Category: Class 12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 3 p-Block Elements – II Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements – II

12th Chemistry Guide p-Block Elements – II Text Book Questions and Answers

Part – I – Text Book Evaluation

I. Choose the correct answer

1. In which of the following, NH₃ is not used?
a) Nessler’s reagent
b) Reagent for the analysis of IV group basic radical
c) Reagent for the analysis of III group basic radical
d) Tollen’s reagent
Answer:
a) Nessler’s reagent

2. Which is true regarding nitrogen?
a) least electronegative element
b) has low ionisation enthalpy than oxygen
c) d – orbitals available
d) ability to form pπ -pπ bonds with itself
Answer:
d) ability to form pπ -pπ bonds with itself

3. An element belongs to group 15 and 3rd period of the periodic table, its electronic configuration would be
a) Is² 2s² 2p⁴
b) Is² 2s² 2p³
c) Is² 2s² 2p⁶ 3s2² 3p²
d) Is² 2s² 2p⁶ 3s² 3p³
Answer:
d) Is² 2s² 2p⁶ 3s² 3p³

4. Solid (A) reacts with strong aqueous NaOH liberating a foul smelling gas(B) which spontaneously burn in air giving smoky rings. A and B are respectively
a) P₄(red) and PH₃
b) P₄ (white) and PH₃
c) S₈ and H₂S
d) P₄(white) and H₂S
Answer:
b) P₄ (white) and Ph₃

5. On hydrolysis, PCl₃ gives
a) H₃PO₃
b) PH₃
c) H₃PO₄
d) POCl₃
Answer:
a) H₃PO₃

6. P₄O₆ reacts with cold water to give
a) H₃PO₃
b) H₄P₂O₇
c) HPO₃
d) H₃PO₄
Answer:
a) H₃PO₃

7. The basicity of pyrophosphorous acid (H₄P₂O₃) is
a) 4
b) 2
c) 3
d) 5
Answer:
b) 2

8. The molarity of given orthophosphoric acid solution is 2M. Its normality is
a) 6N
b) 4N
c) 2N
d) none of these
Answer:
a) 6N
Normality = M x basicity =2×3 = 6

9. Assertion : bond dissociation energy of fluorine is greater than chlorine gas
Reason : chlorine has more electronic repulsion than flourine
a) Both assertion and reason are true and reason is the correct explanation of assertion.
b) Both assertion and reason are true bu t reason is not the correct explanation of assertion
c) Assertion is true bu t reason is false
d) Both assertion and reason are false
Answer:
d) Both assertion and reason are false The converse is true

10. Among the following, which is the strongest oxidizing agent?
a) Cl₂
b) F₂
C) Br₂
d) l₂
Answer:
b) F₂

11. The correct order of the thermal stability of hydrogen halide is (PTA – 4)
a) Hl > HBr > HCl > HF
b) HF > HCl > HBr > HI
c) HCl > HF > HBr > HI
d) HI > HCl > HF > HBr
Answer:
b) HF > HCl > HBr > HI

12. Which one of the following compounds is not formed?
a) XeOF₄
b) XeO₃
c) XeF₂
d) NeF₂
Answer:
d) NeF₂

13. Most easily liquefiable gas is
a) Ar
b) Ne
c) He
d) Kr
Answer:
c) He

14. XeFg on complete hydrolysis produces
a) XeOF₄
b) XeO₂F₂
c) XeO₃
d) XeO₂
Answer:
c) XeO₃

15. Which of the following is strongest acid among all?
a) HI
b) HF
c) HBr
d) HCl
Answer:
a) HI
Reason : H-I bond is weakest.

16. Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules? (NEET)
a) Br₂ > I₂ > F₂ > Cl₂
b) F₂ > Cl₂ > Br₂ > I₂
c) I₂ > Br₂ > Cl₂ > F₂
d) Cl₂ > Br₂ > F₂ > I₂
Answer:
d) Cl₂ > Br₂ > F₂ > I₂

17. Among the following the correct order of acidity is (NEET)
a) HClO₂ < HClO < HClO₃ < HClO₄
b) HClO₄ < HClO₂ < HClO < HClO₃
c) HClO₃ < HClO₄ < HClO₂ < HClO
d) HClO < HClO₂ < HClO₃ < HClO₄
Answer:
d) HClO < HClO₂ < HClO₃ < HClO₄

18. When copper is heated with cone HNOs it produces
a) Cu (NO₃)₂, NO and NO₂
b) Cu (NO₃)₂ and N₂O
c) Cu (NO₃)₂ and NO₂
d) Cu (NO₃)₂ and NO
Answer:
c) Cu (NO₃)₂ and NO₂

II. Answer the following questions.

Question 1.
What is the inert pair effect?
Answer:
In p-block elements, as we go down the group, two electrons present in the valence s-orbital become inert and are not available for bonding (only p-orbital involves chemical bonding). This is called inert pair effect.

Question 2.
Chalcogens belong to p-block.
Answer:
Give reason.

• Chalcogens belong to p-block elements.
• Because their outer electronic configuration is ns² np⁴.
• In these elements the last electron enters np orbital.
• Hence they belong to p-block elements.

Question 3.
Explain why fluorine always exhibits an oxidation state of -1?
Answer:
Fluorine the most electronegative element than other halogens and cannot exhibit any positive oxidation state. Fluorine does not have a d-orbital while other halogens have d-orbitals. Therefore fluorine always exhibits an oxidation state of-1 and others in the halogen family shows +1, +3, +5 and +7 oxidation states.

Question 4.
Give the oxidation state of halogen in the following
Answer:
a) OF₂
b) O₂F₂
c) Cl₂O₃
d) I₂O₄

Question 5.
What are interhalogen compounds? Give examples
Answer:
Each halogen combines with other halogens to form a series of compounds called interhalogen compounds. For example, Fluorine reacts readily with oxygen and forms difluorine oxide (F₂O) and difluorine dioxide (F₂O₂).

Question 6.
Why fluorine is more reactive than other halogens? (PTA – 1, 3)
Answer:
Fluorine is the most reactive element among halogens. This is due to the low value of F – F bond dissociation energy.

Question 7.
Give the uses of helium. (PTA – 2)
Answer:

1. Helium and oxygen mixture is used by divers in place of air oxygen mixture. This prevents the painful dangerous condition called bends.
2. Helium is used to provide an inert atmosphere in the electric arc welding of metals
3. Helium has the lowest boiling point hence used in cryogenics (low-temperature science).
4. It is much less denser than air and hence used for filling air balloons.

8. What is the hybridisation of iodine in IF₇? Give its structure. (PTA – 5)
Answer:

9. Give the balanced equation for the reaction between chlorine with cold NaOH and hot NaOH
Answer:
Chlorine reacts with cold NaOH to give sodium hypochlorite

Chlorine reacts with hot NaOH to give sodium chlorate

10. How will you prepare chlorine in the laboratory? (PTA – 2)
Answer:
1. Chlorine is prepared by the action of cone, sulphuric acid on chlorides in presence of manganese dioxide.
4NaCl + MnO₂ + 4H₂SO₄ → Cl₂ + MnCl₂ + 4NaHSO₄ + 2H₂O

2. It can also be prepared by oxidising hydrochloric acid using various oxidising agents such as manganese dioxide, lead dioxide, potassium permanganate or dichromate.
PbO₂ + 4HCl → PbCl₂ + 2H₂O + Cl₂
MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂
2KMnO₄ + 16HCl → 2KCl + 2MnCl + 8H₂O + 5Cl₂
K₂Cr₂O₇ + 14HCl → 2KCl + 2CrCl₃ + 7H₂O + 3Cl₂

3. When bleaching powder is treated with mineral acids chlorine is liberated
CaOCl₂ + 2HCl → CaCl₂ + H₂O + Cl₂
CaOCl₂ + H₂SO₄ → CaSO₄ + H₂O + Cl₂

11. Give the uses of sulphuric acid.
Answer:
Sulphuric acid is used

• In the manufacture of fertilisers, ammonium sulphate and superphosphates.
• In the manufacture of other chemicals such as hydrochloric acid, nitric acid etc.,
• as a drying agent.
• in the preparation of pigments, explosives, etc.,

12. Give a reason to support that sulphuric acid is a dehydrating agent. (PTA – 1)
Answer:

• Sulphuric acid is highly soluble in water.
• It has a strong affinity towards water.
• Hence it can be used as a dehydrating agent.
• When dissolved in water, it forms mono (H₂SO₄.H₂O) and dihydrates (H₂SO₄.2H₂O) and the reaction is exothermic.
  ex
  C₁₂H₂₂O₁₁ + H₂SO₄ → 12C + H₂SO₄.11H₂O
  HCOOH + H₂SO₄ → CO + H₂SO₄.H₂O

13. Write the reason for the anomalous behaviour of Nitrogen.
Answer:
1. Due to its small size, high electronegativity, high ionisation enthalpy and absence of d-orbitals.

2. N, has a unique ability to form pπ – pπ multiple bond whereas the heavier members of this group (15) do not form pπ – pπ bond, because their atomic orbitals are so large and diffused that they cannot have effective overlapping.

3. Nitrogen exists a diatomic molecule with triple bond between the two atoms whereas other elements form single bond in the elemental state.

4. N cannot form dπ – pπ bond due to the absence of d – orbitals whereas other elements can.

14. Write the molecular formula and structural formula for the following molecules
Answer:
a) Nitric add
b) dinitrogen pentoxide
c) phosphoric acid
d) phosphine

15. Give the uses of argon.
Answer:
Argon prevents the oxidation of hot filament and prolongs the life in filament bulbs.

16. Write the valence shell electronic configuration of group -15 elements.
Answer:
Valence shell electronic configuration of group 15 elements is ns²np³

17. Give two equations to illustrate the chemical behaviour of phosphine.
Answer:
Basic Nature :
Phosphine is weakly basic and forms phosphonium salts.
PH₃ +HI → PH₄ I
PH₄I + H₂O \(\underrightarrow { \triangle } \) PH₃ + H₃O⁺ + I^–
It react with halogen to give phosphorous penta halidae
PH₃ +4Cl₂ → PCl₅ + 3HCl

Combustion:
When phosphine is heated with air or oxygen it burns to give metaphosphoric acid

18. Give a reaction between nitric acid and a basic oxide.
Answer:
Nitric acid rects with a basic oxide to form salt and water.
3 FeO + 10HNO₃ → Fe (NO₃)₃ + NO + 5H₂O

19. What happens when PCl₅ is heated?
Answer:
On heating Phoshorous pentachloride decomposes into phosphorus trichloride and chlorine

20. Suggest a reason why HF is a weak acid, whereas binary acids of all other halogens are strong acids.
Answer:

• HF is only slightly ionised, hence it is a weak acid
• Other halogen acids are almost completely ionised, hence they are strong acids.
• Among halogen acids, the electronegativity difference is maximum (1.9) in HF acid.
• Hence the bond between H and F is stronger and the acid HF is weaker.

21. Deduce the oxidation number of oxygen in hypofluorous acid – HOF.
Answer:
Oxidation number of F = -1
Oxidation number of H = +1
Oxidation number of O in HOF = x
(+1) + x + (-1) = 0
x = 0
Oxidation number of O in HOF = 0

22. What type of hybridisation occur in (PTA – 5)
a) BrF₅
b) BrF₃

23. Complete the following reactions
1. NaCl+ MnO₂ + H₂SO₄ →
2. NaNO₂ + HCl →
3. P₄ + Na0H + H₂O →
4. AgNO₃ + PH₃ →
5. Mg + HNO₃ →

8. Sb + Cl₂ →
9. HBr + H₂SO₄ →
10. XeF₆ + H₂O →
11. XeO₆^4- + Mn²⁺ + H⁺ →
12. XeOF₄ + SiO₂ →

Answer:

III. Evaluate yourself

1. Write the products formed in the reaction of nitric acid (both dilute and concentrated) with zinc.

12th Chemistry Guide p-Block Elements – II Additional Questions and Answers

Part II – Additional Questions

I. Match the following

1.

Answer:

2.

Answer:

3.

Answer:

4.

Answer:

II. Assertion and Reason

i) Both A and R are correct, R explains A.
ii) A is correct, R is wrong
iii) A is wrong, R is correct
iv) Both A and R are correct, but R does not explain A.

1. Assertion (A) : Aqueous solution of potash Alum is acidic
Reason (R) : Aluminium sulphate undergo hydrolysis. (PTA – 2)
Answer:
i) Both A and R are correct, R is explanation of A

2. Assertion (A) : Elements belonging to group 16 are called chalcogens
Reason (R) : Group 16 elements are saltforming elements
Answer:
ii) A is correct, R is wrong
Correct Reason : Group 16 elements are ore forming elements

3. Assertion (A) : Among halogen acids, HF has low melting and boiling points
Reason (R) : In HF hydrogen bond is present.
Answer:
iii) A is wrong, R is correct.
Correct A : Among halogen acids HF has high melting and boiling points.

4. Assertion (A) : A small piece of Zinc dissolved in dilute nitric acid but hydrogen gas is not evolved. (PTA – 3)
Reason (R) : HNO₃ is an oxidising agent and this oxidizes hydrogen.
Answer:
ii) A is correct but R is wrong.

III. Pick out the Correct statement

1. i) Oxygen is diamagnetic
ii) Oxygen forms hydrogen bonds
iii) Oxygen exists in two allotropic forms
iv) Oxygen exists as a triatomic gas
a) (i) &(ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)
Correct statement: (i) Oxygen is paramagnetic (iv) Oxygen exists as a diatomic gas

2. i) Sulphur exists in crystalline as well as an amorphous form
ii) Rhombic sulphur has a characteristic yellow colour and composed of Sg molecules.
iii) When heated slowly above % C monoclinic sulphur is converted into Rhombic sulphur
iv) At around 140°C Rhombic sulphur melts to form mobile pale yellow liquid called X sulphur.
a) (i) &(ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
a) (i) & (ii)
Correct statement : (iii) When heated slowly above 96°C, Rhombic sulphur is converted into monoclinic sulphur
(iv) At around 140°C the monoclinic sulphur melts to form mobile pale yellow liquid called X sulphur

3. i) H₂SO₄ is a dibasic acid
ii) H₃PO₃ is a tribasic acid
iii) H₃PO₄ is a dibasic acid
iv) H₃PO₂ is a monobasic acid
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
d) (i) & (iv)
Correct statement:
(ii) H₃PO₃ is a dibasic acid
(iii) H₃PO₄ is a tribasic acid

4. i) Krypton is used in cryogenics.
ii) Neon is used in high-speed electronic flashbulbs used by photographers.
iii) Helium is used to provide an inert atmosphere in electric arc welding of metals.
iv) Radon is used as a source of gamma rays.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
c) (iii) & (iv)
Correct statement: (i) Helium is used in cryogenics
(ii) Xenon is used in high speed electronic flash bulbs used by photographers.

IV. Pick out the incorrect statement

1. i) In inter halogen compounds the central atom will be the smaller halogen
ii) Interhalogen compounds can be formed only between two halogen atoms.
iii) Flourine can act as a central atom.
iv) Interhalogens are strong oxidising agents
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (i) & (iv)
Answer:
b) (i) & (iii)
Correct statements:
(i) In interhalogen compounds the central atom will be the larger halogen
(iii) Flourine cannot act as a central atom.

2. i) Nitrogen reacts with group 2 metals to form ionic nitrides.
ii) Ammonia is less soluble in water.
iii) Liquid nitrogen is used in biological preservation.
iv) In the conversion of metal oxides to metal ammonia acts as an oxidising agent.
a) (i) & (ii)
b) (ii) & (iii)
c) (i) & (iii)
d) (ii) & (iv)
Answer:
d) (ii) & (iv)
Correct statements:
(ii) Ammonia is extremely soluble in water.
(iv) In the conversion of metal oxides to metal ammonia acts as a reducing agent

3. i) When reacted with metals, nitric acid liberates hydrogen
ii) Chromium when reacted with nitric acid becomes passive due to the formation of nitrate on its surface.
iii) In most of the reactions nitric acid acts as an oxidising agent
iv) Fuming nitric acid contains oxides of nitrogen
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (i) & (iv)
Answer:
a) (i) & (ii)
Correct statement : (i) When reacted with metals, nitric acid does not liberate hydrogen
(ii) Chromium when reacted with nitric acid becomes passive due to the formation of oxide on its surface

4. The rate of decomposition of ozone increases sharply in alkaline solution.
ii) In acidic solution ozone exceeds the oxidising power of fluorine and atomic oxygen
iii) Considerable amount of ozone is formed in the upper atmosphere by the action of UV light
iv) The shape of the Ozone molecule is linear
a) (i) &(ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
d) (i) & (iv)
Correct statement : (i) The rate of decomposition of ozone drops sharply in alkaline solution.
(iv) The shape of the ozone molecule is bent.

V. Pick out the odd man out

1. w.r.t oxidation number pick the odd man out.
a) HPO₃
b) H₃PO₃
c) H₃PO₄
d) H₄P₂O₇
Answer:
b) H₃PO₃ – O.N is +3 while in others the O.N of phosphorous is +5

2. w.r.t the reaction with sulphuric acid pick the odd man out
a) Gold
b) Silver
c) Platinum
d) Copper
Answer:
d) Copper – copper reacts with sulphuric acid while others do not

3. w.r.t reactivity pick the odd man out
a) F₂
b) Cl₂
c) Br₂
d) I₂
Answer:
a) F₂ – F₂ is more reactive than other halogens

4. w.r.t the ability to form oxoacids pick the odd man out
a) fluorine
b) chlorine
c) bromine
d) iodine
Answer:
a) Flour – Fluorine forms only one oxoacid where as other halogens form more than one oxoacid.

VI. Choose the best answer.

1. The principal gas present in atmosphere is
a) O₂
b) N₂
C) H₂
d) CO₂
Answer:
b) N₂

2. The basicity of hypophosphorous acid is (PTA – 2)
a) 1
b) 2
c) 3
d) 4
Answer:
a) 1

3. Chile salt petre is
a) NaNO₂
b) NaNO₃
c) KNO₂
d) KNO₃
Answer:
b) NaNO₃

4. Indian salt petre is
a) NaNO₂
b) NaNO₃
c) KNO₂
d) K.NO₃
Answer:
d) KNO₃

5. Inert character of nitrogen is due to its
a) high electronegativity
b) low electro negativity
c) high bonding energy
d) low bonding energv
Answer:
c) high bonding energy

6. Which of the following is caro’s acid?
a) H₂S₂O₈
b) H₂S₂O₇
c) H₂SO₃
d) H₂SO₃
Answer:
c H₂SO₅

7. Haber’s process is used for the synthesis of
a) NO₂
b) HNO₃
c) NH₃
d) N₂O
Answer:
a) NH₃

8. The substance used in cryosurgery for producing low temperature is
a) liquid oxygen
b) liquid nitrogen
c) liquid hydrogen
d) liquid ammonia
Answer:
b) liquid nitrogen

9. Urea on hydrolysis gives
a) NO₂
b) HNO₃
c) NH₃
d) N₂O
Answer:
a) NH₃

10. The catalyst used in Haber’s process is
a) Ni
b) Fe
c) Co
d) Pt
Answer:
b) Fe

11. The smell of ammonia is
a) rotten egg
b) rotten fish
c) pungent
d) garlic
Answer:
c) pungent

12. Like water, ammonia is a fairly good ionising solvent, because its dielectric constant is
a) considerably low
b) considerably high
c) equal to zero
d) equal to one
Answer:
b) considerably high

13. The process used for the manufacture of nitric acid is known as
a) Haber’s process
b) Deacon’s process
c) contact process
d) Ostwald’s process
Answer:
d) Ostwald’s process

14. With excess of chlorine, ammonia reacts to give an explosive substance
a) N₂
b) NH₄NO₃
c) NH₄Cl
d) NCl₃
Answer:
d) NCl₃

15. The deep blue colour compound formed when excess of ammonia is added to aqueous solution of copper sulphate is
a) [Cu(NO₃)₂]
b) [Cu(NH₃)₂]²⁺
c) [Cu(NH₃)₄]²⁺
d) [Cu(NH₃)₂]⁺
Answer:
c) [Cu(NH₃)₄]²⁺

16. The shape of ammonia molecule is
a) tetrahedral
b) pyramidal
c) square planar
d) octahedral
Answer:
b) pyramidal

17. The bond angle in ammonia is
a) 104°
b) 104°28′
c) 107°
d) 180°
Answer:
c) 107°

18. The colour of Pure nitric acid is
a) colourless
b) brown
c) pale green
d) green
Answer:
a) colourless

19. Fuming nitric acid contains oxides of
a) sulphur
b) hydrogen
c) nitrogen
d) carbon
Answer:
c) nitrogen

20. Nitric acid can act as
a) an acid
b) an oxidising agent
c) nitrating agent
d) all of the above
Answer:
d) all the above

21. The formula of hyponitrous acid is (MARCH 2020)
a) H₂N₂O₂
b) H₄N₂O₄
c) HOONO
d) HNO₂
Answer:
a) H₂N₂O₂

22. The oxidising power of oxo acids follows the order
a) HOX > HXO₂ > HXO₃ > HXO₄
b) HXO₄ > HXO₃ > HXO₂ > HOX
c) HXO₃ > HXO₄ > HXO₂ > HOX
d) HOX > HXO₄ > HXO₃ > HXO₂
Answer:
a) HOX > HXO₂ > HXO₃ > HXO₄

23. White phosphorous is kept under
a) kerosene
b) water
c) alcohol
d) ether
Answer:
b) water

24. White phosphorous becomes yellow phosphorous due to
a) hydrolysis
b) reduction
c) oxidation
d) displacement
Answer:
c) oxidation

25. In the conversion of yellow phosphorous into phosphine, phosphorous acts as
a) oxidising agent
b) reducing agent
c) catalyst
d) hydrolysing agent
Answer:
b) reducing agent

26. In the conversion of phosphorous into orthophosphoric acid, the catalyst used is
a) Cl₂
b) Br₂
c) I₂
d) F₂
Answer:
c) I₂

27. Which is used in match boxes?
a) White phosphorous
b) Red phosphorous
c) Black phosphorous
d) Scarlet phosphorous
Answer:
b) Red Phosphorous

28. The acid having O-O bond in its structure (PTA – 6)
a) H₂SO₃
b) H₂S₂O₆
c) H₂S₂O₈
d) H₂S₄O₆
Answer:
c) H₂S₂O₈

29. The smell of phosphine is
a) rotten egg
b) rotten fish
c) pungent
d) garlic
Answer:
b) rotten fish

30. The compounds used in Holme’s signal are
a) CaC₂ & Ca₃P₂
b) AlP & Ca₃P₂
c) CaC₂ & P₄
d) AlP & P₄
Answer:
a) CaC₂ & Ca₃P₂

31. The gases liberated in Holme’s signal are
a) C₂H₂ & CH₄
b) C₂H₂ & Ph₃
c) C₂H₄ & PH₃
d) CH₄ & Ph₃
Answer:
b) C₂H₂ & Ph₃

32. The formula of pyrophosphoric acid is
a) H₄P₂O₆
b) H₄P₂O₇
c) H₃PO₂
d) H₃PO₃
Answer:
b) H₄P₂O₇

33. Thermodynamically stable allotrophic form of sulphur is
a) Rhombic sulphur
b) Monoclinic sulphur
c) Plastic sulphur
d) Colloidal sulphur
Answer:
a) Rhombic sulphur

34. The gas found in volcanic eruptions is
a) NO₂
b) NO
c) SO₂
d) SO₃
Answer:
c) SO₂

35. The hybridisation of sulphur in SO₂ is
a) sp
b) sp²
c) sp³
d) dsp²
Answer:
b) sp²

36. The gas liberated when dilute sulphuric acid reacts with metals is
a) SO₂
b) SO₃
c) H₂
d) O₂
Answer:
c) H₂

37. The gas liberated when cone, sulphuric acid reacts with metals is
a) SO₂
b) SO₃
c) H₂
d) O₂
Answer:
a) SO₂

38. When sulphuric acid reacts with barium chloride solution, the white precipitate formed is
a) PbSO₄
b) BaSO₄
c) (CH₃COO)₂SO₄
d) PbCl₂
Answer:
b) BaSO₄

39. The halogen which exists as a liquid is
a) flourine
b) chlorine
c) bromine
d) iodine
Answer:
c) bromine

40. The halogen which exists as a solid is
a) flourine
b) chlorine
c) bromine
d) iodine
Answer:
d) iodine

41. Chlorine is manufactured by
a) Haber’s process
b) Deacon’s process
c) Contact process
d) Ostwald’s process
Answer:
b) Deacon’s process

42. The colour of chlorine gas is
a) colourless
b) brown
c) greenish yellow
d) pale green
Answer:
c) greenish yellow

43. Aqua regia is a mixture of cone. HCl and cone. HNO₃ in the ratio
a) 1 : 3
b) 3 : 1
c) 2 : 3
d) 3 : 2
Answer:
b) 3 :1

44. The halogen acid which forms hydrogen bond is
a) HF
b) HCl
c) HBr
d) HI
Answer:
a) HF

45. Among halogen acids, the strongest bond is present in
a) HF
b) HCl
c) HBr
d) HI
Answer:
a) HF

46. Among halogen acids, the weakest bond is present in
a) HF
b) HCl
c) HBr
d) HI
Answer:
d) HI

47. Among halogen acids, the strongest acid is
a) HF
b) HCl
c) HBr
d) HI
Answer:
d) HI

48. Among halogen acids, the weakest acid is
a) HF
b) HCl
c) HBr
d) HI
Answer:
a) HF

49. The correct order of acid strength is
a) HF > HCl > HBr > HI
b) HF < HCl < HBr < HI
c) HF > HCl < HBr > HI
d) HF < HCl > HBr < HI
Answer:
b) HF < HCl < HBr < HI

50. Which is more reactive towards hydrogen?
a) flourine
b) chlorine
c) bromine
d) iodine
Answer:
a) flourine

51. The number of bond pair and lone pair of electrons present in the interhalogen compound BrF₃ is
a) 1 & 3
b) 3 & 2
C) 5 & 1
d) 7 & 0
Answer:
b) 3 & 2

52. The oxidation number of oxygen in F₂O is
a) -2
b) -1
c) +2
d) +1
Answer:
c) +2

53. The oxidation number of chlorine in Cl₂O₇ is
a) +1
b) +4
c) +6
d) +7
Answer:
d) +7

54. The strongest oxidising agent among the following is
a) chlorous acid
b) chloricacid
c) hypochlorous acid
d) perchloric acid
Answer:
c) hypochlorous acid

55. The first ionisation energy of noble gases is in the order
a) He < Ne < Ar < Kr
b) He > Ne > Ar > Kr
c) He < Ne > Ar < Kr
d) He > Ne < Ar > Kr
Answer:
b) He > Ne > Ar > Kr

56. Among noble gases, chemical reactivity is shown by
a) He & Ne
b) Ar & Kr
c) Kr & Xe
d) Xe & Rn
Answer:
c) Kr & Xe

57. Which among the following is used in cryogenics?
a) He
b) Ne
c) Ar
d) Kr
Answer:
a) He

58. Which is used for filling air balloons?
a) He
b) Ne
c) Ar
d) Kr
Answer:
a) He

59. Which is used in advertisement sign boards?
a) He
b) Ne
c) Ar
d) Kr
Answer:
b) Ne

60. Lamps used in airports as approaching lights is filled with
a) He
b) Ne
c) Ar
d) Kr
Answer:
d) Kr

VII. Two Mark Questions

Question 1.
How is pure nitrogen gas prepared?
Answer:
Pure nitrogen gas is prepared by the thermal decomposition of sodium azide at about 575 K
2NaN₃ \(\underrightarrow { 573K } \) 2Na + 3N₂

Question 2.
Nitrogen does not form any penta halides like phosphorus, why?
Answer:
Nitrogen does not form pentahalide although it exhibit +5 oxidation state. Due to the absence of d-orbitals.
It cannot undergo sp3d hybridization and hence cannot form pentahalides.

Question 3.
What is Haber’s process?
Answer:
The synthesis of ammonia from nitrogen and hydrogen at high pressure and optimum temperature in presence of iron catalyst is known as Haber’s process.

∆H_f = -46.2 Kjmol^-1

Question 4.
Write the uses of nitrogen
Answer:
Nitrogen is used
In the manufacture of ammonia, nitric acid, and calcium cyanamide etc.
Liquid nitrogen is used for producing low temperature required in cryosurgery and so used in biological preservation.

Question 5.
How is ammonia prepared in the laboratory?
Answer:
Ammonia is prepared in the laboratory by heating an ammonium salt with a base
2NH₄Cl + CaO → CaCl₂ + 2NH₃ + H₂O

Question 6.
Write about the reducing property of ammonia.
Answer:
When passed over heated metallic oxides Ammonia reduces metal oxides into metal
3PbO + 2NH₃ → 3Pb + N₂ + 3H₂O

Question 7.
What happens when ammonia reacts with excess of chlorine?
Answer:
With excess of chlorine ammonia reacts to give an explosive substance nitrogen trichloride
2NH₃ + 6Cl₂ → 2NO₃ + 6HO

Question 8.
On standing nitric acid becomes yellow in colour why?
Answer:

• Pure nitric acid is colourless
• Fuming nitric acid contains oxides of nitrogen
• It decomposes on exposure to sunlight or on being heated into nitrogendioxide, water and oxygen.
  4HNO₃ → 4NO₂ + 2H₂O + O₂
• Due to this reaction, pure nitric acid or its concentrated solution becomes yellow on standing

Question 9.
Prove that nitric acid is an oxidising agent.
Answer:
Non metals like carbon, sulphur are oxidised by nitric acid.
C + 4HNO₃ → CO₂ + 4NO₂ + 2H₂O
S + 2HNO₃ → H₂SO₄ + 2NO

Question 10.
Prove that nitric acid is a nitrating agent.
Answer:
Nitric acid replaces hydrogen atom from organic compounds with nitronium ion NO₂⁺. This is called nitration.

Question 11.
Write the uses of nitric acid is used
Answer:

• as an oxidising agent.
• in the preparation of aqua regia.
• Salts of nitric acid are used in photography (AgNO₃) and gunpowder for fire arms (NaNO₃)

Question 12.
How is nitrous oxide prepared?
Answer:
By Heating ammonium nitrate nitrous oxide is prepared
NH₄NO₃ → N₂O + 2H₂O

Question 13.
How is nitrous acid prepared?
Answer:
By treating nitrite salt with acids, nitrous acid is prepared
Ba(NO₂)₂ + H₂SO₄ → 2HNO₂ + BaSO₄

Question 14.
What is phosphorescence?
Answer:
White phosphorous undergoes spontaneous slow oxidation in air giving a greenish yellow glow which is visible in the dark. This is known as phosphorescence. The main product of this slow oxidation is P₂O₃.

Question 15.
How is phosphine prepared?
Answer:
Phosphine is prepared by the action of sodium hydroxide with white phosphorous in an inert atmosphere of carbon dioxide

Question 16.
How is orthophosphoric acid prepared in ‘ the laboratory?
When phosphorous is treated with cone, nitric acid in the presence of iodine catalyst, it is oxidised to orthophosphoric acid.

Question 17.
Write the uses of phosphorous Phosphorous is used
Answer:

• in match boxes
• For the production of certain alloys such as phosphor bronze.

Question 18.
What happens when phosphine is heated in the absence of air?
Answer:
Phosphine decomposes into its elements when heated in the absence of air at 317 K
4PH₃ \(\underrightarrow { 317K } \) P₄ + 6H₂

Question 19.
Write about the reducing property of phosphine?
Answer:
Phosphine reduces silver nitrate into silver
PH₃ + 6AgNO₃ + 3H₂O → 6Ag + 6HNO₃ + H₃PO₃

Question 20.
How is phosphorous trichloride prepared?
Answer:

• When a slow stream of chlorine is passed over white phosphorous, PCl₃ is obtained.
• It is also prepared by treating white phosphorous with thionyl chloride.
  P₄ + 8SOCl₂ → 4PCl₃ + 4SO₂ + 2S₂Cl₂

Question 21.
Ozone (O₃) acts as a powerful oxidizing agent why? (PTA – 5)
Answer:
Ozone is not a very stable compound under normal conditions and decomposes readily on heating to give a molecule of oxygen and nascent oxygen.
Nascent oxygen, being a free radical, is very reactive
O₃ \(\underrightarrow { \triangle } \) O₂ + [O]

Question 22.
Write the uses of oxygen
Answer:

• Oxygen is one of the essential components for the survival of living organisms.
• Oxygen is used in oxyacetylene welding.
• Liquid oxygen is used as a rocket fuel.

Question 23.
How is sulphur dioxide prepared in the laboratory?
Answer:
SO₂ is prepared in the laboratory by treating a metal or metal sulphite with sulphuric acid
Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O
SO₃^2- + 2H⁺ → SO₂ + H₂O

Question 24.
Illustrate the oxidising property of SO₂.
Answer:
SO₂ oxidises hydrogen sulphide to sulphur and magnesium to magnesium oxide.
2H₂S + SO₂ → 3S + 2H₂O
2Mg + SO₂ → 2MgO + S

Question 25.
Write about contact process.
Answer:

• In contact process SO₂ is oxidised to SO₃
• It is used in the manufacture of sulphuric acid.
  

Question 26.
Write the uses of sulphurdioxide.
Answer:

• SO₂ is used in bleaching hair, silk, wool etc.
• SO₂ is used for disinfecting crops and plants in agriculture

Question 27.
Write about the structure of sulphr dioxide.
Answer:

• Sulphur undergoes sp² hybridisation.
• A double bond arises between S and O due to pπ – dπ overlapping

Question 28.
Illustrate the dehydrating property of sulphuric acid.
Answer:

Question 29.
Show that sulphuric acid is a dibasic acid.
Answer:
H₂SO₄ forms two types of salts with NaOH Hence it is dibasic.

Question 30.
How is chlorine is manufactured by Deacon’s process?
Answer:

• A mixture of air and hydrochloric acid is passed up a chamber containing a number of shelves, containing pumice stones soaked in cuprous chloride.
• Hot gases at about 723 K are passed through a jacket that surrounds the chamber.
  
• Chlorine obtained is dilute and used for the manufacture of bleaching powder

Question 31.
Write about the bleaching action of chlorine.
Answer:
Chlorine is a strong oxidising and bleaching agent since it produces nascent oxygen.
H₂O + Cl₂ → HCl + HOCl (Hypochlorous acid)
HOCl → HCl +[0]
Colouring matter + Nascent oxygen → Colourless oxidation product.

The bleaching of chlorine is permanent.

Question 32.
How is bleaching powder prepared? (MARCH 2020)
Answer:
Bleaching powder is prepared by passing chlorine gas through dry slaked lime (calcium hydroxide)
Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O

Question 33.
Write the uses of chlorine Chlorine is used
Answer:

• In the purification of drinking water.
• In the bleaching of cotton textiles, paper, and rayon.
• In the extraction of gold and platinum.

Question 34.
How is hydrochloric and prepared in the laboratory?
Answer:
Hydrochloric add is prepared by the action of sodium chloride and cone, sulphuric acid
NaCl + H₂SO₄ → NaHSO₄. + HCl
NaHSO₄ + NaCl → Na₂SO₄. + HCl
Dry hydrochloric acid is obtained by passing the gas through cone, sulphuric acid

Question 35.
How is xenon trioxide prepared?
Answer:
2XeF₆ + SiO₂ \(\underrightarrow { 50°C } \) 2XeOF₄ + SiF₄
2XeOF₄ + SiO₂ → 2XeO₂F₂ + SiF₄
2XeO₂F₂ + SiO₂ → 2XeO₃ + SiF₄
(or)
XeF₆ + 3H₂ → XeO₃ + 6HF

Question 36.
How is sodium per xenate obtained?
Answer:
When XeF₆ reacts with 2.5 M NaOH, sodium per xenate is obtained.
2XeF₆ + 16NaOH → Na₄XeO₆ + Xe + O₂ + 12NaF + 8H₂O

Question 37.
Show that sodium per xenate is a strong oxidising agent
Answer:
Sodium per xenate oxidises manganese (II) ion into permanganate ion even in the
absence of a catalyst
5XeO^4-₆ + 2Mn²⁺ + 14H⁺ → 2MnO^–₄ + 5XeO₃ + 7H₂O

Question 38.
Give reason: ICl is more reactive than l₂ (PTA – 3)
Answer:

• This is because inter-halogen compounds are in general more reactive than halogens due to w eaker inter-halogen X-X bond than X-X bond.
• So, I₂ is more stable and less reactive than ICl.

VIII. Three Mark Questions

Question 1.
Write about the structure of ammonia.
Answer:

• Ammonia molecule is pyramidal in shape.
• Hybridisation of nitrogen is sp³.
• The shape must be tetrahedral but in one of the tetrahedral positions a lone pair of electrons from the nitrogen atom is present, hence it is pyramidal.
• The N – H bond distance is 1.016 A0.
• The H – H bond distance is 1.645 A°.
• The bond angle is 107°
  

Question 2.
How does red phosphorous react with oxygen?
Answer:
Red phosphorous reacts with oxygen on heating to give phosphorous trioxide and phosphorous pentoxide.
P₄ + 3O₂ → P₄O₆
P₄ + 5O₂ → P₄O₁₀

Question 3.
How is pure phosphine prepared?
Answer:
Pure phosphine is prepared by heating phosphorous acid

A pure sample of phosphine is prepared by heating phosphonium iodide with caustic soda solution.

Question 4.
What happens when phosphine is heated with air?
Answer:
When phosphine is heated with air or oxygen, it undergoes combustion to give meta phosphoric acid.

Question 5.
Write about Holmes signal
Answer:

• In a ship during distress, a container with calcium carbide and calcium phosphide mixture is pierced and thrown into the sea.
• The mixture reacts with seawater liberating acetylene and phosphine gases.
• The liberated phosphine catches fire and ignites acetylene.
• These burning gases with lot of smoke serves as a signal to the approaching ships.
• This is known as Holme’s signal.

Question 6.
Write about the structure of phosphine
Answer:

• Phosphorous shows sp³ hybridisation.
• Three orbitals are occupied by bond pair electrons.
• Fourth orbital is occupied by lone pair of electrons.
• Hence instead of tetrahedral, PH₃ has a pyramidal shape.
• Bond angle is reduced to 94°
  

Question 7.
How is oxygen prepared in the laboratory?
Answer:
Oxygen is prepared in the laboratory by the decomposition of hydrogen peroxide in presence of Mn02 catalyst or by the oxidation of potassium permanganate.

5H2O₂ + 2MnO₄^– + 6H⁺ → 5O₂ + 8H₂O + 2Mn²⁺

Question 8.
Write about ozone
Answer:

• Ozone is an allotropic form of oxygen
• Ozone is triatomic gas
• Although negligible amounts of ozone occurs at sea level, it is formed in the upper atmosphere by the action of UV light.
• In the laboratory ozone is prepared by passing electrical discharge through oxygen.
• At a potential of 20,000 V about 10% of oxygen is converted into ozone, it gives a mixture known as ozonised oxygen.
• Pure ozone is obtained as a pale blue gas by the fractional distillation of liquefied ozonised oxygen.
  
• Ozone molecule has a bent shape and symmetrical with delocalised bonding between the oxygen atoms.

Question 9.
Write about the reducing property of sulphur dioxide
Answer:

SO₂ reduces chlorine into hydrochloric acid
SO₂ + 2H₂O + Cl₂ → H₂SO₄ + 2HCl

SO₂ reduces potassium permanganate into manganese sulphate (Mn²⁺).
2KMnO₄ + 5SO₂ + 2H₂O → K₂SO₄ + 2MnSO₄ + 2H₂SO₄

SO₂ reduces potassium dichromate into chromic sulphate (Cr³⁺)
K₂Cr₂O₇ + 3SO₂ + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + H₂O

Question 10.
Write about the bleaching action of sulphur dioxide.
Answer:
In presence of water, sulphur dioxide bleaches coloured wool, silk, sponges and straw into colourless due to its reducing property

When the bleached product (Colourless) is allowed to stand in air, it is reoxidised by atmospheric oxygen to its original colour.

Hence bleaching action of sulphurdioxide is temporary

Question 11.
Explain the manufacture of sulphuric acid
Answer:

• Sulphur dioxide is produced by burning sulphur or iron pyrites in oxygen / air
  S + O₂ → SO₂
  4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
• SO₂ is oxidised to SO₃ by air in presence of V₂O₅ or platinised absestos
  
• SO₃ is absorbed in cone H₂SO₄ and oleum is produced
  SO₃ + H₂SO₄ H₂S₂O₇
• Oleum is converted into sulphuric acid by diluting it with water.
  H₂S₂O₇ + H₂O → 2H₂SO₄
• To maximize the yield the plant is operated at 2 bar pressure and 720 K.
  96% pure H₂SO₄ is obtained.

Question 12.
Show that sulphuric acid is an oxidising agent
Answer:
Sulphuric acid is an oxidising agent as it produces nascent oxygen
H₂SO₄ → H₂O + SO₂ + [O] (Nascert oxygen)

Sulphuric acid oxidises carbon into carbon dioxide
C + 2H₂SO₄ → 2SO₂ + 2H₂O + CO₂

Sulphuric acid oxidises phosphorous into orthophosphoric acid
P₄ + 10H₂SO₄ → 4H₃PO₄ + 10SO₂ + 4H₂0

Sulphuric acid oxidises iodide into iodine.
H₂SO₄ + 2HI → 2SO₂ + 2H₂O + I₂

Question 13.
What is the action of sulphuric acid on metals?
Answer:
Dilute sulphuric acid reacts with metals liberating hydrogen gas.
Zn + H₂SO₄ ZnSO₄ + H₂ ↑

Hot cone. Sulphuric acid reacts with metals to give sulphates and sulphur dioxide
Cu + 2H₂SO₄ → CuSO₄ + 2H₂O + SO₂ ↑

Sulphuric acid does not react with noble metals like gold, silver and platinum.

Question 14.
Give the test for sulphate / sulphuric acid
Dilute solution of sulphuric acid / Sulphates react with barium chloride or lead acetate solution to give a white precipitate

Question 15.
How is chlorine manufactured by the electrolytic process?
Answer:
When brine solution (NaCl) is electrolyzed, Na⁺ and Cl^– ions are formed.
Na⁺ ions react with OH^– ions of water forming sodium hydroxide.
Hydrogen and chlorine are liberated as gases.
NaCl → Na+ +Cl^–
H₂O → H+ +OH^–
Na⁺ + OH^– → NaOH
At cathode : H⁺ + e^– → H
H + H → H₂
At anode : Cl^– → Cl + e^–
Cl + Cl → Cl₂

Question 16.
What is aqua regia? What is its action on gold?
Answer:
Aqua regia is a mixture of three parts of cone, hydrochloric acid and one part of cone, nitric acid.
This is used for dissolving gold, platinum etc.
AU + 4H⁺ + NO₃^– + 4Cl^– → AuCl₄^– +NO + 2H₂O

Question 17.
HF acid is a weaker acid at low concentration, but becomes stronger as the concentration increases why?
Answer:
0.1 M Solution HF is 10% ionised, hence it is a weak acid.
But 5 M, 15 M solution of HF is stronger due to the equilibrium.

At high concentration, the equilibrium involves the removal of flouride ions and increases the hydrogen ion concentration, HF becomes stronger acid as the concentration increases.

Question 18.
HF acid is not stored in glass bottles why? (MARCH 2020)
Answer:
HF attacks silica and silicates present in glass bottles.

Hence HF is not stored in glass bottles. But HF is stored in Teflon bottles.
SiO₂ + 4HF → SiF₄ + 2H₂O
Na₂SiO₃ + 6HF → Na₂SiF₆ + 3H₂O

Question 19.
Mention the characteristic of interhalogen compounds (PTA – 2)
Answer:

• The central atom will be the larger halogen.
• It can be formed only between two halogens and not more than two halogens.
• Fluorine can’t act as a central atom because it is the smallest among halogens and highly electronegative.
• They are strong oxidising agents and undergo auto ionization.

Question 20.
Give the preparation of xenon fluorides
Answer:

Question 21.
What is the hybridisation in XeOF₂? Give its structure. (PTA – 1)
Answer:

IX. Five Mark Questions

Question 1.
How is nitric acid manufactured using Ostwald’s process?

• Ammonia prepared by Haber’s process is mixed about 10 times of air.
• This mixture is preheated and passed into the catalyst chamber where they come in contact with platinum gauze.
• The temperature rises about 1275 K.
• The metallic gauze brings about the rapid catalytic oxidation of ammonia resulting in the formation of NO.
• NO is oxidized to NO₂
  4NH₃ +5O₂ → 4NO + 6H₂O +120 KJ
  2NO + O₂ → 2NO₂
• NO₂ produced is passed through a series of adsorption towers.
• NO₂ reacts with water to give nitric acid.
• Nitric acid formed is bleached by blowing air.
  6NO₂ + 3H₂O → 4HNO₃ + 2NO + H₂O

Question 2.
Explain the action of nitric acid on metals with one example.
Primary reaction:
Metal nitrate is formed with the release of nascent hydrogen.
3Cu + 6HNO₃ → 3CU(NO₃)₂ + 6(H)

Secondary reaction:
Nascent hydrogen produces the reduction products of nitric acid
6(H) + 3HNO₃ → 3HNO₂ + 3H₂O

Tertiary reaction:
With dilute acid, the secondary products decompose to give final products.
3HNO₂ → HNO₃ + 2NO + H₂O
Hence overall reaction is
3Cu + 8HNO₃ → 3CU(NO₃)₂ + 2NO + 4H₂O

With concentrated acid the secondary products react to give the final products.
HNO₂ + HNO₃ → 2NO₂ + H₂O
Hence overall reaction is
Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O

Question 3.
Write the preparation of nitrogen oxides.
Answer:

Question 4.
Write the preparation of oxoacids of nitrogen
Answer:

Question 5.
Explain the structure of oxides of phosphorus
Answer:
Phosphorus trioxide :

• In P₄O₆, four phosphorous atoms lie at the corners of a tetrahedron and six oxygen atoms along the edges.
  The P – O bond distance is 165.6 pm which is shorter than the single bond distance of the P-O bond (184 pm)
• This is due to Pπ – dπ bonding
• This results in the considerable double bond character
  

Phosphorous Pentoxide:

• In P₄O₁₀ each P atom forms a single bond with three oxygen atoms and a coordinate bond with one oxygen atom.
• Terminal coordinate P-O bond length is 143 pm
• This is less than the expected single bond distance
• This may be due to lateral overlap of filled
  p – Orbitals of an oxygen atom with empty
  d – Orbital on phosphorous.

Question 6.
Write the structure of and basicity oxoacids of phosphorous. (PTA – 3)

Answer:

Question 7.
Write the preparation of oxoacids of phosphorous
Answer:

Question 8.
Write the structure of oxo acids of sulphur.
Answer:

Question 9.
List any five compounds of Xenon and mention the type of hybridization and structure of the compounds (PTA – 6)
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.9 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.9

Question 1.
Find the asymptotes of the following curves:
(i) f(x) = \(\frac { x^2 }{ x^2-1 }\)
(ii) f(x) = \(\frac { x^2 }{ x+1 }\)
(iii) f(x) = \(\frac { 3x }{ \sqrt{x^2+2} }\)
(iv) f(x) = \(\frac { x^2-6x-1 }{ x+3 }\)
(v) f(x) = \(\frac { x^2+6x-4 }{ 3x-6 }\)
Solution:
(i) f(x) = \(\frac { x^2 }{ x^2-1 }\)
The function becomes undefined when x = 1 and x = -1
∴ x = 1 and x = -1 are the vertical asymptotes

As ‘x’ gets larger (Positive or negative) the function, the function attaining the value 1.
∴ y = 1 is horizontal asymptote

(ii) f(x) = \(\frac { x^2 }{ x+1 }\)

The function becomes undefined when x = -1
∴ Vertical asymptote is x = -1 and there is no horizontal asymptote.
No horizontal asymptote exists for the curve. Oblique asymptote can be obtained by polynomial long division method.

Oblique (or) slant asymptote is y = x – 1

(iii) f(x) = \(\frac { 3x }{ \sqrt{x^2+2} }\)
No vertical asymptotes
Horizontal asymptotes RHL (Right Hand Limit)


y = 3 and y = -3 are the Horizontal asymptotes
Slant asymptotes’. No such slant asymptotes exist for the given curve.

(iv) f(x) = \(\frac { x^2-6x-1 }{ x+3 }\)

When x = -3, the function becomes undefined.
∴ x = -3 is the vertical asymptote.
No Horizontal asymptote exist for the curve.
Oblique asymptote can be obtained by polynomial long division method

∴ y = x – 9 is the slant (or) oblique asymptote.

(v) f(x) = \(\frac { x^2+6x-4 }{ 3x-6 }\)

The function becomes undefined when x = 2.
∴ x = 2 is the vertical asymptote.
No Horizontal asymptote exist for the given curve.
Oblique asymptote can be obtained by polynomial long division method.
∴ y = \(\frac { x }{ 3 }\) + \(\frac { 8 }{ 3 }\) (or) 3y = x + 8 is the slant asymptote.

Question 2.
Sketch the graphs of the following functions
(i) y = –\(\frac { 1 }{ 3 }\) (x³ – 3x + 2)
(ii) y = x \(\sqrt { 4-x }\)
(iii) y = \(\frac { x^2+1 }{ x^2-4 }\)
(iv) y = \(\frac { 1 }{ 1+e^{-x} }\)
(v) y = \(\frac { x^3 }{ 24 }\) – log x
Solution:
(i) y = –\(\frac { 1 }{ 3 }\) (x³ – 3x + 2)

Factorizing we get
y = –\(\frac { 1 }{ 3 }\) (x – 1)² (x + 2) = f(x)
The domain and the range of the given function f(x) are the entire real line.
Putting y = 0, we get x = 1, 1, – 2. Hence the x-intercepts are (1, 0) and (- 2, 0) and by putting x = 0. We get y = –\(\frac { 2 }{ 3 }\). Therefore, the y-intercept is (0, –\(\frac { 2 }{ 3 }\))
f'(x) = \(\frac { (3x^2-3) }{ 3 }\) = -(x² – 1) = 1 – x²
f'(x) = 0 ⇒ 1 – x² = 0 ⇒ x = ±1
The critical points of the curve occur at x = ± 1 .
f”(x) = -2x
f”(1) = – 2 < 0, ∴ f(x) is maximum at x = 1 and the local maximum is f(1) = o
f”(-1) = 2 > 0, ∴ f(x) is minimum at x = -1 and the local minimum is
f(-1) = –\(\frac { 4 }{ 3 }\)
f”(x) = – 2x < 0 ∀ x > 0, ∴ The function is concave downward in the positive real line.
f”(x) = 2x > 0 ∀ x < 0, ∴ The function is concave upward in the negative real line.
Since f”(x) = 0 at x = 0 and f”(x) changes its sign when passing through x = 0.
Hence the point of inflection is (0, –\(\frac { 2 }{ 3 }\))
The curve has no asymptotes.

(ii) y = x\(\sqrt { 4-x }\)

y = x\(\sqrt { 4-x }\) = f(x)
where x > 4 the curve does not exist and it exists for x ≤ 4
∴ The domain is (-∞, 4] and the Range is (-∞, \(\frac { 16 }{ 3√3 }\) ]
The curve passes through the origin. The curve intersects x-axis at (4, 0).
f'(x) = –\(\frac { x }{ 2 \sqrt{4-x} }\) + \(\sqrt { 4-x }\) = \(\frac { 8-3x }{ 2 \sqrt{4-x} }\)
f'(x) = 0 ⇒ 8 – 3x = 0 ⇒ x = \(\frac { 8 }{ 3 }\)
∴ Critical point of the curve occur at x = \(\frac { 8 }{ 3 }\)
f”(x) = \(\frac { 3x-16 }{ 4(4-x)^{\frac{3}{2}} }\)
f”(\(\frac { 8 }{ 3 }\)) = –\(\frac { 3√3 }{ 4 }\) < 0
∴ f(x) is maximum at x = \(\frac { 8 }{ 3 }\) and the local maximum f(\(\frac { 8 }{ 3 }\)) = \(\frac { 16 }{ 3√3 }\) and local minimum is 0 at x = 4 (from the graph)
f”(x) = \(\frac { 3x-16 }{ 4(4-x)^{\frac{3}{2}} }\) < 0 ∀ x < 4
∴ The curve is concave downward in the negative real line.
No point of inflection exists.
As x → ∞, y → ±∞ , and hence the curve does not have any asymptotes.

(iii) y = \(\frac { x^2+1 }{ x^2-4 }\)

The domain of the given function f(x) is (-∞, -2) ∪ (-2, 2) ∪ (2, ∞)
ie. x < -2 (or) -2 < x < 2 (or) x > 2.
Range of f(x) is (-∞, –\(\frac { 1 }{ 4 }\)) ∪ (1, ∞)
i.,e. f(x) ≤ –\(\frac { 1 }{ 4 }\) (or) f(x) > 1.
Putting y = 0, x is unreal. Hence, there is no ‘x’ intercept.
By putting x = 0, we get y = –\(\frac { 1 }{ 4 }\).
∴ y intercept is (0, –\(\frac { 1 }{ 4 }\))
f'(x) = –\(\frac { 10x }{ (x^2-4)^2 }\)
f'(x) = 0 ⇒ x = 0,
∴ The critical point is at x = 0
f'(x) = \(\frac { 10(x^2-4)(3x^+4) }{ (x^2-4)^4 }\)
f'(0) = –\(\frac { 5 }{ 8 }\) < 0,
∴ f(x) is maximum at
x = 0. Hence the local maximum is f(0) = –\(\frac { 1 }{ 4 }\)
No points of inflection exist for the curve.
When x = ± 2, y = ∞
∴ Vertical asymptotes are x = 2 and x = -2 and Horizontal asymptote is y = 1.

(iv) y = \(\frac { 1 }{ 1+e^{-x} }\)

The Domain of the function f(x) is the entire real line.
ie., (-∞, ∞) ⇒ -∞ < x < ∞ and the range is (0, 1) ie., 0 < f(x) < 1
No ‘x’ intercept for f(x) and when x = 0
y = \(\frac { 1 }{ 2 }\)
∴ The ‘y’ intercept is (0, \(\frac { 1 }{ 2 }\))
f'(x) = \(\frac { e^{-x} }{ (1+e^{-x})^2 }\)
f'(x) = 0 ⇒ which is absurd. Hence there is no extremum.
No vertical asymptote for the curve exist and the Horizontal asymptotes are y = 1 and y = 0.

(v) y = \(\frac { x^3 }{ 24 }\) – log x

The curve exists only for positive values of ‘x’ (x > 0) ie., domain is (0, ∞) and
The range is (\(\frac { 1 }{ 3 }\) – log e², ∞)
No intersection points are possible
f'(x) = \(\frac { x^2 }{ 8 }\) – \(\frac { 1 }{ x }\)
f'(x) = 0 ⇒ x³ – 8 = 0 ⇒ x = 2
∴ Critical point occur at x = 2
f'(x) = \(\frac { x }{ 4 }\) + \(\frac { 1 }{ x^2 }\)
f”(2) = \(\frac { 3 }{ 4 }\) > 0,
∴ f(x) is mimmum at x = 2 and the local minimum is f(2) = \(\frac { 1 }{ 3 }\) – log e²
No point of inflection exists.
No Horizontal asymptotes are possible, but the vertical asymptote is x = 0 (y-axis).

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Zoology Guide Pdf Chapter 11 Biodiversity and its Conservation Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 11 Biodiversity and its Conservation

12th Bio Zoology Guide Biodiversity and its Conservation Text Book Back Questions and Answers

Question 1.
Which of the following region has maximum biodiversity?
(a) Taiga
(b) Tropical forest
(c) Temperate rain forest
(d) Mangroves
Answer:
(b) Tropical forest

Question 2.
Conservation of biodiversity within their natural habitat is ……………….
(a) In-situ conservation
(b) Ex-situ conservation
(c) In vivo conservation
(d) In vitro conservation
Answer:
(a) In-situ conservation

Question 3.
Which one of the following is not coming under in-situ conservation?
(a) Sanctuaries
(b) Natural parks
(c) Zoological park
(d) Biosphere reserve
Answer:
(c) Zoological park

Question 4.
Which of the following is considered a hotspot of biodiversity in India?
(a) Western ghats
(b) Indo-gangetic plain
(c) Eastern Himalayas
(d) A and C
Answer:
(d) A and C

Question 5.
The organization which published the red list of species is ………………
(a) WWF
(b) IUCN
(c) ZSI
(d) UNEP
Answer:
(b) IUCN

Question 6.
Who introduced the term biodiversity?
(a) Edward Wilson
(b) Walter Rosen
(c) Norman Myers
(d) Alice Norman
Answer:
(b) Walter Rosen

Question 7.
Which of the following forests is known as the lungs of the planet Earth?
(a) Tundra forest
(b) Rain forest of north east India
(c) Taiga forest
(d) Amazon rain forest
Answer:
(d) Amazon rain forest

Question 8.
Which one of the following are at high risk extinction due to habitat destruction?
(a) Mammals
(b) Birds
(c) Amphibians
(d) Echinoderms
Answer:
(c) Amphibians

Question 9.
Assertion: The Environmental conditions of the tropics are favourable for speciation and diversity of organisms.
Reason: The climate seasons, temperature, humidity and photoperiod are more or less stable and congenial.
(a) Both Assertion and Reason are true and Reason explains Assertion correctly.
(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) Assertion is true, but Reason is false.
(d) Both Assertion and Reason are false.
Answers:
(a) Both Assertion and Reason are true and Reason explains Assertion correctly.

Question 10.
Define endemism.
Answer:
Endemism: A species or a taxon which is confined to a specific area
E..g: Nilgiri Thar

Question 11.
How many hotspots are there in India? Name them.
Answer:
India encloses 4 biodiversity hotspots. They are

• Himalayan
• Indo-Burma
• Western Ghats
• Sunderland

Question 12.
What are the three levels of biodiversity?
Answer:

• Genetic Diversity
• Species Diversity
• Community / Ecosystem Diversity

Question 13.
Name the active chemical found in the medicinal plant Rauwolfia vomitoria. What type of diversity does it belong to?
Answer:
Rauwolfia vomitoria can be cited as an example of genetic diversity. Reserpine is an active chemical extracted from Rauwolfia vomitoria.

Question 14.
“Amazon forest is considered to be the lungs of the planet”-Justify this statement.
Answer:
Amazon rain forest is known as the “lungs of the planet” since a great proportion of CO₂ released due to anthropogenic activities are uptaken by their dense tropical forests, which is _ homologous to the functioning of human beings except for the difference in gases.

Question 15.
‘Red data book’-What do you know about it?
Answer:
Red Data Book or Red list is a catalogue of taxa facing the risk of extinction. IUCN – International Union of Conservation of Nature and Natural Resources, which is renamed as WCU – World Conservation Union (Morges Switzerland) maintains the Red Databook. The concept of Red list was mooted in 1963. The purpose of preparation of Red List are:

• To create awareness on the degree of threat to biodiversity
• Identification and documentation of species at high risk of extinction
• Provide global index on declining biodiversity
• Preparing conservation priorities and help in conservation of action

Information on international agreements on conservation of biological diversity Red list has eight categories of species.

1. • Extinct
   • Extinct in wild
   • Critically Endangered
   • Endangered
   • Vulnerable
   • Lower risk
   • Data deficiency
   • Not evaluated

Question 16.
Extinction of a keystone species led to loss of biodiversity – Justify.
Answer:
A keystone species is an organism that helps define an entire ecosystem. Without the keystone species, a particular ecosystem would be dramatically disturbed or even ceased. Keystone species either directly or indirectly affects every species in a particular ecosystem. If a keystone species is lost or removed no other organism would compensate its ecological niche.

E.g. Jaguar is a keystone species. As a top predator, it plays a crucial role in the ecosystem. Without jaguar, there is an exponential increase in herbivoral population that would decimate the plants of the ecosystem. At one point even the herbivore populations also get declined due to the lack of vegetation. Thus jaguar acts a keystone species.

Question 17.
Compare and Contrast the insitu and exsitu conservation.
Answer:

Question 18.
What are called endangered species? Explain with examples.
Answer:
A species which has been categorized as very likely to become extinct. E.g., Bengal tiger, Polar bears.

Question 19.
Why do we find a decrease in biodiversity distribution, if we move from the tropics towards the poles?
Answer:
There is a decrease in biodiversity as we move from tropics towards poles due to a drop in temperature which makes the condition unfavourable for the majority of organisms to survive.

Question 20.
What are the factors that drive habitat loss?
Answer:
Natural habitats are destroyed for the purpose of settlement, agriculture, mining, industries, and construction of highways. As a result, species are forced to adapt to the changes in the environment or move to other places. If not, they become victim to predation, starvation, disease and eventually die or results in human animal conflict.

Question 21.
Where are biodiversity hotspots normally located? Why?
Answer:
Hotspots are areas characterized with high concentration of endemic species experiencing unusual rapid rate of habitat modification loss. Norman Myers defined hot spots as “regions that harbour a great diversity of endemic species and at the same time, have been significantly impacted and altered by human activities.”

Question 22.
Why is biodiversity so important and worthy of protection?
Answer:
Biodiversity reflects the number of different organisms and their relative frequencies in an ecological system and constitutes the most important functional component of a natural ecosystem. It helps to maintain ecological processes, create soil, recycle nutrients, influence climate, degrade waste, and control diseases. It provides an index of the health of an ecosystem. The survival of the human race depends on the existence and well-being of all life forms (plants and animals) in the biosphere.

Question 23.
Why do animals have greater diversification than plant diversity?
Answer:
The distribution of plants and animals is not uniform around the world. Organisms require different sets of conditions for their optimum metabolism and growth. Plants in general have the ability to withstand and overcome environmental fluctuations. Moreover, majority of plants are autotrophs so they can synthesize their own food, hence they show minimal modifications.
In case of animals, they have to tolerate climatic fluctuations, migrate to other areas in search of food, or adapt themselves or their body parts according food they consume thus varying from ancestors leading to evolution of new species. Hence it is understood that climatic condition food source, predation, competition and other natural forces lead to much diversification among animals than in plants.

Question 24.
Alien species invasion is a threat to endemic species – substantiate this statement.
Answer:
Exotic species are organisms often introduced unintentionally or deliberately for commercial purpose, as biological control agents and other uses. They often become invasive and drive away the local species and is considered as the second major cause for extinction of species. Tilapia fish (Jilabi kendai) (Oreochromis mosambicus) introduced from east coast of South Africa in 1952 for its high productivity into Kerala’s inland waters, became invasive, due to which the native species such as Puntius dubius and Labeo kontius face local extinction.
Amazon sailfin catfish is responsible for destroying the fish population in the wetlands of Kolkata. The introduction of the Nile Perch, a predatory fish into Lake Victoria in East Africa led to the extinction of an ecologically unique assemblage of more than 200 nature species of cichlid fish in the lake.

Question 25.
Mention the major threats to biodiversity caused by human activities. Explain.
Answer:
Human activities, both directly and indirectly are today’s main reason for habitat loss and biodiversity loss. Fragmentation and degradation due to agricultural practices, extraction (mining, fishing, logging and harvesting) and development (settlements, industrial and associated infrastructures) leads to habitat loss and fragmentation leads to the formation of isolated, small and scattered populations and as endangered species.

Some of the other threats include specialised diet, specialized habitat requirement, large size, small population size, limited geographic distribution and high economic or commercial value. Large mammals by virtue of their size require larger areas to obtain the necessities of life – food, cover and mates than do smaller mammals.

The individual home range of Lion can be . “ about 100 square Km. Mammals have specialized dietary needs such as carnivores, frugivores and the need to forage over much larger areas than general dietary herbivores and omnivores. Mammals also have low reproductive output other than small rodents.

Question 26.
What is mass extinction? Will you encounter one such extinction in the near future. Enumerate the steps to be taken to prevent it.
Answer:
The Earth has experienced quite a few mass extinctions due to environmental catastrophes. A mass extinction occurred about 225 million years ago during the Permian, where 90% of shallow-water marine invertebrates disappeared.

Question 27.
In northeastern states, the jhum culture is a major threat to biodiversity – substantiate.
Answer:
In shifting cultivation, plots of natural tree vegetation are burnt away and the cleared patches 1 are farmed for 2-3 seasons, after which their fertility reduces to a point where crop production is no longer profitable.

The farmer then abandons this patch and cuts down a new patch of forest trees elsewhere for crop production. This system is practiced in the north-eastern regions of India. When vast areas are cleared and burnt, it results in loss of forest cover, pollution, and discharge of CO₂, which in turn attributes to a loss of habitat and climate change which has an impact on the faunal diversity of that region.

Question 28.
List out the various causes for biodiversity losses.
Answer:
The major causes of biodiversity decline are:

• Habitat loss, fragmentation, and destruction (affects about 73% of all species),
• Pollution and pollutants (smog, pesticides, herbicides, oil slicks, and GHGs).
• Climate change.
• Introduction of alien/exotic species.
• Overexploitation of resources (poaching, indiscriminate cutting of trees, overfishing, hunting, and mining).
• Intensive agriculture and aquacultural practices.
• Hybridization between native and non-native species and loss of native species
• Natural disasters (Tsunami, forest fire, earthquake,s, and volcanoes).
• Industrialization, Urbanization, infrastructure development, Transport – Road and Shipping activity, communication towers, dam construction, unregulated tourism, and monoculture are a common area of specific threats.
• Co-extinction

Question 29.
How can we contribute to promoting biodiversity conservation?
Answer:

• identify and protect all threatened species
• identify and conserve in protected areas the wild relatives of all the economically important organisms
• identify and protect critical habitats for feeding, breeding, nursing, resting of each species
• resting, feeding, and breeding places of the organisms should be identified and protected.
• Air, water, and soil should be conserved on a priority basis
• Wildlife Protection Act should be implemented

Question 30.
‘Stability of a community depends upon its species diversity’ – Justify the statement.
Answer:
Species diversity leads to a stable community because an area with more species diversity always leads to higher productivity thus maintains a stable community.

Question 31.
Write a note on
(i) Protected areas
(ii) Wildlife sanctuaries
(iii) WWF.
Answer:
(i) Protected areas are biogeographical areas, where biological diversity along with natural and cultural resources is protected, maintained, and managed through legal measures. Protected areas include national parks, wildlife sanctuaries, community reserves, and biosphere reserves.

(ii) Any area other than the area comprised of any reserve forest or the territorial waters can be notified by the State Government to constitute as a sanctuary if such area is of adequate ecological, faunal, floral, geomorphological, natural or zoological significance. This is for the purpose of protecting, endangered factual species. Some restricted human activities are allowed inside the sanctuary area. Ecotourism is permitted, as long as animal life is undisturbed.

(iii) WWF stands for World Wide Fund for nature is an international NGO working in the field of wildlife conservation.

12th Bio Zoology Guide Biodiversity and its Conservation Additional Important Questions and Answers

12th Bio Zoology Guide Biodiversity and its Conservation One Mark Questions and Answers

Question 1.
Who coined the term Bio-diversity?
Answer:
Walter Rosen.

Question 2.
Which is not an indices of species diversity?
(a) Alpha diversity
(b) Beta diversity
(c) Delta diversity
(d) Gamma diversity
Answer:
(c) Delta diversity

Question 3.
The total number of mega biodiversity countries in the world is …………….
(a) Twelve
(b) Fifteen
(c) Seventeen
(d) Nineteen
Answer:
(c) Seventeen

Question 4.
How many biogeographic zones are there in India?
(a) Twelve
(b) Seventeen
(c) Ten
(d) Fifteen
Answer:
(c) Ten

Question 5.
The most important pattern of biodiversity is ……………………..
(a) Longitudinal gradient in diversity
(b) Latitudinal gradient in diversity
(c) Polar gradient diversity
(d) Equatorial gradient in diversity
Answer:
(b) Latitudinal gradient in diversity

Question 6.
Which of the following denotation is correct regarding increasing diversity?
(a) Poles < Equator
(b) Equator < Pole
(c) Pole = Equator
(d) Latitude = Longitude
Answer:
(a) Poles < Equator

Question 7.
Select the proper sequence indicating the increasing order of biodiversity.
(a) Polar, Temperate and Polar
(b) Tropics, Temperate and Polar
(c) Temperate, Tropic and Polar
(d) Polar, Tropic and Temperate
Answer:
(a) Polar, Temperate and Polar

Question 8.
Select the correct linear equation describing the species-area relationship.
(a) loc C = log S + Z log A
(b) Z log A = log S + log C
(c) log S = log C + Z log A
(d) log C = log S ± Z log A
Answer:
(c) log S = log C + Z log A

Question 9.
If meat-eating animals are called carnivores, how do you call the animals that thrive mostly on fruits?
Answer:
Frugivore

Question 10.
The wild ass is endemic to ………………..
(a) the Western Ghats
(b) Deccan Peninsula
(c) the Himalayas
(d) Indian desert
Answer:
(d) Indian desert

Question 11.
Which is considered as the Biogeographical Gateway of India?
(a) the Himalayas
(b) Andaman & Nicobar
(c) North – East India
(d) Mumbai
Answer:
(c) North – East India

Question 12.
Species introduced deliberately in an area are referred to as …………………
(a) Endemic species
(b) Vulnerable species
(c) Exotic species
(d) Extinct species
Answer:
(c) Exotic species

Question 13.
Tilapia fish (Oreochromis Mozambique) is an exotic breed from ……………………
(a) Mexico
(b) South Africa
(c) Canada
(d) Central America
Answer:
(b) South Africa

Question 14.
Mention the correct number of biodiversity hotspots identified throughout the world.
(a) 29
(b) 16
(c) 34
(d) 46
Answer:
(c) 34

Question 15.
Which is not an accepted biodiversity hotspot of India?
(a) Indian Himalayas
(b) Western Ghats
(c) Indo-Burma
(d) Deccan Plateau
Answer:
(d) Deccan Plateau

Question 16.
A species is considered extinct
(a) When its member is confined to a particular area
(b) When its member is maintained in a non-native area
(c) When none of its members is alive in the native area
(d) When none of its members alive anywhere in the world
Answer:
(d) When none of its members are alive anywhere in the world.

Question 17.
The concept of Red list was noted in …………………
(a) 1953
(b) 1963
(c) 1973
(d) 2003
Answer:
(6) 1963

Question 18.
Match the following
(a) Tiger reserves in India – (i) 4
(b) Hotspots in India – (ii) 104
(c) Biosphere reserves in India – (iii) 27
(d) National parks in India – (iv) 18
Answer:
a – iii, b – i, c – iv, d – ii

Question 19.
Statement 1: Biodiversity is the assemblage of different life form.
Statement 2: The term biodiversity was introduced by Edward Wilson.
(a) Statement 1 is correct, statement 2 in incorrect
(b) Statement 1 is incorrect, statement 2 in correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(fl) Statement 1 is correct, statement 2 is incorrect

Question 20.
Statement 1: India is the seventh-largest country in the world in terms of area.
Statement 2: It includes ten biogeographic areas.
(a) Statement 1 is correct, statement 2 is incorrect
(b) Statement 1 is incorrect, statement 2 is correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(c) Both the statements are correct

Question 21.
Statement 1: the Western Ghats extend from South Gujarat to Karnataka.
Statement 2: Wild ass is an endemic species of Western Ghats
(a) Statement 1 is correct, statement 2 is incorrect
(b) Statement 1 is incorrect, statement 2 is correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(d) Both the statements are incorrect

Question 22.
Statement 1: Exotic species are a non-native organisms.
Statement 2: Sailfin catfish is an exotic species in India.
(a) Statement 1 is correct, statement 2 in incorrect
(b) Statement 1 is incorrect, statement 2 in correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(c) Both the statements are correct

12th Bio Zoology Guide Biodiversity and its Conservation Two Marks Questions and Answers

Question 1.
Define biodiversity.
Answer:
Biodiversity is the variability among living organisms from all sources, including terrestrial, marine, and other aquatic ecosystems and the ecological complexes of which they are part. This includes diversity within species, between species and ecosystems of a region.

Question 2.
Who coined the term biodiversity? Name the levels of biodiversity.
Answer:
The term biodiversity was coined by Walter Rosen (1986). The levels of biodiversity are – Genetic diversity, Species diversity, and ecosystem diversity.

Question 3.
What is species richness? Give example.
Answer:
The number of species per unit area at a specific time is called species richness, which denotes the measure of species diversity.
E.g., Western ghats have more amphibian species than Eastern ghats.

Question 4.
Enlist the factors that determine the distribution of biodiversity.
Answer:
Temperature, precipitation, distance from the equator (latitudinal gradient), altitude from sea level (altitudinal gradient) are some of the factors that determine biodiversity distribution patterns.

Question 5.
What are the most important causes of biodiversity loss?
Answer:
Habitat loss, fragmentation, and destruction.

Question 6.
Name any two alien animal species introduced in India.
Answer:
(a) Tilapia fish (Oreochromis Mozambique)
(b) African Apple snail (Achatina Fulica)

Question 7.
Name any four biogeographic zones in India.
Answer:
(a) the Himalayas
(b) Indian deserts
(c) Trans-Himalayan region
(d) Western ghats.

Question 8.
What do you mean by the term co-extinction?
Answer:
Coextinction of a species is the loss of a species as a consequence of the extinction of another.
E.g., orchid bees and forest trees by cross-pollination.

Question 9.
What are the natural causes of biodiversity loss?
Answer:
Natural threats include spontaneous jungle fires, tree falls, landslides, defoliation by insects or locust attacks.

Question 10.
Define hotspots.
Answer:
Hotspots are areas characterized with high concentration of endemic species experiencing unusual rapid rate of habitat modification loss.

Question 11.
What will be the consequences of the loss of biodiversity?
Answer:
Loss of biodiversity has a immense impact on plant and animal life. The loss of diversity leads to,

• Extinction of species
• Dramatic influence on the food chain and food web in.
• It will lead to immediate danger for food necessity

Question 12.
Name the four biodiversity hotspots in India.
Answer:
(a) the Himalayas
(b) Western Ghats
(c) Indo-Burma region
(d) Sundaland

Question 13.
What does IUCN stand for?
Answer:
IUCN – International Union for Conservation of Nature and Natural resources.

Question 14.
Define species extinction. What makes a species to become extinct?
Answer:
A species is considered extinct when none of its members are alive anywhere in the world. Environmental changes and population characteristics are the two major reasons for species extinction.

Question 15.
When a species is considered as locally extinct?
Answer:
A species is considered to be locally extinct when it is no longer found in an area it once inhabited but is still found elsewhere in the wild.

Question 16.
State the mission of IUCN.
Answer:
IUCN’s mission is to influence, encourage, and assist societies throughout the world to conserve nature and to ensure that any use of natural resources is equitable and ecologically sustainable.

Question 17.
What is the Red list? How many categories of species are mentioned in the Red List?
Answer:
Red List or Red databook is a catalogue of taxa facing a risk of extinction. It has 8 categories of species.

Question 18.
Mention any four categories of species mentioned in the Red data book.
Answer:

• Extinct
• Endangered
• Extinct in wild
• Vulnerable

Question 19.
How a national park can be defined?
Answer:
The national park is a natural habitat that is notified by the state government to be constituted as a National Park due to its ecological, faunal, floral, geomorphological, or zoological association of importance. No human activity is permitted inside the national park except the activities permitted by the Chief Wildlife Warden of the state.

Question 20.
Name any two species that are extinct due to human activities.
Answer:

• Dodo of Mauritius
• Steller’s cow of Russia

Question 21.
Define in-situ conservation.
Answer:
Conservation of animals in their natural habitat is called in-site conservation.
E.g., National parks.

Question 22.
What is the goal of “Project Tiger”?
Answer:
The project ensures a viable population of Bengal tigers in their natural habitats, protecting them from extinction and preserving areas of biological importance as a natural heritage.

Question 23.
Give the number of national parks in India. Name any two of them in Tamil Nadu.
Answer:
India has 104 National Parks. Guindy National Park (Chennai) and Mudumalai National Park (Nilgiris) are located in Tamil Nadu.

Question 24.
State the role of Biosphere Reserve.
Answer:
Biosphere Reserves are designated to deal with the conservation of biodiversity, economic and social development, and maintenance of associated cultural values.

Question 25.
Name few endangered species protected in Arignar Anna Zoological Park.
Answer:
Royal Bengal Tiger, Lion Tailed Macaque, Nilgiri Langur, and Gray Wolf.

Question 26.
Give the names of two methods of in-situ conservation.
Answer:

• Wildlife Sanctuaries
• Biosphere reserve

12th Bio Zoology Guide Biodiversity and its Conservation Three Marks Questions and Answers

Question 27.
Point out the biosphere reserves in Tamil Nadu.
Answer:

• Nilgiris (Tamil Nadu – Kerala)
• Agasthya malai (Tamil Nadu – Kerala – Karnataka)
• Gulf of Mannar (Tamil Nadu)

Question 28.
Write a note on Sacred Groves.
Answer:
A sacred grove or sacred woods are any groves of trees that are of special religious importance to a particular culture. Sacred groves feature in various cultures throughout the world.

Question 29.
What is ex-situ conservation?
Answer:
Ex-situ conservation of selected rare plants/ animals in places outside their natural homes. It includes offsite collections and gene banks.

Question 30.
Why Red list is prepared periodically?
Answer:
The purpose of preparation of Red List are:

• To create awareness on the degree of threat to biodiversity
• Identification and documentation of species at high risk of extinction
• Provide global index on declining biodiversity
• Preparing conservation priorities and help in conservation of action
• Information on international agreements on the conservation of biological diversity

Question 31.
Name the types of extinctions.
Answer:

• Natural Extinction
• Mass Extinction
• Anthropogenic Extinction

Question 32.
Point out the human activities that threaten biodiversity.
Answer:
Direct and indirect human activities have a detrimental effect on biodiversity. Direct human, activities like change in local land use, species introduction or removal, harvesting, pollution, and climate change contribute a greater pressure on the loss of biodiversity. Indirect human drivers include demographic, economic, technological, cultural, and religious factors.

Question 33.
Extinction of Dodo bird led to the danger of Calvaria tree – Justify,
Answer:
Another example for co-extinction is the connection between Calvaria tree and the extinct bird of Mauritius Island, the Dodo. The Calvaria tree is dependent on the Dodo bird for completion of its life cycle. The mutualistic association is that the tough horny endocarp of the seeds of Calvaria tree are made permeable by the actions of the large stones in birds gizzard and digestive juices thereby facilitating easier germination. The extinction of the Dodo bird led to the imminent danger of the Calvaria tree coextinction.

Question 34.
Give an account on slash and burn agriculture.
Answer:
In shifting cultivation, plots of natural tree vegetation are burnt away and the cleared patches are farmed for 2-3 seasons, after which their fertility reduces to a point where crop production is no longer profitable. The farmer then abandons this patch and cuts down a new patch of forest trees elsewhere for crop production.

This system is practiced in the north-eastern regions of India. When vast areas are cleared and burnt, it results in loss of forest cover, pollution and discharge of CO₂ which in turn attributes to a loss of habitat and climate change which has an impact on the faunal diversity of that regions.

Question 35.
Impact of Industrialization on Biodiversity – Comment.
Answer:
Industrialization is a major contributor to climate change and a major threat to biodiversity. Energy drives our industries, which is provided by burning of fossil fuels. This increases the emission of CO₂, a GHG, leading to climate change. Due to large scale deforestation, the emitted CO₂ cannot be absorbed fully, and its concentration in the air increases.

Climate change increases land and ocean temperature changes precipitation patterns and raises the sea level. This, in turn, results in melting of glaciers, water inundation, less predictability of weather patterns, extreme weather conditions, outbreak of squalor diseases, migration of animals and loss of trees in forest. Thus, climate change is an imminent danger to the existing biodiversity.

Question 36.
What are exotic species? Explain with example.
Answer:
Exotic species are organisms often introduced unintentionally or deliberately for commercial purpose, as biological control agents and other uses. They often become invasive and drive away the local species and is considered as the second major cause for extinction of species. Exotic species have proved harmful to both aquatic and terrestrial ecosystems.
Tilapia fish (Jilabi kendai) (Oreochromis mosambicus) introduced from east coast of South Africa in 1952 for its high productivity into Kerala’s inland waters, became invasive, due to which the native species such as Puntius dubius and Labeo kontius face local extinction.

Question 37.
Write a brief note on Habitat fragmentation.
Answer:
Habitat fragmentation is the process where a large, continuous area of habitat is both, reduced in area and divided into two or more fragments. Fragmentation of habitats like forest land into croplands, orchard lands, plantations, urban areas, industrial estates, transport, and transit systems has resulted in the destruction of complex interactions amongst species, (food chain and webs) destruction of species in the cleared regions, annihilation of species restricted to these habitats (endemic) and decreased biodiversity in the habitat fragments.

Animals requiring large territories such as mammals and birds are severely affected. The elephant corridors and migratory routes are highly vulnerable. The dwindling of many well-known birds (sparrows) and animals can be attributed to this.

Question 38.
Write a note on the biogeographic area – the Gangetic plains.
Answer:
Gangetic Plains: These plains are relatively homogeneously defined by the Ganges river system and occupy about 11% of the country’s landmass. This region is very fertile and extends up to the Himalayan foothills. Fauna includes rhinoceros, elephant, buffalo, swamp deer, hog-deer.

Question 39.
Compare Alpha diversity with Beta diversity.
Answer:
(i) Alpha diversity: It is measured by counting the number of taxa (usually species) within a particular area, community or ecosystem.
(ii) Beta diversity: It is species diversity between two adjacent ecosystems and is obtaining by comparing the number of species unique to each of the ecosystem.

Question 40.
What is species diversity?
Answer:
Species diversity refers to the variety in number and richness of the species in any habitat. The number of species per unit area at a specific time is called species richness, which denotes the measure of species diversity. The Western Ghats have greater amphibian species diversity than the Eastern Ghats. The more the number of species in an area the more is the species richness. The three indices of diversity are – Alpha, Beta and Gamma diversity.

Question 41.
State the principle of Stockholm declaration – 1972.
Answer:
The natural resources of the Earth, including air, water, land, flora, and fauna of natural ecosystems must be safeguarded for the benefit of the present and future generations through careful planning and management, as appropriate – Principle of the Stockholm Declaration. 1972.

12th Bio Zoology Guide Biodiversity and its Conservation Five Marks Questions and Answers

Question 42.
Give an account on genetic diversity and community diversity.
Answer:
Genetic diversity refers to the differences in genetic make-up (number and types of genes) between distinct species and to the genetic variation within a single species; also covers genetic variation between distinct populations of the same species. Genetic diversity can be measured using a variety of molecular techniques. India has more than 50,000 genetic variants of Paddy and 1000 variants of Mango. Variation of genes of a species increases with diversity in size and habitat. It results in the formation of different races, varieties and subspecies. Rauwolfia vomitaria, a medicinal plant growing in different ranges of the Himalayas shows differences in the potency and concentration of the active ingredient reserpine due to genetic diversity. Genetic diversity helps in developing adaptations to changing environmental conditions.

Community/Ecosystem diversity is the variety of habitats, biotic communities, and ecological processes in the biosphere. It is the diversity at ecosystem level due to diversity of niches, trophic levels and ecological processes like nutrient cycles, food webs, energy flow, and several biotic interactions. India with its alpine meadows, rain forests, mangroves, coral reefs, grasslands, and deserts has one of the greatest ecosystem diversity on Earth.

Question 43.
Why Tropical regions are rich in biodiversity?
Answer:
The reasons for the richness of biodiversity in the Tropics are:

• Warm tropical regions between the tropic of Cancer and Capricorn on either side of the equator possess congenial habitats for living organisms.
• Environmental conditions of the tropics are favourable not only for speciation but also for supporting both variety and number of organisms.
• The temperatures vary between 25°C to 35°C, a range in which most metabolic activities of living organisms occur with ease and efficiency.
• The average rainfall is often more than 200 mm per year.
• Climate, seasons, temperature, humidity, photoperiods are more or less stable and encourage both variety and numbers.
• Rich resource and nutrient availability.

Question 44.
What is the significance of the slope of regression in a species-area relationship?
Answer:
German Naturalist and Geographer Alexander von Humboldt explored the wilderness of South American jungles and found that within a region the species richness increased with the increasing area but upto a certain limit. The relationship between species richness and area for a wide variety of taxa (angiosperm plants, birds, bats and freshwater fishes) turned out to be the rectangular hyperbola. On a logarithmic scale, the relationship is a straight line described by the equation.
log S = log C + Z log A
where S = Species richness
A = Area
Z = Slope of the line
(regression coefficient)
C = Y-intercept

Regression coefficient Z generally has a value of 0.1-0.2 regardless of taxonomic group or region. However, in case of the species-area relationship in very large areas like entire continents, the slope of the line appears to be much steeper (Z-value in the range of 0.6-1.2). For example, in the case of the fruit eating (frugivorous) birds and mammals in the tropical forests of different continents, the slope is found to be a steeper line.

Question 45.
Point out any 5 functional attributions of biodiversity.
Answer:
The major functional attributes are:

• continuity of nutrient cycles or biogeochemical cycles (N₂, C, H₂O, P, S cycles)
• soil formation, conditioning or maintenance of soil health (fertility) by soil microbial diversity along with the different trophic members
• increases ecosystem productivity and provide food resources
• act as water traps, filters, water flow regulators, and water purifiers (forest cover and vegetation)
• climate stability (forests are essential for rainfall, temperature regulation, C02 absorption, which in turn regulate the density and type of vegetation)
• forest resource management and sustainable development

Question 46.
Explain in detail about various types of extinctions.
Answer:
There are three types of Extinctions
i. Natural extinction: It is a slow process of replacement of existing species with better adapted species due to changes in environmental conditions, evolutionary changes, predators and diseases. A small population can get extinct sooner than a large population due to inbreeding depression (less adaptivity and variation)

ii. Mass extinction: The Earth has experienced quite a few mass extinctions due to environmental catastrophes. Amass extinction occurred about 225 million years ago during the Permian, where 90% of shallow-water marine invertebrates disappeared.

iii. Anthropogenic extinctions: These are abetted by human activities like hunting, habitat destruction, overexploitation, urbanization and industrialization. Some examples of extinctions are Dodo of Mauritius and Steller’s sea cow of Russia. Amphibians seem to be at higher risk of extinction because of habitat destruction. The most serious aspect of the loss of biodiversity is the extinction of species. The unique information contained in its genetic material (DNA) and the niche it possesses are lost forever.

Question 47.
Give a comparative account on ex-situ conservation.
Answer:
Ex-Situ Conservation: It is conservation of selected rare plants/ animals in places outside their natural homes. It includes offsite collections and gene banks.

Offsite Collections: They are live collections of wild and domesticated species in Botanical gardens, Zoological parks, Wildlife safari parks, Arborata (gardens with trees and shrubs). The organisms are well maintained for captive breeding programmes. As a result, many animals which have become extinct in the world continue to be maintained in Zoological Parks. As the number increases in captive breeding, the individuals are selectively released in the wild. In this way, the Indian crocodile and Gangetic dolphin have been saved from extinction.

Gene Banks: Gene banks are a type of biorepository which preserves genetic materials. Seeds of different genetic strains of commercially important plants can be stored for long periods in seed banks, gametes of threatened species can be preserved in viable and fertile condition for long periods using cryopreservation techniques. However, it is not economically feasible to conserve all biological wealth and all the ecosystems. The number of species required to be saved from extinction far exceeds the conservation efforts.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
What will be the result, if the relationship between species richness and area for wide variety of taxa are plotted on a long scale?
Answer:
A rectangular hyperbola.

Question 2.
What may be the reasons for the entry of wild .lives into the agriculatural lands or towns?
Answer:

• Habitat lose / Habitat fragmentation
• Lack of food or water source

Question 3.
When does a species is categorized as endangered?
Answer:
A species that has been categorized as very likely to become extinct is an endangered species.

Question 4.
Give any two examples of anthropogenic extinction.
Answer:
Dodo of Mauritius Steller’s cow of Russia

Question 5.
Mention any two species that had become extinct very recently.
Answer:

• George, the tree snail (Achatinella apexfulva)
• Sudan – Northern white rhinoceros (Ceratotherium simum)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Zoology Guide Pdf Chapter 4 Principles of Inheritance and Variation Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation

12th Bio Zoology Guide Principles of Inheritance and Variation Text Book Back Questions and Answers

Question 1.
Haemophilia is more common in males because it is a………………
(a) Recessive character carried by Y-chromosome
(b) Dominant character carried by Y-chromosome
(c) Dominant trait carried by X-chromosome
(d) Recessive trait carried by X-chromosome
Answer:
(d) Recessive trait carried by X-chromosome

Question 2.
ABO blood group in man is controlled by …………………….
(a) Multiple alleles
(b) Lethal genes
(c) Sex linked genes
(d) Y-linked genes
Answer:
(a) Multiple alleles

Question 3.
Three children of a family have blood groups A, AB and B. What could be the genotypes of their parents?
(a) I^AI^B and ii
(b) I^A1^O and I^BI^O
(c) I^B I^B and I^AI^A
(d) I^AI^A and ii
Answer:
(b) I^A1^O and I^BI^O

Question 4.
Which of the following is not correct?
(a) Three or more alleles of a trait in the population are called multiple alleles.
(b) A normal gene undergoes mutations to form many alleles
(c) Multiple alleles map at different loci of a chromosome
(d) A diploid organism has only two alleles out of many in the population
Answer:
(c) Multiple alleles map at different loci of a chromosome

Question 5.
Which of the following phenotypes in the progeny are possible from the parental combination
(a)AandB only
(b) A,B andAB only
(c) AB only
(d) A,B,AB and O
Answer:
(d) A,B,AB and O

Question 6.
Which of the following phenotypes is not possible in the progeny of the parental genotypic combination I^AI^O x l^Al^B ?
(a) AB
(b) O
(c) A
(d) B
Answer:
(b) O

Question 7.
Which of the following is true about Rh factor in the offspring of a parental combination DdXDd (both Rh positive)?
(a) All will be Rh positive
(b) Half will be Rh positive
(c) About 3/4 will be Rh negative
(d) About one fourth will be Rh negative
Answer:
(d) About one fourth will be Rh negative

Question 8.
What can be the blood group of offspring when both parents have AB blood group?
(a) AB only
(b) A, B and AB
(c) A, B, AB and O
(d) A and B only
Answer:
(b) A, B and AB

Question 9.
If the childs blood group is ‘O’ and fathers blood group is ‘A’ and mother’s blood group is ‘B’ the genotype of the parents will be …………….
(a) I^AI^A and I^BI^O
(b) I^AI^O and I^BI^O
(c) I^AI^O and I^OI^O
(d) I^OI^O and I^BI^B
Answer:
(b) I^AI^O and I^BI^O

Question 10.
XO type of sex determination and XY type of sex determination are examples of …………………
(a) Male heterogamety
(b) Female heterogamety
(c) Male homogamety
(d) Both (b) and (c)
Answer:
(a) Male heterogamety

Question 11.
In an accident there is great loss of blood and there is no time to analyse the blood group Question which blood can be safely transferred?
(a) ‘O’ and Rh negative
(b) ‘O’ and Rh positive
(c) ‘B’ and Rh negative
(d) ‘AB’ and Rh positive
Answer:
(a) ‘O’ and Rh negative

Question 12.
Father of a child is colourblind and mother is carrier for colourblindness, the probability of the child being colourblind is ………………
(a) 25%
(b) 50%
(c) 100%
(d) 75%
Answer:
(b) 50%

Question 13.
A marriage between a colourblind man and a normal woman produces
(a) All carrier daughters and normal sons
(b) 50% carrier daughters, 50% normal daughters
(c) 50% colourblind sons, 50% normal sons
(d) All carrier offsprings
Answer:
(a) All carrier daughters and normal sons

Question 14.
Mangolism is a genetic disorder which is caused by the presence of an extra chromosome number.
(a) 20
(b) 21
(c) 4
(d) 23
Answer:
(b) 21

Question 15.
Klinefelters’ syndrome is characterized by a karyotype of………………
(a) XYY
(b) XO
(c) XXX
(d) XXY
Answer:
(d) XXY

Question 16.
Females with Turners’syndrome have………………
(a) Small uterus
(b) Rudimentary ovaries
(c) Underdeveloped breasts
(d) All of these
Answer:
(d) All of these

Question 17.
Pataus’ syndrome is also referred to as………………
(a) 13-Trisomy
(b) 18-Trisomy
(c) 21-Trisomy
(d) None of these
Answer:
(a) 13-Trisomy

Question 18.
Who is the founder of Modem Eugenics movement?
(a) Mendel
(b) Darwin
(c) Fransis Galton
(d) Karl pearson
Answer:
(c) Fransis Galton

Question 19.
Improvement of human race by encouraging the healthy persons to marry early and produce large number of children is called………………
(a) Positive eugenics
(b) Negative eugenics
(c) Positive euthenics
(d) Positive euphenics
Answer:
(a) Positive eugenics

Question 20.
The ……………… deals with the control of several inherited human diseases especially inborn errors of metabolism.
(a) Euphenics
(b) Eugenics
(c) Euthenics
(d) All of these
Answer:
(a) Euphenics

Question 21.
“Universal Donor” and “Universal Recipients” blood group are ……………… and ………………  respectively.
(a) AB, O
(b) O, AB
(c) A, B
(d) B, A
Answer:
(b) O, AB

Question 22.
ZW-ZZ system of sex determination occurs in………………
(a) Fishes
(b) Reptiles
(c) Birds
(d) All of these
Answer:
(d) All of these

Question 23.
A co-dominant blood group is
(a) A
(b) AB
(c) B
(d) O
Answer:
(b) AB

Question 24.
Which of the following is incorrect regarding ZW-ZZ type of sex determination?
(a) It occurs in birds and some reptiles
(b) Females are homogametic and males are heterogametic
(e) Male produce two types of gametes
(d) It occurs in gypsy moth
Answer:
(b) Females are homogametic and males are heterogametic

Question 25.
What is haplodiploidy?
Answer:
In haplodiploidy, the sex of the offspring is determined by the number of sets of chromosomes it receives. Fertilized eggs develop into females (Queen or Worker) and unfertilized eggs develop into males (drones) by parthenogenesis. It means that the males have half the number of chromosomes (haploid) and the females have double the number (diploid).

Question 26.
Distinguish between heterogametic and homogametic sex determination systems.
Answer:
Heterogametic Sex :

1. Organisms producing two different types of gametes.
2. Example: Human male.
   Sperm with X chromosome
   Sperm with Y chromosome

Homogametic Sex :

1. Organisms producing only one type of gametes.
2. Example: Human female.
   Every egg produced contain X chromosomes.

Question 27.
What is Lyonisation?
Answer:
Lyonisation is a process of inactivation of one of the X chromosomes in some females.

Question 28.
What is criss-cross inheritance?
Answer:
Inheritance of genes from a male parent to female child and then to male grandchild or female parent to male child and then to female grandchild. E.g., X-linked gene inheritance.

Question 29.
Why are sex-linked recessive characters more common in male human beings?
Answer:
Sex linked inherited traits are more common in males than females because, males are hemizygous and therefore express the trait when they inherit one mutant allele.

Question 30.
What are holandric genes?
Answer:
The genes present in the differential region of the Y chromosome are called Y- linked or holandric genes. The Y linked genes have no corresponding allele in X chromosome.

Question 31.
Mention the symptoms of Phenylketonuria.
Answer:
Severe mental retardation, light pigmentation of skin and hair. Phenylpyruvic acid is excreted in the urine.

Question 32.
Mention the symptoms of Down’s syndrome.
Answer:
Severe mental retardation, defective development of the central nervous system, increased separation between the eyes, flattened nose, ears are malformed, mouth is constantly open and the tongue protrudes.

Question 33.
Differentiate Intersexes from Supersexes.
Answer:Intersexes:
Intersexes refers to the individuals having the characteristics of both female and male sexes and their sexual anatomy does not seem to fit the typical definition of male or female.

Supersexes:
Supersexes ar formed as a result of an abnormal combination of sex chromosomes.
Example: Super males in humans human beings have 44+XYY chromosomes.

Question 34.
Explain the genetic basis of ABO blood grouping in man.
Answer:
Multiple allele inheritance of ABO blood groups
Blood differs chemically from person to person. When two different incompatible blood types are mixed, agglutination (clumping together) of erythrocytes (RBC) occurs. The basis of these chemical differences is due to the presence of antigens (surface antigens) on the membrane of RBC and epithelial cells. Karl Landsteiner discovered two kinds of antigens called antigen ‘A’ and antigen ‘B’ on the surface of RBC’s of human blood. Based on the presence or absence of these antigens three kinds of blood groups, type ‘A’, type ‘B’, and type ‘O’ (universal donor) were recognized. The fourth and the rarest blood group ‘AB’ (universal recipient) was discovered in 1902 by two of Landsteiner’s students Von De Castelle and Sturli.

Bernstein in 1925 discovered that the inheritance of different blood groups in human beings is determined by a number of multiple allelic series. The three autosomal alleles located on chromosome 9 are concerned with the determination of blood group in any person. The gene controlling blood type has been labeled as ‘L’ (after the name of the discoverer, Landsteiner) or I (from isoagglutination). The I gene exists in three allelic forms, I^A, I^B and I^O. I^A specifies A antigen. I^B allele determines B antigen and I^O allele specifies no antigen. Individuals who possess these antigens in their fluids such as the saliva are called secretors.

Each allele (I^A and I^B) produces a transferase enzyme. I^A allele produces N-acetyl galactose transferase and can add N-acetyl galactosamine (NAG) and I^B allele encodes for the enzyme galactose transferase that adds galactose to the precursor (i.e. H substances). In the case of I^O/I^O allele no terminal transferase enzyme is produced and therefore called “null” allele and hence cannot add NAG or galactose to the precursor.

From the phenotypic combinations it is evident that the alleles I^A and I^B are dominant to 1°, but co-dominant to each other (I^A = I^B). Their dominance hierarchy can be given as (I^A=I^B> 1^O). A child receives one of three alleles from each parent, giving rise to six possible genotypes and four possible blood types (phenotypes). The genotypes are I^AI^A , I^AI^O, I^BI^B, I^BI^O, I^AI^B and I^OI^O.

Question 35.
How is sex determined in human
Answer:
Genes determining sex in human beings are located on two sex chromosomes, called allosomes. In mammals, sex determination is associated with chromosomal differences between the two sexes, typically XX females and XY males. 23 pairs of human chromosomes include 22 pairs of autosomes (44A) and one pair of sex chromosomes (XX or XY). Females are homogametic producing only one type of gametes (egg), each containing one X chromosome while the males are heterogametic producing two types of sperms with X and Y chromosomes. An independently evolved XX: XY system of sex chromosomes also exist in Drosophila.

Question 36.
Explain male heterogamety.
Answer:
Male heterogamety (XY males) is a type of sex determination in which males produce two different types of gametes. For example, human males produce two kinds of sperms that is sperm with X-chromosome and sperms with Y-chromosome.

Question 37.
Brief about female heterogamety.
Answer:
Female heterogamety (ZO females) refers to the condition, where female produces two types of egg cells. Some with Z chromosome and some without Z chromosome.

Question 38.
Give an account of genetic control of Rh factor?
Answer:
Genetic control of Rh factor
Fisher and Race hypothesis: Rh factor involves three different pairs of alleles located on three different closely linked loci on the chromosome pair. This system is more commonly in use today, and uses the ‘Cde’ nomenclature.
In the given figure, three pairs of Rh alleles (Cc, Dd and Ee) occur at 3 different loci on homologous chromosome pair-1. The possible genotypes will be one C or c, one D or d, one E or e from each chromosome. For e.g. CDE/cde; CdE/cDe; cde/cde; CDe/CdE etc. All genotypes carrying a dominant ‘D’ allele will produce Rh+positive phenotype and double recessive genotype ‘dd’ will give rise to Rh negative phenotype.

Wiener Hypothesis
Wiener proposed the existence of eight alleles (R¹, R², R⁰, R^z, r, r¹, r¹¹, r^y) at a single Rh locus. All genotypes carrying a dominant ‘R allele’ (R¹, R² ,R⁰ ,R^z) will produce ‘Rh-positive’ ^phenotype and double recessive genotypes (rr, rr¹, rr¹¹, rr^y) will give rise to Rh-negative phenotype.

Question 39.
Explain the mode of sex determination in honeybees.
Answer:
In hymenopteran insects such as honeybees, ants and wasps, a mechanism of sex determination called haplodiploidy mechanism of sex determination is common. In this system, the sex of the offspring is determined by the number of sets of chromosomes it receives. Fertilized eggs develop into females (Queen or Worker) and unfertilized eggs develop into males (drones) by parthenogenesis. It means that the males have half the number of chromosomes (haploid) and the females have double the number (diploid), hence the name haplodiploid for this system of sex determination.

This mode of sex determination facilitates the evolution of sociality in which only one diploid female becomes a queen and lays the eggs for the colony. All other females which are diploid having developed from fertilized eggs help to raise the queen’s eggs and so contribute to the queen’s reproductive success and indirectly to their own, a phenomenon known as Kin Selection. The queen constructs their social environment by releasing a hormone that suppresses fertility of the workers.

Question 40.
Discuss the genic balance mechanism of sex determination with reference to Drosophila?
Answer:
XX-XY type (Lygaeus Type) sex determination is seen in Drosophila. The females are homogametic with XX chromosomes, while the males are heterogametic with X and Y chromosomes. Homogametic females produce only one kind of egg, each with one X chromosome, while the heterogametic males produce two kinds of sperms some with X chromosome and some with Y chromosome.

The sex of the embryo depends on the fertilizing sperm. An egg fertilized by an ‘X’ bearing sperm produces a female, if fertilized by a ‘Y’ bearing sperm, a male is produced.

Question 41.
What are the applications of Karyotyping?
Answer:

• Karyotyping helps in gender identification.
• It is used to detect the chromosomal aberrations like deletion, duplication, translocation, non-disjunction of chromosomes.
• It helps to identify the abnormalities of chromosomes like aneuploidy.
• It is also used in predicting the evolutionary relationships between species.
• Genetic diseases in human beings can be detected by this technique.

Question 42.
Explain the inheritance of sex linked characters in human being.
Answer:
Haemophilia is commonly known as bleeder’s disease, which is more common in men than women. This hereditary disease was first reported by John Cotto in 1803. Haemophilia is caused by a recessive X-linked gene. A person with a recessive gene for haemophilia lacks a normal clotting substance (thromboplastin) in blood, hence minor injuries cause continuous ’bleeding, leading to death. The females are carriers of the disease and would transmit the disease to 50% of their sons even if the male parent is normal. Haemophilia follows the characteristic criss-cross pattern of inheritaitce.

Question 43.
What is extra chromosomal inheritance? Explain with an example.
Answer:
The cytoplasmic extra nuclear genes have a characteristic pattern of inheritance which does not resemble genes of nuclear chromosomes and are known as Extrachromosomal/ Cytoplasmic inheritance.

Question 44.
Comment on the methods of Eugenics.
Answer:
Eugenics refers to the study of the possibility of improving the qualities of human population.

Methods of Eugenics:

• Sex-education in school and public forums.
• Promoting the uses of contraception.
• Compulsory sterilization for mentally retarded and criminals.
• Egg donation.
• Artificial insemination by donors.
• Prenatal diagnosis of genetic disorders and performing MTP
• Gene therapy
• Cloning
• Egg/sperm donation of healthy individuals.

12th Bio Zoology Guide Principles of Inheritance and Variation Additional Important Questions and Answers

12th Bio Zoology Guide Principles of Inheritance and Variation One Mark Questions and Answers

Question 1
If a colorblind female marries a normal male, their sons will be ………………
(a) All normal visioned
(b) All color blinded
(c) One half normal visioned other half colorblind
(d) Three fourth colorblind one fourth normal
Answer:
(c) One half normal visioned other half colorblind

Question 2
Excess hair growth on pinna is a feature noticed only in males because ……………
(a) Males produce more testosterone
(b) gene responsible for the character is located in Y-chromosome
(c) Estrogen suppresses the character in females
(d) females act only as a carriers for this character
Answer:
(b) gene responsible for the character is located in Y-chromosome

Question 3.
ABO blood group is a classical example for ………………..
(a) Multiple allelism
(b) Pleotropism
(c) Incomplete dominance
(d) Polygenic mechanism
Answer:
(a) Multiple allelism

Question 4
Unit of heredity is ……………….
(a) allele
(b) allelomorph
(c) trait
(d) gene
Answer:
(d) gene

Question 5.
Identify the proper dominance hierarchy.
(a) I^A = I^O > I^B
(b) I^A = I^B > O
(b) I^A = I^B > O
(d) I^B = I^A > O
Answer:

Question 6
Haemophilia is more common in human males than human females. The reason is due to
(а) X-linked dominant gene
(b) X-linked recessive gene
(c) Y-linked recessive gene
(d) Allosomal abnormality
Answer:
(b) X-linked recessive gene

Question 7.
Identify the correct statement.
(a) Homozygous sex chromosome (XX) produce males in Drosophila
(b) Homozygous sex chromosome (ZZ) determine female sex in birds
(c) Heterozygous sex chromosome (XO) determine male sex in grasshopper
(d) Heterozygous sex chromosome (ZW) determine male sex in gypsy moth
Answer:
(c) Heterozygous sex chromosome (XO) determine male sex in grasshopper

Question
Which blood group doesnot possess antibodies?
(a) I^AI^B
(b) I^OI^O
(c) I^AO
(d) I^BI^B
Answer:
(a) I^AI^B

Question 9.
Assertion (A): On diagnosis, Ramu is reported to have underdeveloped testis and gynecomastia.
Reason (R): His karyotype reveals XXY condition.
(а) A is right but R is wrong
(b) R explains A
(c) Both A and R are wrong
(d) Both and R are right but R is not the correct explanation of A
Answer:
(b) R explains A

Question 10.
Pick out the odd man.
(a) Klinefelter’s syndrome
(b) Turner’s syndrome
(c) Huntington’s chorea
(d) 13-Trisomy
Answer:
(c) Huntington’s chorea

Question 11.
Pick the odd one out regarding Mendelian disorder.
(a) Thalassemia
(b) phenylketonuria
(c) Albinism
(d) Huntington’s chorea
Answer:
(d) Huntington’s chorea

Question 12.
Match the following:

(a) A – iv, B – ii, C – i, D – iii
(b)A – ii, B – iv, C – iii, D – i
(a)A – iii, B – i, C – ii, D – iv
(c) A – i, B – iv, C – iii, D – iii
Answer:
(a) A – iv, B – ii, C – i, D – iii

Question 13.
Identify the proper ratio of normal visioned individuals against colorblind individuals, if colorblind carrier female marries a normal male.
(a) 1 : 1
(b) 3:1
(c) 1 : 3
(d) All four are normal visioned
Answer:
(c) 1 : 3

Question 14.
Pick out the correct statement.
(i) Karyotyping helps in gender identification
(ii) Holandric genes are located on X-chromosome
(iii) Trisomy-21 is an allosomal abnormality
(iv) Cooley’s anemia is an autosomal recessive disorder
(a) i, iii, iv are correct
(c) i and iv are correct
Answer:
(c) i and iv are correct

Question 15.
DOPA stands for ……………….
(a) 3,4- dihydroxy phenylacetate
(b) 3, 4 – dihydroxy phenylalanine
(c) 3,4- dihydroxy phenyl aspartate
(d) 3, 4- dihydroxy phenol aldehyde
Answer:
(b) 3,4 – dihydroxy phenylalanine

Question 16.
The type of antibody generated against Rh antigen is ….
(a)IgE
(b) IgG
(c) IgA
(d) IgB
Answer:
(b) IgG

Question 17.
Which of the following symbol is used in the pedigree analysis to represent unspecified sex?

Answer:

Question 18:
A colorblind man marries a woman with normal sight who has no history of color blindness in her family. What is the probability of their grandson being colorblind?
(a) 1/4
(b) 3/4
(c) 2/4
(d) 4/4
Answer:
(a) 1/4

Question 19.
Multiple alleles are located……………………
(a) at different loci on homologous chromosome
(b) at same locus on homologous chromosome
(c) at different loci on non-homologous chromosome
(d) at different chromosomes
Answer:
(b) at same locus on homologous chromosome

Question 20.
Identify the incorrect statement regarding haplodiploidy.
(g) Haplodiploidy is noticed in honeybees and drosophila
(b) Unfertilized eggs develop into drones
(c) Fertilized eggs develop into queen and worker bees
(d) Males have half the total chromosomal number
Answer:
(a) Haplodiploidy is noticed in honeybees and drosophila

Question 21.
I^A and I^B genes of ABO blood group are
(a) Co-dominant
(b) Pleotropic
(c) Dominant and recessive
(d) Epistatic
Answer:
(a) Co-dominant

Question 22.
Which one of the following crosses show 3 : 1 ratio of normal visioned versus carrier blind?
(a) X^CX^C x X⁺Y
(b) X⁺ X^C x X^C Y^–
(c) X⁺X^C x X⁺Y^–
(d) X⁺X⁺ x X^CY^–
Answer:
(c) X⁺X^C x X⁺Y^–

12th Bio Zoology Principles of Inheritance and Variation Two Marks Questions and Answers

Question 1.
Define multiple allelism.
Answer:
When three or more alleles of a gene that control a particular trait occupy the same locus on the homologous chromosome of an organism, they are called multiple alleles and their inheritance is called multiple allelism.

Question 2.
Name the discoverers of antigen A, B and AB.
Answer:
Antigens A and Antigen B was discovered by Karl Landsteiner. Antigen AB was discovered by Von De Castelle and Sturli.

Question 3.
What happens if type A blood is injected to a person having B blood group? Explain the reason.
Answer:
When two different incompatible blood types are mixed, agglutination (clumping together) of erythrocytes (RBC) occurs. The basis of these chemical differences is due to the presence of antigens (surface antigens) on the membrane of RBC and epithelial cells.

Question 4.
State the allelic forms of I gene and mention its chromosomal location.
Answer:
The I gene exists in three forms: I^A, I^B and I^O. The alleles are located on chromosome 9.

Question 5.
Write the possible genotypes for a person having a B-blood group.
Answer:
The possible genotypes of a B-blood group person are I^BI^B or I^BI^O.

Question 6.
State Wiener Hypothesis on Rh-factor.
Wiener proposed the existence of eight alleles (R¹, R², R⁰, R^z, r, r¹, r¹¹, r^y) at a single Rh locus. All genotypes carrying a dominant ‘R allele’ (R¹, R² ,R⁰ ,R^z) will produce ‘Rh-positive’ phenotype and double recessive genotypes (rr, rr¹, rr¹¹, rr^y) will give rise to Rh-negative phenotype.

Question 7.
Distinguish between homogametic and heterogametic condition with example.
Answer:
Homogametic organism:

1. Organism producing only one type of gametes.
2. e.g. Human female (Only X)

Heterogametic organism :

1. Organism producing two different types of gametes.
2. e.g. Human Male (X and Y)

Question 8.
Name any four organism expressing ZW-ZZ type of sex determination.
Answer:
Gypsy moth, fishes, reptiles and birds.

Question 9.
Expand (a) SRY (b) TDF
Answer:
SRY – Sex Determining region Y
TDF – Testes Determining Factor

Question 10.
Define Barr body.
Answer:
In 1949, Barr and Bertram first observed a condensed body in the nerve cells of female cat which was absent in the male. This condensed body was called sex chromatin by them and was later referred as Barr body.

Question 11.
Based on Lyon’s hypothesis, mention the number of Barr bodies in XXY males, XO females.
Answer:
XXY males – One Barr body.
XO females – No Barr body.

Question 12.
State Lyon’s hypothesis.
Answer:
Lyon’s hypothesis states that in mammals the necessary dosage compensation is accomplished by the inactivation of one of the X chromosomes in females so that both males and females have only one functional X chromosome per cell.
Mary Lyon suggested that Barr bodies represented an inactive chromosome, which in females becomes tightly coiled into a heterochromatin, a condensed and visible form of chromatin (Lyon’s hypothesis). The number of Barr bodies observed in cell was one less than the number of X-Chromosome. XO females have no Barr body, whereas XXY males have one Barr body.

Question 13.
Mention few X-linked inherited diseases.
Answer:
Red-green colour blindness or daltonism, haemophilia and Duchenne’s muscular dystrophy.

Question 14.
Define Karyotyping.
Answer:
Karyotyping is a technique through which a complete set of chromosomes is separated from a cell and the chromosomes are arranged in pairs. An idiogram refers to a diagrammatic representation of chromosomes.

Question 15.
Explain the inheritance pattern of Y-linked genes for example.
Answer:
Genes in the non-homologous region of the Y-chromosome are inherited directly from male to male. In humans, the Y-linked or holandric genes for hypertrichosis (excessive development of hairs on pinna of the ear) are transmitted directly from father to son, because males inherit the Y chromosome from the father. Female inherits only X chromosome from the father and are not affected.

Question 16.
Observe the symbol used in pedigree analysis and give the proper terms they represent.
Answer:

Question 17.
Write a brief note on pedigree analysis.
Answer:
Pedigree is a “family tree”, drawn with standard genetic symbols, showing the inheritance pathway for specific phenotypic characters. Pedigree analysis is the study of traits as they have appeared in a given family line for several past generations.

Question 18.
What do you mean by ‘Mendelian disorder’.
Answer:
Alteration or mutation in a single gene causes Mendelian disorders. These disorders are transmitted to the offsprings on the same line as the Mendelian pattern of inheritance. E.g., Thalassemia.

Question 19.
Name any four Mendelian disorders.
Answer:
(a) Thalassemia (b) Albinism (c) sickle cell anaemia (d) Huntington’s chorea

Question 20.
What is the phenotype of (a) I^AI^O (b) I^OI^O
Answer:
(a) I^AI^O – A blood group person
(b) I^OI^O – O blood group person

Question 21.
On which chromosomes does HBA1 gene and HBB genes are located?
Answer:
HBA1 gene is located on chromosome 16.
HBB gene is located on chromosome 11.

Question 22.
Complete the equation.

Answer:
(a) A = Phenylalanine hydroxylase
(b) B = Tyrosinase

Question 23.
Write a note on Huntington’s chorea.
Answer:
Huntington’s chorea is inherited as an autosomal dominant lethal gene in man. It is characterized by involuntary jerking of the body and progressive degeneration of the nervous system, accompanied by gradual mental and physical deterioration. The patients with this disease usually die between the age of 35 and 40.

Question 24.
Comment on Trisomy-21.
Answer:
Trisomic condition of chromosome – 21 results in Down’s syndrome. It is characterized by severe mental retardation, defective development of the central nervous system, increased separation between the eyes, flattened nose, ears are malformed, mouth is constantly open and the tongue protrudes.

Question 25.
Mention the genetic makeup of Turner’s syndrome person and Klinefelter’s syndrome , person.
Answer:
Klinefelter’s syndrome – 44AA+XXY
Turner’s syndrome – 44AA+XO

Question 26.
List out any four clinical symptoms of Klinefelter’s syndrome.
Answer:
Gynaecomastia, high pitched voice, under developed genetalia and tall with long limbs.

12th Bio Zoology Principles of Inheritance and Variation Three Marks Questions and Answers

Question 27.
Write the types of sex-determination mechanisms does the following crosses as shown. Give an example for each.
(a) Female XX with Male XO (6) Female ZW with Male ZZ
Answer:
(a) Male heterogamety. e.g., Human beings.
(b) Female heterogamety. e.g., Birds.

Question 28.
What are the enzymes encoded by the alleles I^A, I^B and I^O?
Answer:
IA allele produces N-acetyl galactose transferase and can add N-acetyl galactosamine (NAG) and I^B allele encodes for the enzyme galactose transferase that adds galactose to the precursor (i.e. H substances). In the case of I^O/I^O allele no terminal transferase enzyme is produced and therefore called “null” allele and hence cannot add NAG or galactose to the precursor.

Question 29.
Draw a tabular column representing various types of blood group in human beings, their phenotypes, genotypes, antigens and respective antibodies.
Answer:
Genetic basis of the human ABO blood groups:

Question 30.
Give an account on Rhesus factor.
Answer:
Rhesus or Rh – Factor: The Rh factor or Rh antigen is found on the surface of erythrocytes. It was discovered in 1940 by Karl Landsteiner and Alexander Wiener in the blood of rhesus monkey, Macaca rhesus and later in human beings. The term ‘Rh factor’ refers to “immunogenic D antigen of the Rh blood group system. An individual having D antigen are Rh D positive (Rh+) and those without D antigen are Rh D negative (Rh”)”. Rhesus factor in the blood is inherited as a dominant trait.

Naturally occurring Anti D antibodies are absent in the plasma of any normal individual. However if an Rh” (Rh negative) person is exposed to Rh+ (Rh positive) blood cells (erythrocytes) for the first time, anti D antibodies are formed in the blood of that individual. On the other hand, when an Rh positive person receives Rh-negative blood no effect is seen.

Question 31.
How Erythroblastosis foetalis can be prevented?
Answer:
If thefmother is Rh negative and foetus is Rh positive, anti D antibodies should be administered to the mother at 28th and 34th week of gestation as a prophylactic measure. If the Rh-negative mother delivers Rh positive child then anti D antibodies should be administered to the mother soon after delivery. This develops passive immunity and prevents the formation of anti D antibodies in the mothers blood by destroying the Rh foetal RBC before the mother’s immune system is sensitized. This has to be done whenever the woman attains pregnancy.

Question 32.
Explain XX-XO type of sex determination.
Answer:
XX-XO method of sex determination is seen in bugs, some insects such as cockroaches and grasshoppers. Pi The female with two X chromosomes are homogametic Gametes (XX) while the males with only one X chromosome are heterogametic (XO). The presence of unpaired X chromosomes determines the male sex. The males PI Generation with unpaired ‘X’ chromosome produce two types of sperms, one half with X chromosome and other half without X chromosome. The sex of the offspring depends upon the sperm that fertilizes the egg.

Question 33.
Name the type of sex-determination mechanism of the following organisms.
(a) Gypsy moth (A) Human beings (c) Butterflies
Answer:
(a) Gypsy moth -ZW – ZZ type (ZW-females, ZZ – males)
(b) Human beings – XX – XY type (XX-females, XY – males)
(c) Butterflies – ZO – ZZ type (ZO-females, ZZ – males)

Question 34.
Complete the following cross.

Answer:

Question 35.
Role of Y- chromosome is crucial for maleness – Justify.
Answer:
Current analysis of Y chromosomes has revealed numerous genes and regions with potential genetic function; some genes with or without homologous counterparts are seen on the X. Present at both ends of the Y chromosome are the pseudoautosomal regions (PARs) that are similar with regions on the X chromosome which synapse and recombine during meiosis.

The remaining 95% of the Y chromosome is referred as the Non-combining Region of the Y (NRY). The NRY is divided equally into functional genes (euchromatic) and non-functional genes (heterochromatic). Within the euchromatin regions, is a gene called Sex determining region Y (SRY). In humans, absence of Y chromosome inevitably leads to female development and this SRY gene is absent in X chromosome. The gene product of SRY is the testes determining factor (TDF) present in the adult male testis.

Question 36.
Color blindness is a perfect example for criss-cross of inheritance – Justify the statement.
Answer:
A marriage between a colour blind man and a normal visioned woman will produce normal visioned male and female individuals in F1 generation but the females are carriers. The marriage between a F1 normal visioned carrier woman and a normal visioned male will produce one normal visioned female, one carrier female, one normal visioned male and one colour blind male. Colour blind trait is inherited from the male parent to his grandson through carrier daughter, which is an example of criss-cross pattern of inheritance.

Question 37.
How the Karyotype of lymphocytes was prepared by Tjio and Levan?
Answer:
Preparation of Karyotype Tjio and Levan (1960) described a simple method of culturing lymphocytes from the human blood. Mitosis is induced followed by addition of colchicine to arrest cell division at metaphase stage and the suitable spread of metaphase chromosomes is photographed. The individual chromosomes are cut from the photograph and are arranged in an orderly fashion in homologous pairs. This arrangement is called a karyotype. Chromosome banding permits structural definitions and differentiation of chromosomes.

Question 38.
What is a genetic disorder? Mention its types?
Answer:
A genetic disorder is a disease or syndrome that is caused by an abnormality in an individual -DNA. Abnormalities can range from a small mutation in a single gene to the addition or subtraction of an entire chromosome or even a set of chromosomes. Genetic disorders are of two types namely, Mendelian disorders and chromosomal disorders.

Question 39.
Explain the genetic basis of Phenylketonuria.
Answer:
Phenylketonuria is an inborn error of Phenylalanine metabolism caused due to a pair of autosomal recessive genes. It is caused due to mutation in the gene PAH (phenylalanine hydroxylase gene) located on chromosome 12 for the hepatic enzyme “phenylalanine hydroxylase”.

This enzyme is essential for the conversion of phenylalanine to tyrosine. Affected individual lacks this enzyme, so phenylalanine accumulates and gets converted to phenylpyruvic acid and other derivatives. It is characterized by severe mental retardation, light pigmentation of skin and hair. Phenylpyruvic acid is excreted in the urine.

Question 40.
Give an account of Patau’s syndrome.
Answer:
Trisomic condition of chromosome 13 results in Patau’s syndrome. Meiotic non-disjunction is thought to be the cause for this chromosomal abnormality. It is characterized by multiple and severe body malformations as well as profound mental deficiency. Small head with small eyes, cleft palate, malformation of the brain and internal organs are some of the symptoms of this syndrome.

Question 41.
Define aneuploidy.
Answer:
Failure of chromatids to segregate during cell division resulting in the gain or loss of one or more chromosomes is called aneuploidy. It is caused by the non-disjunction of chromosomes.

12th Bio Zoology Principles of Inheritance and Variation Five Marks Questions and Answers

Question 42.
What do you mean by “syndrome”? Give two examples.
Answer:
Group of signs and symptoms that occur together and characterize a particular abnormality is called a syndrome, e.g., Down’s syndrome and Turner’s syndrome.

Question 42.
Explain in detail about Erythroblastosis foetalis.
Answer:
Rh incompatability has great significance in childbirth. If a woman is Rh-negative and the man is Rh positive, the foetus may be Rh positive having inherited the factor from its father. The Rh negative mother becomes sensitized by carrying Rh positive foetus within her body. Due to damage of blood vessels, during child birth, the mother’s immune system recognizes the Rh antigens and gets sensitized. The sensitized mother produces Rh antibodies. The antibodies are IgG type which are small and can cross placenta and enter the foetal circulation. By the time the mother gets sensitized and produce anti ‘D’ antibodies, the child is delivered.

Usually no effects are associated with exposure of the mother to Rh positive antigen during the first child birth, subsequent Rh positive children carried by the same mother, may be exposed to antibodies produced by the mother against Rh antigen, which are carried across the placenta into the foetal blood circulation. This causes haemolysis of foetal RBCs resulting in haemolytic jaundice and anaemia. This condition is known as Erythroblastosis foetalis or Haemolytic disease of the new bom (HDN).

Question 43.
Decribe female heterogamy and its types.
Answer:
Heterogametic Females:
In this method of sex determination, the homogametic male possesses two ‘X’ chromosomes as in certain insects and certain vertebrates like fishes, reptiles and birds producing a single type of gamete; while females produce dissimilar gametes. The female sex consists of a single ‘X’ chromosome or one ‘X’ and one ‘Y’ chromosome. Thus the females are heterogametic and produce two types of eggs. Heterogametic females are of two types, ZO-ZZ type and ZW-ZZ type. ,

ZO-ZZ Type
This method of sex determination is seen in certain moths, butterflies and domestic chickens. In this type, the female possesses single ‘Z’ chromosome in its body cells and is heterogametic (ZO) producing two kinds of eggs some with ‘Z’ chromosome and some without ‘Z’ chromosome, while the male possesses two ‘Z’ chromosomes and is homogametic (ZZ).

ZW-ZZ type
This method of sex determination occurs in certain insects (gypsy moth) and in vertebrates such as fishes, reptiles and birds. In this method the female has one ‘Z’ and one ‘ W’ chromosome (ZW) producing two types of eggs, some carrying the Z chromosomes and some carry the W chromosome. The male sex has two ‘Z’ chromosomes and is homogametic (ZZ) producing a single type of sperm.

Question 44.
Write elaborately about the following Mendelian disorders.
(a) Thalassemia (b) Albinism
Answer:
(a) Thalassemia
Thalassemia is an autosomal recessive disorder. It is caused by gene mutation resulting in excessive destruction of RBC’s due to the formation of abnormal haemoglobin molecules. Normally haemoglobin is composed of four polypeptide chains, two alpha and two beta globin chains. Thalassemia patients have defects in either the alpha or beta globin chain causing the production of abnormal haemoglobin molecules resulting in anaemia.

Thalassemia is classified into alpha and beta based on which chain of haemoglobin molecule _ is affected. It is controlled by two closely linked genes HBA1 and HBA2 on chromosome 16. Mutation or deletion of one or more of the four alpha gene alleles causes Alpha Thalassemia. In Beta Thalassemia, production of beta globin chain is affected. It is controlled by a single gene (HBB) on chromosome 11. It is the most common type of Thalassemia and is also known as Cooley’s anaemia. In this disorder, the alpha chain production is increased and damages the membranes of RBC.

(b) Albinism
Albinism is an inborn error of metabolism, caused due to an autosomal recessive gene. Melanin pigment is responsible for skin colour. Absence of melanin results in a condition called albinism. A person with the recessive allele lacks the tyrosinase enzyme system, which is required for the conversion of dihydroxyphenylalanine (DOPA) into melanin pigment inside the melanocytes. In an albino, melanocytes are present in normal numbers in their skin, hair, iris, etc., but lack melanin pigment.

Question 45.
Discuss any two Allosomal anomalies in human.
Answer:
Allosomal abnormalities in human beings
Mitotic or meiotic non-disjunction of sex chromosomes causes allosomal abnormalities. Several sex chromosomal abnormalities have been detected. E.g. Klinefelter’s syndrome and Turner’s syndrome.

1. Klinefelter’s Syndrome (XXY Males)
This genetic disorder is due to the presence of an additional copy of the X chromosome resulting in a karyotype of 47,XXY. Persons with this syndrome have 47 chromosomes (44AA+XXY). They are usually sterile males, tall, obese, with long limbs, high pitched voice, under developed genetalia and have feeble breast (gynaecomastia) development.

2. Turner’s Syndrome (XO Females)
This genetic disorder is due to the loss of a X chromosome resulting in a karyotype of 45,X. Persons with this syndrome have 45 chromosomes (44 autosomes and one X chromosome) (44AA+XO) and are sterile females. Low stature, webbed neck, underdeveloped breast, rudimentary gonads lack of menstrual cycle during puberty, are the main symptoms of this syndrome.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
On analysis, a person’s karyotype reveals an extra one chromosome of the twenty-first pair. What does this condition represent? which type of symptoms can be noticed in the person?
Answer:
Trisomy-21 or Down’s syndrome.
Symptoms – Mental retardation, malformed ears, protruded tongue, mouth is constantly open etc.

Question 2.
A female whose blood group is AB^– got conceived and later it is diagnoised that her – foetus possess B+. What measures would be taken to prevent the foetus from Haemolytic disease of Newborn (HDN)
Answer:
If the mother is Rh-negative and foetus is Rh-positive, anti D antibodies should be administered to the mother at 28th and 34th week of gestation as a prophylactic measure. If the Rh-negative mother delivers a Rh-positive child then anti D antibodies should be administered to the mother soon after delivery. This develops passive immunity and prevents the formation of anti D antibodies in the mother’s blood by destroying the Rh foetal RBC before the mother’s immune system is sensitized. This has to be done whenever the woman attains pregnancy.

Question 3.
The following table shows the genotypes for ABO blood grouping and these phenotypes. Complete the table by filling the gaps.

Answer:
2) I^AI^O
3) I^AI^B
4) O

Question 4.
Give one example for each of the following group of drugs, (a) Stimulants (b) Analgesic (c) Hallucinogens
Answer:
(a) Stimulants – Eg.: Nicotine
(b) Analgesic – Eg.: Opium
(c) Hallucinogens – Phencyclidine

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.8 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8

Question 1.
Find two positive numbers whose sum is 12 and their product is maximum.
Solution:
Let the two numbers be x, 12 – x.
Their product p = x (12 – x) = 12x – x²
To find the maximum product.
p'(x) = 12 – 2x
p”(x) = -2
p'(x) = 0 ⇒ 12 – 2x = 0 ⇒ 2x = 12
⇒ x = 6
at x = 6, p”(x) = -2 = -ve
⇒ p is maximum at x = 6
when x = 6, 12 – x = 12 – 6 = 6
So the two numbers are 6, 6

Question 2.
Find two positive numbers whose product is 20 and their sum is minimum.
Solution: Let the two positive numbers be ‘x’ and ‘y’
Given product is 20 ⇒ xy = 20 ⇒ y = \(\frac { 20 }{ x }\)
Sum S = x + y
S = x + \(\frac { 20 }{ x }\)
\(\frac { dS }{ dx }\) = 1 – \(\frac { 20 }{ x^2 }\)
For maximum or minimum, \(\frac { dS }{ dx }\) = 0
x² – 20 = 0 x² = 20
x = ±2√5
[x= -2√5 is not possible
\(\frac { d^2S }{ dx^2 }\) = \(\frac { 40 }{ x^3 }\)
at x = 2√5, \(\frac { d^2S }{ dx^2 }\) > 0
∴ Sum ‘S’ is minimum when x = 2√5
y = \(\frac { 20 }{ 2√5 }\) = 2√5
Minimum sum = 2√5 + 2√5 = 4√5

Question 3.
Find the smallest possible value of x² + y² given that x + y = 10.
Solution:
Given x + y = 10 ⇒ y = 10 – x
Let A = x² + y²
A = x² + (10 – x)²
\(\frac { dA }{ dx }\) = 2x + 2(10 – x)(-1)
For maximum or minimum,
\(\frac { dA }{ dx }\) = 0 ⇒ 2(2x – 10) = 0
x = 5
\(\frac { d^2A }{ dx^2 }\) = 4
at x = 5, \(\frac { d^2A }{ dx^2 }\) > 0
∴ A is minimum when x = 5
y = 10 – 5 = 5
∴ The smallest possible value of x² + y² is
(5)² + (5)² = 25 + 25 = 50

Question 4.
A garden is to be laid out in a rectangular area and protected by a wire fence. What is the largest possible area of the fenced garden with 40 meters of wire?
Solution:

Perimeter = 40 m
2(l + b) = 40 ⇒ l + b = 20
Let l = x m
b = (20 – x)m
Area = l × b = x(20 – x) = 20x – x²
To find the maximum area
A(x) = 20x – x²
A'(x) = 20 – 2x
A”(x) = -2
A'(x) = 0 ⇒ 20 – 2x = 0
⇒ x = 10
x = 10 is a maximum point
:. Maximum Area = x (20 – x)
= 10(20 – 10)
= 10 × 10 = 100 sq.m.

Question 5.
A rectangular page is to contain 24 cm² of print. The margins at the top and bottom of the page are 1.5 cm and the margins at the other sides of the page are 1 cm. What should be the dimensions’ of the page so that the area of the paper used is minimum?
Solution:
Let the width of the printed part be ‘x’ cm
Let the height Of the printed part be ‘y’ cm
Given, Area of the printed part = 24 cm²
i.e., xy = 24
y = \(\frac { 24 }{ x }\)

From the given data,
Width of the page
= x + 2(1) = x + 2 cm
Height of the page
= y + 2(1.5) = y + 3 cm
∴ Area of the paper
‘A’ = (x + 2) (y + 3)
= (x + 2) (\(\frac { 24 }{ x }\) + 3)
A = 24 + 3x + \(\frac { 48 }{ x }\) + 6
\(\frac { dA }{ dx }\) = 3 – \(\frac { 48 }{ x^2 }\)
For maximum or minimum,
\(\frac { dA }{ dx }\) = 0
3x² – 48 = 0
x² = 16
x = ±4 [∵ x cannot be negative
∴ x = 4
Now, \(\frac { d^2A }{ dx^2 }\) = \(\frac { 96 }{ x^3 }\)
at x = 4, \(\frac { d^2A }{ dx^2 }\) > 0
∴ Area is minimum when x = 4
y = \(\frac { 24 }{ 4 }\) = 6
∴ Dimensions of the page:
Width of the page = x + 2 = 4 + 2 = 6 cm
Height of the page = y + 3 = 6 + 3 = 9 cm

Question 6.
A farmer plans to fence a rectangular pasture adjacent to a river. The pasture must contain 1,80,000 sq. mtrs in order to provide enough grass for herds. No fencing is needed along the river. What is the length of the minimum needed fencing material?
Solution:

Let the length of the pasture be ‘x’ m
Let the breadth of the pasture be ‘y’ m
Given Area = 1,80,000
xy = 1,80,000
y = 1,80,000
For fencing, we need 2y + x
(one side is River)
Let P = 2y + x
P = 2(\(\frac { 180000 }{ 2 }\)) + x = \(\frac { 360000 }{ x }\) + x
\(\frac { dP }{ dx }\) = –\(\frac { 360000 }{ x^2 }\) + 1
For maximum or minimum,
\(\frac { dP }{ dx }\) = 0
⇒ – 360000 + x² = 0
x² = 360000
x = ±600
[x = -600 is not possible]
∴ x = 600
Now, \(\frac { d^2P }{ dx^2 }\) = \(\frac { 720000 }{ x^3 }\)
at x = 600, \(\frac { d^2P }{ dx^2 }\) > 0
∴ P is minimum when x = 600
y = \(\frac { 180000 }{ 600 }\) = 300
∴ Length of the minimum needed fencing material = 2y + x = 2(300) + 600 = 1200 m

Question 7.
Find the dimensions of the rectangle with the maximum area that can be inscribed in a circle of a radius of 10 cm.
Solution:
Let the length of the rectangle be ‘x’ cm
The breadth of the rectangle be ‘y’ cm

From the figure,
x² + y² = (20)² [Pythagoras Theorem
y² = 400 – x²
[∵ radius of the circle is 10 cm
y = \(\sqrt { 400-x^2 }\)
Now, Axea of the rectangle A = xy

For maximum or minimum,
\(\frac { dA }{ dx }\) = 0 ⇒ \(\frac { -2x^2+400 }{ \sqrt{400-x^2} }\)
x² = 200
x = ±10√2
x = -10√2 is not possible
∴ x = 10√2

Area of the rectangle is maximum When x = 10√2
y = \(\sqrt { 400-200 }\) = \(\sqrt { 200 }\) = 10√2
∴ x = y = 10√2
Length of the rectangle = 10√2 cm
Breadth of the rectangle = 10√2 cm
(Note: A largest rectangle that can be inscribed in a circle is a square)

Question 8.
Prove that among all the rectangles of the given perimeter, the square has the maximum area.
Solution:
Let x, y be the length and breadth of a rectangle and given perimeter is P (say)
ie. 2(x + y) = P
y = \(\frac { P }{ 2 }\) – x
Area of a rectangle ‘A’ = xy
A = x(\(\frac { P }{ 2 }\) – x) = \(\frac { P }{ 2 }\) x – x²
\(\frac { dA }{ dx }\) = \(\frac { P }{ 2 }\) – 2x
For maximum or minimum,
\(\frac { dA }{ dx }\) = 0 ⇒ \(\frac { P }{ 2 }\) – 2x = 0
x = \(\frac { P }{ 4 }\)
Now, \(\frac { d^2A }{ dx^2 }\) = -2
at x = \(\frac { P }{ 4 }\), \(\frac { d^2A }{ dx^2 }\) < 0
∴ Area of the rectangle is maximum when x = \(\frac { P }{ 4 }\)
Now, y = \(\frac { P }{ 2 }\) – x = \(\frac { P }{ 2 }\) – \(\frac { P }{ 4 }\) = \(\frac { P }{ 4 }\)
∴ Length of a rectangle = \(\frac { P }{ 4 }\)
Breadth of a rectangle = \(\frac { P }{ 4 }\)
Since Length = Breadth, the rectangle is a square.
Hence Proved.

Question 9.
Find the dimensions of the largest rectangle that can be inscribed in a semi-circle of radius r cm.
Solution:

Given radius of the semi-circle = ‘r’ cm
Let the length of the rectangle be ‘x’ cm
Let the breadth of the rectangle be ‘y’ cm
From the figure,

For maximum or minimum,
\(\frac { dA }{ dx }\) = 0
⇒ \(\frac { 1 }{ 2 }\) [ \(\frac { 4r^2-2x^2 }{ \sqrt{4x^2-x^2} }\) ] = 0
4r² – 2x² = 0
x² = 2r²
x = ± √2 r
x = -√2 r is not possible
∴ x = √2 r

∴ Length of the rectangle is √2 r cm
Breadth of the rectangle is \(\frac { r }{ √2 }\) cm

Question 10.
A manufacturer wants to design an open box having a square base and a surface area of 108 sq. cm. Determine the dimensions of the box for the maximum volume.
Solution:
Let ‘x’ be the length Of the box.
‘y’ be the height of the box.
Given, surface area = 108 sq.cm
i,e. 4(xy) + x² = 108
⇒ y = \(\frac { 108-x^2 }{ 4x }\)
Volume of the box V = x²y

maximum or minimum, \(\frac { dV }{ dx }\) = 0
⇒ 108 – 3x² = 0
x² = 36
x = ± 6 [x = – 6 is not possible
∴ x = 6
Now, \(\frac { d^2V }{ dx^2 }\) = –\(\frac { 6x }{ 4 }\) = –\(\frac { 3x }{ 2 }\)
at x = 6, \(\frac { d^2V }{ dx^2 }\) < 0
Volume of theboxis maximum when x = 6
y = \(\frac { 108-36 }{ 24 }\) = \(\frac { 72 }{ 24 }\) = 3
∴ Length of the box = 6 cm
Breadth of the box = 6 cm
Height of the box = 3 cm

Question 11.
The volume of a cylinder is given by the formula V = πr²h. Find the greatest and least values of V if r + h = 6.
Solution:
Given r + h = 6
⇒ r = 6 – h
Volume V = πr²h
V = π(6 – h)²h
\(\frac { dV }{ dh }\) = π [(6 – h)² (1) + 2h(6 – h) (-1)] = π(6 – h)[6 – 3h]
For maximum or minimum,
\(\frac { dV }{ dh }\) = 0
⇒ π (6 – h) (6 – 3h) = 0
⇒ h = 6, h = 2
h = 6 is not possible as r + h = 6
∴ h = 2
\(\frac { d^2V }{ dh^2 }\) = π [(6 – h)(-3) + (6 – 3h)(-1)] = π [6h – 24]
at h = 2, \(\frac { d^2V }{ dh^2 }\) < 0
∴ Volume of the cylinder is maximum when h = 2 and r = 6 – 2 = 4
greatest value of V = π(4)² (2) = 32 π
least value of V = 0

Question 12.
A hollow cone with a base radius of a cm and’ height of b cm is placed on a table. Show that) the volume of the largest cylinder that can be hidden underneath is \(\frac { 4 }{ 9 }\) times the volume of the cone.
Solution:

Cone
heigthof the cone = b cm
base radius = a cm
Cylinder
Let the base radius be ‘r’ cm
height be ‘h’ cm
From the figure, \(\frac { h }{ a-r }\) = \(\frac { b }{ a }\)
(using similar triangles property
⇒ h = \(\frac { b }{ a }\) (a – r)
= b – \(\frac { b }{ a }\) r
Volume of cylinder V = πr²h

For maximum or minimum,
\(\frac { dV }{ dr }\) = 0
⇒ br(2a – 3r) = 0
r = 0 and r = \(\frac { 2a }{ 3 }\)
r = 0 is not possible
∴ r = \(\frac { 2a }{ 3 }\)

Hence proved.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Zoology Guide Pdf Chapter 3 Reproductive Health Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 3 Reproductive Health

12th Bio Zoology Guide Reproductive Health Text Book Back Questions and Answers

Question 1.
Which of the following is correct regarding HIV, hepatitis B, gonorrhoea and trichomoniasis?
(a) Gonorrhoea is a STD whereas others are not.
(b) Trichomoniasis is a viral disease whereas others are bacterial.
(c) HIV is a pathogen whereas others are diseases.
(d) Hepatitis B is eradicated completely whereas others are not.
Answer:
(c) HIV is a pathogen whereas others are diseases.

Question 2.
Which one of the following groups includes sexually transmitted diseases caused by bacteria only?
(a) Syphilis, gonorrhoea and candidiasis
(b) Syphilis, chlamydiasis and gonorrhoea
(c) Syphilis, gonorrhoea and trichomoniasis
(d) Syphilis, trichomoniasis and pediculosis
Answer:
(b) Syphilis, chlamydiasis and gonorrhoea

Question 3.
Identify the correct statements from the following:
(a) Chlamydiasis is a viral disease.
(b) Gonorrhoea is caused by a spirochaete bacterium, Treponema palladium.
(c) The incubation period for syphilis is 2 to 14 days in males and 7 to 21 days in females.
(d) Both syphilis and gonorrhoea are easily cured with antibiotics.
Answer:
(d) Both syphilis and gonorrhoea are easily cured with antibiotics

Question 4.
A contraceptive pill prevents ovulation by …………….
(a) blocking fallopian tube
(b) inhibiting release of FSH and LH
(c) stimulating release of FSH and LH
(d) causing immediate degeneration of released ovum.
Answer:
(b) inhibiting release of FSH and LH

Question 5.
The approach which does not give the defined action of contraceptive is

Answer:
(b) Vasectomy – Prevents spermatogenesis

Question 6.
Read the given statements and select the correct option.
Statement 1: Diaphragms, cervical caps, and vaults are made of rubber and are inserted into the female reproductive tract to cover the cervix before coitus.
Statement 2: They are chemical barriers of conception and are reusable.
(a) Both statements 1 and 2 are correct and statement 2 is the correct explanation of statement 1.
(b) Both statements 1 and 2 are correct but statement 2 is not the correct explanation of statement 1.
(c) Statement 1 is correct but statement 2 is incorrect.
(d) Both statements 1 and 2 are incorrect.
Answer:
(c) Statement 1 is correct but statement 2 is incorrect.

Question 7.
Match column I with column II and select the correct option from the codes given below.

(a) A – (iv), B – (ii), C – (i), D-(iii)
(b) A – (iv), B – (i), C – (iii), D – (ii)
(c) A – (i), B – (iv), C – (ii), D – (iii)
(d) A – (iv), B – (i), C – (ii), D – (iii)
Answer:
(d) A-(iv), B-(i), C-(ii), D-(iii)

Question 8.
Select the incorrect action of hormonal contraceptive pills from the following
(a) Inhibition of spermatogenesis.
(b) Inhibition of ovulation.
(c) Changes in cervical mucus impairing its ability to allow passage and transport of sperms.
(d) Alteration in uterine endometrium to make it unsuitable for implantation.
Answer:
(a) Inhibition of spermatogenesis.

Question 9.
What is amniocentesis? Why a statutory ban is imposed on this technique?
Answer:
Amniocentesis is a prenatal technique used to detect any chromosomal abnormalities in the foetus and it is being often misused to determine the sex of the foetus. Once the sex of the foetus is known, there may be a chance of female foeticide. Hence, a statutory ban on amniocentesis is imposed.

Question 10.
Select the correct term from the bracket and complete the given branching tree
Answer:

Question 11.
Correct the following statements

1. Transfer of an ovum collected from donor into the fallopian tube is called ZIFT.
2. Transfering of an embryo with more than 8 blastomeres into uterus is called GIFT,
3. Multiload 375 is a hormone-releasing IUD.

Answer:

1. Transfer of 8 celled blastomeres collected from donor into the fallopian tube is called ZIFT.
2. Transfering of an embryo with more than 8 blastomeres into uterus is called IUT.
3. Multiload 375 is a copper-releasing IUD.

Question 12.
Which method do you suggest the couple to have a baby, if the male partner fails to inseminate the female or due to very low sperm count in the ejaculate?
Answer:
Micro-testicular sperm extraction (TESE):
Microsurgical sperm retrieval from the testicle involves a small midline incision in the scrotum, through which one or both testicles can be seen. Under the microscope, the seminiferous tubules are dilated and small amount of testicular tissue in areas of active sperm production are removed and improved for sperm yield compared to traditional biopsy techniques.

Question 13.
Expand the following:

1. ZIFT
2. ICSI

Answer:

1. Zygote intra-fallopian transfer
2. Intracytoplasmic sperm injection

Question 14.
What are the strategies to be implemented in India to attain total reproductive health?
Answer:

• Creating awareness and providing medical assistance to build a healthy society.
• Introducing sex education in schools to provide information about adolescence and adolescence related changes.
• Educating couples and those in the marriageable age groups about the available birth control methods and family planning norms.
• Creating awareness about care for pregnant women, post-natal care of mother and child, and the importance of breastfeeding.
• Encouraging and supporting governmental and non-governmental agencies to identify new methods and/or to improve upon the existing methods of birth control.

Question 15.
Differentiate foeticide and infanticide.
Answer:
Female foeticide refers to ‘aborting the female in the mother’s womb’.
Female infanticide is ‘killing the female child after her birth’.

Question 16.
Describe the major STDs and their symptoms.
Answer:

Cervical cancer:
Cervical cancer is caused by a sexually transmitted virus called Human Papillomavirus (HPV). HPV may cause abnormal growth of cervical cells or cervical dysplasia. The most common symptoms and signs of cervical cancer are pelvic pain, increased vaginal discharge and abnormal vaginal bleeding. The risk factors for cervical cancer include

1. Having multiple sexual partners
2. Prolonged use of contraceptive pills

Gervical cancer can be diagnosed by a Papanicolaou smear (PAP smear) combined with an HPV test. X-Ray, CT scan, MRI and a PET scan may also be used to determine the stage of cancer. The treatment options for cervical cancer include radiation therapy, surgery and chemotherapy.

Modem screening techniques can detect precancerous changes in the cervix. Therefore screening is recommended for women above 30 years once in a year. Cervical cancer can be prevented with vaccination. Primary prevention begins with HPV vaccination of girls aged 9-13 years before they become sexually active. Modification in lifestyle can also help in preventing cervical cancer. A healthy diet, avoiding tobacco usage, preventing early marriages, practicing monogamy and regular exercise minimize the risk of cervical cancer.

Question 17.
How are STDs transmitted?
Answer:
Sexually transmitted diseases (STD) are called as Sexually transmitted infections (STI). Normally STI are transmitted from person to person during intimate sexual contact with an infected partner. Infections like Hepatitis-B and HIV are transmitted sexually as well as by sharing of infusion needles, surgical instruments, etc with infected people, blood transfusion or from infected mother to baby.

Question 18.
Write the preventive measures of STDs.
Answer:
Prevention of STDs:

• Avoid sex with unknown partner/ multiple partners
• use condoms
• In case of doubt, consult a doctor for diagnosis and get complete treatment.

Question 19.
The procedure of GIFT involves the transfer of female gametes into the fallopain tube, can gametes be transferred to the uterus to achieve the same result? Explain.
Answer:
Gamete intra-fallopian transfer (GIFT): Transfer of an ovum collected from a donor into the fallopian tube. In this the eggs are collected from the ovaries and placed with the sperms in one of the fallopian tubes. The zygote travels toward the uterus and gets implanted in the inner lining of the uterus.

Question 20.
Amniocentesis, the foetal sex determination test, is banned in our country. Is it necessary? Comment.
Answer:
Amniocentesis is a prenatal technique used to detect any chromosomal abnormalities in the foetus and it is being often misused to determine the sex of the foetus. Once the sex of the foetus is known, there may be a chance of female foeticide. Hence, a statutory ban on amniocentesis is imposed.

Question 21.
Open Book Assessment ‘Healthy reproduction, legally checked birth control measures and proper family planning programmes are essential for the survival of mankind’ Justify.
Answer:
Reproductive health and proper family planning programmes are highly essential for the survival of mankind. Reproductive health refers to the state of physical, psychological and social well-being merely in absence of illness in all matters related to the reproductive system and its proper functioning. Family planning has the following health and social benefits.

1. Protecting the health of women by reducing frequent pregnancies.
2. Reducing abortions.
3. Providing stable population growth.
4. Assuming the infants well being by providing adequate time lapse between successive pregnancies.

12th Bio Zoology Guide Reproductive Health Additional Important Questions and Answers

12th Bio Zoology Guide Reproductive Health One Mark Questions and Answers

Question 1.
Which of the following is Not a natural contraceptive?
(a) Rhythm method
(b) Lactational amenorrhoea
(c) Progestasert
(d) Continuous abstinence
Answer:
(c) Progestasert

Question 2.
Identify the fungal STD(s) ………………
(i) Trichomoniasis
(ii) Genital herpes
(iii) Candidiasis
(a) Only (z)
(b) Only (z’zz)
(c) Only (zv)
Answer:
(A) Only (iii)

Question 3.
Match the following.

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

Question 4.
Pick out the incorrect statement regarding the character of a good contraceptive.
(a) It should be user friendly
(b) should not affect sexual drive
(c) side effects must be least
(d) should not be easily available
Answer:
(d) should not be easily available

Question 5.
Select the proper hormonal composition of oral contraceptive pills
(a) FSH & Prolactin
(b) Prolactin & TSH
(c) TSH & FSH
(d) FSH & LH
Answer:’
(d) FSH & LH

Question 6.
Assertion (A): IUD’s are inserted into the ovary.
Reason (R): IUD’s Increases phagocytosis of the sperm.
(a) Both A and R are correct
(b) Both A and R are incorrect
(c) A is correct R is incorrect
(d) A is incorrect R is correct
Answer:
(d) A is incorrect R is correct

Question 7.
Identify the mismatched pair.

Answer:
(c) Candidiasis (iii) Albugo Candida

Question 8.
In India, a family planning programme was initiated in …………………
(a) 1961
(b) 1981
(c) 1951
(d) 1971
Answer:
(c) 1951

Question 9.
Assertion (A): Amniocentesis helps to diagnose the chromosomal aberrations in foetus.
Reason (R): Amniocentesis is legalized is our country.
(a) Both A and R are wrong
(b) A is right and R is wrong
(c) R explains A
(d) A is the wrong R is right
Answer:
(b) A is right and R is wrong

Question 10.
The legalized marriageable age of female in India is …………….
(a) 19 years
(b) 20 years
(c) 18 years
(d) 21 years
Answer:
(c) 18 years

Question 11.
Identify the correct statement.
(a) Lactational amenorrhea is a permanent birth control method
(b) Condoms are made of polyethylene glycol and lambskin
(c) LNG -20 is a copper – releasing IUD
(d) Diaphragm covers the cervix thereby preventing sperm entry
Answer:
(d) Diaphragm covers the cervix thereby preventing sperm entry

Question 12.
According to WHO, India is the…………largest HIV affected country.
(a) first
(b) second
(c) third
(d) seventh
Answer:
(c) third

Question 13.
Identify the correct statement.
(a) MTP is the voluntary killing of an infant.
(b) MTP is legalized in India in 1974.
(c) Performing MTP during the second trimester is riskier.
(d) It is a surgical – based abortion.
Answer:
(c) Performing MTP during the second trimester is riskier.

Question 14.
Saheli contains a non-steroidal preparation called ………………
Answer:
Centchroman

Question 15.
The average foetal heartbeat rate is between ……………… beats per minute.
Answer:
120-160

Question 16.
World AIDS Day is observed on ………………
Answer:
11th July

Question 17.
Indian Government legalized MTP in ………………
Answer:
1971

Question 18.
In the chorionic villus sampling test, the tissue sample is taken from ………………
(a) amniotic fluid
(b) placental tissue
(c) Intestinal villi
(d) foetal liver
Answer:
(b) placental tissue

Question 19.
In ZIFT technique, the zygote is transferred at the stage of ………………
(a) 16blastomere
(b) morula
(c) 12 blastomere
(d) 8 blastomere
Answer:
(d) 8 blastomere

Question 20.
Given below are the basic steps in IVF treatment cycle. Select the proper sequence.
(i) Ovarian stimulation
(ii)Egg retrieval
(iii) fertilization
(iv) Embryo culture
(v) Embryo transfer
(a) (ii) → (iv) → (v) → (i) → (iii)
(b) (i) → (iii)) → (ii) → (v) → (iv)
(c) (i) → (ii) → (iii) → (iv) → (v)
(d) (ii) → (1) → (iii) → (v) → (iv)
Answer:
(c) (i) → (ii) → (iii) → (iv) → (v)

Question 21.
Enactment of ……………………..banned the identification of sex and to prevent prenatal abortion.
(a) POGSOAct
(b) POTAAct
(c) PCPNDTAct
(d) GOONDAAct
Answer:
(c) PCPNDT Act

Question 22.
Which is NOT a national health care programme?
(a) Pradhan Mantri Surakshit Matritva Abhiyan
(b) Pradhan Mantri Fiscal BhimaYojana
(c) RMNCH + A approach
(d) Janani Shishu Suraksha Karyakaram
Answer:
(b) Pradhan Mantri Fiscal Bhima Yojana

Question 23.
…………… is Known as an anti-sterility vitamin.
Answer:
Vitamin-E

12th Bio Zoology Guide Reproductive Health Two Marks Questions and Answers

Question 1.
Name any four national-level health care programmes run by the Indian Government.
Answer:

1. Janani Suraksha Yojana
2. Shishu Suraksha Karyakaram
3. RMNCH+A Programme
4. Pradhan Mantri Surakshit Matritva Abhiyan

Question 2.
Comment on ‘Saheli’
Answer:
Saheli is an oral contraceptive pill provided by Central Drug Research Institute in Lucknow, India. It contains a non-steroidal hormone preparation called centchroman.

Question 3.
What is Mayer – Rokitansky Syndrome?
Answer:
All women are bom with ovaries, but some do not have functional uterus. This condition is called Mayer-Rokitansky syndrome.

Question 4.
Point out any four STD’s caused by viruses.
Answer:

1. Genital herpes
2. Hepatitis-B
3. Genital warts
4. AIDS

Question 5.
Define Surrogacy.
Answer:
Surrogacy is a method of assisted reproduction or agreement whereby a woman agrees to carry a pregnancy for another person, who will become the newborn child’s parent after birth. Through In Vitro Fertilization (IVF), embryos are created in a lab and are transferred into the surrogate mother’s uterus.

Question 6.
Name the causative organism of (a) Syphilis (b) Genital warts
Answer:
(a) Syphilis is caused by Treponema palladium
(b) Genital warts caused by Human Papilloma vims.

Question 7.
Define Infertility.
Answer:
Inability to conceive or produce children even after unprotected sexual cohabitation is called infertility. That is, the inability of a man to produce sufficient numbers or quality of sperm to impregnate a woman or inability of a woman to become pregnant or maintain a pregnancy.

Question 8.
Expand the acronyms:

1. GIFT
2. ICSI

Answer:

1. GIFT – Gamete Intra – Fallopian Transfer.
2. ICSI – Intra Cytoplasmic Sperm Injection.

Question 9.
What is amniocentesis?
Answer:
Amniocentesis involves taking a small sample of the amniotic fluid that surrounds the foetus to diagnose chromosomal abnormalities.

Question 10.
How copper IUD’s provide contraception?
Answer:
Copper IUD’s release free copper and copper salts into the uterus and suppress the sperm motility. They remain in uterus for 5-10 years. Eg: NovaT, Cu T-380.

Question 11.
What do you mean by the term – coitus interruptus?
Answer:
Coitus interruptus is a oldest family planning method, where the male partner withdraws his penis before ejaculation, there by preventing the deposition of semen into vagina.

Question 12.
What is MTP?
Answer:
Medical Termination of Pregnancy (MTP) : Medical method of abortion is a voluntary or intentional termination of pregnancy in a non-surgical or non-invasive way. Early medical termination is extremely safe upto 12 weeks (the first trimester) of pregnancy and generally has no impact on a women’s fertility. Abortion during the second trimester is more risky as the foetus becomes intimately associated with the maternal tissue.

Question 13.
Which type of women are benefited by In Vitro Fertilization?
Answer:
IVF is used to treat women with blocked, damaged or absent fallopian tubes.

12th Bio Zoology Guide Reproductive Health Three Marks Questions and Answers

Question 1.
What are the steps taken by Government to overcome population explosion?
Answer:
To overcome the problem of population explosion, birth control is the only available solution.
People should be motivated to have smaller families by using various contraceptive devices. Advertisements by the Government in the media as well as posters/bills, etc., with a slogan Naam iruvar namakku iruvar (we two, ours two) and Naam iruvar namakku oruvar (we two, ours one) have also motivated to control population growth in Tamil Nadu. Statutory rising of marriageable age of the female to 18 years and that of males to 21 years and incentives given to pouples with small families are the other measures taken to control population growth in our country.

Question 2.
Lactational amenorrhoea – Comment.
Answer:
Menstrual cycles resume as early as 6 to 8 weeks from parturition. However, the reappearance of normal ovarian cycles may be delayed for six months during breastfeeding. This delay in ovarian cycles is called lactational amenorrhoea. It serves as a natural, but an unreliable form of birth control.

Question 3.
Write a note on sterilization procedure in the male.
Answer:
Vasectomy is the surgical procedure for male sterilisation. In this procedure, both vas deferens are cut and tied through a small incision on the scrotum to prevent the entry of sperm into the urethra. Vasectomy prevents sperm from heading off to penis as the discharge has no sperms in it.

Question 4.
Define Tubectomy.
Answer:
Tubectomy is the surgical sterilisation in women. In this procedure, a small portion of both fallopian tubes are cut and tied up through a small incision in the abdomen or through vagina. This prevents fertilization as well as the entry of the egg into the uterus.

Question 5.
Complete the table by filling the gaps.

Answer:
A – Neisseria Gonorrhoea
B – AIDS
C- Jaundice

Question 6.
List various natural methods of birth control.
Answer:

1. Periodic abstinence
2. Coitus interruptus
3. Lactational amenorrhea

Question 7.
What does ICSI stand for? Describe the technique.
Answer:
ICSI stands for Intra-Cytoplasmic Sperm Injection. In this method, a sperm is carefully injected into the cytoplasm of the egg and the zygote formed is allowed to divide till 8 celled stages and transferred to the uterus. Fertilization occurs in 75-85% of eggs injected with sperm.

Question 8.
How LNG-20 act as a contraceptive?
Answer:
LNG-20 is an Intra-Uterine System of contraceptive method. It increases the viscosity of cervical mucus and thereby preventing the sperms from entering the cervix.

Question 9.
MTP is legalized in our country. Yes or No? Why?
Answer:
Yes. The government of India legalized MTP in 1971 for medical necessity and social consequences ’with certain restrictions like sex discrimination and illegal female foeticides to avoid its misuse. MTP performed illegally by unqualified quacks is unsafe and could be fatal. MTP of the first conception may have serious psychological consequences.

Question 10.
Write the hormonal composition of oral contraceptive pills. Also explain their action mode.
Answer:
Oral contraceptive pills are enriched with synthetic progesterone and oestrogen hormones. These pills prevent the ovulation by inhibiting the secretion of LH and FSH hormones.

Question 11.
Suggest some methods to help infertile couples to have children.
Answer:

1. Invitro fertilization
2. Zygote intra-fallopian transfer (ZIFT)
3. Gamete intra-fallopian transfer (GIFT)
4. Surrogacy

Question 13.
What are the characteristics of a good contraceptive?
Answer:
An ideal contraceptive should be user friendly, easily available, with the least side effects, and should not interfere with sexual drive.

Question 14.
Write is a note on a foetoscope.
Answer:
Foetoscope is used to monitor the foetal heart rate and other functions during late pregnancy and labour. The average foetal heart rate is between 120 and 160 beats per minute. An abnormal foetal heart rate or pattern may mean that the foetus is not getting enough oxygen and it indicates other problems. A hand-held Doppler device is often used during prenatal visits to count the foetal heart rate. During labour, continuous electronic foetal monitoring is often used.

Question 15.
How the technique of amniocentesis is performed?
Answer:
Amniocentesis is generally performed in a pregnant woman between the 15th and 20th weeks of pregnancy by inserting a long, thin needle through the abdomen into the amniotic sac to withdraw a small sample of amniotic fluid. The amniotic fluid contains cells shed from the foetus.

Question 16.
Mention the role of prolactin in lactational amenorrhoea.
Answer:
Suckling by the baby during breastfeeding stimulates the pituitary to secrete increased prolactin hormone in order to increase milk production. This high prolactin concentration in the mother’s blood may prevent the menstrual cycle by suppressing the release of GnRH (Gonadotropin-Releasing Hormone) from hypothalamus and gonadotropin secretion from the pituitary.

Question 17.
Name anyone

1. Fungal STI
2. Bacterial STI
3. Protozoan STI

Answer:

1. Fungal STI – Candidiasis
2. Bacterial STI -Gonorrhoea
3. Protozoan STI – Trichomoniasis

Question 18.
Why ultrasonography is performed for carrying women?
Answer:
Ultrasonography is usually performed in the first trimester for dating, determination of the number of foetuses, and for assessment of early pregnancy complications.

Question 19.
Suggest any two simple precautions to avoid contracting RTFs.
Answer:

1. Avoiding coitus with unknown/multiple partners.
2. Use of condoms during coitus.

Question 20.
Name the most effective long-acting reversible contraceptive methods.
Answer:
Intrauterine devices and contraceptive implants.

12th Bio Zoology Guide Reproductive Health Five Marks Questions and Answers

Question 1.
Give an detailed account on various natural methods of contraception.
Answer:
Natural method is used to prevent meeting of sperm with ovum, i.e., Rhythm method (safe period), coitus interruptus, continuous abstinence and lactational amenorrhoea.

a. Periodic abstinence/rhythm method: Ovulation occurs at about the 14th day of the menstrual cycle. Ovum survives for about two days and sperm remains alive for about 72 hours in the female reproductive tract. Coitus is to be avoided during this time.

b. Continuous abstinence is the simplest and most reliable way to avoid pregnancy is not to have coitus for a defined period that facilitates conception.

c. Coitus interruptus is the oldest family planning method. The male partner withdraws his penis before ejaculation, thereby preventing deposition of semen into the vagina.

d. Lactational amenorrhoea : Menstrual cycles resume as early as 6 to 8 weeks from parturition. However, the reappearance of normal ovarian cycles may be delayed for six months during breast-feeding. This delay in ovarian cycles is called lactational amenorrhoea. It serves as a natural, but unreliable form of birth control.

Suckling by the baby during breastfeeding stimulates the pituitary to secrete increased prolactin hormone in order to increase milk production. This high prolactin concentration in the mother’s blood may prevent menstrual cycle by suppressing the release of GnRH (Gonadotropin-Releasing Hormone) from hypothalamus and gonadotropin secretion from the pituitary.

Question 2.
What are IUD’s? Explain its way of functioning. Also, describe their types.
Answer:
Intrauterine Devices (IUDs) are inserted by medical experts in the uterus through the vagina. These devices are available as copper releasing IUDs, hormone releasing IUDs, and non-medicated IUDs. IUDs increase the phagocytosis of sperm within the uterus. IUDs are the ideal contraceptives for females who want to delay pregnancy. It is one of the popular methods of contraception in India and has a success rate of 95 to 99%.

Copper releasing IUDs differ from each other by the amount of copper. Copper IUDs such as Cu T-380 A, Nova T, Cu 7, Cu T 380 Ag, Multiload 375, etc. release free copper and copper salts into the uterus and suppress sperm motility. They can remain in the uterus for five to ten years.

Hormone-releasing IUDs such as Progestasert and LNG – 20 are often called as intrauterine systems (IUS). They increase the viscosity of the cervical mucus and thereby prevent sperms from entering the cervix. Non-medicated IUDs are made of plastic or stainless steel. Lippes loop is a double S-shaped ‘.plastic device.

Question 3.
Write in detail about cervical cancer.
Answer:
Cervical cancer is caused by a sexually transmitted virus called Human Papillomavirus (HPV). HPV may cause abnormal growth of cervical cells or cervical dysplasia.
The most common symptoms and signs of cervical cancer are pelvic pain, increased vaginal discharge and abnormal vaginal bleeding. The risk factors for cervical cancer include

1. Having multiple sexual partners
2. Prolonged use of contraceptive pills

Cervical cancer can be diagnosed by a Papanicolaou smear (PAP smear) combined with an HPV test. X-Ray, CT scan, MRI, and PET scan may also be used to determine the stage of cancer. The treatment options for cervical cancer include radiation therapy, surgery, and chemotherapy.

Modem screening techniques can detect precancerous changes in the cervix. Therefore screening is recommended for women above 30 years once a year. Cervical cancer can be prevented with vaccination. Primary prevention begins with HPV vaccination of girls aged 9-13 years, before they become sexually active. Modification in lifestyle can also help in preventing cervical cancer. A healthy diet, avoiding tobacco usage, preventing early marriages, practicing monogamy and regular exercise minimize the risk of cervical cancer.

Question 4.
List out the causes of fertility in humans.
Answer:
Causes for male infertility:

• Undescended tests and swollen veins in the scrotum
• Underdeveloped testes
• Tight clothing increases the temperature in the scrotum and affects sperm production.
• Autoimmune response against own sperm.
• Usage of alcohol, tobacco, marijuana drugs etc.

Causes for female infertility:

• Malformation of the cervix or fallopian tubes.
• Inadequate nutrition at puberty.
• Low body fat (anorexia = psychological eating disorder due to fear of gaining weight)
•  Pelvic Inflammatory Disease (PID), uterine disorders, endometriosis.
• underdeveloped ovaries.
• Developing antibodies against sperm.

Common Causes for both the sexes:

• Tumours in pituitary or sex organs
• Inherited mutation of hormone synthesizing genes
• Long term stress
• Ingestion of toxins (Cadmium) & drugs
• Injuries to gonads
• Ageing

Higher Order Thinking Skills (HOTs) Questions

Question 1.
Dr. Sheela is a famous Gynaecologist at Poes garden. However illegally she performed amniocentesis for several pregnant illiterates and did MTP if identified as female foetus, on request. Under which act will she get arrested, if a complaint is filed against her.
Answer:
Dr. Sheela will be arrested under the PCPNDT Act – preconception and prenatal diagnostic technique Act-1994.

Question 2.
Identify the picture and describe the surgical procedure.
Answer:
The picture describes tubectomy.

1. Tubectomy is the surgical sterilization in women.
2. In this a portion of the oviduct is cut and tied through vagina or minor incision in the abdomen.
3. It prevents fertilization and also the entry of egg to uterus.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Botany Guide Pdf Chapter 8 Environmental Issues Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Botany Solutions Chapter 8 Environmental Issues

12th Bio Botany Guide Environmental Issues Text Book Back Questions and Answers

I. Choose the correct answer
Question 1.
Which of the following would most likely help to slow down the greenhouse effect.
a) Converting tropical forests in to grazing land for cattle.
b) Ensuring that all excess paper packing is buried to ashes.
c) Redesigning landfill dumps to allow methane to be collected.
a) Promoting the use of private rather than puplic transport.
Answer:
c) Redesigning landfill dumps to allow methane to be collected.

Question 2.
With respect to Eichhornia.
Statement A: It drains off oxygen from water and is seen growing in standing water.
Statement B: It is an indigenous spicies of our country.
a) Statement A is correct and Statement B is wrong.
b) Both Statements A and B are correct.
c) Statement A is correct and Statement B is wrong.
d) Both Statements A and Bare wrong.
Answer:
a) Statement A is correct and Statement B is wrong.

Question 3.
Find the wrongly matched pair.
a) Endemism – Species confined to a region and not found anywhere else.
b) Hotspots – Western ghats.
c) Ex-situ Conservation – Zoological parks.
d) Sacred groves – Saintri hills of Rajasthan.
e) Alien sp. Of India – Water hyacinth.
Answer:
e) Alien sp. Of India – Water hyacinth.

Question 4.
Depletion of which gas in the atmosphere can lead to an increased incidence of skin cancer?
a) Ammonia
b) Methane
c) Nitrous oxide
d) Ozone
Answer:
d) Ozone.

Question 5.
One green house gas contributes 14% of total global warming and another contributes 6%. These are respectively identified as
a) N₂O and CO₂
b) CFC_S and N₂O
c) CH₄ and CO₂
d)CH₄ and CFCS
Answer:
b) CFC_S and N₂O.

Question 6.
One of the chief reasons among the following for the depletion in the number of species making endangered is.
a) over hunting and poaching
b) green house effect
c) competition and predation
d) habitat destruction
Answer:
d) habitat destruction.

Question 7.
Deforestation means
a) growing plants and trees in an area where there is no forest
b) growing plants and trees in an area where the forest is removed
c) growing plants and trees in a pond
d) removal of plants and trees
Answer:
d) removal of plants and trees.

Question 8.
Deforestation does not lead to
a) Quick nutrient cycling
b) soil erosion
c) alternation of local weather conditions
d) Destruction of natural habitat weather conditions
Answer:
a) Quick nutrient cycling.

Question 9.
The unit for measuring ozone thickness
a) Joule
b) Kilos
c) Dobson
d) Watt
Answer:
c) Dobson

Question 10.
People’s movement for the protection of environment in sirsi of Karnataka is
a) Chipko movement
b) Amirtha Devi Bishwas movement
c) Appiko movement
d) None of the above
Answer:
c) Appiko movement.

Question 11.
The plants which are grown in silivpasture system are
a) Sesbania and Acacia
b) Solenum and Crotalaria
c) Clitoria and Begonia
d) Teak and sandal
Answer:
a) Sesbania and Acacia.

Question 12.
What is the ozone hole?
Answer:
The decline in the thickness of the ozone layer over restricted area is called Ozone hole.

Question 13.
Give four examples of plants cultivated in commercial agroforestry.
Answer:
Commercial Agroforestry includes Casuarina, Eucalyptus, Malai vembu, Teak and Kadambu trees.

Question 14.
Expand CCS.
Answer:
CCS – Carbon capture and storage.

• Carbon capture and storage is a technology of capturing carbon dioxide and inject it deep into the underground rocks at a depth of 1 km (or) more.
• It is an approach to mitigate global warming.

Example:

• It is capturing CO₂ released from industries and power plants.
• Such as declining oil fields, gas fields saline aquifers and unmineable coal have been suggested as storage sites.
• Various safe sites have been selected for permanent storage.
• liquid storage in ocean and solid storage by reduction of CO₂ with metal oxide to produce stable carbonates.
• It is also known as Geological sequestration.

Question 15.
How do forests help in maintaining the climate?
Answer:

• Forest regulate co2 levels in the atmosphere.
• If greater the forest area more CO₂ is removed and the impact of global climate change is decreased.
• They regulate ecosystem, protect biodiversity, play an integral part in the carbon cycle.
• In the tropics water evaporates naturally from trees, increasing cloud cover and keeping temperature cooling.

Question 16.
How do sacred groves help in the conservation of biodiversity?
Answer:
These are the patches or grove of cultivated trees which are community protected and are based on strong religious belief systems which usually have a significant religious connotation for protecting the community. Each grove is an abode of a deity mostly village God Or Goddesses like Aiyanar or Amman.

448 grooves were documented throughout Tamil Nadu, of which 6 groves (Banagudi shola, Thirukurungudi and Udaiyankudikadu, Sittannnavasal, Puthupet and Devadanam) were taken up for detailed floristic and faunistic studies. These groves provide a number of ecosystem services to the neighbourhood like protecting watersheds, fodder, medicinal plants, and microclimate control.

Question 17.
Which one gas is most abundant out of the four commonest greenhouse gases? Discuss the effect of this gas on the growth of plants?
Answer:
CO₂ Carbon-di-oxide is the most abundant among greenhouse gases.

• Low agricultural productivity in tropics.
• Frequent heat waves (weeds, pests, fungi, need warmer temperature)
• increase of vectors and epidemics
• strong storms and intense flood damage.
• Water crisis and decreased irrigation
• Change in flowing seasons and pollinators
• Change in species distributional ranges.
• Speakes extinction.

Question 18.
Suggest a solution to water crisis and explain its advantages.
Answer:
Rainwater harvesting is the accumulation and storage of rain water for reuse in-site rather than allowing it to run off. Rainwater can be collected from rivers, roof tops and the water collected is directed to a deep pit. The water percolates and gets stored in the pit. RWH is a sustainable water management practice implemented not only in urban area but also in agricultural fields, which is an important economical cost effective method for the future. Environmental benefits of Rain Water Harvesting:

1. Promotes adequacy of underground water and water conservation.
2. Mitigates the effect of drought.
3. Reduces soil erosion as surface run-off is reduced.
4. Reduces flood hazards.
5. Improves groundwater quality and water table / decreases salinity.
6. No land is wasted for storage purpose and no population displacement is involved.
7. Storing water underground is an eco-friendly measure and a part of sustainable water storage strategy for local communities.

Question 19.
Explain afforestation with case studies.
Answer:
Afforestation:

• Afforestation is planting of trees where there was no previous tree coverage and the conversion of non-forested lands into forests by planting suitable trees to retrieve the vegetation.
• Example: Slopes of dams afforested to reduce water run-off, erosion and siltation. It can also provide a range of environmental services including carbon sequestration, water retention.

The Man who Single-Handedly Created a Dense Forest

• Jadav “Molai” Payeng (born 1963) is an environmental activist has single-handedly planted a forest in the middle of a barren wasteland.
• This Forest Man of India has transformed the world’s largest river island, Majuli, located on one of India’s major rivers, the Brahmaputra, into a dense forest, home to rhinos, deers, elephants, tigers and birds. And today his forest is larger than Central Park.
• Former vice-chancellor of Jawahar Lai Nehru University, Sudhir Kumar Sopory named Jadav Payeng as Forest Man of India, in the month of October 2013.
• He was honoured at the Indian Institute of Forest Management during their annual event ‘Coalescence1.
• In 2015, he was honoured with Padma Shri, the fourth highest civilian award in India.
• He received honorary doctorate degree from Assam Agricultural University and Kaziranga University for his contributions.

Tamil Nadu Afforestation project TAP I:

TaP I
objectives:

• It aimed to uplight the quality and life of villagers, abutting forest areas.
• It is resolve the degraded forests in Tamil Nadu

Tap II
objectives

• To restore the ecological equilibrium of the forests, watersheds and adjacent villages of Tamil Nadu.
• To improve the quality of the life of inhabitants through reforestation. Water conservation and sustained community action.

Question 20.
What are the effects of deforestation and benefits of agroforestry?
Answer:

Effects of deforestation:

1. Burning of forest wood release stored carbon, a negative impact just opposite of carbon sequestration.
2. Trees and plants bind the soil particles. The removal of forest cover increases soil erosion and decreases soil fertility. Deforestation in dry areas leads to the formation of deserts.
3. The amount of runoff water increases soil erosion and also creates flash flooding, thus reducing moisture and humidity.
4. The alteration of local precipitation patterns leading to drought conditions in many regions. It triggers adverse climatic conditions and alters water cycle in ecosystem.
5. It decreases the bio-diversity significantly as their habitats are disturbed and disruption of natural cycles.
6. Loss of livelihood for forest dwellers and rural people.
7. Increased global warming and account for one-third of total CO₂ emission.
8. Loss of life support resources, fuel, medicinal herbs and wild edible fruits.

Benefits of agroforestry:

1. It is an answer to the problem of soil and water conservation and also to stabilise the soil (salinity and water table) reduce landslide and water run-off problem.
2. Nutrient cycling between species improves and organic matter is maintained.
3. Trees provide micro climate for crops and maintain CO₂ balanced, atmospheric temperature and relative humidity.
4. Suitable for dry land where rainfall is minimum and hence it is a good system for alternate land use pattern.
5. Multipurpose tree varieties like Acacia are used for wood pulp, tanning, paper and firewood – industries.
6. Agro-forestry is recommended for the following purposes. It can be used as Farm Forestry for the extension of forests, mixed forestry, shelter belts and linear strip plantation.

12th Bio Botany Guide Environmental Issues Additional Important Questions and Answers

I. Match
Question 1.
Match the column I with column II and select correct option

Option:
a) A – 4,B – 3, C – 1, D – 2
b) A – 3, B – 2, C – 1, D – 4.
c) A – 2, B – 1, C – 3, D – 4.
d) A – 1, B – 3, C – 2, D – 4.
Answer:
a) A – 4,B – 3, C – 1, D – 2

Question 2.
Match the column I with column II and select correct option

Option:
a) A – 1,B – 2, C – 3, D – 4
b) A – 2, B – 1, C – 4, D – 3.
c) A – 2, B – 3, C – 4, D – 1
d) A – 3, B – 2, C – 1, D – 2.
Answer:
b) A – 2, B – 1, C – 4, D – 3

Question 3.
Match the column I with column II and select correct option

Option:
a) A – 3, B – 1, C – 4, D – 2.
b) A – 2, B – 3, C – 4, D – 1.
c) A – 1, B – 2, C – 3, D – 4.
d) A – 3, B – 2, C – 1, D – 4.
Answer:
a) A – 3, B – 1, C – 4, D – 2

II. State True or False and choose the correct option

Question 1.
A – Eichhornia crassiper decreases the oxygen content of water bodies.
B – Prosopis juliflora enrich soil nutrient and important local species growth.
C – Petunia and chrysanthemum are referred as nitrate phytoindicators.
D – Robinia pseudoacocia is a indicator of heavy metal contamination.
Option:

Answer:
b) A – T, B – F, C – T, D – T

Question 2.
A) Jadav “molai payeng” is an forest man of India.
B) The world largest river island, Majuli located on major rivers the Brahmaputra.
C) In 2015 he was honoured with padma shri award in India.
D) He was a student of forest management student.
Option:

Answer:
b) A – T, B – T, C – T, D – F.

III. Choose the incorrect statement

Question 1.
Choose the incorrect statement related to effects of ozone depletion.
a) Juvenile mortality of animal, Increased incidence of mutations.
b) Increases the incidence of cataract, throat and lung irritation, emphysema, skin cancer.
c) Flood/ drought, sea water rise, Imbalance in ecosystem affecting flora and fauna.
d) Diminishing the functioning of immune system is not related to ozone depletion.
Answer:
d) Diminishing the functioning of immune system is not related to ozone depletion.

Question 2.
Choose the incorrect statement related to forestry.
a) The tank foreshore plantations have been a major source of firewood in Tamilnadu.
b) The production of woody plants combined with pasture is referred to silvopasture system.
c) Trees provide micro climate for crops and maintain 0₂ – CO₂ balance.
d) Agro-forestry is an integration of trees, animals, water bodies and humans.
Answer:
d) Agro-forestry is an integration of trees, animals, water bodies and humans

IV. Choose the correct statement

Question 1.
Choose the correct statement from the following.
a) Agricultural drones are animals used to do heavy agricultural works.
b) CARTOSAT-2 is used to watch border surveillance.
c) The production of flowering plants combined with pasture is reffered to silvopasture system.
d) GTS is a satellite navigation system used to determine the ground position of an object
Answer:
b) and d)

V. Pick out the odd one out and give Reason

Question 1.
In-situ, Ex-situ, National parks, Biosphere Reserves, Remote seming.
a) Remote seming, while other are related to biodiversity conservation.
b) National parks, while others are man made project.
c) Biosphere reserves, detecting and monitoring the physical characteristic of an area.
d) None of the above.
Answer:
a) Remote seming, while other are related to biodiversity conservation.

Question 2.
Reduces flood hazards, decreases salinity Reduces soil erosion, carbon sink.
a) Reduces soil erision. while others are related to Rainwater harvesting system.
b) Carbon sink, while others are related to benefits of Rain water harvesting.
c) Decreases salinity, while others are related to rain water harvesting.
d) Reduces soil erosion, while others are eco friendly method.
Answer:
b) Carbon sink, while others are related to benefits of Rain water harvesting

Question 3.
Lichens, Ficus, Pinus, Rose, Gladiolus.
a) Lichens, while others are SO₂ pollution.
b) Gladiolus is the heavy metal indicator, while others are related to SO₂ pollution.
c) Gladiolus. while others are indicator for SO₂ pollution.
d) Pinus is the nitrate indicator, while others are SO₂pollution.
Answer:
c) Gladiolus. while others are indicator for SO₂ pollution

VI. Pictorial questions

Question 1.
Relative contribution of green house gas are shown below. Which one of the given option is correct?

a) A – others, B – CH₄, C – CFC, D – O₂
b) A – CH₄, B – CO₂, C – O₂,D – CFC
c) A – CO₂, B – CH₄, C – CFC, D – O₂
d) A – others, B – CH₄, C – CFC, D – CO₂
Answer:
d) A – others, B – CH₄, C – CFC, D – CO₂

Question 2.
Examine the diagram Which is showing the percentage. Find out the Green house gas related to its percentage.

Answer:
a) CO₂ CH₄ CFC others

VII. Assertion and Reason

Question 1.
Assertion: Sacred groves and sacred lakes are community protected Bio-diversity conservation.
Reason: Which are based on strong religious belief system,
a) A is correct R is wrong.
b) A is correct but R does not explains A.
c) A is correct and R is the correct explanation for A
d) A and R are wrong.
Answer:
c) A is correct and R is the correct explanation for A.

Question 2.
Assertion : Appiko movement started in Gubbi Gadde village sirsi in karnataka by panduranga Hegde.
Reason: It is started to protest against felling of trees, monoculture, forest policy and deforestation
a) Both are wrong.
b) A and R is correct
c) A is correct R- does not explains A.
d) A is wrong R is correct.
Answer:
b) A and R is correct

VIII. Spot the error

Question 1.
Biosphere Reserves, National parks and wildlife sanctuaries are community protected Bio-diversity conservation.
Answer:
Biosphere Reserves,National parks and wild life sanctuaries are Goverment protected Bio-diversity conservation.

IX. Choose the incorrect pair

Question 1.
Choose the incorrect pair.
A) Protein Bank-Fodder production.
B) Livefence, foddertree – Erythrina spp.
C) Agro forestry – Extension of forest, mixed forestry.
D) Social forestry-Jadav,Molaipayeng.
Answer:
D) Social forestry-Jadav,Molai payerg

X. Read the following statement with two blanks A and B select the correct option for blank A and B.

Question 1.
Eichhornia crassipes is an invasive weed native to south America. It affects the growth of A and finally leads to B.

Answer:
a) Phytoplankton – 1) eutrophication

XI. Fill in the blanks Answers

1. ………………. is another long term method to store carbon.
Answer:
Biochar

2. …………………. is the total amount of green house gases produced by human activated.
Answer:
Carbon foot print

3. Eating indigenous fruits and products are reduce ………………
Answer:
Carbon foot print

4. The forest, soil, ocean are …………………. and landfills are ………….. sinks.
Answer:
Natural, artificia
Dobson unit

5. The thickness of the ozone column of air is measured in terms of ………………
Answer:
Chloro fluro carban

6. …………………. is the anthropogenic greenhouse gas.
Answer:
Tank Foreshore

7. _____ plantations have been a major source of fire wood in Tamilnadu.
Answer:
Plantations

8. World ozone Day is celebrated on ……………….
September 16

XII Choose the correct option

Question 1.
Which creates a breeding habitat for disease causing mosquito Anopheles?
a) Eichhornia crassipes
b) Lantana camara
c) Prosopis juliflora
d) Parthenium hysterophorus
Answer:
a) Eichhornia crassipes

Question 2.
Which one the activities is replacement of conventional electrification project solar panels or other energy efficient boilers?
a) Clean Development Mechanism (CDM)
b) Chloro Fluoro Carbon (CFC)
c) Certified Emission Reduction (CER)
d) TamilNadu Afforestation Project (TAP)
Answer:
a) Clean Development Mechanism (CDM)

Question 3.
Some of the major species cultivated in Agroforestry for commercial use:
a) Erythrina, Albizzia
b) Malaivembu, Kadambu
c) Acacia, Azadirachta Indica
d) Sesbania, Acacia
Answer:
b) Malaivembu, Kadambu

Question 4.
Which one of the following is not a carbon sequestration method?
a) Forest conservation and soil conservation
b) Carbon foot print
c) Biochar
d) Increasing the number of animals.
Answer:
d) Increasing the number of animals.

Question 5.
Chlorella, sceuedesmus, chroococcus and chlamydomonas are used globally for
a) conservation movement
b) micro climate.
c) carbon sequestration
d) Biochar preparation.
Answer:
c) Carbon sequestration

Question 6.
Botanical garden, zoological park, in-vitro conservation, cryo preservation, seedling, tissue culture and DNA banks are ………………..
a) Sacred groves
b) In-situ conservation.
c) Ex-situ conservation
d) Appiko movement
Ans:
c) Ex-situ conservation

Question 7.
…………….. grooves were documented through out Tamil Nadu
a) 446
b) 447
c) 448
d) 449
Answer:
c) 448

Question 8.
Species which is present in some part of continent or present in single island is.
a) Endemic
b) Epidemic
c) Pandamic
d) Sporadic
Answer:
a) Endemic

Question 9.
Approximately one third of India flora have been identified in Indian Himalayas, Peniris India, and Andaman nicobar island it is ……………… species.
a) pandamic
b) epidemic
c) endemic
d) sporadic
Answer:
c) endemic.

Question 10.
Bentinckia condappana tree, which is endemic to………………… of Tamil Nadu and kerala
a) Western ghats
b) Peninsular
c) Coastal
d) Slope
Answer:
a) Western ghats.

Question 11.
Lianas, Nepenthes khasiyana is endemic to …………….. of meghalaya
a) Western ghat
b) Peninsular
c) Meghalaya
d) Khasi hills
Answer:
d) Khasi hills

Question 12.
Macroalgae and Maine grasses and Mangloves have ability to mitigate ………………..
a) N₂O
b) CFC
c) CO₂
d) CH₄
Answer:
c) CO₂

Question 13.
Trees like Eugenia Caryo phyllata, Tecomastans cinnamomum verum have high capacity to sequester …………….
a) N₂O
b) Carbon
c) Methane
d) CFC
Answer:
b) Carbon

Question 14.
Which one of the following is an alien invasive species?
a) Mangifera indica
b)Eichhornia crassipes
c) Solanum nigrum
d) Zizipus jujupa
Answer:
b) Eichhornia crassipes

Question 15.
Environmental management tool is an …………………
a) Biodiversity Impact assessment
b) Environmental Impact assessment
c) Bio monitoring
d) G I S
Answer:
b) Environmental Impact assessment

Question 16.
……………. system is used in mining, Aviation, surveying agricultural and marie ecosystem.
a) GIS
b) GPS
c) BIA
d) EIA
Answer:
b) GPS

Question 17.
InSAT3DR Satellites used in …………………….
a) Earth observation
b) Communication
c) Disaster management
d) Weather forecasting
Answer:
c) Disaster management

Question 18.
Now a days, scientists suggest carbon sequestration is a solution for global warming. The reason is
a) The balance between photosynthesis and respiration is disturbed
b) The absorption process of plants from the soil is disturbed
c) Due to high intensity of light respiration process is disturbed
d) Carbon sequestration is not the solution for global warming.
Answer:
d) Carbon sequestration is not the solution for global warming.

Question 19.
Reforestation refers to
a) Chipko movement
b) Development of forest in an area which was already subjects to deforestation
c) Development of forest through cultivabel land
d) None of there
Answer:
b) Development of forest in an area which was already subjects to deforestation.

Question 20.
World’s must problematic aquatic weed is
a) Clitoria
b) Parthenium
c) Eichhornia crassipes
d) sesbania
Answer:
c) Eichhornia crassipes.

Question 21.
Which of the following gas related to release cars with catalytic converter and buring of organic matter.
a) CO₂
b) Methane
c) N₂O
d) CFC
Answer:
c) N₂O

Question 22.
The Ozone layer of troposphere is called
a) Middle Ozone
b) Ozone Shield
c) Bad Ozone
d) Good Ozone
Answer:
c) Bad Ozone

Question 23.
Which one of the following is a livefence of fodder?
a) Gliricidia sepium
b) Nerium
c) Aloevera
d) Chrysanthemum
Answer:
a) Gliricidia sepium

Question 24.
Coral bleeching observed in Gulf of mannar, Tamil Nadu due to ………………..
a) Decreases of fresh water
b) Low rainfall
c) Green house
d) Dust particles
Answer:
c) green house

Question 25.
Read the following statement and fill it with correct answer in the blank A and B the ozone layer of the troposphere is called A and the ozone layer of startosphere is known as B ……………..
a) important, useful
b) bad and lower
C) bad zone, good zone
d) good zone,bad zone
Answer:
c) bad zone, good zone

Question 26.
The purple and blue colours of ozone picture indicate ……………. ozone.
a) Least
b) High
c) Medium
d) Large
Answer:
a) Least

Question 27.
The yellow and red colours of ozone picture indicate ……………….. ozone.
a) more
b) medium
c) low
d) least
Answer:
a) more

Question 28.
The objective of clean development mechanism are/is
a) Prevention of dangerous climate change
b) Reduction of emission of green house gases
c) Both (a) and (b)
d) Reduction of electricity generation and its need.
Answer:
c) Both (a) and (b)

Question 29.
Remote sensing is .
a) Mapping ocean bottom and its resources
b) Mapping forest fire
c) Mapping species distribution
d) all the above
Answer:
d) all the above

Question 30.
Bentinckia and Baccaurea are …………….. plants.
a) Invasive species
b) Endemic
c) Silvopasture
d) None of these
Answer:
b) Endemic

Question 31.
Eichornia. prosopis are plants ………………….
a) endemic
b) Invasive
c) Silvopasture
d) None of these
Answer:
b) Invasive

Question 32.
The management of forest and afforestation on barren lands is …………………..
a) agroforestry
b) Silvopasture
c) Social forestry
d) Afforestation
Answer:
c) Social forestry

Question 33.
Which is fast growing and more adopted species?
a) Invasive species
b) epidemic species
c) Pandamic species
d) Wild species
Answer:
a) Invasive species

Question 34.
The plant species (or) community acts as a measure of enviornmental conditions is referred as …………….
a) Both b and d
b) Plant indicators
c) Protocol
d) Biological indicators
Answer:
b) Plant indicators

Question 35.
The international treaty called …………….. (1987) was held in Canada on substances that deplete ozone layer.
a) Montreal Protocol
b) Kyoto protocol
c) CDM
d) CER
Answer:
a) Montreal Protocol

XIII. Two Marks

Question 1.
Why the green house gases cause global warming
Answer:

• Green House effect is a process by which radiant heat from the sun is captured by gas in the atmosphere that increase the temperature of the earth.
• The gases that capture the heat are called Green Hosue gases includes CO₂, CH₄, N₂O and CFC.

Question 2.
Draw the pie diagram which shows Relaitive contribution of green house gases.
Answer:

• Green House effect is a process by which radiant heat from the sun is captured by gas in the atmosphere that increase the temperature of the earth.
• The gases that capture the heat are called Green House gases includes CO₂, CH₄4, N₂Oand CFC
  

Question 3.
Why dust and humid night than dust free nights? (or) Does clouds and dust particles cause global warming?
Answer:

• Yes, clouds and dust particles can also produce Green House effect.
• That is why clouds dust and humid nights are warmer than clear dust free dry nights.

Global warming

• The increase in mean global temperature due to increased concentration of green house gases is called global warming.

Question 4.
Differentiate bad ozone from good ozone.
Answer:

Question 5.
Which is indicating the least ozone part and more ozone part of the atmosphere?
Answer:

• The colour view of total ozone indicating the least ozone part and more ozone part.
• The purple and blue colours are where there is the least ozone.
• The yellows and reads are where there is more ozone.

Question 6.
What would be the result of ozone depletion on living organisms?
Answer:

• UVB radiation destroys biomolecules (skin aging) and damaging living tissues.
• UV radiation causing DNA damage, enhancing skin cancer.

Question 7.
Is there any hole in the ozone?
Answer:

• No. If the ozone chield is damaged by the chlorofluoro carbon widely used in refrigeration, aerosol, chemicals used as cleaners in industries.
• The decline in the thickness of the ozone layer over restricted area is called ozone hole.

Question 8.
What are the main objectives of Montreal protocol?
Answer:
The main goal of it is gradually eliminating the production and consumption of ozone depleting substances and to limit their damage on the earth’s ozone layer.

Question 9.
What is Agroforestry?
Answer:

• Agroforestry is an integration of trees crops and livestock on the same plot of land.
• The main objective is on the interaction among them.

Question 10.
What is silvopasture? (or) What is of degraded forests recreation forestry?
Answer:

• The production of wood plants combined with pasture is referred to silvopasture system.
• The trees and shrubs may be used primarily to produce fodder for livestock (or) grown for timber, fuel wood and fruit (or) to improve the soil.

Question 11.
Differentiate Agroforestry and Social forestry.
Answer:

Question 12.
What could be the causes for deforestation?
Answer:

• The conversion of forest into agricultural plantation and livestock ranching is a major cause of deforestation.
• Developmental activities like road construction electric tower lines and dams.

Question 13.
Who is called as Forest Man of India? Why?
Answer:

• Former vice Chancellor of Jawaharlal Nehru University Sudhir Kumar named Jadav payeng as Forest Man of India.
• He has transformed the world’s largest river island Majuli (located on river of Brahmaputra) into dense forest, home to rhinos, deers, elephants, tigers and birds.

Question 14.
What is invasive species?
Answer:
A non – native species to the ecosystem (or) country that spreads naturally, interferes with native species, poses a serious threat to the ecosystem and cause economic loss.

Question 15.
Conservation movement – What does it signifies.
Answer:
A community level participation can help in preservation and conservation of our environment.

Question 16.
What is Five F’s? Which is related to what?
Answer:

• > Main aim of chipko movement was to give a slogan of Five – Fs. food, fodder, fuel, fibre and fertilizer.
• > It make the communities self sufficiency in their basic needs.

Question 17.
Each grove is an abode of a diety mostule Village God (or) Goddesses, like Aiyanar (or) Amman – What does it states?
Answer:

• There are called scared groves.
• The patches (or) grove of cultivated trees which are community protected and based on religeous belief system.
• There are 448 groves were documented through out Tamilnadu.

Question 18.
What is Biochar?
Answer:

• It is another Long term method to store carbon.
• Plants are partly burnt such as crop waste woods to become carbon rich slow decomposing substances called Biochar.

Question 19.
What is carbon foot print?
Answer:
C.F.P is the total amount of green house gases produced by human activities such as agriculture,Industries, deforestation, waste disposal burning fossil fuel directly (or) indrectly.

Question 20.
What are the benefits related to environmental impact assessment?
Answer:

• Benefit are healthier environment.
• Maintenance of bio diversity
• Decreased resource usage.
• Reduction in gas emission and environment damage.

Question 21.
What is Biomonitoring?
Answer:
The act of observing and assessing the current state ongoing
a) changes in ecosystem
b) bio diversity components
c) landscape including natural habitat
d) population and species.

Question 22.
What is Agricultural drone?
Answer:

• Agricultural drone is an unmanned aerial vehicle to help increased crop production and monitor crop, growth.
• Farmers can see their fields from the sky.

Question 23.
What are the uses of agricultural drone (or) Bio monitoring?
Answer:

• This bird’s – eye – view instrument can reveal many issues such as irrigation problems, soil variation pest and fungal infestations
• It is also used for cost effective safe method of spraying pesticides and fertilizers.

Question 24.
What Geographic Information system?
Answer:

• GIS is a computer system for capturing storing, checking, and displaying data related to positions on earths surface.
• Also manipulate, analyse, manage and present special (or) geographic data.

Question 25.
What are the scope of GPS satellite? (or) (Global Positioning System)
Answer:

• Global positioning system is satellite navigation system used to determine the ground position of an object.
• A constellation of approximately 30 well spaced Satellites that orbit the earth and make it possible geographic location.
  eg:- Mining, surveying Agricultural and marine ecosystem.

Question 26.
Which is providing exact picture and data on identification of even a single tree to large area and wild lifor for classification?
Answer:
Remote sensing is the process of datelining and monitoring the physical characteristics of an area by measuring its reflected and emitted radiation at a distance from the target area.

Question 27.
What are green house gases?
Answer:

• The gases that capture heat are called Green house gases.
• Which includes CO₂ , CH₄ , Nitrous oxide (N₂O) chioro fluoro carbon.

Question 28.
If you buy imported fruit like Kilvi indirectly it increases (GFP) carbon foot print. How?
Answer:
The fruit has travelled a long distance in shipping (or) airlines thus emitting tons of cO₂.

Question 29.
Which is names of Alien invasive species?
Answer:

• Eichhornia crassipes
• Prosopisjuliflora
• parthenium hysterophorus

Question 30.
Define carbon sink. Give an example
Answer:
Any system having the capacity to accumulate more atmospheric carbon during a given time interval than releasing CO₂
Example :- forest, soil, clean are natural sinks, Landfills are artificial sinks.

XIV. Three Marks

Question 1.
What are the effects of Green House Gases? Give example.
Answer:

• Increase in green house gases leads to irreversible changes in ecosystem and climatic patterns.
• eg:- Coral bleaching observed in Gulf of mannen Tamilnadu. [coral system is affected by increase in temperature]

Question 2.
Why we want to control global warming? Write the (or) effects of global warming.
Answer:

• Rise in global temperature which causes sea levels to rise as polar ice caps and glaciers begin to melt causing submergence of many coastal cities in many parts of the world.
• There will be drastic change in whether patterns bringing more floods (or) droughts in some areas
• Biological diversity may get modified.

Question 3.
What are the reasons for global warming?
Answer:

• Drastic increase in population resulted in demand for more productivity of food, fibres fuels.
• Which led to many environmental issues in agriculture, land use modification resulting in loss of biodiversity, land degradation reducing in fresh water availability resulting man mode global warming.

Question 4.
Global warming is a threatening problem nowadays. What are the stratifies to deal with it?
Answer:

• Increasing the vegetation cover, grow more trees.
• Reducing the use of fossil fuels and green house gases.
• Minimising use of nitrogenous fertilizers and aerosols.

Question 5.
Why ozone layer is known as ozone shield?
Answer:
Ozone layer is a region of earth’s stratosphere that absorbs most of the sun’s ultra violet radiation. So it is called as ozone shield.

Question 6.
What is Dobson unit?
Answer:

• DU is the unit of measurement for total ozone.
• One DU (0.001 atm.cm) is the number of molecules of ozone that would be required to create a layer of pure ozone 0.01 mm thick at a temperature of 0°C and a pressure of atmosphere.
• Total ozone layer over the earth surface is 0.3 centimetres (3mm) thick and is writtern can we ozone

Question 7.
Can we see ozone layer?
Answer:

• It is not very strong layer but it will contribute blue colour to the sky even at the very low concentration.
• We can visualize by using satellites.

Question 8.
Why was Montreal protocol signed?
Answer:

• During 1970’s research findings indicated that man – made chlorofluoro carbons reduce and convert ozone molecules in the atmosphere.
• Vienna conference provided the frame works necessary to create regulative measures in the form of Montreal protocol to climate production and consumption of ozone depleting substances.

Question 9.
What is Montreal protocol? What is its aim?
Answer:

• The international treaty called the Montreal protocol (1987) was held in Canada on substances that deplete ozone layer.
• The main goal of it is gradually eliminating the production and consumption of ozone depleting substances and to limit their damage on the earth ozone layer.

Question 10.
What is clean development mechanism (or) Kyoto protocol?
Answer:

• CDP (or) protocol (2007) provides project based mechanisms with two objectives.
• To prevent dangerous climate change and to reduce green house gas emissions.
• It help the countries to reduce (or) limit emission and stimulate sustainable development.
  eg:- Replacement of conventional electrification projects with solar panels (or) other energy efficient boilers.

Question 11.
Differentiate protein Bank from live fence of fodder trees and hedges.
Answer:

Question 12.
What is social forestry?
Answer:

• It refers to the sustainable management of forests by local communities.
• de-pollution, deforestation, forest restoration and providing indirect employment.
• Its refers to the management of forests and afforestation of barren lands.

Question 13.
Why forestry extension centres are important? (or) What ae the major activities of forestry extension centre?
Answer:

• Training on tree growing methods.
• Publicity and information regarding tree growing
• Raising and supply of seedlings on subsidy.
• Awareness creation among school children and youth about the importance of forests through training camps.

Question 14.
Forest, soil, ocean are called natural sink why? (or) What is carbon sink?
Answer:
Any system having the capacity to accumulate more atmospheric carbon during a given time interval than releasing CO₂.
eg: Landfills are artificial sinks forest, soil, ocean are natural sinks.

Question 15.
Why Eichhorinia crassipes is called Terror of Bengal?
Answer:

• Its widespread growth affects the growth of phytoplanktons and finally changing the aquatic ecosystem.
• It also decreases the oxygen content of the water bodies which leads to eutrophication.
• It poses a threat to human health because it creates a breeding habitat for disease causing mosquitoes and snails.

Question 16.
Differentiate Chipko movement and Appiko movement.
Answer:

Question 17.
Differentiate in – Situ conservation and ex-site conservation.
Answer:

Question 18.
What is Geological sequent ration?
Answer:
Various safe sites have been selected for permanent storage, liquid storage in the ocean and solid storage by reduction of CO₂ called Geological sequestration.

Question 19.
Write the benefits of environmental impact assessment.
Answer:

• A healthier environment.
• Maintenance of biodiversity
• Decrased resource usage.
• Reduction in gas emission and environment damage.

Question 20.
What is environmental impact assessment?
Answer:

• It is an environmental management tool.
• It helps to regulate and recommend optimal use of natural resources with minimum impact on ecosystem and biotic communication.

Question 21.
What is Bio – diversity impact Assessment?
Answer:
It can be defined as a decision supporting tool to help biodiversity development planning and implementation.

Question 22.
How Bio – diversity impacts can be assessed?
Answer:

• Change in land use and cover
• Fragmentation and isolation
• External inputs such as emissions, effluents and chemicals impact on endemic and threatened flora and faura.

Question 23.
Write a short note on production of methane?
Answer:
Methane is 20 times as effective as CO ₂ at trapping heat in the atmosphere.
It sources are attributed paddy cultivation, cattle rearing, bacteria in water bodies, fossil fuel production. Ocean, non – wetland soils and forest/wild fires.

Question 24.
Write a note on (N₂O) Nitrous Oxide?
Answer:

• It is naturally produced in oceans from biological sources of soil and water due to microbial actions and rainforests.
• Man-made sources include nylon and nitric acid production, use of fertilizers in agriculture manures cars with catalytic converter and burning of organic matter.

Question 25.
Write the significances of lakes.
Answer:

• Water bodies like lakes, ponds not only provide us a number of environmental benefits but they strengthen our economy as well as our quality of life like health.
• Lakes as a storage of rain water provides drinking water, improves ground water level and preserve the fresh water bio-diversity and habitat of the area where in occurs
• In terms of services lakes offer sustainable solutions to key issues of water management and climatic influences and benefits like nutrient retention, influencing local rainfall removal of pollutants, phosphorous and nitrogen and carbon sequestration.

XV. Five Marks

Question 1.
Define Global warming. Write the reasons for it what are green hours Gases? What are the human activities lead to produce green house effect?
Answer:
Global Warming

• The increase in mean global temperature due to increased concentration of green house gases is called global warming.

Reasons for global warming

• Drastic increase in population resulted in demand for more productivity of foof, fibres fuels.
• Which led to many environmental issues in agriculture, land use modification resulting in loss of biodiversily, land degradation reduction in fresh water availanility resulting man made global warming.

green house gases

• The gases that capture heat are called Green house Gases.
• Which includes CO₂, CH₄, Nitrous oxide (N₂O) and chlorofluoro carbon.

Human activities lead to produce the green house effect.
– Human activities lead to produce the green house effect by

• Buring fossil, which release CO₂ and CH₄
• Way of Agriculture and animal husbandry practice.
• Electrical gadgets like refrigerator and air conditioners release chloro fluoro carbons.
• The fertilizers used in Agriculture which release N₂O
• The emissions from automoblies.

Question 2.
Write about Natural and anthropogenic sources of Green House Gases Emission.
Answer:
CO₂ (Carbon dioxide)

• Coal based power plants, by the burning of fossil fuels for electricity generation.
• Combustion of fuels in the engines of automoniles, commercial vehicles and air plances contribute the most of global warming.
• Agricultural practices like stubble burning result in emission of CO₂
• Natural from organic matter, volcanoes, warm oceans and sediments.

Methane

• Methane is 20 times as effective as CO₂ at trapping heat in the atomosphere.
• Its sources are paddv cultivations field cattle rearing, bacteria in water bodies , fossil fuel production, ocean, non-wetland soils and forest/ wild fires.

N₂O (Nitrous oxide)

• It is naturally produced in Oceans from biological sources of soil and water due to microbial actions and rainforest.
• Man-made sources include nylon and nitric acid production, use of fertilizers in agriculture, manures, cars with catalytic com orter and burning of organic matter.

Question 3.
What are the effects of Global warming on plants? (or) What are the changes may occur on plants and climate due to global warming?
Answer:
Global Warming Effects on Plants

• Low agriclutural productivity in tropics.
• Frequent heat waves (Weeds, pests, fungineed warmer temperature).
• Increase of vectors and epidemics.
• Strong storms and intense flood damage
• Water crisis and decreased irrigation.
• Change in flowering seasons and pollinators.
• Change in Species distributional ranges.
• Species extinction.

Question 4.
The presence (or) absence of certain plants indicates the state of environment by its response. What does it called ? Give some examples.
Answer:
Plant indicators

• The presence (or) absence of certain plants indicate the state environment by their response.
• The plant species or plant community acts as a measure of environmental conditions, it is referred as biological indicatores or phytoindicators or plant indicators.

Examples

Question 5.
If the ozone layer is affected, U-V radiations on the sun will reach the earth surface and sure many damage.
Can you list out some effects of ozone depletion?
Answer:
Effects of Ozone depletion.
The main ozone depletion are:

• Increases the incidence of cataract, throat and lung irritation and aggravation of asthma or emphysema, skin cancer and diminishing the functioning of immune system in human beings.
• Juvenile mortality of animals.
• Increased incidence of mutations.
• In plants, photosynthetic chemicals will be affected and therefore photosynthesis will be inhibited. Decreased photosyntheses is will result in increased atmospheric CO₂ resulting in global warming and also shortage of food leading to food crisis.
• Increase in temperature changes the climate and rainfall pattern which may result in flood/drought, sea water rise, imbalance in ecosystems affecting flora and fauna.

Question 6.
Write an essay on ozone.
Answer:
Ozone shield

• Ozone layer is a region of earth’s stratosphere that absorbs most of the sun’s ultra violet radiation. So it is called as ozone shield.

Ozone HOLE

• No. If the ozone shield is damaged by the cholrofluorocarbons widely used in refrigeration, aerosol, chemicals used as cleanes in industries.
• The decline in the thickness of the ozone layer over restricted area is called ozone hole.

Montreal protocol

• During 1970s research findings indicated that man-made chlorofluoro carbons reduce and convert ozone molecules in the atmosphere.
• Vienna conference provided the frame works necessary to create regulative measures in the form of montreal protocol to elimate production and consumption of ozone depleting substances.
• The international treaty called the montreal protocol (1987) was held in Canada on substances that deplete ozone layer.
• The main goal of it is gradually eliminating the production and consumption of ozone depleting substances and to limit their damage on the earths ozone layer.

Kyoto protocol

• C D M (or) protocol (2007) provides project based mechanisms with two objectives.
• To prevent dangerous climate change and to reduce green house gas emissions.
• It help the contries to reduce (or) limit emission and stimulate sustainable development
• eg: Replacement of conventional electrification projects with solar panels (or) other energy efficient boilers.

Question 7.
Write the objectives and achievements of Afforestation.
Answer:
Afforestation Objectives

• To increase forest cover, planting more trees, increases
• O₂
• production and air quality.
• Rehabilitation of degraded forests to increase carbon fixation and reducing CO₂ from atmosphere.
• Raising bamboo plantations.
• Mixed plantations of minor forest produce and medicinal plants.
• Regeneration of indigenous herbs/ shrubs. Awareness creation, monitoring and evaluation.

Achievements

• Degraded forests were restored.
• Community assets like overhead tanks bore- wells, hand pumps, community halls, libraries, etc were established.
• Environmental and ecological stability was maintained.
• Conserved bio-diversity, wildlife and genetic resources.
• Involvement of community especially women in forest management.

Question 8.
What is Bio-Diversity conservation movement.
Answer:
Conservation movement

• A community level participation can help in preservation and conservation of our environement.
• Our environment is a common treasure for all the living organisms on earth. Every individual should be aware of this and participate actively in the programs meant for the conservation of the local environment.
• Indian histroy has witnessed many people movements for the protection of environment.

Chipko Movement

• The tribal women of Himalayas protested against the exploitation of forests in 1972. Later on it transformed into Chipkon Movement by Sundarlal Bahuguna in , Mandle village of Chamoli district in 1974.
• People protested by hugging trees together which were felled by a sports goods company.

Features of chipko Movement

• This movement remained non political
• It was a voluntary movement based on Gandhian thought.
• It was concerned with the ecological balance of nature.
• Main aim of Chipko movement was to give a slogan of five F’s _ Food, Fodder, Fuel, Fibre and Fertilizer, to make the communities self sufficient in all their basic needs.

Appiko Movement

• The famous Chipko Andolen in the Himalayas inspired the villagers of Uttar Karnataka to launch a similar movement to save their forests.
• This, movement started in Gubbi Gadde a small village near Sirsi in Karnataka by Panduranga Hegde.
• This movement started to protest against felling of trees, monoculture, forest policy and deforestation.

Question 9.
Write about endemic centre’s and endemic plants.
Answer:

• Endemic Centres and Endemic Plants.
• Endemic species ar plants and animals that exist only in one geographic region.
• It may be due to various reasons such as isolation, interspecific interactions, seeds dispersal problems.
• There are 3 Megacentres of endmism and 27 microendemic centres in India.
• That is ¡nid an Himalayas,Peninsular India and Andaman nicobar islands.
• A large percentage of Endemic Poaceae. Apiaceae, Asteraceae and Orchidaceae.

Question 10.
Write about two approach to mitigate global warming.
Answer:
Carbon Capture and Storage (CCS)

• Carbon capture and storage is a téchnology of capturing carbondioxide and injects it deep into
  the underground rocks into a depth of 1 km
• Various safe sites have been selected for permanent storage i.n various deep geological
  formations, liquid storage in the Ocean and solid storage by reduction of CO ₂ with metal oxide to
  produce stable carbonates. It is also known as Geological sequestration.

Carbon Sequestration

• Carbon sequestration is the process of capturing and storing CO ₂ which reduces the amount of
  CO ₂ in the atmosphere with a goal of reducing global climate change.
• Carbon sequestration occurs naturally by plants and in ocean.
• Terrestrial sequestration is typically accomplished through forest and soil conservation practices that enhance the storage carbon.
• As an example microalgae such as species of Chiorella, Scenedesmus, Chroococcus and Chiamydomonas are used globally for CO ₂sequestration.
• Macroalgae and marine grasses and mangroves are also have ability to mitigate carbon – di – oxide.

Question 11.
What is Carbon foot print ? How will you reduce this CFP?
Answer:
Carbon Foot Print (CPF)

• Every human activity leaves a mark just like our footprint. This Carbon foot print is the total amount of green house gases produced by human activities such as agriculture, industries, deforestation, waste disposal, buring fossil etc.

To Reduce carbon foot print

• Eating indigenous fruits and products.
• Reduce useofyour electronic devices.
• Reduce travelling
• Do not buy fast and preserved, processed, packed foods.
• Plant a garden
• Less consumption of meat and sea food. Poultry requires little space, nutrients and less pollution comparing cattle farming.
• reduce use of Laptops (when used for 8 hours, it releases nearly 2 kg, of CO₂ annually).
• Line dry your clothes.

Question 12.
Write an essay on Environmental impact Assesment and its benefits.
Answer:
Environmental Impact Assessment (EIA)

• Environment Impact Assessment is an environment management tool.
• It helps to regulate and recommend optimal use of natural resources with minimum impact on ecosystem and biotic communities.
• It is used to predict the environmental consequences of future, (example : river, projects, dams, highway projects.)
• It reduces environmental stress thus helping utilization of natural resources and disposal of wastes to avoid environemental degradation.

The benefits of EIA to society

• A healthier environment
• Maintenance of biodiversity
• Decreased resource usage
• Reduction in gas emission and environment demage.

Question 13.
Write about Bio-diversity impact Assesment and How will you anses its impact.
Answer:
Biodiversity Impact Assessment (BI A)

• Biodiversity Impact Assessment can be defined as a decision supporting tool to help biodiversity inclusive of development, planning and implementation.

Bio-diversity Impacts can be assessed by

• Change in land use and cover.
• Fragmentation and isolation.
• Extraction
• External inputs such as emissions, effluents and chemicals.
• Introduction of invasive, alien or genetically modified species.
• Impact on endemic and threatened flora and fauna.

Question 14.
What is Geographic Information system? What are its importance?
Answer:
Geographic Information System

• GIS is a computer system for capturing, storing checking and displaying data related to positions on Earth’s surface. Also to manipulate, analyse, manage and present spacial or geographic data.
• GPS is a satellite navigation system used to determine the ground position of an object. It is a constellation of approximately 30 well spaced satellites that orbit the earth and make it possible geographic location.

Importance of GIS

• Environmental impact assessment.
• Disaster management.
• Zoning of landslide hazard
• Determination of land cove and land use
• Estimation of flood damage.
• Management of natural resources.
• Soil mapping
• Wetland mapping
• Irrigation management and identification of volcanic hazard.
• Vegatation studies and mapping of threatened and endemic species.

Question 15.
What is Remote sensing and what are its uses?
Answer:
Remote Sensing

• Remote Sensing is the process of detecting and monitoring the physical characteristics of an area by measuring its reflected and emitted radiation at a distance from the targeted area. It is an tool used in conservation practices by giving exact picture and data on identification of even a single tree to large area of vegetation and wild life.
• Mapping of forest fire and species distribution.
• Mapping ocean bottom and its resources.

Applications of Satellites

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Zoology Guide Pdf Chapter 2 Human Reproduction Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction

12th Bio Zoology Guide Human Reproduction Text Book Back Questions and Answers

Question 1.
The mature sperms are stored in the ………………..
(a) Seminiferous tubules
(b) Vas deferens
(c) Epididymis
(d) Seminal vesicle
Answer:
(c) Epididymis

Question 2.
The male sex hormone testosterone is secreted from ………………….
(a) Sertoli cells
(b) Leydig cell
(c) Epididymis
(d) Prostate gland
Answer:
(b) Leydigcell

Question 3.
The glandular accessory organ which produces the largest proportion of semen is …………………
(a) Seminal vesicle
(b) Bulbourethral gland
(c) Prostate gland
(d) Mucous gland
Answer:
(a) Seminal vesicle

Question 4.
The male homologue of the female clitoris is …………………
(a) Scrotum
(b) Penis
(c) Urethra
(d) Testis
Answer:
(b) Penis

Question 5.
The site of embryo implantation is the …………………
(a) Uterus
(b) Peritoneal cavity
(c) Vagina
(d) Fallopian tube
Answer:
(a) Uterus

Question 6.
The foetal membrane that forms the basis of the umbilical cord is ………………..
(a) Allantois
(b) Amnion
(c) Chorion
(d) Yolk sac
Answer:
(a) Allantois

Question 7.
The most important hormone in initiating and maintaining lactation after birth is .. ………………….
(a) Oestrogen
(b) FSH
(c) Prolactin
(d) Oxytocin
Answer:
(c) Prolactin

Question 8.
Mammalian egg is …………………
(a) Mesolecithal and non-cleidoic
(b) Microlecithal and non-cleidoic
(c) Alecithal and non-cleidoic
(d) Alecithal and cleidoic
Answer:
(c) Alecithal and non-cleidoic

Question 9.
The process which the sperm undergoes before penetrating the ovum is …………………..
(a) Spermiation
(b) Cortical reaction
(c) Spermiogenesis
(d) Capacitation
Answer:
(d) Capacitation

Question 10.
The milk secreted by the mammary glands soon after child birth is called ………………….
(a) Mucous
(b) Colostrum
(c) Lactose
(d) Sucrose
Answer:
(b) Colostrum

Question 11.
Colostrum is rich in …………………
(a) Ig E
(b) Ig A
(c) Ig D
(d) Ig M
Answer:
(b) Ig A

Question 12.
The Androgen Binding Protein (ABP) is produced by…………………
(a) Leydig cells
(b) Hypothalamus
(c) Sertoli cells
(d) Pituitary gland
Answer:
(c) Sertoli cells

Question 13.
Which one of the following menstrual irregularities is correctly matched?
(a) Menorrhagia – excessive menstruation
(b) Amenorrhoea – absence of menstruation
(c) Dysmenorrhoea – irregularity of menstruation
(d) Oligomenorrhoea – painful menstruation
Answer:
(b) Amenorrhoea – absence of menstruation

Question 14.
Find the wrongly matched pair
(a) Bleeding phase – fall in oestrogen and progesterone
(b) Follicular phase – rise in oestrogen
(c) Luteal phase – rise in FSH level
(d) Ovulatory phase – LH surge
Answer:
(c) Luteal phase – rise in FSH level

Answer the following type of questions:

Assertion (A) and Reason (R)
(a) A and R are true, R is the correct explanation of A
(b) A and R are true, R is not the correct explanation of A
(c) A is true, R is false
(d) Both A and R are false

Question 15.
A – In human male, testes are extra abdominal and lie in scrotal sacs.
R – Scrotum acts as thermoregulator and keeps temperature lower by 2°C for normal sperm production.
Answer:
(a) A and R are true, R is the correct explanation of A

Question 16.
A – Ovulation is the release of ovum from the Graafian follicle.
R – It occurs during the follicular phase of the menstrual cycle.
Answer:
(c) A is true, R is false

Question 17.
A – Head of the sperm consists of acrosome and mitochondria.
R – Acrosome contains spiral rows of mitochondria.
(d) Both A and R are false
Answer:
(d) Both A and R are false

Question 18.
Mention the differences between spermiogenesis and spermatogenesis.
Answer:

1. Spermiogenesis: Transformation of spermatids into mature sperm.
2. Spermatogenesis: Spermatogenesis is the sequence of events in the seminiferous tubules of testes that produces male gametes, the sperms.

Question 19.
At what stage of development are the gametes formed in newborn males and females?
Answer:
In males, at puberty, the spermatogonia (sperm mother cells) begin to undergo meiotic division and produces sperms throughout life, whereas in females during the stage of foetal development, the germinal epithelial cells undergo mitosis and produce oogonia (egg mother cells) and they further enter prophase-I of meiosis-I forming primary oocytes and get arrested. No more oogonia is formed further. At puberty, out of million eggs (prime oocytes) produced at birth only 300-400 will ovulate till menopause.

Question 20.
Expand the acronyms

1. FSH
2. LH
3. hCG
4. hPL

Answer:

1. FSH – Follicle Stimulating Hormone
2. LH – Luteinizing Hormone
3. HCG – Human Chorionic Gonadotropin
4. HPL – human Placental Lactogen

Question 21.
How is polyspermy avoided in humans?
Answer:
Once fertilization is accomplished, cortical granules from the cytoplasm of the ovum form a barrier called the fertilization membrane around the ovum preventing further penetration of other sperms. Thus polyspermy is prevented.

Question 22.
What is colostrum? Write its significance.
Answer:
The mammary glands secrete a yellowish fluid called colostrum during the initial few days after parturition. It has less lactose than milk and almost no fat, but it contains more proteins, vitamin A and minerals. Colostrum is also rich in antibodies. This helps to protect the infant’s digestive tract against bacterial infection.

Question 23.
Placenta is an endocrine tissue. Justify.
Answer:
During pregnancy, the placenta acts as a temporary endocrine gland and produces large quantities of human Chorionic Gonadotropin (hCG), human Chorionic Somatomammotropin (hCS) or human Placental Lactogen (hPL), oestrogens and progesterone which are essential For a normal pregnancy. A hormone called relaxin is also secreted during the later phase of pregnancy which helps in relaxation of the pelvic ligaments at the time of parturition.

Question 24.
Draw a labeled sketch of a spermatozoan.
Answer:

Question 25.
What is inhibin? State its functions.
Answer:
inhibin is a hormone secreted by Sertoli cells of the testes which are involved in the negative feedback control of sperm production.

Question 26.
Mention the importance of the position of the testes in humans.
Answer:
The testes are positioned in such a way hanging out from the body in scrotal sac that provides optimal temperature 2°C to 3°C lower than internal body temperature for effective sperm production.

Question 27.
What is the composition of semen?
Answer:
Semen or seminal fluid is a milky white fluid which contains sperms and seminal plasma, which is secreted from the seminal vesicles, prostate gland, and bulbourethral glands.

Question 28.
Name the hormones produced from the placenta during pregnancy.
Answer:
human Chorionic Gonadotropin (hCG)
human Placental Lactogen (hPL)
Relaxin.

Question 29.
Define gametogenesis.
Answer:
Gametogenesis is the process of formation of gametes i.e., sperms and ovary from the primary sex organs in all sexually reproducing organisms. Meiosis plays the most significant role in the process of gametogenesis.

Question 30.
Describe the structure of the human ovum with a neat labelled diagram.
Answer:
Human ovum is non-cleidoic, alecithal and microscopic in nature. Its cytoplasm called ooplasm contains a large nucleus called the germinal vesicle. The ovum is surrounded by three coverings namely an inner thin transparent vitelline membrane, middle thick zona pellucida and outer thick coat of follicular cells called corona radiata. Between the vitelline membrane and zona pellucida is a narrow perivitelline space.

Question 31.
Give a schematic representation of spermatogenesis and oogenesis in humans.
Answer:

Question 32.
Explain the various phases of the menstrual cycle.
Answer:
Menstrual cycle: The menstrual or ovarian cycle occurs approximately once in every 28/29 days during the reproductive life of the female from menarche (puberty) to menopause except during pregnancy. The cycle of events starting from one menstrual period till the next one is called the menstrual cycle during which cyclic changes occur in the endometrium every month. Cyclic menstruation is an indicator of normal reproductive phase.

Menstrual cycle comprises of the following phases:

1. Menstrual phase
2. Follicular or proliferative phase
3. Ovulatory phase
4. Luteal or secretory phase

1. Menstrual phase: The cycle starts with the menstrual phase when menstrual flow occurs and lasts for 3-5 days. Menstrual flow is due to the breakdown of endometrial lining of the uterus, and its blood vessels due to decline in the level of progesterone and oestrogen. Menstruation occurs only if the released ovum is not fertilized. Absence of menstruation may be an indicator of pregnancy. However it could also be due to stress, hormonal disorder and anaemia.

2. Follicular or proliferative phase: The follicular phase extends from the 5th day of the cycle until the time of ovulation. During this phase, the primary follicle in the ovary grows to become a fully mature Graafian follicle and simultaneously, the endometrium regenerates through proliferation. These changes in the ovary and the uterus are induced by the secretion of gonadotropins like FSH and LH, which increase gradually during the follicular phase. It stimulates follicular development and secretion of oestrogen by the follicle cells.

3. Ovulatory phase: Both LH and FSH attain peak level in the middle of the cycle (about the 14th day). Maximum secretion of LH during the mid cycle called LH surge induces the rupture of the Graafian follicle and the release of the ovum (secondary oocyte) from the ovary wall into the peritoneal cavity. This process is called as ovulation.

4. Luteal or secretory phase: During luteal phase, the remaining part of the Graafian follicle is transformed into a transitory endocrine gland called corpus luteum. The corpus luteum secretes large amount of progesterone which is essential for the maintenance of the endometrium. If fertilization takes place, it paves way for the implantation of the fertilized ovum.

5. The uterine wall secretes nutritious fluid in the uterus for the foetus. So, this phase is also called as secretory phase. During pregnancy all events of menstrual cycle stop and there is no menstruation. In the absence of fertilization, the corpus luteum degenerates completely and leaves a scar tissue called corpus albicans. It also initiates the disintegration of the endometrium leading to menstruation, marking the next cycle.

Question 33.
Explain the role of oxytocin and relaxin in parturition and lactation.
Answer:

1. Relaxin is the hormone secreted by the placenta that causes the contraction of pelvic joints and promotes parturition (childbirth).
2. Oxytocin causes the Let-down reflex – the actual ejection of milk from the alveoli of mammary glands. Oxytocin also stimulates the uterus to regain its pre-pregnancy size after childbirth.

Question 34.
Identify the given image and label its parts marked as a, b, c and d.
Answer:

The given image is the diagram of human egg cell or ovum,
a-vitelline membrane; b- Nucleus; c- Zona pellucida; d- Corona radiata

Question 35.
The following is the illustration of the sequence of ovarian events (a-i) in a human female.

(a) Identify the figure that illustrates ovulation and mention the stage of oogenesis it represents.
(b) Name the ovarian hormone and the pituitary hormone that have caused the above- mentioned events.
(c) Explain the changes that occurs in the uterus simultaneously in anticipation.
(d) Write the difference between C and H.
Answer:
(a) A- Primordial follicle; B- Primary follicle; C- Secondary follicle; D-Tertiary follicle; E- Mature graafian follicle; F- Ovulation (release of egg); G- Empty Graafian follicle; H- Corpus luteum; I – Corpus albicans.

(b) Pituitary hormones: Follicle Stimulating Hormones (FSH) and Lutenizing Hormone (LH). Ovarian hormones: Estrogen and Progesterone.

(c) At the start of menstrual cycle, the endometrium of uterus starts regenerating through proliferation of cells induced by FSH and CH. After ovulation, the progesterone secreted by corpus luteum prepares the endometrium (uterine wall) to receive the egg if it is fertilized.

(d) C- Secondary follicle
H – Corpus luteum
During development of ovum, the primary follicle gets surrounded by many layers of granular Cells and forms a new layer called secondary follicle.

Corpus luteum is the empty graafian follicle that remains after ovulation. It acts as a transitory endocrine gland secreting progesterone to maintain pregnancy.

12th Bio Zoology Guide Human Reproduction Additional Important Questions and Answers

12th Bio Zoology Guide Human Reproduction One Mark Questions and Answers

Question 1.
The developing spermatoza are nourished by………………..
(a) Leydig cells
(b) Sertoli cells
(c) Follicular cells
(d) Epididymis
Answer:
(b) Sertoli cells

Question 2.
Identify the correct sequence of reproductive events in human beings.
(a) Insemination, Implantation, Fertilization, Parturition and Placentation.
(b) Implantation, Fertilization, Insemination, Placentation and Parturition.
(c) Implantation, Insemination, Fertilization, Parturition and Placentation.
(d) Insemination, Fertilization, Implantation, Placentation and Parturition.
(d) Insemination, Fertilization, Implantation, Placentation and Parturition.
Answer:
(d) Insemination, Fertilization, Implantation, Placentation and Parturition.

Question 3.
Expulsion of baby from the mother’s womb is referred as ………………
(a) Ejection
(b) Relaxation
(c) Parturition
(d) Implantation
Answer:
(c) Parturition

Question 4.
Match the Column I with Column II

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

Question 5.
Which of the following statement is not correct?
(i) Interstitial cells are seen surrounding the seminiferous tubule.
(ii) Nurse cells secrete inhibin.
(iii)Males have single prostate gland which encircles the urethra.
(iv) Insemination, Fertilization, Implantation, Placentation and Parturition.
(a) i and ii
(b) iii only
(c) iii and iv
(d) iv only
Answer:
(d) iv only

Question 6.
Assertion (A): In scrotum, the temperature is maintained 2 – 3°C lower than body temperature.
Reason (R): Reduced temperature results in efficient sperm production.
(a) R explains A.
(b) A is right R is wrong.
(c) A and R are right. R does not explains A.
(d) Both A and R are wrong.
Answer:
(a) R explains A.

Question 7.
Assertion (A): The acrosome of the sperm cell contains sperm lysin.
Reason (R): Sperm lysin destroys the deformed sperm cells.
(a) R explains A.
(b) A is right, R is wrong.
(c) A and R are right. R does not explains A.
(d) Both A and R are wrong.
Answer:
(A) A is right, R is wrong.

Question 8.
Assertion (A): Human ovum is non – cieidoic
Reason (R): Human does not contain yolk.
(a) R explains A.
(b) A is right, R is wrong.
(c) A and R are right. R does not explains A.
(d) Both A and R are wrong.
Answer:
(c) A and R are right. R does not explains A.

Question 9.
Assertion (A): Menopause refers to the absence of menstruation during pregnancy.
Reason (R): Ovulation occurs during menstrual phase.
(a) R explains A.
(b) A is right, R is wrong.
(c) A and R are right. R does not explains A.
(d) Both A and R are wrong.
Answer:
(d) Both A and R are wrong.

Question 10.
Assertion (A): Cervix is common site of ectopic pregnancies
Reason (R): Implantation of fertilized ovum outside uterus.
(a) A is wrong, R is right.
(b) A is right, R is wrong.
(c) A and R are right. R does not explains A.
(d) Both A and R are wrong.
Answer:
(a) A is wrong, R is right.

Question 11.
Which of the following contributes to the seminal plasma?
(i) Cowper’s gland
(ii) Seminal vesicles
(iii) Prostate gland
(iv) Bulbourethral gland

(a) ii, iii and ii
(b) i, ii, and iii
(c) i, iii and iv
(d) all the above
Answer:
(d) all the above

Question 12.
Organ of copulation in human female is ……………
(a) Cevix
(b) Fundus
(c) Vagina
(d) Uterus
Answer:
(c) Vagina

Question 13.
Identify the gland which is homologous to the Cowper’s glands of male.
(a) Bartholin’s gland
(b) Bulbourethral gland
(c) Prostate gland
(d) Skene’s gland
Answer:
(a) Bartholin’s gland

Question 14.
Find out the proper sequence representing the parts of female reproductive system.
(a) Vagina → Ovary → Uterus → Cervix →Infundibulum → Oviduct
(b) Vagina → Ovary → Oviduct → Infundibulum → Cervix → Uterus
(c) Ovary → Infundibulum → Oviduct → Uterus → Cervix → Vagina
(d) Oviduct → Ovary → Uterus → Infundibulum Vagina → Cervix
Answer:
(c) Ovary → Infundibulum → Oviduct → Uterus → Cervix → Vagina

Question 15.

(a) Spermatogenesis
(b) Spermiation
(c) Spermiogenesis
(d) Gametogenesis
Answer:
(c) Spermiogenesis

Question 16.
An adult male produces an average of…………….. sperms per day
(a) 200 million
(b) 300 million
(c) 300 billion
(d) 120 million
Answer:
(a) 200 million

Question 17.
Statement (1): During spermiation, the sperms are released into the cavity of I seminiferous tubule.
Statement (2): During spermiogenesis, the spermatids get mature into sperms.
(a) Statement 1 is correct; statement 2 is incorrect.
(b) Statement 1 is incorrect; statement 2 is correct.
(c) Both the statements 1 and 2 are correct.
(d) Both the statements 1 and 2 are incorrect.
Answer:
(c) Both the statements 1 and 2 are correct.

Question 18.
Statement (1): Siamese twins are conjoined twins who are joined during birth. Statement (2): Dizygotic twins will be of same sex.
(a) Statement 1 is correct; statement 2 is incorrect.
(b) Statement 1 is incorrect; statement 2 is correct.
(c) Both the statements 1 and 2 are correct.
(d) Both the statements 1 and 2 are incorrect.
Answer:
(a) Statment 1 is correct; statement 2 is incorrect.

Question 19.
Statement (1): The endometrium acts as transitory endocrine gland secreting progesterone
Statement (2): Progesterone maintain pregnancy
(а) Statement 1 is correct; statement 2 is incorrect.
(б) Statement 1 is incorrect; statement 2 is correct.
(c) Both the statements 1 and 2 are correct.
(d) Both the statements 1 and 2 are incorrect.
Answer:
(b) Statement 1 is incorrect; statement 2 is correct

Question 20.
Statement (1): Human pregnancy lasts for 35 weeks.
Statement (2): During gestation, embryo’s heat develops during 12th week.
(a) Statement 1 is correct; statement 2 is incorrect.
(b) Statement 1 is incorrect; statement 2 is correct.
(c) Both the statements 1 and 2 are correct.
(d) Both the statements 1 and 2 are incorrect.
Answer:
(d) Both the statements 1 and 2 are incorrect.

Question 21.
Statement (1): Menstrual cycle occurs once in every 29 days.
Statement (2): The average age of menopause is 45-50 years.
(a) Statement 1 is correct; statement 2 is incorrect.
(b) Statement 1 is incorrect; statement 2 is correct.
(c) Both the statements 1 and 2 are correct.
(d) Both the statements 1 and 2 are incorrect.
Answer:
(c) Both the statements 1 and 2 are correct.

Question 22.
The first ejaculation of the semen in male is called as ………………..
Answer:
Spermarche

Question 23.
Identify the mismatched pair.
(a) Castration – Orchidectomy
(b) Spermiogenesis – Release of sperms into the cavity of seminiferous tubule
(c) Ovulation – Release of egg from ovary
(d) Capacitation – Process enabling the sperm to penetrate the egg.
Answer:
(b) Spermiogenesis – Release of sperms into the cavity of seminiferous tuble.

Question 24.
Given below are the extra embryonic membranes of which identify the outermost membrane.
(a) Amnion
(b) Chorion
(c) Yolk sac
(d) Allantois
Answer:
(b) Chorion

Question 25.
Identify the given figure and select the correct option representing X, Y and Z.


Answer:
(b) Acrosome, Nucleus and Mitochondria

Question 26.
The entire process of spermatogenesis takes about days
(a) 60 days
(b) 44 days
(c) 64 days
(d) 50 days
Answer:
(c) 64 days

Question 27.
Observe the diagram and select the correct option denoting the proper sequence of parts.

Answer:
(b) Fimbriae, Infundibulum, Uterus and Isthmus

Question 28.
Pick out the incorrect statements.
(a) The upper rounded portion of uterus is fundus.
(b) Uterus open into vagina through narrow cervix.
(c) Cervix is the organ of copulation in female.
(d) Vagina extends from the cervix and opens to exterior.
Answer:
(c) Cervix is the organ of copulation in female.

Question 29.
What is the role of fimbriae?
(a) Secretion of oestrogen and prolactin.
(b) Helps in the collection of the ovum after ovulation.
(c) Attaches the ovary to the abdominal cavity.
(d) Connects oviduct with ovary.
Answer:
(b) Helps in the collection of the ovum after ovulation.

Question 30.
Name the enzyme found in the acrosomal tip of sperm cell.
Answer:
Hyaluronidase

Question 31.
Which is not a correct statement regarding Oogenesis?
(i) During foetal development, cells in germinal epithelium of foetal ovary undergo , mitosis and produce oogonia.
(ii) Oogonial cell divide and enter into prophase I of meiosis I and from primary oocytes.
(iii)Primary oocytes later develop into primary follicles.
(iv) No oogonia is formed or added after the foetal birth.
(a) Only i
(b) ii and iii
(c) iv only
(d) None of the above
Answer:
(d) None of the above .

Question 32.
In embryo development of human beings, how long does it takes for a zygote to convert into morula?
(a) 24hrs
(b) 36hrs
(c) 48hrs
(d) 72hrs
Answer:
(d) 72 hrs

Question 33.
Identify the hormone which is produced only during the time of pregnancy
(a) Relaxin
(b) Oxytocin
(c) Progesterone
(d) Cortisol
Answer:
(a) Relaxin

Question 34.
The type of antibodies present in colostrum.
(a)IgE
(b) IgM
(c) IgA
(d) IgB
Answer:
(c) IgA

12th Bio Zoology Guide Human Reproduction Two Marks Questions and Answers

Question 1.
Enumerate the functions of reproductive system.
Answer:
The reproductive system has four main functions namely,

• to produce the gametes namely sperms and ova
• to transport and sustain these gametes
• to nurture the developing offspring
• to produce hormones

Question 2.
Define the terms (a) Insemination (6) Fertilization.
Answer:
(a) Insemination: Transfer of sperms by the male into the female genital tract.
(b) Fertilization: Fusion of male and female gametes to form zygote, called fertilization.

Question 3.
What are seminiferous tubules? Mention its role.
Answer:
Seminiferous tubules are highly coiled tubules seen in the lobules of testis. They occupy 80% – of testicular substance. They are the site for sperm production.

Question 4.
Name the cells noticed in the epithelial layer of seminiferous tubule.
Answer:
(a) Sertoli cells or Nurse cells
(b) Spermatogonic cells or male germ cells.

Question 5.
Mention the role epididymis.
Answer:

• Epididymis is a temporary store house for sperms.
•  Sperms undergo physiological maturation, increased motility and fertilizing capacity inside epididymis.

Question 6.
Seminal plasma is acidic or alkaline. Write its composition.
Answer:

• Seminal plasma is alkaline in nature.
• It contains fructose, ascorbic acid, prostaglandins and a coagulating enzyme called vesiculase.

Question 7.
Define Semen.
Answer:
Semen or seminal fluid is a milky white fluid which contains sperms and the seminal plasma, which is secreted from the seminal vesicles, prostate gland and the bulbourethral glands.

Question 8.
Why do males have Penis?
Answer:

• Penis is the male external genitalia.
• It functions as both excretory and copulatory organ.
• It is made of special tissue that erects the penis to facilitate insemination.

Question 9.
Point out the female accessory organs.
Answer:
Fallopian tubes, Uterus and Vagina.

Question 10.
Define the nature of uterus.
Answer:
The uterus or womb is a hollow, thick-walled, muscular, highly vascular and inverted pear shaped structure lying in the pelvic cavity between the urinary bladder and rectum.

Question 11.
What are the components that make up external genitalia female?
Answer:
Labia Majora, Labia Minora, Hymen and Clitoris.

Question 12.
Name the accessory reproductive glands in female which are homologous to (a) Cowper’s gland and (b) Prostate gland.
Answer:
In female, Bartholin’s gland is homologous to Cowper’s gland and Skene’s gland is homologous to prostate gland.

Question 13.
Define Gametogenesis.
Answer:
Gametogenesis is the process of formation of gametes i.e., sperms and ovary from the primary sex organs in all sexually reproducing organisms. Meiosis plays the most significant role in the process of gametogenesis.

Question 14.
Define the terms (a) Spermiogenesis (b) Spermiation
Answer:
(a) Spermiogenesis : Transformation of spermatids into mature sperm.
(b) Spermiation : Release of mature sperm into the lumen of seminiferous tubule.

Question 15.
What do you mean by ‘Sperm lysin’? Mention its function.
Answer:

• Sperm lysin is a proteolytic enzyme secreted in the acrosome of sperm.
• It helps to penetrate the ovum during fertilization.
• It is also called as hyaluronidase.

Question 16.
Name the four phases of menstrual cycle.
Answer:

1. Menstrual phase
2. Follicular or proliferative phase
3. Ovulatory phase
4. Luteal or secretory phase

Question 17.
What is corpus albicans?
Answer:
In the absence of fertilization, the corpus luteum degenerates completely and leaves a scar tissue called corpus albicans. It also initiates the disintegration of the endometrium leading to menstruation, marking the next cycle.

Question 18.
Define menopause.
Answer:
Menopause is the phase in a women’s life when ovulation and menstruation stops. The average age of menopause is 45-50 years. It indicates the permanent cessation of the primary functions 1 of the ovaries.

Question 19.
When does capacitation occurs? Define it.
Answer:
The sperms deposited in the female reproductive tract undergo capacitation. It is a bio chemical event that makes the sperm to penetrate and fertilize the egg.

Question 20.
Write a brief note on ectopic pregnancy?
Answer:
If the fertilized ovum is implanted outside the uterus it results in ectopic pregnancy. About 95 percent of ectopic pregnancies occur in the fallopian tube. The growth of the embryo may cause internal bleeding, infection and in some cases even death due to rupture of the fallopian tube.

Question 21.
Point out the extra embryonic membranes of human embryo.
Answer:
(a) amnion
(b) Chorion
(c) allantois
(d) Yolk nac

Question 22.
What is placenta?
Answer:
Placenta is a temporary endocrine organ formed during pregnancy and it connects the foetus to the uterine wall through the umbilical cord. It is the organ by which the nutritive, respiratory and excretory functions are fulfilled.

Question 23.
Name the organs developed from embryonic ectoderm.
Answer:
Brain and spinal cord (CNS), peripheral nervous system (PNS), epidermis and its derivatives and mammary glands.

Question 24.
Mention the hormones secreted by the placenta during pregnancy.
Answer:

• human Chorionic Gonadotropin (hCG)
• human Chorionic Somatomammotropin (hCS)
• human Placental Lactogen (hPL)
• Oestrogen and progesterone and relaxin.

Question 25.
Name the hormones that are secreted in human only during pregnancy.
Answer:

• human Chorionic Gonadotropin (hCG)
• human Chorionic Somatomammotropin (hCS)
• relaxin

Question 26.
State the role of relaxin.
Answer:
Relaxin is an hormone secreted by the placenta during the later phase of pregnancy. It helps in relaxation of pelvis during child birth.

Question 27.
Define parturition and labour.
Answer:
Parturition is the completion of pregnancy and giving birth to the baby. The series of events that expels the infant from the uterus is collectively called “labour”.

Question 28.
What do you mean by ‘false labour’?
Answer:
Throughout pregnancy the uterus undergoes periodic episodes of weak and strong contractions, fhese contractions called Braxter-Hick’s contractions lead to false labour.

Question 29.
Explain the term C-section.
Answer:
When normal vaginal delivery is not possible due to factors like position of the baby and nature of the placenta, the baby is delivered through a surgical incision in the woman’s abdomen and uterus. It is also termed as abdominal delivery or Caesarean Section or ‘C’ Section.

12th Bio Zoology Guide Human Reproduction Three Marks Questions and Answers

Question 1.
Compare gametogenesis with organogenesis.
Answer:

Question 2.
What are primary reproductive organs? What role does they play in organisms?
Answer:
The primary reproductive organs namely the ovary and testis are responsible for producing the ova and sperms respectively. Hormones secreted by the pituitary gland and the gonads help in the development of the secondary sexual characteristics, maturation of the reproductive system and regulation of normal functioning of the reproductive system.

Question 3.
Scrotum acts as a thermoregulator – Justify.
Answer:
The scrotum is a sac of skin that hangs outside the abdominal cavity. Since viable sperms cannot be produced at normal body temperature, the scrotum is placed outside the abdominal cavity to provide a temperature 2-3°C lower than the normal internal body temperature. Thus, the scrotum acts as a thermoregulator for spermatogenesis.

Question 4.
Write any three statements on Sertoli cells.
Answer:

• Sertoli cells are elongated and pyramidal cells.
• They provide nourishment to sperm till maturation. ‘
• They secrete a hormone called inhibin which is involved in negative feedback control of sperm production.

Question 5.
Give a brief account on leydig cells.
Answer:

• Interstitial cells of Leydig are seen embedded in the soft connective tissue surrounding the seminiferous tubules.
• These cells are endocrine natured and produce testosterone (androgen).
• These cells are characteristic to mammalian testes.

Question 6.
Name the accessory glands of male reproductive system.
Answer:

• A pair of seminal vesicles.
• A pair of bulbourethral gland (Cowper’s gland).
• A single prostate gland.

Question 7.
State the location and secretion of prostate gland.
Answer:
The prostate encircles the urethra and is just below the urinary bladder and secretes a slightly acidic fluid that contains citrate, several enzymes and prostate specific antigens.

Question 8.
Write a note on hymen.
Answer:
The external opening of the vagina is partially closed by a thin ring of tissue called the hymen. The hymen is often tom during the first coitus (physical union). However in some women it remains intact. It can be stretched or tom due to a sudden fall or jolt and also during strenuous physical activities such as cycling and horseback riding, etc., and therefore cannot be considered as an indicator of a woman’s virginity.

Question 9.
“Role of hormones in spermatogenesis” – comment on the statement.
Answer:
Spermatogenesis starts at the age of puberty and is initiated due to the increase in the release of Gonadotropin Releasing Hormone (GnRH) by the hypothalamus. GnRH acts on the anterior pituitary gland and stimulates the secretion of two gonadotropins namely Follicle Stimulating Hormone (FSH) and Lutenizing Hormone (LH). FSH stimulates testicular growth and enhances the production of Androgen Binding Protein (ABP) by the Sertoli cells and helps in the process of spermiogenesis. LH acts on the Leydig cells and stimulates the synthesis of testosterone which in turn stimulates the process of spermatogenesis.