine barriers) like mucous membrane / Nostril hairs / skin / Antibodies / Interferons.

Question 6.
Arrange the following diseases in the following columns in a meaningful order.  (MARCH-2012)
Typhoid, Ringworms, Amoebiasis, AIDS, Malaria, Pneumonia, Common cold.

Answer:

Question 7.
In a classroom discussion a student argues that allergic diseases are more common in children of metrocities than in villages. ,  (MARCH-2012)
a) Do you agree with this statement.
b) Which type of immunoglobulin is responsible for allergic reactions?
c) Suggest two drugs which reduce allergic symptoms.
Answer:
a) Yes
b) IgE
c) Antilistamine and Adrenalin drugs

Question 8.
Note the relation between first two terms and suggest a suitable term for the fourth place. (MAY-2012)
a) Erythroxylum coca : cocaine :
Papaversomniferum : _______
b) Salmonella typhi: typhoid :
Plasmodium falciparum: ______
Answer:
a) morphine
b) malaria

Question 9.
Nature has as many varieties of plants which give drugs for abuse, as there are medicinal plants which give medicines. Substantiate with two examples. (MARCH-2013)
Answer:
a) Cocaine is obtained from coca plant Erythroxylum coca. It affects the transport of the neuro-transmitter dopamine.
b) Opioids are the drugs, receptors present in our central nervous system and gastrointestinal tract Heroin (smack the depressant,it is prepared by the acetylation of morphine which is extracted from the latex of poppy plant Papaversomniferum.

Question 10.
“Prevention is better than cure”. This statement is true in the case of AIDS as well as immunisation. Substantiate. (MAY-2013)
Answer:
Diseases like AIDS are not curable even after the treatment for long period. This is the immuno deficiency disease affect the resistant power of the body. Medicine for these diseases are not discovered so far. Likewise immunization is done in earlier period of growth is good .Hence prevention is better than cure.

Question 11.
Classify the diseases given in the box as two groups based on their causative organisms. Specify the type of causative organism for each group. (MARCH-2014)

Answer:
protozoans – Malaria and amoebiasis Bacterial — diphtheria, typhoid and pneumonia

Question 12.
Prepare a pamphlet for an awareness programme in your school about the measures to prevent and control alcohol and drug abuse in adolescents. (MARCH-2014)
Answer:
Education and counseling, moral education, Avoid undue peer pressure, visual publicity through TV, seeking professional and medical help etc.

Question 13.
Briefly describe the characteristics of cancer cells. (MAY-2014)
Answer:
1. lose of contact inhibition- Normal cells are contact with other cells inhibits their uncontrolled growth. Cancer cells lost this property.
2. Metastasis- Cells sloughed from tumors reach distant sites through blood, and wherever they get lodged in the body, they start a new tumor there.

Question 14.
It is said that “Chikunguinea” once affected will not affect a person in the next half of his life. Justify this statement. (MAY-2014)
Answer:
This is due to acquired immunity. Due to infection body first time produces a response called primary response which is of low intensity. Subsequent encounter with the same pathogen a highly intensified secondary response. This is appears to have memory of the first encounter. The primary and secondary immune responses are carried out with the help of two special types of lymphocytes present in our blood, i.e. B-lymphocytes and T- lymphocytes.

Question 15.
Mother’s milk is considered essential for new born infants. (MARCH-2015)
a) Name the fluid secreted by mother from breast during the initial days of lactation.
b) What type of immunity it provides?
Answer:
a) Colostrum
b) IgA provides passive immunity

Question 16.
Cancer is one of the most dreaded diseases of human beings, and is a major cause of death all over the globe. (MARCH-2015)
a) What are the causes of cancer?
b) What are the methods of detection of cancer?
c) What are the types of treatment for cancer?
Answer:
a) Ionising radiations like X – rays and gamma rays and non-ionizing radiations like UV cause DNA damage leading to neoplastic transformation. The chemical carcinogens present in tobacco smoke have been identified as a major cause of lung cancer.

b) Biopsy and histopathological studies of the tissue and blood and bone marrow tests for increased cell counts in the case of leukemias Radiography (use of X-rays), CT (computed tomography) and MRI (magnetic resonance imaging) are very useful to detect cancers of the internal organs

c) 1. Surgery, radiation therapy and immuno therapy
2. Chemotherapeutic drugs are used to kill cancerous cells

Question 17.
Match the terms given in the three columns of the table correctly: (MAY-2015)

Answer:
Haemophilus Influenzae – Bacteria – pneumonia
Plasmodium Vivax – protozoa – malaria Wuchereria
Bancrofli – flat worm – filarisis
Trichophyton – fungus – ringworm

Question 18.
Identify the disease shown in the following figure and write the causative organism of the disease.(MARCH-2016)

Answer:
Elephantiasis orfilariasis

Question 19.
“Blood of a man is tested positive for cannabinoid.” (MARCH-2016)
a) What are these?
b) From where these are extracted naturally?
c) Which part of the body is affected by these?
Answer:
a) Drug obtained from plants
b) Natural cannabinoids are obtained from the inflorescences of the plant Cannabis sativa
c) Cardiovascular system of the body

Question 20.
Breast feeding during initial period of infant growth is necessary to develop immunity of new born babies. Why? (MARCH-2016)
Answer:
Initial days of lactation contains colostrum which is rich in antibodies and provide immunity to new borne babies.

Question 21.
Answer the questions about the given figure: (MAY-2016)

a) Identify the parts X and Y.
b) Name any two types of this molecule.
Answer:
a) X-antigen binding site
Y-Heavy chain
b) lgA, lgG, lg E

Question 22.
Select the odd one out and justify your selection. (MAY-2016)
Malaria, Gonorrhea, Amoebiasis, Filariasis
Answer:
Gonorrhoea – It is the only sexually transmitted disease

Question 23.
Complete the table by filling a, b, c and d. (MAY-2016)

Answer:
a) pneumonia
b) Rhino virus
c) malaria
d) Inflammation of lymphatic vessels of lower limb/ genital organs

Question 24.
Feeding _______ in the first few days is essential for preventing infections in a newly born baby. (MARCH-2017)
Answer:
colostrum

Question 25.
Morphine is said to be an abused drug. Discriminate the terms ‘use’ and ‘abuse’ of drugs based on this example. (MARCH-2017)
Answer:
For medical purpose it is used as painkiller or sedative but it is abused by some individuals as narcotic drug.

Question 26.
Differentiate Active immunity from Passive immunity. Give an example for Passive immunity. (MARCH-2017)
Answer:
When the antigens are coming in the form of living or dead microbes, the body of organism produce antibodies. This type of immunity is called active immunity.
When ready – made antibodies are directly injected to protect the body against foreign agents, it is called passive immunity.
Eg-Anti-venom.

Question 27.
Prepare a brief note to be presented in an awareness programme for adolescents about AIDS, their causes and preventive measures. (MAY-2017)
Answer:
AIDS-it is acquired immuno deficiency syndrome caused by virus called HIV It is mainly prevented by

Avoid sex with multiple partners
Use disposable needles and syringes
Ensure safe blood transfusion

Question 28.
Fill the boxes A, B, C and D. (MAY-2017)

Answer:
A – Acquired immunity
B – Physiological barrier
C -Cytokin barrier
D-passive immunity

Kerala State Board New Syllabus Plus Two Zoology Chapter Wise Previous Questions and Answers Chapter 5 Evolution.

Kerala Plus Two Zoology Chapter Wise Previous Questions Chapter 5 Evolution

Question 1.
Arrange the following in a hierarchical manner based on the period of their evolution. (MARCH-2010)
Homoerectus, Ramapithicus, Australopithicines, Homosapiens, Neanderthal man.
Answer:
Ramapithicus -> Australopithicines -> Homoerectus -> Neanderthalman -> Homosapiens

Question 2.
Fill the columns A and B using the items given below: (MARCH-2010)

(Lamarkism, Evolution of anthropogenic action, Geneflow by chance, varieties of marsupials in Australia, De Vries)
Answer:

Question 3.
The diagrams shown below represents the operation of natural selection on different traits. Observe the diagrams and answer the following : (MAY-2010)

a) Why graph C shows a marked difference from graph A.
b) What is the evolutionary significance of Directional Selection?
c) Mention the factors affecting gene frequency.
d) What is meant by founder effect?
Answer:
a) A is stabilizing selection and C is disruptive selection.
b) Peak shifted to one direction.
c) Gene migration
Genetic drift
Mutation
Genetic recombination
Natural selection
d) The effect where original drifted population becomes founders is called founder effect.

Question 4.
a) Arrange the given chemical compound in the sequential order as per the concept of origin of life. (Ammonia, Hydrogen, Protein, Nucleic acid, Amino acid) (MARCH-2011)
b) Correlate Miller’s experiment with this.
Answer:
a) Hydrogen -> Ammonia -> Amino acid -> Protein Nucleic acid
OR
b) Brief account of Miller’s Experiment / Labelled diagram of the experimental setup.

Question 5.
Note the relationship between the first pair and complete the second pair. (MARCH-2012)
a) Natural selection : Drawing ; Inheritance of acquired characters : ________.
b) Heart of vertebrates: homologous organs; Flippers of Penguin and Dolphin : _______.
Answer:
a) Lamarck
b) Analogous organs

Question 6.
An evolutionary process occurred in the evolution of marsupial mammals in Australia is given below. (MARCH-2012)

a) Name this evolutionary process.
b) Suggest another example for this phenomenon.
Answer:
a) Adaptive radiation
b) Darwins finches

Question 7.
Arrange the following examples under two heads viz- homologous organs and analogous organs. (MARCH-2013)
Forelimb of whale and bat,
Wings of butterfly and bat,
Heart of man and cheetah,
Eyes of Octopus and mammals
Answer:
Homologous organs
Forelimb of whale and bat Heart of man and cheetah
Analogous organs
Wings of butterfly and bat Eyes of octopus and mammals

Question 8.
Theory of chemical evolution is a version of theory of abiogenesis. Analyze the statement. (MARCH-2013)
Answer:
Oparin and Haldane proposed that the first form of life that arose from preexisting non-living organic molecules (e.g. RNA, protein, etc.) and it is followed by chemical evolution

Question 9.
A specific rat population was controlled for about a decade by a poison. After apopulation decline for about 10 years, the rat population was increased stabilized. (MAY-2013)

Resistance to poison is governed by a dominant autosomal gene ‘R’. In 1975 majority of the resistant animals are heterozygous at this locus (Rr).
a) What was the major genotype of the rat population before 1961 ?
(A) RR, (B) Rr, (C) rr, (D) R is absent as it is produced by a mutation.
b) What explanation you give for the development of resistance against poison in these rats?
c) This illustration can be utilized to explain a theory of Evolution” Substantiate.
Answer:
a) R is absent as it is produced by mutation
b) Mutation leads to genetic variation and which in turn lead to evolution
c) Evolution by anthropogenic action

Question 10.
Given below is the diagrammatic representation of the operation of Natural Selection on different traits. (MARCH-2014)

a) Identify the type of natural selection A, B and C with explanation of each.
b) Define Hardy – Weinberg Principle.
Answer:
a) A- Stabilising selection-More individuals acquire mean character value
B – Directional selection – More individuals ac-quire value other than mean character
C – Disruptive selection – More individuals acquire peripheral character value at both ends.
b) According to Hardy-Weinberg principle allele frequencies in a population are stable and is con-stant from generation to generation. This is called genetic equilibrium. Sum total of all the allelic frequencies is 1.
Hence,p2 + 2pq + q2 = 1 Disturbance in Hardy- Weinberg equilibrium, i.e., change of frequency of alleles in a population affected by five factors These are gene migration or gene flow, genetic drift, mutation, genetic recombination and natural selection.

Question 11.
Arrange the following in a hierarchical manner in ascending order based on the period of their evolution.  (MAY-2014)Homoerectus, Ramapithecus, Australopithecus, Homo sapiens.
Answer:
Ramapithecus —> Australopithecus—>Homoerectus —> Homo sapiens

Question 12.
a) The diagram given below shows a particular type of evolutionary process in Australian marsupials. Identify the’ evolutionary phenomenon and comment on. (MAY-2014)

b) Give another example for such a type of evolutionary process and explain.
Answer:
a) convergent evolution
Number of marsupials are different from the other evolved from an ancestral stock, but all within the Australian island continent. When more than one adaptive radiation appeared to have occurred in an isolated geographical area(representing different habitats), it is called as convergent evolution.
b) Placental mammals in Australia also exhibit adaptive radiation in evolving into varieties of such placental mammals each of which appears to be ‘similar’ to a corresponding marsupial (e.g., Placental wolf and Tasmanian wolf marsupial).

Question 13.
Match the following: (MARCH-2015)

Answer:

Question 14.

The above shown pictures are beaks of a particular type of bird seen in an island during Darwin’s journey.  (MARCH-2015)
a) Identify the bird and name the island.
b) Write the significance of this process in evolution.
Answer:
a) Darwin’s Finches, Galapagos Islands
b) From the original seed-eating features, many other forms with altered beaks arose, enabling them to become insectivorous and vegetarian finches.

Question 15.
Four groups of organs are given below:  (MAY-2015)
Read them carefully and answer the questions:
A. Thorns of bougainvilla and Tendrils of cucurbita
B. Eyes of octopus and mammals
C. Flippers of penguin and dolphin
D. Forelimbs of cheetah and man
(a) Categorise the four groups of organs as homologous organs and analogous organs.
(b) Based on each group of organs differentiate convergent evolution and divergent evolution.
(c) Illustrate homologous and analogous organs as evidences for evolution.
OR
Observe the diagrammatic representation and answer the questions:

a) Explain the phenomenon shown in the figure.
b) How can it be considered as an evidence of evolution?
c) Write any other example for this phenomenon. Explain.
Answer:
a) Homologous organs
Thorns of bougainvilla and tendrils of cucurbits , fore limbs of cheetah and man Analogous organs: Eye of octopus and mammals, flippers of penguin and dolphin
b) Convergent evolution – different structures evolving for the same function
Divergent evolution – same structure developed for different function
c) In convergent evolution, the similar habitat that has resulted in selection of similar adaptive fea-tures in different groups of organisms but toward the same function:
In divergent evolution ,same structure developed along different directions due to adaptations to different needs
OR
a) Adaptive radiation
b) A number of marsupials, each different from the other evolved from an ancestral stock, but all within the Australian island continent. When more than one adaptive radiation appeared to have occurred in an isolated geographical area (representing different habitats), it represents convergent evolution.
c) Darwin’s finches represent one of the best examples of this phenomenon. From the original seed-eating features, many other forms with altered beaks arose, it helps them to become insectivorous and vegetarian finches.

Question 16.
Which theory talks about the huge explosion that leads to origin of universe?  (MARCH-2016)
Answer:
Big Bang theory

Question 17.
Read the principle and answer the questions: “Allele frequencies in a population are stable and constant from generation to generation called genetic equilibrium.’ (MARCH-2016)
a) Name the principle mentioned here.
b) Mention any two factors affecting the equilibrium.
c) What is the significance of disturbances occur in the genetic equilibrium?
OR
’Natural selection can lead to stabilisation, directional change and disruptive changes.’
Explain the terms stabilization, directional change and disruptive change mentioned above.
Answer:
a) Hardy-Weinberg principle
b) gene flow and genetic drift
c) they become a different species
OR
Natural selection can lead to 1 stabilisation (in which more individuals acquire mean character value)
2. directional change (more individuals acquire value other than the mean character value)
3. disruption (more individuals acquire peripheral character value at both ends of the distribution curve)

Question 18.
Observe the diagram and answer the questions below: (MAY-2016)

a) Identify the types of evolution in the concept diagrams A and B.
b) Write example pair each for homologous and analogous organs.
Answer:
A – Divergent evolution
B – Convergent evolution
Homologous organs- Cheetah and human – bones of forelimbs, thorn of Bougainvillea and tendrils of cucurbits.
Analogous organs – eye of the octopus and of mammals and flippers of Penguins and Dolphins.

Question 19.
Statement below show the features of some human fossils. Read carefully and identify the fossil. (MAY-2016)
a) Human like beings with brains capacities between 650 – 800 cc
b) Lived in East and Central Asia with brain capacity of 1400 cc.
Answer:
a) Homohabilus
b) Neaderthal man

Question 20.
A population of 208 people of MN blood group was sampled and it was found that 119 were MM group, 76 MN group and 13 NN group. Answer the following questions: (MARCH-2017)
a) Determine the gene frequencies of M and N alleles in the population.
b) How does the above frequencies affect evolution?
OR
Examine the pictures of Darwin’s Finches given below and answer the following questions:
a) What phenomenon in evolution is represented in the picture?
b) Explain the phenomenon with the help of an additional example.

Answer:
a) Adaptive radiation –In this evolution starting from a point and radiating to other areas of geography
b) A number of different marsupials evolved from an ancestral stock within the Australian island.

Question 21.
Which of the following sets of gases were used in Mliller’s experiment? (MARCH-2017)
1) \(\mathrm{CH}_{4}, \mathrm{NO}_{2}, \mathrm{H}_{2} \mathrm{O}, \mathrm{CO}_{2}\)
2) \(\mathrm{NH}_{3}, \mathrm{CH}_{3}, \mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2}\)
3) \(\mathrm{NH}_{3}, \mathrm{CH}_{3}, \mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2}\)
4) \(\mathrm{H}_{2} \mathrm{O}, \mathrm{N}, \mathrm{CH}_{4}, \mathrm{H}_{2}\)
Answer:
2) \(\mathrm{NH}_{3}, \mathrm{CH}_{3}, \mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2}\)

Question 22.
Diagrammatic representation of the operation of Natural Selection on different traits is given. Observe it and answer the questions: (MAY-2017)

a) What do B and C represent?
b) Explain the process shown in B and C.
Answer:
a) B is directional selection C is disruptive selection
b) B natural selection leads to directional change in which more individuals acquire value other than the mean character value.
C natural selection leads to disruption in which more individuals acquire peripheral character value at both ends of the distribution curve.

Question 23.
Rearrange the following in the order of their evolution period (MAY-2017)
Australopithecines
Neanderthal man
Homo sapiens
Homo erectus
Dryopithicus
Answer:
Dryopithecus —> australopithecus —> Homo erectus —> neanderthal man —> Homo sapiens

Kerala State Board New Syllabus Plus Two Zoology Chapter Wise Previous Questions and Answers Chapter 4 Molecular Basis of Inheritance.

Kerala Plus Two Zoology Chapter Wise Previous Questions Chapter 4 Molecular Basis of Inheritance

Question 1.
Observe the diagram below: (MAY-2010)

Which among the following is the complimentary sequence of the DNA fragment shown above?
a) 5′ -> ATTCG -> 3′
b) 3′ -> ATTCG -> 5′
c) 3′ -> TAAGC -> 5′
d) 3′ -> CGAAT -> 5′
Answer:
3′ -> TAAGC -> 5′

Question 2.
DNA sequence is provided below:   (MAY-2010)
TACGAGTTATATATACAT
a) Write down the triplet codon- it codes for.
b) If a nitrogen base is added in between 4th & 5th nitrogen bases, what will be its effect on transcription?
Answer:
a) AUG / CUC / AAU / AUA / UAU / GUA
OR
TAC / GAG / TTA / TAT / ATA / CAT
b) Frame shift mutation / point mutation/Mutation / Entire amino acid sequence changed / incorrect protein produced / change in protein structure / incorrect transcription and translation /abnormal protein.

Question 3.
In a paternity dispute, the VNTR DNA samples of parent and child were DNA finger printed. The diagrammatic representation of the DNA fingerprint is shown below.  (MAY-2010)

a) What is your opinion about the paternity of child? Substantiate your opinion.
b) List down any four major steps of molecular biological procedures adopted for this.
Answer:
a) Paternity is confirmed
VNTR bands / finger print bands / bands are similar.
b) Steps of DNA finger printing techniques are
i) Isolation of DNA
ii) Digestion of DNA by restriction endo nuclease
iii) Seperation of DNA fragment by Electrophoresis,
iv) Plotting of DNA fragment into Nitro cellulose
v) Ditection of hybridised DNA by Auto radiography

Question 4.
A transcriptional unit is given below. Observe it and answer the questions.  (MARCH-2012)

a) How can you identify the coding strand?
b) Write the sequence of RNA formed from this unit?
c) What would happen if both strands of the DNA act as templates for transcription?
Answer:
a) Coding strand has 5′ end at the promoter or 5’TCAGTACA3′
b) 5′ UCAGUACA 3′
c) The two RNA will be complimentary and may form double stranded RNA and it prevents the translation.

Question 5.
In E.coli lactose catabolism is controlled by Lac operon. Lac operon in the absence of inducer (lactose) is given below. (MARCH-2012)

a) What is ‘P’?
b) Name the enzymes produced by the structural genes ‘Z’, ‘Y’ and ‘a’?
c) Redraw the diagram in the presence of an inducer.
Answer:
a) Promoter
Z-Bgalactosidase
b) Y – Permease
a-Transacetylase
c)

Question 6.
Following are the first two steps in Griffith transformation experiment: (MAY-2012)
1) S strain 2 Inject into mice 2 mice live
2) R strain 2 Inject into mice 2 mice die
a) If there is any mistake correct it.
b) Write the remaining steps.
Answer:
a) 1. S strain -> inject into mice -> mice die
2. R strain inject into mice -> mice live
b) When Griffith was injected heat-killed S strain into mice, bacteria did not kill them. But he injected a mixture of heat-killed S and live R bacteria, the mice died .

Question 7.
DNA is better genetic material than RNA. Do you agree with this statement ? Substantiate. (MAY-2012)
Answer:
The 2-OH group present at the nucleotide in RNA is a reactive group and makes RNA labile and easily degradable. Hence DNA is less reactive and structurally more stable when compared to RNA. Therefore, among the two nucleic acids, the DNA is a better genetic material.

Question 8.
Given below is the diagrammatic representation of first stage of a process in bacteria. (MAY-2012)

a) Identify the process.
b) Name the enzyme catalyses this process.
c) What are the additional complexities in Eukaryotes for this process?
Answer:
a) Prokaryotic transcription
b) RNA polymerase
c) In eukaryotes hnRNA undergo additional processing called as splicing, capping and tailing.
In splicing the introns are removed and exons are joined together. In capping methyl guanosine triphosphate is added to the 5-end of hnRNA. In tailing, adenylate residues are added at 3-end in a template.The fully processed hnRNA, called as mRNA, that is transported out of the nucleus for translation.

Question 9.
The flow of genetic information is shown below. (MARCH-2013)

Name the processes a, b, c
Answer:
a) Replication
b) Transcription
c) Translation
d) Reverse transcription

Question 10.
RNA is not an ideal molecule as genetic material because ________. (MARCH-2013)
a) 2’OH group of ribose is reactive and make it labile
b) It is catalytic and hence reactive
c) Both (a) and (b)
d) None of the above
Answer:
c) Both a and b

Question 11.
Presence of lactose enhances the production of (3 galactosidase and other enzymes in bacteria, how will you explain this phenomenon? (MAY-2013)
Answer:
In lac operon,lactose is the inducer.Inducer binds with repressor protein but repressor cannot bind to op-erator gene,the free operator gene induces the RNA polymerase to bind with promoter and initiates tran-scription Three structural genes synthesise three mRNAs to produce enzymes.

Question 12.
A DNA sequence needed for coding a peptide is given below. (MAY-2013)
CAAGTAAATTGAGGACTC
(Hint: Codons and Aminoacids)
CAAGTAAATTGAGGACTC
UUA-Leu
CCU – Pro
CAU – His
ACU-Thr
GUU-Val
GAG-Glu
a) Write the complementary mRNA coding sequence for this.
b) Find out the amonoacid sequence of the peptide chain using the codons given in the hints.
c) If a mutation cause a change in the sixth codon CTC to CAC. It leads to a mendelian disorder. Identify the disease and write the specific characteristics of the diseasie.
Answer:
a) GUUCAUUUAACUCCUGAG
b) Val,His,Leu,Thr,Pro,Glu
c) Sickle cell anaemia, RBC become sickle shaped and unable to take oxygen .These are destroyed more rapidly leading to anaemia.

Question 13.
Draw a flow chart showing the steps of southern blot hybridization using radio labelled VNTR. (MAY-2013)
Answer:

Question 14.
“Prediction of the sequence of amino acids from the nucleotide sequence in mRNA is very easy, but the exact prediction of the nucleotide sequence in m RNA from the sequence of amino aids coded by mRNA is difficult”.(MARCH-2014)
a) Which properties of the genetic code is the reason for the above condition? Explain.
b) Which are the stop codons in DNA replication?
Answer:
a) Degeneracy – A single aminoacid is represented by many codons (degenerate codons)
b) Stop codons are UAA,UAG,UGA

Question 15.
Diagrammatic representation of Central dogma’ is given below: Observe the diagram carefully and redraw it making appropriate corrections. (MARCH-2014)

Answer:

Question 16.
Explain the phenomenon shown in the following figure and the reason for difference in the production of  II recombinants in Cross A and Cross B as explained by morgan. (MARCH-2014)

Difference in chromosome number of some human beings A, B. C and D are given below:
A) 22 pairs of Autosomes
B) 22 pairs of Autosomes+XO
C) 22 pairs of Autosomes + 1 Autosome
D) 22 pairs of Autosomes+XXY
OR
a) Identify the person who suffers from Klinefelter’s syndrome. Write its symptoms.
b) Differentiate between aneuploidy and polyploidy.
Answer:
The percentage of cross over depends upon the distance between the genes.
The genes are closer in cross A,so the less number of recombinants are produced. This is called link-age.
In cross B, genes are distantly located so more number of recombinants are produced.
OR
a) 22 pairs of autosomes+XXY, Sterile male, poorly developed testis and mental retardation
b) Aneuploidy-failure of separation of chromosomes during cell division results in-the gain or loss of chromosome.
Polyploidy – Failure of cytokinesis aftertelophase results in the increase in a whole set of chromosome

Question 17.
Correct the amino acid sequence of sickle cell haemoglobin. (MAY-2014)

Answer:

Question 18.
Diagram of components of DNA are given below: Identify and differentiate the two diagrams I and II. (MAY-2014)

Answer:
I- Nucleotide II- Nucleoside

Question 19.
a) Identify the diagram and explain. (MAY-2014)

b) In same cases DNA is produced from RNA. Name this process and give example.
Answer:
a) Central Dogma in molecular biology
b) Reverse transcription

Question 20.
a) Define mutation. (MAY-2014)
b) What are the different types of mutation?
Answer:
a) Mutation is a phenomenon which results in alteration of DNA sequences and consequently results in changes in the genotype and the phenotype of an organism
b) physical mutation and chemical mutation

Question 21.
a) Paternity or maternity can be determined by certain scientific methods. What is it? Define. (MAY-2014)
b) Briefly write the* methodology involved in the technique.
c) Comment on its other applications.
Answer:
a) DNA fingerprinting. DNA fingerprinting involves identifying differences in some specific regions in DNA sequence
b) i) isolation of DNA
ii) digestion of DNA by restriction endonucleases
iii) separation of DNA fragments by electrophoresis
iv) transferring (blotting) of separated DNA fragments to synthetic membranes, such as nitrocellulose or nylon
v) hybridisation using labelled VNTR probe
vi) detection of hybridised DNA fragments by autoradiography
c) It is also used in forensic science

Question 22.
Fill in the blanks: (MARCH-2015)
a) _________is a metabolic disorder that occurs due to the lack of an enzyme, that converts phenylalanine to tyrosine.
b) _______is a disease caused by the substitution of Glutamic acid by valine at 6th position.

Answer:
a) Phenylketonuria
b) Sickle-cell anaemia

Question 23.
The flow of genetic information is shown below. Name the process of (a) and (b). (MARCH-2015)

Answer:
a) Trancription
b) Translation

Question 24.
Explain transcription. A transcription unit in DNA is defined primarily by three regions. Write the names of any two regions. (MARCH-2015)
Answer:
Information contained in DNA is copied down to mRNA is called Transcription – Promoter and structural gene.

Question 25.
Observe the diagram and answer the questions: (MAY-2015)

a) What is the difference in the replication processes in A strand and B strand?
b) What is the role of DNA ligase in the replication process in B strand?
c) What is meant by Replication fork?
Answer:
a) In A, replication is continuous and form leading strand but in B replication is discontinuous and form lagging strand.
b) In B, short stretch of Okazaki fragments are connected with the help of DNA ligase and form continuous strand
c) The replication occur within a small opening of the DNA helix, referred to as replication fork. It appears as Y shaped forked structure.

Question 26.
Observe the following diagram and answer the questions: (MAY-2015)

a) Diagrammatically represent the changes take place when lactose is added to the medium.
b) What is the role of z, y and a genes in this metabolic pathway?
Answer:
a)

b) The z gene codes for beta-galactosidase which is responsible for the hydrolysis of the disaccharide, lactose into galactose and glucose. The y gene codes for permease, which increases permeability of the cell to beta-galactosides.The a gene encodes a transacetylase

Question 27.
Read carefully the sequence of codons in the mRNA unit and answer the questions. (MARCH-2016)

a) What change in needed in the first codon to start the translation process?
b) If translation starts by that change, till which codon it can continuous? Why?
Answer:
a) AUG is needed
b) UGA, Termination of translation occurs

Question 28.
Schematic representation of DNA fingerprints are shown below: (MARCH-2016)

[Hints : C is a sample taken from a crime scene, A and B from two suspected individuals]
a) Which one of the suspected individual may involved in the crime?
b) Write any other use of DNA fingerprinting.
Answer:
a) suspected person is B
b) disputed parentage determining population and genetic diversities.

Question 29.
Observe the figure of mRNA and answer the questions: (MAY-2016)

a) Find the start and stop codons.
b) How many amino acids will be present in the protein translated from this mRNA?
c) The additional sequences that are not translated in mRNA are called _______.
Answer:
a) Start codon – AUG
Stop codon – UAG
b) 4
UTR- Untranslated region

Question 30.
a) The hints of the lac operon is given below: (MAY-2016)
Hints:
Inducer, Repressor,
Structural genes, operator Regulatory gene
i) Which substance is acting as inducer in this operon?
ii) Explain the working of operon in presence of the inducer.
OR
b) With the help of the figure given, explain the processing of hnRNA to mRNA in eukaryotes.

Answer:
a) i) lactose inducer
ii) If lactose is present in a medium, it has to be breakdown into glucose and galactose,for this positive control of operon works and structural genes transcribe. The switch on condition of operator gene is due to the binding of RNA polymerase with promotersite. In +ve control, repressor from regulator gene is inactivated due to the presence of lactose.
OR
b) 1) Splicing-the introns are removed and exons are joined together.
2) capping -methyl guanosine triphosphate is added to the 5-end of hnRNA.
3) Tailing- adenylate residues (200-300) are added at 3-end in a template
After these three process, fully processed mRNA is released from nucleus into cytoplasm for protein synthesis.

Question 31.
Examine the following fragment of beta globin chain in human haemoglobin and identify the hereditary disease with reason. (MARCH-2017)

Answer:
Sickle cell anaemia
The defect is caused by the substitution of Glutamic acid by Valine at the sixth position of the beta globin chain of the haemoglobin molecule The mutant haemoglobin molecule under low oxygen tension causing the change in the shape of the RBC from biconcave disc to elongated sickle like structure.

Question 32.
Which of the following combinations do not apply to DNA? (MARCH-2017)
a) Deoxyribose, Guanine
b) Ribose, Adenine
c) Deoxyribose, Uracil
d) Guanine, Thymine
1) (a) and (b)
2) (b) and (c)
3) (c) and (d)
4) (a) and (d)
Answer:
2) b) and c)

Question 33.
Examine the diagram of mRNA given below. Mark the ‘5’ and ‘3’ ends of the mRNA by giving reasons. (MARCH-2017)

Answer:
Polyadynilation always at 3′ end of m RNA ,so the other must be 5′ end.

Question 34.
A small fragment of skin of a different person was extracted from the nails of a murdered person. This fragment of skin led the crime investigators to the murderer. Based on this incident answer the following questions: (MARCH-2017)
1) What technique was used by the investigators?
2) What is the procedure involved in this technique?
OR
In an E. coli culture lactose is used as food instead of glucose. If so, answer the following questions:
1) How do the bacteria respond to the above situation at genetic level?
2) If lactose is removed from the medium what will happen?
Answer:
1) DNA finger printing
2) The technique, is based on following procedure
i) isolation of DNA,
ii) digestion of DNA by restriction endonucleases,
iii) separation of DNA fragments by electrophoresis,
iv) transferring (blotting) of separated DNA fragments to synthetic membranes, such as nitrocellulose or nylon,
v) hybridisation using labelled VNTR probe, and vi) detection of hybridised DNA fragments by autoradiography
OR
In this closely associated genes function as unit called operon If lactose is present in a medium, it has to be breakdown into glucose and galactose ,for this positive control of operon works and structural genes transcribe with the help of switch on condition of operator gene. This is due to binding of RNA polymerase on promoter site of DNA.

Question 35.
Find the odd one and write the common feature of others. (MAY-2017)
Cytidine, Adenine, Thymine, Guanine.
Answer:
Cytidine, All others are nitrogen bases.

Question 36.
Observe the diagram:  (MAY-2017)

a) Redraw the diagram correctly if any mistake is there.
b) What does the diagram indicate?
c) What is the function of DNA ligase in this process?
Answer:
a)

b) Replication fork.
c) Discontinuously synthesised fragments of DNA are joined by DNA ligase.

Question 37.
Read the codon sequence in the mRNA unit which is undergoing translation.  (MAY-2017)

a) What will happen if the nitrogen base ‘U’ in the sixth position is replaced by ‘A’ by point mutation?
b) Name and define this type of mutation.
c) Draw the base sequence in the coding DNA strand from which the above mRNA is transcribed.
Answer:
a) Translation stop as UAA is a stop codon.
b) In this point mutation one nitrogen base is deleted.
c)

Kerala State Board New Syllabus Plus Two Zoology Chapter Wise Previous Questions and Answers Chapter 3 Principles of Inheritance and Variation.

Kerala Plus Two Zoology Chapter Wise Previous Questions Chapter 3 Principles of Inheritance and Variation

Question 1.
Polypeptide chains of two haemoglobin molecules are shown below. One of the chains shows an abnormality. Observe the diagram and answer the following questions. (MARCH-2010)

a) Which of the polypeptide chain in haemoglobin is abnormal leading to a disease?
b) What is the reason for this abnormality?
c) What will be the effect of this change in polypeptide chain?
Answer:
a) A
b) Substitution of glutamic acid by valine in the 6th position of polypeptide chain.
c) The RBC become sickle shaped causing a disease sickle cell anaemia / Affect the o₂ carrying capacity of RBC.

Question 2.
To find out the unknown genotype of a violet flowered pea plant a researcher done the following cross. Observe the diagram and answer the following questions: (Hint: Violet flower colourin pea plant is dominant over white). (MARCH-2010)

a) What would be the above cross called?
b) Can you determine the unknown genotype of violet flowered parent by drawing Punnet square?
Answer:
a) Test cross
b)

Genotype of unknown parent Ww/ Heterozygous.

Question 3.
Some genetic abnormalities, their genotypes and features are distributed in columns A, B and C respectively. Match them correctly. (MAY-2010)

Answer:

Question 4.
The flow charts A and B given below represents the inheritance of normal haemoglobin and sickle cell haemoglobin: (MAY-2010)

a) Observe flow chart A and complete the flow chart B.
b) Note down the genotype of a sickel cell anaemia patient and mention the symptom of the disease.
c) Mention the peculiarity of Hb^A Hb^s phenotype.

Answer:
a)

b) Sickle cell Anemia genotype is homozygous for Hb^s (Hb^sHb^s).
Symptoms
SevereAnemia
Oxygen shortage etc.
Haemoglobin becomes sickle shaped.
c) Hb^A Hb^s indicate heterozygous individuals, who are the carrier of sickle cell Anemia, but unaffected ones.

Question 5.
After analyzing’the karyotype of a short statured round headed person with mental retardation, a general physician noticed an addition of autosomal chromosome. (MARCH-2011)
Answer the following questions.
a) Addition or deletion of chromosome generally results in ______
b) What may be the possible syndrome or disorder of the above person should suspected to be?
c) Suggest two / more morphological peculiarities to confirm the chromosome disorder in that person.
Answer:
a) Aneuploidy / Chromosomal disorders/ Chromo-somal abberrations/Chromosomal mutations.
b) Down’s syndrome / Mongolism / Trisomy 21st / 45A + xx / xy.
c) Furrowed tongue, partially opened mouth, Broad palm, Skin fold at the corner of the eye.
Two correct points.

Question 6.
The frequency of occurrence of Royal disease or . haemophilia is high in the pedigree of royal families of Queen Victoria. Which of the following cannot be generally inferred from this? (MARCH-2011)
a) Queen Victoria was not homozygous forthe disease.
b) Many heterozygous females were there in the Royal families.
c) Non-Royal families were not affected with haemophilia.
d) There is less possibility to become a female diseased.
e) Generally a diseased female cannot survive after the first menstruation.
f) Pedigreee analysis is the study of inheritance patterns of traits in human females.
Answer:
‘C’ and ‘f ’ c₁ non royal families were not affected with haemophilia.
f₂ Pedegree analysis is the study of inheritance pattern of traits in human females.

Question 7.
A couple have two daughters. The blood group of husband and wife is ‘O’ (MARCH-2011)
a) What is the possible blood groups of the children should have?
b) Whether any change in blood group will occur if they have two sons instead of daughters.
Substantiate your answer.
Answer:
a) O group
b) No.
There is no sex specificity in blood group alleles / Co-dominance / Homozygous recessive. Not related to sex chromosomes or autosomes /

Question 8.
Complete the table using suitable terms. (MARCH-2012)
Turner’s syndrome (a) ______ Sterile female
(b) __________44A + XXY (c) _________
(d) __________ Trisomy 21 Mental retardation
Answer:
a) 44A + X0
b) Klinefelters Syndrome
c) Sterile male
d) Down’s Syndrom

Question 9.
In pea plant the gene for yellow seed colour is dominant over green and round seed shape is dominant over wrinkled. Write the four types of gametes formed in a heterozygous pea plant with yellow and round seeds. (Yy Rr).
Answer:
YR Yr yR yr

Question 10.
The first child of a couple is affected with phenylketonuria. During the second pregnancy they visited a genetic counsellor and he prepared a pedigree chart of their family. (MARCH-2012)
a) What is pedigree analysis?
b) Draw the symbols for
i) Affected female.
ii) Sex unspecified.
iii) Consanguineous mating.
Answer:
a) It is the study of inheritance of a trait for several generation of a family.
b)

Question 11.
Diagrammatic representation of chromosome map of Drosophila is given below: (MARCH-2012)

Y – Yellow
W – White
M – Miniature
a) Which genes are more linked?
b) Who mapped the chromosome firstly?
c) Tightly linked genes show low recombination. Why?
Answer:
a) y and w
b) Alfred Sturtevant
c) Crossing over rarely takes between genes

Question 12.
Work of a student is given below:  (MAY-2012)

a) From the above give an example for genotype and phenotype.
b) Complete the work’using punnett square and find out the phenotypic ratio in the F2 generation.
Answer:
a) First filial generation Genotype-RrYy,Phenotype- round yellow

Question 13.
Identify the traits from the pedigree chart. Give one example each.  (MAY-2012)

Answer:
a) Autosomal dominant trait
Eg. Muscular dystrophy
b) Autosomal recessive trait
Eg. Sickle cell anaemia

Question 14.
A poultry farm manager was cursing his hens for producting lion share of cocks in its progeny. Hearing this, Kumar – farm attender starts to blame his wife for delivering consecutive girl children. Analyze the situtations scientifically and state whether you agree with kumar.  (MARCH-2013)
Answer:
No, During spermatogenesis two types of gametes are produced of which 50 per cent carry the X- chromosome and the rest 50 per cent has Y- chromosome besides the autosomes. Females produce only one type of ovum with an X- chromosome.
In case the ovum fertilises with a sperm carrying X- chromosome the zygote develops into a female (XX) and with Y-chromosome results into a male offspring. Thus the genetic makeup of the sperm that determines the sex of the child.

Question 15.
In the given pedigree the shaded figures denote individuals expressing a specific trait. (MARCH-2013)

Which of the following is the most probable mode of inheritance of this trait.
Answer:
Principles of inheritance & Variation
A – Simple Mendelian Recessive
B – Co dominant relationship of a single pair of alleles.
C- X – linked recessive transmission
D – X- linked dominant transmission
E – Polygenic inheritance.
X-Linked recessive transmission

Question 16.
“Gopalan argues that if father is of A’ blood group, mother is of ‘B’ blood group. Their children can only be A’ group, B group or AB’ group.” (MARCH-2014)
a) Do you agree with Gopalan’s Argument ?
b) Give reason for your answer.
Answer:
a) No ,All groups are possible including O group
b) 

Question 17.
Identify the syndrome from the genotype given below: (MAY-2014)
1) 44 Autosomes + XXY
2) 44 Autosomes + XO
Answer:
1) Klinefelters syndrome
2) Turners syndrome

Question 18.
The family of Queen Victoria shows a number of haemophilic descendants as she was the carrier of the disease. Name the pattern of inheritance of this royal disease. (MAY-2014)
Answer:
Sex linked inheritance .
The heterozygous female (carrier) for haemophilia may transmit the disease to sons.
(Criss cross inheritance)

Question 19.

a) Identify the syndrome from the diagram, and write the genotype. (MARCH-2015)
b) It occurs in both sexes (male and female)? Write the reason.
Answer:
a) Downs syndrome , 45 A+XX or 45A+XY
b) The cause of this genetic disorder is the presence of an additional copy of the chromosome number 21 (trisomy of 21).

Question 20.
It is evident that, it is the genetic make up of the sperm that determine the sex of the child in human beings. Substantiate. (MARCH-2015)
Answer:
During spermatogenesis among males, two types of gametes are produced. 50% of the total sperm produced carry the X-chromosome and the rest 50 per cent has Y-chromosome besides the autosomes. In case the ovum fertilises with a sperm carrying X- chromosome the zygote develops into a female (XX) and the fertilisation of ovum with Y-chromosome carrying sperm results into a male offspring. Hence the genetic makeup of the sperm that determines the sex of the child.

Question 21.
Diagrammatic representation of the pedigree analysis of the inheritance of sickle cell anaemia is shown below: (MAY-2015)

a) Name the type of inheritance shown in the figure.
b) Write the genotype of A and B.
(Hint: Disease is controlled by a pair of alleles Hb^A & Hb^s)
Represent pedigree analysis of an X-linked recessive inheritance diagrammatically.
Answer:
a) Mendalian inheritance
b) A-Hb^A Hb^s B- Hb^A Hb^s

Question 22.
Observe the inheritance shown in A (MAY-2015)

a) Name the type of inheritance shown in A and B.
b) What is the difference between the two types of inheritance.?
Answer:
a) A- Dominance B-Co dominance
b) In dominance , dominant gene mask the expression of recessive gene and tall character in F₁ progenies but in co dominance both dominant gene express together and shows AB blood group.

Question 23.
Results of a famous experiment is given in the figure. Answer the questions. (MARCH-2016)

a) Identify the experiment.
b) Which property of the DNA is proved by this experiment?
Answer:
a) Semiconservative DNA replication experiment
b) During DNA replication one parent strand is conserved

Question 24.
Which of the following is not a Mendelian disorder? (MARCH-2016)
i) Colourblindess
ii) Down’s syndrome
iii) Haemophilia
iv) Thalassemia
Answer:
ii) Down’s syndrome

Question 25.
Study the following cross and answer the questions. (MARCH-2016)

[Hint: ABO blood group in man is controlled by three alleles I^A, I^B, and i]

a) Write the geno types of Father, Mother and Son.
b) The type of dominance of human blood group inheritance is ________.
Answer:
Father-I_A i or i i Mother-I_B or i i Son-i i

Question 26.
Observe the figures and answer the questions. (MARCH-2016)

a) Identify the syndromes A and B.
b) What is the chromosome numbers in A and B?
Answer:
a) A-KlinefelterSyndrome;
B- Turner’s Syndrome
b) A-47 B-45

Question 27.
a) Complete the flow chart of chromosomal disorder by filling the blank boxes (A and B). (MAY-2016)

b) What is an euploidy?
Answer:
a) A-monosomy of sex chromosome B – Klinefelters syndrome
b) Failure of separation of homologous chromosomes during anaphase I of meosis lead to gain or loss of chromosome, it is called aneupoidy.

Question 28.
Observe the figure below and answer the questions following: (MAY-2016)

a) Identify the figure
b) What shows the shaded symbols used?
Answer:
a) Pedigree analysis or Autosomal dominant trait
b) Individuals with genetic disorder (male and female)

Question 29.
The following table shows the F₂ generation of a dihybrid cross, identify the ‘Phenotype’ with homozygous recessive genotype.(MARCH-2017)
Find out A : B : C : D.

Answer:
B , Genotype A: B:C:D: = 3:1:9:3 OR 9:3:3:1

Question 30.
Which of the following do not have similar sex chromosomes? (Homogametic) (MARCH-2017)
a) Human female
b) Drosophila female
c) Bird female
d) Bird male
Answer:
c) Bird female

Question 31.
Observe the following diagram and answer the questions: (Hint: Steps in making a cross in pea plant) (MAY-2017)

a) Name the process marked as A and write its significance.
b) Diagrammatically represent a monohybrid cross between Tall and dwarf pea plants.
Answer:
a) Removal of anthers from female plants
Significance –It prevents self pollination and fertilization

Question 32.
Observe the diagrammatic representation of the following pedigree analysis and answer the questions (MAY-2017)
a) Describe the type of inheritance shown in the diagram.
b) Distinguish between Mendelian disorder and chromosomal disorder with example.

Answer:
a) Sex linked inheritance
b) Mendelian disorder is determined by mutation in the single gene.
These disorders are transmitted to off springs in the same line as the principles of inheritance.

Kerala State Board New Syllabus Plus Two Zoology Chapter Wise Previous Questions and Answers Chapter 2 Reproductive Health.

Kerala Plus Two Zoology Chapter Wise Previous Questions Chapter 2 Reproductive Health

Question 1.
Diagram shown below is a surgical method used for female sterilization. (MARCH-2010)
a) What is the method shown in diagram?
b) Mention any two Intra uterine devices to prevent conception.
c) What is the surgical method of male sterilization called?

Answer:
a) Tubectomy
b) Any two methods (Cu-T, Cu-7, loop, Multibad-375, Hormone producing progestogent, LNG-20 etc)
c) Vasectomy

Question 2.
Select the Assisted Reproductive Technique that uses an early embryo with upto 8 blastomeres.   (MAY-2010)
a) ZIFT
b) IUT
c) GIFT
d) IUI
Answer:
a) ZIFT

Question 3.
One among the contraceptive method is peculiar. Find the odd one. What is common among others?  (MARCH-2011)
a) Periodic abstinence
b) Coitus interruptus
c) Lactational amenorrhea
d) lUD’s
Answer:
d) lUD’s
Others are natural contraceptives/Natural methods/ Others have more chance of conception / IUD is the barrier method.

Question 4.
The treatment facility advertised on the brochure of a private clinic is shown below.  (MARCH-2011)
a) Can you suggest what type of a clinic it is ?
b) Make a brief note on any three of the treatment procedure.

Answer:
a) Infertility clinic / Fertility treatment clinic / Assisted Reproductive clinic or Hospital.
b) IVF-Invitro Fertilization
ZIFT – Zygote Intra Fallopian Transfer
GIFT – Gamete Intra Fallopian Transfer
IUI – Intra Uterine Insemination.

Question 5.
“STDs present a major health concern in both industralized and developing countries.”  (MARCH-2012)
a) What do you mean by STDs?
b) Name two STDs.
c) Suggest two preventive measures.
Answer:
a) Sexually transmitted diseases
b) Gonorrhoea and Syphilis
c) Avoid sex with unknown partners and use condom during coitus.

Question 6.
Find out the odd one from the following, write the reason.  (MAY-2012)
a) CuT
b) Cu7
c) LNG-20
d) Multiload – 375
Answer:
LNG-20

Question 7.
One couple came to know that they have a girl child during fourth month of pregnancy and they decide to do MTP.  (MAY-2012)
a) What is MTP ?
b) At which stage of pregnancy MTP relatively safe?
c) How will you respond to the decision of female foeticide by the couple?
Answer:
a) Medical termination of pregnancy
b) MTPs are safe during the first trimester, i.e., upto 12 weeks of pregnancy.
c) MTPs must be essential where continuation of the pregnancy could be harmful or even fatal either to the mother or to the foetus.

Question 8.
One of your friend argued that anti-retroviral drugs are effective medicine to treat AIDS.  (MAY-2012)
a) What is your opinion about it?
b) How HIV affect out immunity?
Answer:
a) It prevents the growth and multiplication of viruses
b) HIV enters into helper T-lymphocytes (TH), replicates and produce progeny viruses. The progeny viruses released in the blood attack other helper T-lymphocytes, as result the patient becomes immuno-deficient

Question 9.
Most often HIV infection occur due to conscious behaviour patterns. Do you agree with this statement? Substantiate your answer.  (MARCH-2013)
Answer:
No,
1) blood transfusion
2) During birth of child
Yes,
1) sexual contact
2) Intraveinous injection

Question 10.
Suggest the ART which may be successful in the following conditions. (MARCH-2013)
a) A female cannot produce ovum, but can provide suitable environment for fertilization and further development.
b) Male partner is unable to inseminate the female or has very low sperm count.
c) Fusion of gametes and zygote formation does not occur within the body of the female.
Answer:
a) Transfer of an ovum collected from a donor into the fallopian tube (GIFT – gamete intra fallopian transfer) of another female who cannot produce one, but provide suitable environment for fertilisation and further development
b) Intra cytoplasmic sperm injection (ICSI) Here a sperm is directly injected into the ovum. It is the procedure to form an embryo in the laboratory
c) In vitro fertilisation (IVF—fertilisation outside the body in almost similar conditions as that in the body) followed by embryo transfer (ET).

Question 11.
Prepare a pamphlet for Adolescent children to make them aware of alcohol and drug abuse. (MAY-2013)
Answer:
Education and counseling, moral education, Avoid undue peer pressure .visual publicity through TV, seeking professional and medical help etc.

Question 12.
One of your neighbour is suffering from itching, fluid discharge, slight pain and swelling in genital region. (MARCH-2014)
a) What do you think the disease he is suffering from?
b) What measures are to be taken to prevent such diseases?
Answer:
a) Sexually transmitted disease (STD) Eg-gonor- rhoea, syphilis etc.
b) 1) Avoid unknown sexual partner
2) Use condoms during coitus

Question 13.
Expand the following abbreviations which are commonly used in reproductive health: (MARCH-2014)
a) ART
b) ZIFT
Answer:
a) ART – Assisted reproductive techonology
b) ZIFT – Zygote intra fallopian transfer

Question 14.
______ and _____ are two surgical contraceptive methods in males and females respectively. (MAY-2014)
Answer:
Vasectomy,Tubectomy

Question 15.
Sex of the baby is determined by the father, not by the mother. Substantiate. (MAY-2014)
Answer:
ovum fertilises with a sperm carrying X- chromosome the zygote develops into a female (XX) and the fertilisation of ovum with Y-chromosome carrying sperm results into a male offspring. Hence the genetic make up of the sperm that determines the sex of the child.

Question 16.
Amniocentesis for sex determination is banned in our country? Is this ban necessary? Comment. One use of amniocentesis. (MAY-2014)
Answer:
Yes, Ban on amniocentesis for sex-determination to legally check increasing female foeticides. It is a foetal sex determination test based on the chromosomal pattern in the amniotic fluid surrounding the developing embryo.

Question 17.
Identify the diagram and write how it acts? (MARCH-2015)

Answer:
Copper T (CuT), lUDs increase phagocytosis of sperms within the uterus and the Cu ions released suppress sperm motility and the fertilising capacity of sperms.

Question 18.
Foetal sex can be determined by a test based on the chromosomal pattern from the amniotic fluid.  (MARCH-2015)
a) What is this test?
b) Revealing of sex determination through this test is banned. Is this ban necessary? Comment.
c) Invitro fertilization followed by embryo transfer is known as _____.
Answer:
a) Aminocentesis
b) Yes, Ban on amniocentesis for sex- determination to legally check increasing female foeticides.
c) Test tube baby programme

Question 19.
If proper care and attention is not given by adults, adolescents may become addicted to drugs/ alcohol.” What is your opinion about this statement? Substantiate your answer.  (MAY-2015)
Answer:
Adolescence is a bridge linking childhood and adulthood. It is the very vulnerable phase of mental and psychological development of an individual. Repeated use of drugs, the tolerance level of the receptors present in our body increases. Consequently the receptors respond only to higher doses of drugs or alcohol leading to greater intake and addiction.

Question 20.
Some techniques commonly used for infertility treatment are given below. Read them carefully and answer the questions.  (MAY-2015)
ZIFI , GJFF, ICSI, IUI, IVF
a) Which of the above technique is used forthe collection of sperm from the husband or a healthy donor and artificially introduced into the vagina or uterus of the female?
b) Distinguish between ZIFT and GIFT.
c) Write the common term used to denote ‘ tech-niques given above
Answer:
a) IUI
b) ZIFT- Ova from the wife/donor (female) and sperms from the husband/donor (male) are collected and are induced to form zygote under simulated conditions in the laboratory. The zygote or early embryos (with upto 8 blastomeres) are transferred into the fallopian tube
b) GIFT-Transferof an ovum collected from a donor into the fallopian tube of another female who cannot produce it ,but can provide suitable environment for fertilisation
c) Assisted reproductive technologies (ART)

Question 21.
Categorise the given birth control methods into three groups with proper heads.  (MARCH-2016)
Cervical caps, Vasectomy, Cu T, Tubectomy, Diaphragms, Condom, Lippes Loop
Answer:
lUDs- Cu T, LiPPes loop
Barrier method- Diaphrams, Condoms Cervical caps, surgical method – Vasectomy, Tubectomy,

Question 22.
Diagnostic report of two couples having infertility problems are given below:  (MAY-2016)
1) The woman cannot produce ovum.
2) The man has very low sperm count in semen. Suggest a suitable Assisted Reproductive Technologies (ART) for each problem in expanded form
Answer:
1) GIFT -Gamete intra fallopian transfer
2) lUI-Intra uterine insemination/AI-Artificial insemination.

Question 23.
Which of the following pairs of STDs is completely curable?  (MARCH-2017)
1) HIV, Hepatitis-B
2) Hepatitis-B, Gonorrhoea
3) Syphilis, Gonorrhoea
4) Chlamydomonas, genital-herpes
Answer:
3) syphilis, gonorrohea

Question 24.
a) What is ART?  (MARCH-2017)
b) Categorize the following ARTs based on their applications in male sterility and female sterility:
GIFT, Al
Answer:
a) In the case of infertility, the couples could be assisted to have children through certain special techniques commonly known as assisted reproductive technologies (ART),
b) GIFT-Female,
Al – Male

Kerala State Board New Syllabus Plus Two Zoology Chapter Wise Previous Questions and Answers Chapter 1 Human Reproduction.

Kerala Plus Two Zoology Chapter Wise Previous Questions Chapter 1 Human Reproduction

Question 1.
Given below is the diagrammataic representation of human blastocyst. Observe the diagram and answer the following questions: (MARCH-2010)

a) Identify A and B
b) Write the functions of A and B.
Answer:
a) A – Inner cell mass, B – Trophoblast
b) A – Differentiate into embryo germ layers/gastrula B – Attachment to endometrium / form different embryonic membranes/ placenta formation/ nutritional supply.

Question 2.
When the urine sample of a lady is tested, presence of Human Chorionic Gonadotropin (HCG) is detected. (MARCH-2010)
a) What does the presence of HCG indicate?
b) Which is the source of HCG?
Answer:
a) Pregnancy
b) Placenta

Question 3.
The graph shown below shows the levels of LH and FSH at various stages of menstrual cycle. (MAY-2010)

a) Name the source of LH and FSH.
b) The level of LH is maximum during the middle day of the cycle. Mention its effect.
c) Note the function of LH in males.
Answer:
a) Pituitary gland
b) Helps for ovulation
c) Stimulates the synthesis of horrnones (androgens)

Question 4.

The above graph shows the level of ovarian hormones in a normally menstruacing woman during the follicular phase. (MARCH-2011)
a) Name ‘a’ and ‘b’
b) Mention the role of pituitary hormones in maintaining this condition.
c) Reconstruct the graph for Luteal phase.
Answer:
a) i) Estrogens
ii) Progesterone
b) FSH / Follicle stimulating Hormone, LH / Luteinizing Hormone
OR
Gonadotropins
One function each of FSH and LH.
c)

Question 5.
Some stages of the embryonic development are given below. Observe these diagrams and answer the questions. (MAY-2011)

a) What is A & B?
b) Name thetwo types of cells found in the blastocyst.
c) Which layer of blastocyst is attached to the endometrium? And name that process.
Answer:
a) A – Blastomere B – Morula
b) Trophoblast and inner cell mass
c) Trophoblast, Implantation

Question 6.
Observe the diagram provided (Do not copy the picture) (MARCH-2012)

a) Label A and B
b) On which day of menstrual cycle Graafian follicle rupture.
c) Name the process induces the rupture of Graafian follicle.
d) Write the name and function of the structure forming inside the ovary after the rupture of Graafian follicle.
Answer:
a) A – Primary follicle B – Tertiary follicle
b) 14th day
c) The rupture of Graafian follicle and the release of ovum is called ovulation.
d) The corpus luteum secretes large amounts of progesterone which helps in the maintenance of the endometrium.

Question 7.
The following statements compares the process of oogenesis and spermatogenesis. Which one is not true?  (MAY-2012)
a) Production of ovum ceases at certain age, but sperm production continues even in old men.
b) Oogenesis begins in the embryonic stages, but spermatogenesis starts at the onset of puberty.
c) Meiotic arrest occurs both in Oogenesis and spermatogenesis.
d) Polar bodies are formed in oogenesis.
Answer:
a) Replication
b) Transcription
c) Translation
d) Reverse transcription

Question 8.
The diagram represents a process of gametogenesis. Closely observe it and answer the following  (MARCH-2013)

Is it spermatogenesis or Oogenesis?
What does the smaller shaded circle represent? Write down two significance of production of the same.
Answer:
a) oogenesis
b) polar bodies
c) Retention of cytoplasm in ovum Maintaining one functional haploid ovum

Question 9.
Though one ovum is produced from a primary oocyte it can result into a male or female child after fertilization. But in the case of spermatocyte though 4 sperms are produced only two of them can result to a female child after fertilization”. Justify.  (MARCH-2013)
Answer:
Two types of sperms are produced one with x chromosomes and other withy chromosome. Sperm with x chromosome after fertilization results in the formation of female baby

Question 10.
Sterilization and IUDS are effective birth control measures, but lactational amenorrhea may not be so effective.  (MAY-2013)
a) How the sterilization procedure of males differ from that of females in preventing pregnancy?
b) Which part of the female reproductive organ is utilized for the IUD procedure? How this procedure prevents pregnancy?
c) Why the lactational amenorrhea is not so effective?
Answer:
a) In males the sterilization method is called vasectomy but in females it is called Tubectomy
b) uterus, lUDs suppresses sperm motility and fertilizing capacity of sperm or they make uterus unsuitable for implantation.
c) During intense lactation, menstrual cycle does not occur. Therefore the chance of conception is almost nil. But the method is effective only upto six months following delivery

Question 11.
Observe the diagram, and answer the questions: (MARCH-2013)
a) Identify A and B.
b) What is the function of C ?
c) In which of the marked part reduction division takes place? What is the significance of it?

Answer:
a) A – Spermatozoa, B-Primary spermatocyte
b) Sertoli cells provide nutrition to the germ cells.
c) Primary spermatocyte, chromosome number reduced to half

Question 12.
Diagram of a mammalian sperm’is given. Label the parts marked. (MAY-2014)

Answer:
A – Acrosome
B – Nucleus containing chromosomal material

Question 13.
Schematic representation of gametogenesis is given below. Identify A. Write one difference between A& B. (MARCH-2015)

Answer:
A) Spermatogenesis – It results 4 spermatozoa
B) Oogenesis – It results one ovum and polar body

Question 14.
1) In which part of the human reproductive system the following events occur? (MARCH-2015)
a) Fertilisation
b) Implantation
2) Diagram of a human biastocyst is given below. Identify A & B.

Answer:
a) ampullary-isthmicjunction
b) Endometrium of the uterus
A – inner cell mass
B – trophoblast

Question 15.
Choose the odd one from the following and write the common feature of others; (MAY-2015)
a) Estrogen
b) Androgen
c) Relaxin
d) Progesterone
Answer:
Androgen.
Others are ovarian hormones.

Question 16.
Complete the flow chart showing spermatogenesis by filling A and B and answer the questions: (MAY-2015)

a) What is the chromosome number of primary spermatocyte
b) What is the significance of reduction division in spermatocyte?
Answer:
A- Spermatogonia
B- Spermatids
a) Primary spermatocyte-Diploid number, in man it is 23 pair, spermatids-haploid number, in man it is 23
b) Reduction division helps to reduce chromosome number as half in gametes (haploid).
It helps keep chromosome number species as constant for many generations.

Question 17.
Match the columns A and B: (MARCH-2016)

Answer:

Question 18.
The process of fusion of a sperm with ovum is called _______. (MAY-2016)
Answer:
Fertilisation

Question 19.
Match columns A and B.  (MAY-2016)

Answer:

Question 20.
LH and FSH are gonadotrophins. Distinguish their roles in males and females.  (MARCH-2017)
Answer:
LH acts at the Leydig cells and stimulates synthesis and secretion of androgens.
Androgens stimulate the process of spermato-genesis.
FSH acts on the Sertoli cells and stimulates secretion of some factors which help in the process of spermiogenesis.
Female Rapid secretion of LH leading to rupture of Graafian follicle and release ovum (ovulation). FSHhelpsinthe growth and development of ovarian follicle.

Question 21.
Human female possess 44 + XX chromosome number. The chromosome number of secondary oocyte is  (MAY-2017)
a) 44 + X
b) 22 + X
c) 44 +XX
d) 22 +XX
Answer:
b) 22 + X

Question 22.
Observe the diagram and answer the questions:  (MAY-2017)

a) Identify A and B
b) Write the function of B.
Answer:
a) A – peri vitelline space
B- zona pellucida
b) Prevent the entry of further sperms after fertilization

Kerala State Board New Syllabus Plus Two Botany Chapter Wise Previous Questions and Answers Chapter 8 Environmental Issues.

Kerala Plus Two Botany Chapter Wise Previous Questions Chapter 8 Environmental Issues

Question 1.
Destruction of forest leads to the increase of CO, in atmosphere. Recently Govt, of India instituted an award for individuals or communities from rural areas that shown extraordinary courage and dedication in protecting wildlife. (MARCH-2010)
a) Identify the award.
b) Comment on deforestation and reforestation.
Answer:
a) Amrita Devi Bishnoi award
b) Deforestation cutting down of forest trees Reforestation replanting trees which are already destroyed.

Question 2.
Industrial effluents and domestic sewage seriously affect fresh water bodies. For protecting aquatic life Govt, of India recently declared an animal as National aquatic animal. (MARCH-2010)
a) Identify the animal.
b) Distinguish biomagnification from eutrophication.
Answer:
a) Dolphin
b) Bio magnification is the increase in the amount of non biodegradable substance in successive trophic level in a food chain.
Eg: DDT Eutrophication is the overgrowth of aquatic plants as a result of accumulation of nutrients in water body.

Question 3.
In 1990s, Delhi ranked 4th among the most polluted cities of the world. But now air quality of Delhi has significantly improved mainly by switching vehicles from diesel to CNG. (MAY-2010)
a) Expand CNG.
b) CNG is betterthan diesel. Comment.
Answer:
a) Compressed Natural Gas
b) CNG is compressed natural gas. It burns more efficiently than Petrol and diesel, thus brings down the amount of pollutants from automobiles (unbumt hydrocarbons).

Question 4.
Greenhouse effect is a naturally occuring phenomenon that is responsible for heating of earth’s surface and atmosphere. (MAY-2010)
a) Explain greenhouse effect.
b) What will happen if there is absolutely no greenhouse effect over earth’s surface ?
Answer:
a) The greenhouse gases such as CO₂, N₂0, Methane present in the atmosphere will reradiate the reflected infrared radiations. Thus raising the temperature of earth.
b) It will affect the climate of earth as the greenhouse effect causes melting of snow in the polar regions. This will in turn determine the level of water in the sea.

Question 5.
Arrange the following words into suitable categories in the given table. (MARCH-2011)
Algal bloom, Catalytic converter, Eichhornia, Electrostatic precipitator

Answer:

Question 6.
Geetha resides in a city nearby a lake. Water from this lake was used for various domestic purposes earlier. Now-a-days. this water has become turbid and is with an unpleasant odour. (MARCH-2011)
a) What can be the reason for this ?
b) Name the scientific term that explains this effect.
Answer:
a) Nutrient enrichment
b) Eutrophication or Aging of lake

Question 7.
As Head of the Vehicle Department, issue a notice to vehicle owners to observe any two measures to reduce vehicular air pollution and record the merits and demerits of CNG. (MAY-2011)
Answer:
Use of catalytic converter.
Use of lead-free petrol or diesel.
Merits of CNG
1) It burns most efficiently,
2) It is cheaper than petrol or diesel, cannot be siphoned off by thieves and adulterated like petrol or diesel.
Demerits of CNG
It is difficult of laying down pipelines to deliver CNG through distribution points/pumps and ensuring uninterrupted supply.

Question 8.
increase in green house gases. Name two green house gases. (MAY-2011)
Answer:
Green house gases – CO₂ and CFC

Question 9.
The increased use of chemicals like CFCs (Chloro fluro carbons) cause adverse ecological impacts. Why CFCs are considered harmful to the environment? Mil It causes the depletion of ozone layer. (MARCH-2012)
Answer:
CFC is considered as green house gas.lt causes global warming.
The incoming UV rays cuases many diseases like skin cancer,snow blindness etc.

Question 10.
Meena an environmental activist, noticed a gradual decline in the’population of birds in the open agricultural fields near her place. She has heard of the excessive use of pesticides like DDT around that area. (MARCH-2012)
a) What might have led to the decline of bird population in that area?
b) Name the process that has caused this phenomenon.
Answer:
a) It disturbs calcium metabolism in birds due to accumulation of DDT and causes the thinning of egg shell. lt affects the premature breaking of egg.
b) Biomagnification.

Question 11.
Ammu read in the newspaper that, BOD of a water body in a nearby village was high and there is algal bloom. (MAY-2012)
a) What is BOD ?
b) What is algal bloom ?
c) Can you give possible reason for these phenomenon?
Answer:
a) Biological oxygen demand
b) Inceased algal population
c) It is due to the accumulation of inorganic nutrients like phosphate and nitrate in water body

Question 12.
Enviornmentalists usually says: There are many causes for biodiversity losses’. Illustrate four major causes of biodiversity loss. (MARCH-2013)
Answer:
i) Habitat loss and fragmentation
ii) Over-exploitation
iii) Alien species invasions
iv) Co-extinctions

Question 13.
In a study conducted, the concentration of DDT was found to increase in the successive trophic levels, The results of the study is shown below: (MAY-2013)

Answer:
Biomagnification
High concentrations of DDT disturb calcium metabolism in birds, which causes thinning of eggshell and their premature breaking, eventually causing decline in bird populations.

Question 14.
An article in the newspaper reports that ‘Refirigerants like Chlorofluorocarbons (CFCs) pose threat to the environment’. How CFCs are harmful to the environment? (MARCH-2014)
Answer:
CFC (CCI2F2) splits in the presence of UV and release active chlorine. This active chlorine breaks ozone molecule into O₂ and (O) .It causes the thinning of stratospheric good ozone and harmful UV rays reaches earth surface.

Question 15.
Now a days many farmers are interested in organic farming. What is meant by organic farming? Can you suggest any two advantages of organic farming?  (MARCH-2014)
Answer:
It is the cyclical, sustainable and zero-waste procedure.
Advantageous
1) No need of chemical fertilizers.
2) No need of insecticides and pesticides.
3) Never kills microorganisms in the soil.

Question 16.
An aquatic ecosystem having luxurious growth of cyanobacteria (Algal bloom) leads to eutrophication.  (MARCH-2015)
a) What kind of pollutants cause algal bloom to colonize the aquatic ecosystem?
b) What are the consequences of eutrophication?
Answer:
a) Waste water contains contains large quatities of nutrients.
b) It causes ageing of lake and damage to indigenous flora and fauna.

Question 17.
During the past century, the temperature of the earth has increased by 0.6°C, most of it during the last few decades. Rise in temperature causes deleterious changes in the environment, thus leading to increased melting of polar ice caps as well as other places like Himalayan snow caps. Suggest any two control measures that will reduce global warming?  (MARCH-2015)
Answer:
1) Cutting down use of fossil fuel
2) limproving efficiency of energy usage
3) Reducing deforestation
4) Planting trees and slowing down the growth of human population.

Question 18.
Observe the diagram and answer the following:  (MAY-2015)

a) Suggest the reasons for the presence of DDT in the water.
b) Fish eating birds of this area have higher DDT concentration in their body. Justify.
c) What will be the impact of DDT in the birds?
OR
United Nations Framework convention on climate change, an international treaty signed by 194 countries to cooperatively discuss global climate change and its impact.
As a science student,
a) What is global warming?
b) Explain the reasons and give suggestions to control global warming?
Answer:
a) Use of pesticide in agricultural field
b) DDT is accumulated in successive trophic level and finally its concentration is very high fish eating birds.
c) DDT affect calcium metabolism, that causes thinning of eggshell and premature breaking of egg. Hence the bird population is decreased.
OR
a) Increasing the temperature of earth surface due to green house effect is called global warming
b) The incoming radiations of sunlight reaches the earth’s surface re-emits heat in the form of infrared radiation but part of this does not escape into space as atmospheric gases (e.g., carbon dioxide, methane,CFCand nitrogen oxides) absorb a major fraction of it. This cycle is repeated many a times and temperature of earth increases it leads to global warming.
It is reduced by

1. cutting down use of fossil fuel.
2.  improving efficiency of energy usage.
3. reducing deforestation.
4. planting trees and slowing down the growth of human population.

Question 19.
Increase in the concentration of toxicants at successive trophic level is called _____.  (MARCH-2016)
a) BOD
b) Biomagnification
c) Eutrophication
d) Algal Bloom
Answer:
b) Biomagnification

Question 20.
The major pollution in the environment is caused by automobiles. Expand the term CNG. Mention any two of its merits. (MARCH-2016)
Answer:
CNG – compressed natural gas It burn efficiently
It is cheaper than diesel and petrol

Question 21.
Temperature is generally increasing making the earth a hot plate. Mention any two measures to control global warming. (MARCH-2016)
Answer:
Cutting down the use of fossil fuel
Improving the efficiency of energy usage

Question 22.
Quantity of pollutants increase in successive trophic levels. Observe the flowchart regarding biomagnifications of DDT in an aquatic food chain and answerthe following: (MAY-2016)
a) What is biomagnification?
b) What are the consequences of biomagnification?

Answer:
a) Non bio degradable chemicals which accumulate in the body of organism and is passed on to the organisms belonging to the next trophic level, the increase in the concentration of toxicants at successive trophic level and finally concentration become high in last trophic level of the food chain.
b) High concentrations of DDT disturb calcium metabolism in birds, which causes the thinning of eggshell and their premature breaking, causing decrease in bird populations.

Question 23.
Adequate waste management is an environment issue to be considered. Discuss the advantages of Eco-san toilet. (MAY-2016)
Answer:
it is a sustainable system for handling human excreta, using dry composting toilets.
This is a practical, hygienic, efficient and cost- effective solution to human waste disposal.

Question 24.
A common cause of deforestation is slash and burn agriculture. (MARCH-2017)
a) What is the common name attributed to such type of cultivation?
b) Explain how this type of cultivation is practised?
Answer:
a) phytoplankton stage
b) Submerged plant stage
c) Submerged free floating plant Stage
d) Reed swamp stage
f) Marsh – meadow stage
g) Scrub stage
h) Forest stage

Question 25.
Particulate matter in polluted air is removed by the application of electrostatic precipitator. Explain the working principle of lectrostatic precipitator. (MARCH-2017)
Answer:
At high voltage the electrons produced in an instrument are attached to dust particles giving them a net negative charge. These charged dust particles are attracted by collecting plates. Then reducing the velocity of air between the plates which help the dust to fall. Electrostatic precipitator which can remove over 99 per cent particulate matter present in the exhaust from a thermal power plant.

Question 26.
Among the following which one Is used for reducing the emission of poisonous gases from automobiles (MAY-2017)
a) Landfills
b) Catalytic converter
c) Electrostatic precipitator
d) Earmuffs
Answer:
Catalytic converter

Question 27.
Nutrient enrichment in a fresh water lake leads to (MAY-2017)
eutrophication.
a) What happens during eutrophication?
b) How dissolved oxygen level is affected as a result of this?
Answer:
a) Due to accumulation of nitrate and phosphate into the lake, leads the rapid growth of algae, that changes the colour and quality of water.
It depletes the oxygen content of water and finally causes the death of aquatic organs.
b) Due the death and decay of algae, dissolved oxygen content is decreased in lake that causes the mortality of fish and other aquatic organs.

Kerala State Board New Syllabus Plus Two Botany Chapter Wise Previous Questions and Answers Chapter 7 Ecosystem.

Kerala Plus Two Botany Chapter Wise Previous Questions Chapter 7 Ecosystem

Question 1.
During a study tour teacher showed the primary colonisers on the banks of the river ‘Nila’.(MARCH-2010)
a) Identify the succession and justify your answer.
b) List the different stages of the identified succession.
Answer:
Succession taking place in water is called Hydrarch.
b) Phytoplankton stage -> Submerged plant stage -> Submerged free floating plant stage -> Reed swamp stage -> Marsh meadow -> Scrub stage -> Forest stage

Question 2.
While learning trophic levels in class-room, teacher asked you to explain ‘standing crop’ to Raman. Explain. (MARCH-2010)
Answer:
Standing crop – The mass of living material present in each trophic level.

Question 3.
Pond is a self-sustainable unit. Some organisms related to pond ecosystem is listed below, tadpole, fish, water, plants, kingfisher. (MAY-2010)
a) Construct a food chain with the listed organisms.
b) Explain trophic level.
c) Point out trophic level of each organism in the constructed food chain.
d) Name interconnection of food chains in nature.
OR
Two students, Unni and Kannan studied inter specific interactions between different species. They made a table assigning a ‘+’ for beneficial interaction, for detrimental and ‘0’ for neutral interaction. Can you help them by naming the interaction between species in different cases ? Write one example for each interaction.

Answer:
Water plants Tadpole -> Fish -> Kingfisher
b) Specific place of an organism in the food chain based on the source of nutrition or food.
c)

d) Food web
OR
1-Mutualism Eg; Lichen
2-Competition
Eg: Competition for food (Phytoplankton) between Flamingoes and fishes
3 – Predation Eg: Lion and Deer
4 – Commensalism
Eg: Epiphyte, Vanda and Mango tree

Question 4.
Consider pond as an ecosystem showing the number of individuals in the following categories. (MARCH-2011)
Carnivores-2500, Producers-15000, Herbrivores- 5000
a) Draw the pyramid of numbers in this ecosystem.
b) Comment on the energy flow in the ecosystem
Answer:

b) Unidirectional flow of energy from producers to consumers. It decreases in successive trophic levels

Question 5.
The gradual change in the species composition of a given area leading to the formation of climax community is called ecological succession. In a rocky area, (MARCH-2011)
a) What is the expected type of pioneer species?
b) How this pioneer species leads to the establishment of a stable climax community?(2 Scores)
Answer:
a) lichens
b) lichens -> mosses -> herbs -> shrubs ->forest.
These are different stages and leads to stable climax community.

Question 6.
Fill up the blanks with appropriate terms in the given pyramid of trophic level: (MAY-2011)

Answer:
b – Primary consumer a- secondary consumer

Question 7.
Kalyani wrote man, hen, earthworm, mango-tree in her note book. Arrange the terms in a food chain sequence. Explain food chain and name the types of food chain. (MAY-2011)
Answer:
Earth worm—> Hen —> Man —> Mango tree
The transfer of food from the producers through a series of organisms with repeated eating and being eaten is referred to as Food chain Detritus food chain

Question 8.
In a marine ecosystem, a population of phytoplankton (150,000) supports a standing crop of fishes (40,000). (MARCH-2012)
a) Draw the pyramid of biomass and
b) The pyramid of numbers in this ecosystem,
Answer:
a) Inverted Pyramid

b) Upright Pyramid

Question 9.
The gradual and fairly predictable changes in the species composition in an area is called ecological succession. (MARCH-2012)
a) Name the pioneer species in the primary succession in water.
b) Give the sequence of events and climax community in the hydrarch succession.
Answer:
a) Phytoplankton
b) Phytoplankton -> submerged free floating -> Reed swamp marsh meadow -> scrub -> Forest.

Question 10.
Given number of individuals in a grassland ecosystem. (MAY-2012)
Grasshopper – 1500
Grass – 5,842,000
Wolf – 28
Birds – 215
a) Draw a pyramid of numbers showing various trophic levels.
b) Explain trophic level.
Answer:

b) Each step in food chain is called trophic level

Question 11.
Rate of biomass production is called productivity and can be divided into GPP and NPP : (MAY-2012)
a) Define GPP and NPP.
b) How can we relate GPP and NPP?
Answer:
a) GPP – Gross primary productivity NPP – Net primary productivity
b) NPP = GPP-R

Question 12.
Final community that is in near equilibrium with environment in ecological succession is called ________. (MARCH-2013)
Answer:
Climax community

Question 13.
Natural interlinked food chains are called _______. (MARCH-2013)
Answer:
Food web

Question 14.
A list of organisms are given. Place them in different trophic levels. Grass, Man, Fishes, Birds, Lion, Grasshopper, Zooplankton, Trees. (MARCH-2013)
Answer:
First trophic level – trees, grass
Second trophic level – grasshopper, zooplankton
Third trophic level – birds, fishes
Fourth trophic level – lion, man

Question 15.
In the equation, GPP-R = NPP; If NPP = Net primary productivity. (MAY-2013)
Explain GPP – R = NPP.
Answer:
Gross primary productivity minus respiration losses (R) is the net primary productivity (NPP).
That means Net primary productivity is the available biomass for the consumption of heterotrophs (herbivores and decomposers).

Question 16.
The teacher, pointing to a forest said “ Long back, this place was a pond’’. This gradual change is an example of (MAY-2013)
1) Secondary succession
2) Xerarch succession
3) Pioneer species
4) Hydrarch succession
Answer:
Hydrarch succession

Question 17.
The species that invade a nude area are called ________ species. In a primary succession on rocks, the group that invade first are usually. (MARCH-2014)
Answer:
Pioneer species, Lichens

Question 18.
The rate of biomass production in an ecosystem is called productivity. They are of two types, gross primary productivity and net primary productivity, how these two productivities are related? (MARCH-2014)
Answer:
Gross primary productivity of an ecosystem is the rate of production of organic matterduring photosynthesis.
Gross primary productivity minus respiration losses (R), is the net primary productivity (NPP).
i.e: GPP – R = NPP

Question 19.
A list of different organisms in an ecosystem are given below. (MARCH-2014)
Arrange them in 1st, 2nd, 3rd and 4th trophic level,
i) Phytoplankton
ii) Man
iii) Fish
iv) Zooplankton
Answer:

Question 20.
By observing the relationship of the first pair fill up the blanks: (MAY-2014)
a) Grazing food chain consists of producers and consumers whereas Detritus chain comprises dead organic matter and ______.
b) Nitrogen : Gaseous cycle
Sulphur: _______
Answer:
a) Detrivores
b) Sedimentary cycle

Question 21.
Field survey by a team of students recorded the following data related to biomass of the organisms in each tropic level of an ecosystem. Draw, name and explain the pyramid (MAY-2014)
Answer:

Biomass increases in successive trophic levels hence the pyramid given here is inverted.

Question 22.
Gradual, sequential changes of a given area including species composition is known as ecological succession. If so name the first two stages of the succession in hydricarea. (MAY-2014)
Answer:
1) phytoplankton stage
2) submerged plant stage

Question 23.
Primary succession on rocks is known as Xerosere. Answer the following related with Xerosere. (MARCH-2015)
a) Name the pioneer community.
b) Organic acids have important roles in this succession. Justify.
Answer:
a) Lichen
b) It helps in weathering of rocks and soil formation

Question 24.
By observing the relationship of the first pair fills up the blanks. (MAY-2015)
a) Net primary productivity =
Gross primary productivity — Respiration.
Gross primary productivity is ________.
b) Carbon: Gaseous cycle
Phosphorus: ______.
Answer:
a) total organic matter
b) sedimentary cycle

Question 25.
Field survey by a team of students recorded the following data related to number of organisms in an ecosystem and plotted that into a figure shown below: (MAY-2015)

Observe the figure and explain the pyramid.
Answer:
It is the inverted pyramid . In this number of organism increases in the successive trophic levels.

Question 26.
Hydrosere succession stages are given below. Arrange them in order. (MAY-2015)
Scrub stage — forest —submerged free floating — Marsh Meadow —Submerged stage —Reed swamp—Phytopiankton.
Answer:
Phytoplankton- submerged-submerged free floating- reed swamp -Marsh meadow -srub stage – forest.

Question 27.
Nutrients are never lost from the ecosystems and are recycled. Write about. (MARCH-2016)
a) Gaseous cycle
b) Sedimentary cycle
Answer:
a) Gaseous cycle- The reservoir of gaseous type nutrients are present in atmosphere
b) Sedimentary cycle- The reservoir of nutrients are present in earth crust

Question 28.
Ecological pyramids are usually upright. Meanwhile some, pyramid of biomass is inverted. Explain the reason. (MARCH-2016)
Answer:
Pyramid of biomass of sea is inverted because biomass of fishes far exceeds phytoplankton

Question 29.
a) Biogeochemical cycle is an important phenomenon in very ecosystem. Describe phosphorus cycle. (MAY-2016)
OR
b) The plant communities in a given area show successive changes. Mention the stages of succession in a xerosere.
Answer:
a) The cycle consists of following steps.

OR
b) In xerach succession first plant communities appear in bare area are lichens,They secrete
carbonic acid and dissolve rocks. This process forms soil that help in the growth of mosses. Then herbs,shrubs and forest stage appears .this process takes thousands of years and form climax community.

Question 30.
Earthworms are commonly referred as farmers’ friends. Define fragmentation. (MAY-2016)
Answer:
During fragmentation large detritus (dead remains of plants & animals) is converted into into smaller particle by the detritrivores like earthworm, its surface area increases that helps in process, of decomposition.

Question 31.
The different stages of primary succession in water are represented below. Fill the gaps that are unfilled. (MARCH-2017)
a) Phytoplankton
b) _______
c) Submerged free floating plant stage
d) _______
e)________
f) Shrub stage
g) _______
Answer:
a) phytoplankton stage.
b) Submerged plant stage.
c) Submerged free floating plant Stage
d) Reed swamp stage.
f) Marsh – meadow stage
g) Scrub stage.
h) Forest stage

Question 32.
An ecosystem consist of the following population: (MARCH-2017)
Phytoplankton
Man
Fish
Zooplankton
Draw a food chain denoting each trophic level
Answer:

Question 33.
The natural reservoir of phosphorous is rock where it is present in the form of phosphates. How this phosphorous is cycled in ecosystem? (MAY-2017)
Answer:
The reservoir of phosphorus is rock. During weathering process orthophosphates reaches the soil solution. It is taken by plants, that passes through food chain and reaches animals. After the death and decay, the organic form of phosphorus is coming into the soil solution as orthophosphates by the activity of phosphate solubilising bacteria.

Question 34.
Birds represent members in a food chain. (MAY-2017)
Draw a food chain representing each of the above in different tropic levels.
Answer:

Kerala State Board New Syllabus Plus Two Botany Chapter Wise Previous Questions and Answers Chapter 6 Organisms and Populations.

Kerala Plus Two Botany Chapter Wise Previous Questions Chapter 6 Organisms and Populations

Question 1.
Given below is a table which shows the interspecific interaction.’+’ sign indicates beneficial,sign indicates detrimental and ‘0’ indicates neutral. (MARCH-2010)
a) Fill in the blanks.

b) Name the interactions where one species is benefited and the other is detrimental.
Answer:
Amensalism
Commensalism

Question 2.
Small bottle labelled with rDNA insulin. (MARCH-2010)
a) Does it a natural insulin ?
b) Identify the major steps involved in this rDNA insulin production.
Answer:
No.
1. Isolation of desired genes.
2. Insertion of desired Genes into plasmids of E. coli.
3. Introduction of plasmids into E coli cells.
4. Culture of E coli cells.
5. After this, a polypeptide chains A and B are separated and connected together by disulphide linkages.
Thus, genetically engineered insulin is prepared

Question 3.
Given below a schematic representation with circles and squares, which shows four factors/processes influence the population density. (MARCH-2010)
Write the positive factors in circles and negative factors in squares.

Answer:

Question 4.
Snakes change their body temperature with changes in external temperature, but human beings not. Organism may be classed according to above character with explanation. (MAY-2010)
Answer:
Temperature has a significant role in the kinetics of enzymes and thus influence the metabotic activities and physiological functions. Accordingly organisms can be classified into eurythermal (tolerate wide range of temperature) and Stenothermal (restricted to narrow range of temperature)

Question 5.
Density of a population in a given habitat during a given period, fluctuates due to changes in 4 basic processes – Natality, Mortality, Immigration & Emigration. (MAY-2010)
a) Differentiate Natality and Mortality.
b) Differentiate Immigration and Emigration.
Answer:
a) Mortality is the number of death of a population at a given period and Natality is the number of birth during a given period.
b) Immigration is the number of individuals of the same species that have come into the habitat from elsewhere during the time period. Emigration is the number of individuals of the population who left the habitat and gone elsewhere during the time period.

Question 6.

Given above is the bar diagram showing age structure of three different populations. Observe the diagram carefully and answer the following questions. (MARCH-2011)
a) Select the stable population.
b) Compare the nature of population growth in A,B, and C

Answer:
a) ‘B’Stable population
b)

Question 7.
By observing the relationship of the first, fill in the blanks. (MARCH-2011)
a) Unisexual male flower-Staminate ______
Unisexual female flower ______
b) Organisms tolerating a wide range of temperature – eurythermal
Organisms tolerating a narrow range of temperature ______
Answer:
a) pistillate
b) stenothermal

Question 8.
Population interactions: (MAY-2011)

Where’+’ beneficial interaction detrimental interaction ‘0’ neutral interaction.
Observe the interactions of populations of 3 species as shown in the table. Name the interactions
a) Species x and species y in case 1.
b) Species y and species z in case 2.
c) Species x and species z in case 3.
d) Species y and species z in case 1
Answer:
a) Mutualism
b) Predation/Parasitism
c) Competition
d) Commensalism

Question 9.
Mohammed and his family left to Dubai from Kozhikode on March, 2009. In Kozhikode they are referred as after 2009. How it affects Kozhikode population? (MAY-2011)
Answer:
Emigrants, Decrease the size of population

Question 10.
Inter specific interaction from the interaction of populations of two different species. If we assign + for beneficial, — for detrimental and 0 for neutral interactions, copy and complete the following chart. (MARCH-2012)

Answer:

Question 11.
Prakash parked his car in bright sunlight for a few hours, with glass windows fully raised. After sometime inside of the car was very hot. (MAY-2012)
a) Name the phenomenon.
b) How can you correlate this phenomenon with global warming?
Answer:
a) Green house effect
b) The green house effect is due to various green house gases, of which the percentage of carbon dioxide is very high. This causes the increase of temperature on earth called Global warming.

Question 12.
Students involved in nature club activity found some interspecific interactions between organisms in a garden area. They made a table of interaction giving’+’ for beneficial interaction,for detrimental and ‘O’ for neutral interaction. (MAY-2012)

a) Give name of interaction in each case.
b) Explain how parasitism differ from predation.
c) Give the significance of species interaction.
Answer:
i – Mutualism
ii – competition
iii – commensalism
iv- amensalism
b) In parasitism, parasite absorb nutrients from the living host In predation, predator kills and eat the prey.
c) Species interaction is beneficial, detrimental or neutral (neither harm nor benefit) to one of the species or both.

Question 13.
Read the statements below and identify the mode of interaction between the species. (MARCH-2013)
a) Tiger eating deer
b) Butterfly feeding pollen
c) Human liver fluke feed on snail
d) Lice on humans
e) Orchid attached to a tree
f) Mycorrhizal association of fungi and roots of higher plants.
g) Sparrow eating seed
h) Egrets foraging close to cattle
Answer:
a) predation
b) mutualism
c) predation or parasitism
d) parasitism
e) commensalism
f) mutualism
g) predation
h) commensalism

Question 14.
In summer we use air conditioners and in winter we use heaters. Here homeostasis is accomplished by artificial means. Explain four ways by which other living organisms cope with the situation. (MARCH-2013)
Answer:
Hibernation – winter sleeping
Aestivation – Summer sleeping
Migration – Moving into more suitable area
Diapause – inactive in adverse condition

Question 15.
Many desert plant have adaptations to prevent loss of water from their body. Mention any two adaptations to minimise water loss from plant body. (MAY-2013)
Answer:
a) Many desert plants have a thick cuticle on their leaf surfaces and have their stomata arranged in deep pits to minimise water loss through transpiration.
b) They also have a special photosynthetic pathway (CAM) that enables their stomata to remain closed during daytime.

Question 16.
The size of a population is not static. Which of the following leads to decrease in population? (MAY-2013)
1) Natality and Mortality
2) Mortality and Emigration
3) Mortality and immigration
4) Natality and Immigration
Answer:
Mortality and Emigration

Question 17.
Some type of Orchids live on the branches of Mango trees. The relationship between mango tree and Orchid is an example of. (MAY-2013)
1) Mutualism
2) Predation
3) Commensalism
4) Parasitism
Answer:
Commensalism

Question 18.
The density of population in a given habitat increase or decrease due to different reasons. Name two factors responsible for increase in population in a given area. (MARCH-2014)
Answer:
Natality and Immigration

Question 19.
Observe the diagram:(MAY-2014)

Define the following terms:
a) Natality
b) Mortality
c) Emigration
d) Immigration
Answer:
a) Birth rate or total number of live births per 1,000 of a population in a year.
b) Death rate or total number of death per 1,000 of a population in a year.
c) Emigration- movement of individuals out of the population
d) Immigration- movement of individuals into the population

Question 20.
Response of organisms to abiotic stress involves different methods. Explain any two such responses with suitable examples. (MAY-2014)
Answer:
i) Regulate: Organisms are able to maintain constant body temperature and constant osmotic Concentration. Eg- birds and mammals
ii) Conform: Organisms cannot maintain a constant internal environment.
Eg- Majority (99 percent) of animals and all plants.

Question 21.
Suckerfish and shark live in close association, is a classic example of commensalism. What is commensalism? (MARCH-2015)
Answer:
Commensalism -This is the interaction in which one species benefits and the other is neither harmed nor benefited.

Question 22.
Desert plants like Opuntia are able to grow in extreme conditions. Suggest any two adaptations of this plant. (MARCH-2015)
Answer:
1) They have a thick cuticle on their leaf surfaces
and have their stomata arranged in deep pits to minimise water loss through transpiration.
2) They also have a special photosynthetic pathway (CAM) that enables their stomata to remain closed during day time
3) The photosynthetic function is carried out by the flattened stems.

Question 23.
With regard to population growth rate, when responses are limiting the plit is logistic. Verhulst-Pearl Ligstic growth is represented by the equation. (MARCH-2015)
\(\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}=\mathrm{rN} \frac{(\mathrm{K}-\mathrm{N})}{\mathrm{K}}\) what,are
a) r
b) K
Answer:
a) r-lntrinsic rate of natural increase or( b-d)
b) k-Carrying capacity

Question 24.
Observe the equation (MAY-2015)
\(\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}=\mathrm{rN} \frac{(\mathrm{K}-\mathrm{N})}{\mathrm{K}}\)
a) Which type of growth curve does it represents?
b) What do the following notations represent:
a) N b) r c) K
Answer:
a) Logistic growth
b) N-population of size
b) r- intrinsic rate of natural increase
c) K-Carrying capacity

Question 25.
On earth, life exists even in extreme and harsh conditions. Mention any two major biomes in India. (MARCH-2016)
Answer:
Tropical deciduous forest
Rain forest

Question 26.
a) Population interactions may be beneficial or not. Write any three interactions in detail. (MARCH-2016)
OR
b) Organism are influenced by biotic and abiotic factors. Write an account of any three abiotic environmental factors.
Answer:
a) Mutualism- in this both partners are benefitted eg lichen (+,+)
Commensalism- In this one partner is benefitted other partner is neither benefitted nor harmed (-, +) Competition- In this both partners have detrimental effect or negative effect (-, -)
OR
b) Temperature- it affect the enzyme kinetics of reaction. Enzyme works at optimum temperature
Water- it affect productivity and distribution of plants in aquatic ecosystem.
Light- It influence the photoperiodic flowering of plants

Question 27.
Population growth may be exponential or logistic. Differentiate between them. (MAY-2016)
Answer:
When the resources in the habitat are unlimited, each species has the ability to grow in number. Here the population grows in an exponential or geometric fashion. dN/dt = rN
Limited resources leads to competition between individuals and the ‘fittest’ individual will survive and reproduce. This is called logistic growth
\(\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}=\mathrm{rN} \frac{(\mathrm{K}-\mathrm{N})}{\mathrm{K}}\)

Question 28.
Plants are adapted to grow in different habitats. Name any four adaptations of plants in desert habitat. (MAY-2016)
Answer:
Desert plants have a thick cuticle on their leaf surfaces and stomata arranged in deep pits to minimise water loss through transpiration.
They also have CAM pathway in which they open stomata during night and closed during day time.

Question 29.
In a given habitat, the maximum number possible for a species is called _________ of that species in that habitat. (MARCH-2017)
Answer:
Carrying capacity (K)

Question 30.
Different types of population interaction has been observed in a population. (MARCH-2017)
Write the types of interaction observed among the following species:

OR
B) Organisms other than human beings manage or adapt to stressful conditions by adopting different mechanisms. Explain any three mechanisms adopted by them to maintain the internal environment.
Answer:
A) Mutualism or pseudocopulation Commensalism Commensalism Parasitism Parasitism Predation
OR
B) 1) Conform : About 99% of animals and all plants cannot maintain a constant internal environment according to the external environment. They change their body temperature and osmotic concentration of body fluid when external environment changes.
2) Migrate : Some organisms move away temporarily from the stressful habitat to a more hospitable area and return when stressful period is over.
3) Suspend : Some organisms like bacteria, fungi and lower plants produce thick walled spores to tide over unfavourable conditions.
Some organism avoid the stress by escaping in time by method of hibernation during winter (eg. polar bear) or aestivation to avoid summer related problem (eg: snails and shells).

Question 31.
There are four mechanism by which living organisms other than human beings maintain the constancy of internal environment. Name these processes. (MAY-2017)
Answer:
Organisms maintain internal environment as constant by sweating shivering deposition of fat layer below skin and hairy covering on body surface.

Question 32.
Adaptations are the attributes of the organism that enables it to survive and reproduce in its habitat. Give the adaptations of  (MARCH-2014)
a) Cactus plant in desert
b) Kangaroo rat in desert
c) Seals in polar region.
Answer:
a) Desert plants have a thick cuticle on their leaf surfaces and have theirstomata arranged in deep pits to minimise water loss through transpiration
b) kangaroo rat in North American deserts is capable of meeting all its water requirements through its internal fat oxidation
c) In the polar seas aquatic mammals like seals have a thick layer of fat (blubber) below their skin that acts as an insulator and reduces loss of body heat.