Category: Class 12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 2 p-Block Elements – I Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements – I

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

I. Choose the qorrect answer

1. An aqueous solution of borax is __________ .
a) neutral
b) acidic
c) basic
d) amphoteric
Answer:
c) basic

2. Boric acid is an acid because its molecule (NEET)
a) contains replaceable H⁺ ion
b) gives up a proton
c) combines with proton to form water molecule
d) accepts OH^– from water, releasing proton
Answer:
d) accepts OH^– from water, releasing proton

3. Which among the following is not a borane?
a) B₂H₆
b) B₃H₆
c) B₄H₁₀
d) none of these
Answer:
b) B₃H₆

4. Which of the following metals has the largest abundance in the earth’s crust?
a) Aluminium
b) Calcium
b) Magnesium
d) Sodium
Answer:
a) Aluminium

5. In diborane, the number of electrons that accounts for banana bonds is
a) six
b) two
c) four
d) three
Answer:
c) four

6. The element that does not show catenation among the following p-block elements is
a) Carbon
b) Silicon
c) Lead
d) germanium
Answer:
c) Lead

7. Carbon atoms in fullerene with formula C60 have
a) sp³ hybridised
b) sp hybridised
c) sp² hybridised
d) partially sp² and partially sp³ hybridised
Answer:
c) sp2 hybridised

8. Oxidation state of carbon in its hybrides
a) +4
b) -4
c) +3
d) +2
Answer:
a) +4

9. The basic structural unit of silicates is (NEET) (PTA – 1)
a) (SiO₃)^2-
b) (SiO₄)^2-
c) (SiO)^–
d) (SiO₄)^4-
Answer:
d) (SiO₄)^4-

10. The repeating unit in silicone is

11. Which of these is not a monomer for a high molecular mass silicone polymer?
a) Me₃SiCl
b) PhSiCl₃
c) MeSiCl₃
d) Me₂SiCl₂
Answer:
a) Me₃SiCl

12. Which of the following is not sp² hybridised?
a) Graphite
b) graphene
c) Fullerene
d) dry ice
Answer:
d) dry ice

13. The geometry at which carbon atom in diamond are bonded to each other is
a) Tetrahedral
b) hexagonal
c) Octahedral
d) None of these
Answer:
a) Tetrahedral

14. Which of the following statements is not correct?
a) Beryl is a cylic silicate
b) Mg₂SiO₄ is an orthosilicate
c) SiO₄^4- is the basic structural unit of silicates
d) Feldspar is not aluminosilicate
Answer:
d) Feldspar is not aluminosilicate

15. Match items in Column-I with the items of Column-II and assign the correct code.

Answer:
a) 2 1 4 3

16. Duralumin is an alloy of
a) Cu, Mn
b) Cu, AZ, Mg
c) AZ, Mn
d) AZ, Cu, Mn, Mg
Answer:
d) Al, Cu, Mn, Mg

17. The compound that is used in nuclear reactors as protective shields and control rods is
a) Metal borides
b) Metal oxides
c) Metal carbonates
d) Metal carbide
Answer:
a) Metal borides

18. The stability of +1 oxidation state increases in the sequence
a) AZ < Ga < In < TZ
b) TZ < In < Ga < Al
c) In < TZ < Ga < Al
d) Ga < In < AZ < TZ
Answer:
a) Al< Ga < In < TZ

II. Answer the following questions

Question 1.
Write a short note on anamolous properties of the first element of p-block.
Answer:
The following factors are resposible for the anamolous properties of the first elements of p-blick.
1. Small size of the first member
2. High ionisation enthalpy and high electronegativity.
3. Absence of d-orbitals in their valence shell.

Question 2.
Describe briefly allotropiam in p-block elements with specific reference to carbon.
Answer:

• Some elements exist in more than one crystalline or molecular forms in the same physical state.
• This phenomenon is called allotropism.
• The different forms of an element are called allotropes.
• Example: Carbon exists as diamond, graphite, graphene, fullerenes, carbon nanotubes

Question 3.
Give the uses of Borax.
Answer:

1. Borax is used for the identification of coloured metal ions.
2. In the manufacture optical and borosilicate glass, enamels and glazes for pottery.
3. It is also used as a flux in metallurgy and also acts as a good preservative.

Question 4.
What is catenation? Describe briefly the catenation property of carbon. (MARCH 2020)
Answer:
Catenation:
It is the phenomenon of an atom to form a strong covalent bond with the atoms of itself. Carbon shares the property of catenation to the maximum extent because it is small in size and can form pn-pn multiple bonds to itself. The following conditions are necessary for catenation.

1. The valency of element is greater than or equal to two.
2. Element should have the ability to bond with itself.
3. The self-bond must be as strong as its bond with other elements.
4. Kinetic inertness of catenated compound towards other molecules.
5. Carbon possesses all the above properties and forms a wide range of compounds with itself.

Question 5.
Write a note on Fisher Tropsch synthesis. Fischer Tropsch synthesis: (PTA – 4)
Answer:
This is a reaction in which carbon monoxide reacts with hydrogen at a pressure less than 50 atm and temperature 500 – 700 K in presence of metal catalysts to give saturated and unsaturated hydrocarbons.
n CO + (2n+l) H₂ → CnH_2n+2 + nH₂O
n CO + 2n H₂ → C_nH_2n + nH₂O

Question 6.
Give the structure of CO and CO₂.
Answer:
Structure of CO:
Structure is linear.

Structure of CO₂:
Structure is linear.

Question 7.
Give the uses of silicones.
Answer:

1. Used for low temperature lubrication.
2. Used in vacuum pumps.
3. Used in high temperature oil baths.
4. Used for making water proof cloths.
5. Used as insulating material in electrical motor and other applicances.
6. Mixed with paints and enamels to make them resistant towards high temperature, sunlight, dampness and chemicals

Question 8.
Describe the structure of diborane. (PTA – 3)
Answer:

• In diborane two BH₂ units are linked by two bridged hydrogens, rherefore it has eight B-H bonds.
• Diborane has only 12 valence electrons anc are- not sufficient to form normal covalen bonds.
• The four terminal B-H bonds are norma covalent bonds. (2c 2e bond) (Totally 8e-s)
• The remaining four electrons have to be used for the bridged bonds, ie two 3 centred B-H-B bonds utilise two electrons each.
• Hence these bonds are 3c – 2e bonds.
  
• The bridging hydrogen atoms are in a plane.
• In diborane, boron is sp³ hybridised.
• Three sp³ hybridised orbitals contain single electron and the fourth orbital is empty.
• Two half filled sp³ hybridised orbitals of each boron overlap with two hydrogens to form four terminal 2C – 2e bonds.
• One empty and one half filled sp³ hybridised orbital on each boron is left.
• Empty sp³ hybridised orbital of one boron, overlaps with half filled sp³ hybridised orbital of the other boron and Is orbital of hydrogen to form two bridged 3C – 2e B-l 1-B bonds.

Question 9.
Write a short note on hydroboration.
Answer:

• Diborane adds on to alkenes and alkynes in ether solvent at room temperature.
• This reaction is known as hydroboration.
• This is used in synthetic organic chemistry especially for anti Markovnikov addition.
  B₂H₆ + 6 RCH = CHR → 2B (CH₂-CH₂ R)₃

Question 10.
Give one example for each of the following:
Answer:
i) icosogens
ii) tetragen
iii) pnictogen
iv) chalcogen

Question 11.
Write a note on metallic nature of p-block elements.
Answer:

• The tendency of an element to form a cation by losing electrons is known as electro positive or metallic character.
• This character depends on the ionisation energy.
• Generally on moving down a group ionisation energy decreases and hence the metallic character increases.
• In p-block, the elements present in lower left part are metals, while the elements in the upper right part are non metals.

Question 12.
Complete the following reactions:
a) B(OH)₃ + NH₃ →
b) Na₂B₄O₇ + H₂SO₄ + H₂O →
c) B₂H₆ + 2NaOH + 2H₂O →
d) B₂H₆ + CH₃OH →
e) BF₃ + 9H₂O →
f) HCOOH+ H₂SO₄ →
g) SiCl₄ + NH₃ →
h) SiCl₄ + C₂H₅OH →
I) B + NaOH →
j) H₂B₄O₇ \(\underrightarrow { Red\quad hot } \)
Answer:

Question 13.
How will you identify borate radical? (PTA – 5)
Answer:

• When boric acid or borate salt is heated with ethyl alcohol in presence of cone, sulphuric acid, an ester triaikvl borate is formed.
• The vapour of this ester bums with a green edged flame.
• This is ethyl borate test to identify borate radical,
  
  B(OC₂H₅)₃ Ethyl borate (Green edged flame)

Question 14.
Write a note on zeolites. ( PTA – 2)
Answer:

• Zeolites are three dimensional crystalline solids containing aluminium, silicon and oxvgen in their regular three dimensional frame work.
• They are hydrated sodium alumino silicates.
• General formula is
  Na₂O.(Al₂O₃).x(SiO₂).y(H₂O)
  where x = 2 to 10; y = 2 to 6
• Zeolites have porous structure in which the monovalent sodium ions and water molecules are loosely held.
• Si and AI atoms are tetrahedrally coordinated with each other through shared oxygen atoms.
• Zeolites are similar to Clay minerals but they differ in their crystalline structure.
• Zeolites have a three dimensional crystalline structure looks like a honey comb consisting of a network of interconnected tunnels and cages.
• Water molecules move freely In and out of these pores but the zeolite frame work remains rigid.
• Another special aspect of this structure is that the pore/channel sizes are nearly uniform, allowing the crystal to act as a molecular sieve.
• Zeolites are used in the removal of permanent hardness of water.

Question 15.
How will you convert boric acid to boron nitride? (PTA – 3)
Answer:
Fusion of urea with boric acid in an atmosphere of ammonia at 800 -1200 K gives boron nitride.

Question 16.
A hydride of 2nd period alkali metal (A) on reaction with compound of Boron (B) to give a reducting agent (C). Identify (A), (B) and ( C) (PTA – 1)
Answer:
A hydride of 2nd period alkali metal (A) is LiH

Lithium hydride reacts with compound of boron (B) B₂H₆ to give reducing agent (C) lithium boro hydride.
∴ Compound B is diborane
Compound C is lithium boro hydride.

Question 17.
A double salt which contains fourth period alkali metal (A) on heating at 500 K gives (B). Aqueous solution of (B) gives white precipitate with BaCl₂ and gives a red colour compound with alizarin. Identify (A) and (B).
Answer:

•  A double salt which contains fourth period alkali metal (A) is Potash alum
  K₂SO₄. Al₂ (SO₄)₃.24H₂O
• (A) on heating at 500 K gives
  K₂SO₄.Al₂(SO₄)₃ (B) which is burnt alum.

Question 18.
CO is a reducing agent, justify with an example.
Answer:

• CO is a strong reducing agent.
• It reduces metallic oxides inlo melais.
  Example : 3CO + Fe₂CO₃ → 2Fe + 3CO₂

III. Evaluate Yourself

Question 1.
Why group 18 elements are called inert gases? Write the general electronic configuraton of group 18 elements.
Answer:

• These elements are gases.
• Their outer electronic configuration is ns²np⁶ which is stable completely filled configuration.
• So they are more stable and least reactive.
• Hence they are called inert gases.

12th Chemistry Guide Chapter 2 p-Block Elements – I Additional Questions and Answers

Part – II – Additional Questions

I. Choose the correct answer

1. The general electronic configuration of p-block elements is
a) ns¹
b) ns²
c) ns² np^1-6
d) (n-1)s² np^1-6
Answer:
c) ns² np^1-6

2. p-block element consists of the groups
a) 1 & 2
b) 3 – 12
c) 13 – 17
d) 13 – 18
Answer:
d) 13 – 18

3. Group 18 elements are inert because of their
a) unstable incompletely filled orbitals
b) stable completely filled orbitals
c) half filled orbitals
d) stable nucleus
Answer:
b) stable completely filled orbitals

4. As we go down the group ionisation energy
a) decreases
b) increases
c) becomes constant
d) becomes zero
Answer:
a) decreases

5. As we go down the group metallic character
a) decreases
b) increases
c) becomes constant
d) becomes zero
Answer:
b) increases

6. As ionisation energy decreases, the metallic character of elements
a) decreases
b) increases
c) becomes constant
d) becomes zero
Answer:
b) increases

7. In p-block, metals are placed in
a) upper right part
b) middle part
c) lower left part
d) top of the group
Answer:
c) lower left part

8. In p-block, non-metals are placed in
a) upper right part
b) middle part
c) lower left part
d) bottom of the group
Answer:
a) upper right part

9. Which of the following factor is not responsible for the anamolous behaviour of the first member of each group in p-block elements?
a) small size
b) high ionisation enthalpy
c) outer electronic configuration
d) absence of d-orbitals
Answer:
c) outer electronic configuration

10. The correct order of catenation property in group 14 elements is
a) C << Si < Ge = Sn < Pb
b) C >> Si > Ge = Sn > Pb
c) C >> Si < Ge = Sn < Pb
d) C << Si » Ge = Sn > Pb
Answer:
b) C >> Si > Ge = Sn > Pb

11. The elements N, O, F readily forms hydrogen bonds due to their high
a) ionisation energy
b) electron affinity
c) electro negativity
d) atomic radius
Answer:
c) electro negativity

12. The most electro negative element is
a) Flourine
b) Chlorine
c) Bromine
d) Iodine
Answer:
a) Flourine

13. The element with maximum electron affinity is
a) Flourine
b) Chlorine
c) Bromine
d) Iodine
Answer:
b) Chlorine

14. The most reactive element among halogens is
a) Flourine
b) Chlorine
c) Bromine
d) Iodine
Answer:
a) Flourine

15. The strongest oxidising agent among halogens is
a) Flourine
b) Chlorine
c) Bromine
d) Iodine
Answer:
a) Flourine

16. The important property shown by p-block elements is
a) complex formation
b) coloured ion formation
c) inert pair effect
d) metallic character
Answer:
c) inert pair effect

17. In 13th group Tl⁺¹ ion is more stable than Tl³⁺ ion due to
a) high electronegatively
b) inert pair effect
c) high ionisation energy
d) stable electronic configuration
Answer:
b) inert pair effect

18. Diamond and graphite are ______ of carbon.
a) Isotopes
b) Isobars
c) Isomers
d) Allotropes
Answer:
d) Allotropes

19. The formula of Borax is
i) Na₂B₄O₇.10H₂O
ii) Na₂[B₄O₅(OH)₄].8H₂O
iii) Na₂[B₄O₅(OH)₄].2H₂O
a) (i) only
b) (i) & (ii) only
c) (i) & (iii) only
d) (iii) only
Answer:
b) (i) & (ii) only

20. Ortho boric acid on dehydration at 373K produces mainly (PTA – 3)
a) metaboric acid
b) boric anhydride
c) Boron metal and Oxygen
d) tetra boric acid
Answer:
a) metaboric acid

21. The formula of colemanite is
a) Na₂B₄O₇
b) Na₂B₄O₇.10H₂O
c) Ca₂B₆O₁₁
d) NaBO₂
Answer:
c) Ca₂B₆O₁₁

22. Which is used as moderator in nuclear reactors?
a) boron nitride
b) boron
c) borax
d) boric acid
Answer:
b) boron

23. The compound used in eye drops and antiseptics is
a) boron nitride
b) boric acid
c) sodium meta borate
d) boron tri oxide
Answer:
b) boric acid

24. The compound used as a flux in metallurgy is
a) boron nitride
b) boric acid
c) borax
d) boron tri oxide
Answer:
c) borax

25. Boric acid on heating at 413 K gives
a) meta boric acid
b) tetra boric acid
c) boric anhydride
d) borax
Answer:
b) tetra boric acid

26. In ethyl borate test the colour of the flame obtained is
a) red
b) yellow
c) blue
d) green
Answer:
d) green

27. On hydrolysis BF3 gives Boric acid and converted to fluroboric acid. The fluoroboric acid contains the species. (PTA – 6)
a) H⁺, F^– & BF₃
b) H⁺ & [BF₄]
c) [H BF₃]⁺ & F^–
d) H⁺, B³⁺ & F^–
Answer:
b) H⁺ & [BF₄]

28. In organic benzene is
a) diborane
b) borazole
c) borax
d) boric acid
Answer:
b) borazole

29. The formula of Inorganic benzene is
a) B₃N₃
b) B₃N₃H₃
c) B₃N₃H₆
d)B₆N₆H₆
Answer:
c) B₃N₃H₆

30. The most stable form of carbon is
a) graphite
b) diamond
c) fullerene
d) carbon nano tubes
Answer:
a) graphite

31. The formula of buckminster fullerene is
a) C₃₂
b) C₅₀
e) C₆₀
d) C₇₀
Answer:
c) C₆₀

32. The number of six membered and five membered rings fused together respectively in buckminster fullerene is
a) 12 & 20
b) 20 & 12
c) 10 & 22
d) 22 & 10
Answer:
b) 20 & 12

33. Water gas is a mixture of
a) CO₂ + H₂
b) CO + H₂O
c) CO + H₂
d) CO + N₂
Answer:
c) CO + H₂

34. Producer gas is a mixture of
a) CO₂ + H₂
b) CO + H₂O
c) CO + H₂
d) CO + N₂
Answer:
d) CO + N₂

35. In the presence of light carbon monoxide reacts with chlorine to form a poisonous gas called
a) mustard gas
b) phosgene
c) phosphine
d) carbylamine
Answer:
b) phosgene

36. Fischer Tropsch synthesis is used for preparing
a) Silicones
b) Boranes
c) Hydrocarbons
d) Carbonyls
Answer:
c) Hydrocarbons

37. In metal carbonyls the oxidation state of metals is
a) 0
b) +1
c) +2
d) +3
Answer:
a) 0

38. The structure of CO molecule is
a) trigonal
b) tetrahedral
c) linear
d) square planar
Answer:
c) linear

39. The structure of CO₂ molecule is
a) trigonal
b) tetrahedral
c) linear
d) square planar
Answer:
c) linear

40. The critical temperature of CO₂ is
a) 21 °C
b) 31°C
c) 12°C
d) 13°C
Answer:
b) 31°C

41. When CO₂ is dissolved in water, the solution is slightly
a) acidic
b) basic
c) amphoteric
d) neutral
Answer:
a) acidic

42. Which among the following is important for photo synthesis?
a) O₂
b) N₂
C) CO
d) CO₂
Answer:
d) CO₂

43. The water repellant property of silicones is due to the presence of
a) -OH group
b) -Si group
c) -R group
d) -Cl group
Answer:
c) -R group

44. The percentage of silicate minerals and silica present in earth’s crust is
a) 75
b) 85
c) 95
d) 100
Answer:
c) 95

45. The basic unit present in silicates is
a) SiO₂
b) [SiO₃]^–
c) [SiO₄]^2-
d) [SiO₄]^4-
Answer:
d) [SiO₄]^4-

46. Talc is an example of
a) Ino silicates
b) Phyllo silicates
c) Tecto silicates
d) Chain silicates
Answer:
b) Phyllo silicates

47. Quartz is an example of
a) Ino silicates
b) Phyllo silicates
c) Tecto silicates
d) Chain silicates
Answer:
c) Tecto silicates

48. The formula of Spodumene is
a) Sc₂Si₂O₇
b) Li Ai(SiO₃)₂
c) [Be₃ Al₂(SiO₃)₆]
d) Be₂SiO₄
Answer:
b) Li Ai(SiO₃)₂

49. The silicate which is used in the removal of permanent hardness of water is
a) Feldspar
b) Quartz
c) Zeolites
d) Talc
Answer:
c) Zeolites

50. Thermodynamically the most stable form of carbon is (PTA – 4)
a) Diamond
b) Fullerenes
c) graphite
d) Nano tubes
Answer:
c) graphite

II. Pick the odd man out

1. W.r.t. their metallic character pick the odd man out.
a) Ge
b) Ga
c) B
d) As
Answer:
b) Ga – It is a metal while others are metalloids

2. W.r.t. their metallic character pick the odd man out
a) In
b)Pb
c) Cl
d) Bi
Answer:
c) Cl – It is a non metal while others are metals.

3. Pick the odd man out
a) Borax
b) Kernite
c) Colemanite
d) Bauxite
Answer:
d) Bauxite – It is an ore of aluminium others are ores of boron.

4. W.r.t. to hybridisation pick the odd man out.
a) Graphite
b) Diamond
c) Fullerene
d) Graphene
Answer:
b. Diamond – It is sp³ hybridised while others are sp² hybridised.

III. Assertion and Reason

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

1. Assertion (A) : Boron shows non metallic character.
Reason (R) : Atomic radius of boron is small and its nuclear charge is high.
Answer:
(i).Both A and R are correct, R explains A

2. Assertion (A) : As we move down Boron group the elements show less tendency to exhibit +1 oxidation state rather than +3. Reason (R) : As we move down Boron group the elements show inert pair effect.
Ans : (iii).A is wrong but R is correct

3. Assertion (A) : Graphite conducts electricity.
Reason (R) : In Graphite, successive carbon sheets are held together by weak Vander Waals force.
Answer:
(iv). Both A and R are corrrect but R does not explain A

4. Assertion (A) : Silicones are used for making water proofing clothes.
Reason (R) : In silicones the organic side groups which surrounds silicon make the molecule looks like an alkane.
Answer:
(i).Both A and R are correct, R explains A

IV. Choose the correct statement

1. i) Some of the p-block elements show negative oxidation states also.
ii) Halogens gain two electrons to give a stable halide ion.
iii) Inert gases have ns²np⁶ configuration and hence more stable.
iv) p-block elements have a general electronic configuration (n-1)s² np^1-6
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
b) (i) & (iii)
Correction:
ii) Halogens gain one electron to give a stable halide ion.
iv) p-block elements have a general electronic configuration ns² np^1-6

2. i) Boron compounds are electron rich compounds.
ii) Boron does not react directly with hydrogen.
iii) Borax is sodium salt of metaboric acid.
iv) Boric acid is used as an antiseptic,
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (iii) & (iv)
Answer:
c) (ii) & (iv)
Correction:
i) Boron compounds are electron deficient compounds.
iii) Borax is sodium salt of tetraboric acid.

3. i) In graphite carbon atoms are sp³ hybridised.
ii) A single planar sheet of graphite is known as graphene.
iii) In diamond each carbon atom is tetrahedrally surrounded by four other carbon atoms.
iv) Carbon nanotubes do not conduct electricity,
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)
Correction:
i) In graphite carbon atoms are sp² hybridised.
iv) Carbon nanotubes conduct electricity.

4. i) Silicones are organo silicon polymers.
ii) Hydrolysis of R₂SiCl₂ yields complex cross linked polymer.
iii) Silicones are good thermal and electrical conductors.
iv) All silicones are water repellent,
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
d)(i) & (iv)
Correction:
ii) Hydrolysis of R₂SiCl₂ yields a straight chain polymer.
iii) Silicones are good thermal and electrical insulators.

V. Choose the wrong statement

i) Boron is a metal.
ii) Nitrogen is a metalloid.
iii) Oxygen is a non metal.
iv) Antimony is a metalloid.
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
a) (i) & (ii)
Correction:
i) Boron is a metalloid (or) non metal
ii) Nitrogen is a non metal.

2. i) Aluminium chloride is a Lewis acid.
ii) Alum is a double salt of potassium aluminium sulphate.
iii) Aluminium chloride is used as a styptic agent to arrest bleeding.
iv) Alum is used as a catalyst in Friedel Crafts reaction.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
c) (iii) & (iv)
Correction:
iii) Alum is used as a styptic agent to arrest bleeding.
iv) Anhydrous Aluminium chloride is used as a catalyst in Friedel Crafts reaction.

3. Which of the following statement about H₃BO₃ is not correct? (PTA – 5)
a) It is a strong tribasic acid
b) It is prepared by acidifying an aqueous solution of borax.
c) It is a layer structure in which planer BO₃ units are joined by hydrogen bonds.
d) It does not act as proton donor but acts as a Lewis acid by accepting hydroxyl ion.
Answer:
a) It is a strong tribasic acid

4. i) Silicates which contain discrete [SiO₄]^4- units are called neso silicates.
ii) Beryl is an example for amphiboles.
iii) Spodumene is an example for phyllo silicates.
iv) Silicates which contain [Si₇O₇]^6- ions are called Soro silicates,
a) (i) & (ii)
b) (ii) & (iv)
c) (ii) & (iii)
d) (i) & (iv)
Answer:
c) (ii) & (iii)
Correction:
ii) Beryl is an example for cyclic silicates.
iii) Spodumene is an example for chain silicates.

VI. Match the following
1.

Answer:

2.

Answer:

3.

Answer:

4.

Answer:

VII. 2 Marks questions

Question 1.
What are ‘p’-block elements? Write their general outer electronic configuration.
Answer:
The elements in which their last electron enters the ‘p’ orbital are called ‘p’-block elements.

• They are placed in 13 -18 groups.
• General outer electronic configuration is ns²np^1-6.

Question 2.
How are the p-block elements classified.
Answer:

• Based on the outer electronic configuration they are classified as 13 -18 group elements.
• Based on the nature of the elements they are classified as non metals, metalloids and metals.

Question 3.
Aluminium (III) chloride is stable where as Thallium (III) chloride is unstable. Why? (PTA – 2)
Answer:

• Due to inert pair effect, as we move down the 13th group ns² electrons remain inert and np¹ electron takes part in the reaction.
• So Tl³⁺ ion is less stable and Tl⁺¹ ion is more stable.
• Hence AlCl₃ is stable where as TICl₃ is unstable and decomposes into TlCl.

Question 4.
How is boric acid prepared from borax?
Answer:
Boric acid can be extracted from borax by treating with HCl or H₂SO₄.
Na₂B₄O₇ + 2HCl + 5H₂O → 4H₃BO₃ + 2NaCl
Na₂B₄O₇ + H₂SO₄ + 5H₂O → 4H₃BO₃ + 2Na₂SO₄

Question 5.
How is boric acid prepared from Colemanite?
Answer:
When sulphur dioxide is passed through colemanite solution, boric acid is obtained.
Ca₂B₆O₁₁ + 2SO₂ + 9H₂O → 2CaSO₃ + 6H₃BO₃

Question 6.
What is the action of sodium hydroxide on boric acid?
Answer:
Boric acid reacts with sodium hydroxide to form sodium metaborate and sodium tetra borate.

Question 7.
Write the action of water on diborane.
Answer:
Diborane reacts with water to form boric acid.
B₂H₆ + 6H₂O → 2H₃BO₃ + 6H₂

Question 8.
What is the action of NaOH on diborane.
Answer:
Diborane reacts with NaOH to form sodium meta borate.
B₂H₆ + 2NaOH + 2H₂O → 2NaBO₂ + 6H₂

Question 9.
What is the action of air on diborane?
Answer:
At room temperature pure diborane does not react with air or oxygen.

But impure diborane reacts with air or oxygen to giveB203 along with large amount of heat.
B₃H₆ + 3O₂ → B₂O₃ + 3H₂O
∆H =-2165 KJ mol^-1

Question 10.
How does diborane react with methyl alcohol?
Answer:
Diborane reacts with methyl alcohol to give trimethyl borate.
B₂H₆ + 6CH₃OH → 2B(OCH₃)₃ + 6H₂

Question 11.
How does diborane react with metal hydrides?
Answer:
When treated with metal hydrides, diborane forms metal boro hydrides.

Question 12.
How does diborane react with ammonia at low temperature?
Answer:
When treated with excess ammonia at low temperature diborane gives diborane di ammonate.
3B₂H₆ + 6NH₃ \(\underrightarrow { -153K } \) 3B₂H₆.2NH₃

Question 13.
How is inorganic benzene prepared? (PTA – 1)
Answer:

• On heating at higher temperatures with ammonia, diborane forms borazole or borazine.
• Borazole or borazine is called as Inorganic benzene
  

Question 14.
BF₃ acts as a Lewis acid. Give example.
Answer:
BF₃ is an electron deficient compound and accepts electron pairs to form coordinate covalent bonds. Hence BF3 acts as a Lewis acid.
BF₃ + NH₃ → F₃B ← NH₃
BF₃ + H₂O → F₃B ← OH₂

Question 15.
Convert BF3 into hydro fluoro boric acid.
Answer:
On hydrolysis BF3 gives boric acid, which is converted into hydro fluoro boric acid.
4BF₃ + 3H₂O → H₃BO₃ + 3HBF₄
3HBF₄ (Hydro fluoro boric acid)

Question 16.
Write about McAfee process of manufacturing AlCl₃.
Answer:
AlCl₃ is obtained by heating a mixture of alumina and coke in a current of chlorine.
Al₂O₃ + 3C + 3Cl₂ → 2AlCl₃ + 3CO

Question 17.
Write the action of NaOH on AlCl₃
Answer:
With excess of NaOH, AlCl₃ gives sodium alumina te.
AlCl₃ + 4NaOH → NaAlO₂ + 2H₂O + 3NaCl

Question 18.
Write the uses of aluminium chloride.
Answer:
1. Anhydrous AlCl₃ is used as a catalyst in Friedel crafts reaction.
2. AlCl₃ is used for the manufacture of petrol by cracking the mineral oils.
3. AlCl₃ is used as a catalyst in the manufacture of dyes, drugs and perfumes.

Question 19.
What are alums? Give examples.
Answer:
1. Alum is a double salt of potassium aluminium sulphate.
2. Now a days the name alum is used for all the double salts with the formula
M’₂ SO₄ M”2 (SO₄)₃.24H₂O
Where M’ is univalent metal ion or NH₄⁺
M” is trivalent metal ion
Example: K₂SO₄.Al₂(SO₄)₃.24H₂O Potash alum
K₂SO₄.Cr₂(SO₄)₃.24H₂O Chrome alum

Question 20.
Aqueous solution of carbon di oxide is acidic. Why?
Answer:
Aqueous solution of carbon di oxide is slightly acidic as it forms carbonic acid which dissociates to give H⁺ ions.

Question 21.
How is silicon tetra choride prepared?
Answer:
SiCl₄ is prepared by passing dry chlorine over an intimate mixture of silica and carbon heating to 1675 K in a porcelain tube.
SiO₂ + 2C + 2Cl₂ → SiCl₄ + 2CO
SiCl₄ is prepared commercially by the reaction of silicon with hydrogen chloride gas above 600 K.
SiO + 4HCl → SiCl₄ + 2H₂

Question 22.
Write the uses of silicon tetra chloride.
Answer:
Silicon tetra chloride is used
i) In the production of semi conducting silicon.
ii) As a starting material in the synthesis of silica gel, silicic esters, a binder for ceramic materials.

Question 23.
What is water gas equilibrium? (PTA – 5)
Answer:
Water gas equilibrium
The equilibrium involved in the reaction between carbon di oxide and hydrogen, has many industrial applications and is called water gas equilibrium.

VIII. Three Marks questions

Question 1.
How is borax prepared from colemanite?
Answer:
When colemanite ore solution is boiled with sodium carbonate solution borax is obtained.

Question 2.
Write the uses of boron.
Answer:
1. ₅B¹⁰ absorbs neutrons, hence it is used as a moderator in nuclear reactors.
2. Amorphous boron is used as a rocket fuel igniter.
3. Boron is essential for the cell walls of plants.
4. Boric acid and borax are used in eye drops, antiseptics, washing powders.
5. Boric oxide is used in the manufacture of pyrex glass.

Question 3.
Aqueous solution of borax is basic. Why?
Answer:
In hot water borax dissociates into boric acid and sodium hydroxide.
Na₂B₄O₇ + 7H₂O → 4H₃BO₃ + 2NaOH
Boric acid is a weak acid, whereas sodium hydroxide is a strong base.

As a result the resulting solution is basic.

Question 4.
What is the action of heat on borax?
Answer:
On heating borax loses its water of crystallisation first and then decomposes into sodium metaborate and boron trioxide.

Boron trioxide appears as transparent glassy beads.

Question 5.
What is the action of heat on boric acid?
Answer:

Question 6.
Describe the structure of boric acid.
Answer:

• Boric acid has a two dimensional structure.
• It consists of [BO₃]^3- unit.
• These unit are linked to each other by hydrogen bonds.

Question 7.
Write the uses of boric acid.
Answer:
Boric acid is
1. Used in the manufacture of pottery glazes, glass, enamels and pigments.
2. Used as an antiseptic.
3. Used as an eye lotion.
4. Used as a food preservative.

Question 8.
How is diborane prepared?
Answer:

• When sodium boro hydride in diglyme is reacted with iodine diborane is obtained.
  2NaBH₄ + I₂ → B₂H₆ + 2NaI + H₂
• On heating magnesium boride with Hcl, a mixture of volatile boranes are obtained.
  2Mg₃B₂ + 12HCl → 6MgCl₂ + B₄H₁₀ + H₂
  B₄H₁₀ + H₂ → 2B₂H₆

Question 9.
Write the uses of diborane.
Answer:
Diborane is
1. Used as a high energy fuel for propellant.
2. Used as a reducing agent in organic

Question 10.
How is boron trifluoride prepared from boron trioxide?
Answer:
When boron trioxide is treated with calcium fluroide in presence of conc.sulphruic acid, boron trifluoride is obtained.

When boron trioxide is reacted with carbon and fluorine, boron trifluoride is obtained.
B₂O₃ + 3C + 3F₂ → 2BF₃ + 3CO

Question 11.
How is boron trifluoride prepared in the laboratory?
Answer:
In the laboratory pure BF₃ is prepared by the. thermal decomposition of benzene, diozonium tetrafluro borate.

Question 12.
How is potash alum prepared? (PTA – 4)
Answer:
Potash alum is prepared from alunite or alum stone.
When alunite is treated with excess of sulphuric acid, the aluminium hydroxide present is converted into aluminium sulphate.
A calculated quantity of potassium sulphate is added.
The solution is crystallised to obtain potash alum.
It is purified bv recrystallisation.

Question 13.
Write the uses of alum.
Answer:
Alum is used
i) for the purification of water.
ii) for water proofing and textiles.
iii) in dyeing, paper and leather tanning industries.
iv) as a styptic agent to arrest bleeding.

Question 14.
Write the uses of carbon monoxide.
Answer:
Carbon monoxide is used
i) as a reducing agent and can reduce many metal oxides to metal.
ii) as an important ligand and forms metal carbonyls.
iii) a mixture of CO & H₂ is called as water gas and a mixture of CO & N₂ is called as producer gas. Both are used as important industrial fuels.

Question 15.
Write the uses of carbon dioxide.
Answer:
Carbon dioxide is used

• to produce an inert atmosphere for chemical processing.
• by plants in photosynthesis.
• as fire extinguisher.
• as a propellant gas.
• in the production of carbonated beverages.
• in the production of foam.

Question 16.
Write note on Boron Neutron Capture Therapy (BNCT).
Answer:

• The affinity of Boron-10 for neutrons is the bases of this technique BNCT for treating patients suffering from brain tumours.
• It is based on the nuclear reaction which occurs when Boron-10 is irradiated with low- energy thermal neutrons to give high linear energy a-particles and a Li particle.
• Boron compounds are injected into a brain tumour patient and the compounds collect preferentially in the tumour.
• The tumour area is then irradiated with / thermal neutrons and results in the release of an alpha particle.
• This a-particle damages the tissue in the tumour each time a Boron-10 nucleus captures a neutron.
• In this wav damage can be limited preferentially to the tumour, leaving the normal brain tissue less affected.
• BNCT has been studied as a treatment for several other tumours of the head and neck, the breast the prostate, the bladder and the liver.

IX. Five Marks questions

Question 1.
How is higher boranes obtained from diborane.
Answer:
At high temperatures diborane forms higher boranes liberating hydrogen.

Question 2.
Explain the allotropes of carbon.
Answer:

• Carbon exists in many allotropic forms.
• Graphite and diamond are the most common allotropes.
• Graphene, fullerenes and carbon nano tubes are other important allotropes of carbon.

Graphite:

• It is the most stable allotrope of carbon at normal temperature and pressure.
• It is composed of flat two dimensional hexagonal sheets of sp² hybridised carbon atoms.
• C-C bond length is 1.41 Å which is close to the C-C bond distance in benzene (1.40 Å )
• Each carbon atom forms three sbonds with three neighbouring carbon atoms using three of its valence electrons and the fourth electron present in the unhybridised p-orbital forms a p-bond.
• These pelectrons are delocalised over the entire sheet, hence graphite conducts electricity.
• Successive carbon sheets at a distance of 3.40 Ao are held together by weak Vander Waals forces, hence graphite is soft, slippery and used as a lubricant.

Diamond:

• Carbon atoms in diamond are sp3 hybridised.
• Each carbon is bonded tetra hedrally with four other carbon atoms by s-bonds with C-C bond length of 1.54 Å. Hence diamond is hard.
• Since all the four valence electrons of carbon are involved in bonding and there is no free electrons, diamond is not a conductor.
• Being the hardest substance, diamond is used for sharpening hard tools, cutting glasses, making bores and rock drilling.

Fullerenes:

• There are newly synthesised allotropes of carbon.
• Unlike graphite and diamond these are discrete molecules of carbon like C₃₂, C₅₀, C₆₀, C₇₀, C₇₆ …….
• These have cage like structures.
• Buck‘minster fullerene or bucky ball have a soccer ball like structure with the formula C₆₀.
  It has a fused’ ring structure with 20 six membered rings and 12 five membered rings.
• Each carbon is sp² hybridised and forms three π sbonds and one delocalised pbond giving aromatic character.
• C-C bond distance is 1.44 Å and C=C bond distance is 1.38 Å.

Carbon nano tubes:

• This is another recently discovered allotropes of carbon.
• They have graphite like tubes with fullerene ends.
• Along the axis, carbon nano tubes are stronger than steel and conduct electricity.
• They have many applications in nano scale electronics, Catalysis, polymers and medicine.

Graphene:

• It is a single planar sheet of graphite.
• In this sp² hybridised carbon atoms are densely packed in a honey comb crystal lattice.

Question 3.
Write about the preparation and structure of silicones.
Answer:

• Silicones or poly siloxanes are organo silicon polymers.
• Their general empirical formula is (R₂SiO)
• Since their empirical formula is similar to Ketones (R₂CO) they are called as Silicones.
• They may be linear or cross linked.
• Due to their very high thermal stability they are called high-temperature polymers.

Types of Silicones:
i) Linear Silicones:
They are obtained by the hydrolysis and subsequent condensation of dialkyl or diaryl
a) Silicone rubbers:
These are bridged together by methylene or similar groups.
b) Silicone resins:
They are obtained by blending silicones with organic resins such as acrylic esters.

ii) Cyclic Silicones:
These are obtained by the hydrolysis of R₂SiCl₂.

iii) Cross linked Silicones:
These are obtained by the hydrolysis of RSiCl₃.

Preparation:
Vapours of RCl or ArCl are passed over silicon at 570 K with copper catalyst gives R₂SiCl₂ (dialkyl dichloro silanes) or Ar₂SiCl₂ (diaryl dichloro silanes)
2RCl + Si \(\underrightarrow { Cu / 570K } \) R₂SiCl₂

Hydrolysis of R₂SiCl₂ gives a straight chain polymer which grows from both sides.

Hydrolysis of mono alkyl trichloro silanes RSiCl₃ gives a very complex cross linked polymer.

Linear silicones can be converted into cyclic or ring silicones when water molecules are removed from the terminal -OH groups.

Question 4.
Explain various types of silicates.
Answer:
The mineral which contains silicon and oxygen in tetrahedral [SiO₄]^4- units linked together in different patterns are called silicates.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.1 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.1

Question 1.
A particle moves along a straight line in such a way that after t seconds its distance from the origin is s = 2t² + 3t metres.
(i) Find the average velocity between t = 3 and t = 6 seconds.
(ii) Find the instantaneous velocities at t = 3 and t = 6 seconds.
Solution:
s = 2t² + 3t
(i) Average velocity between t = 3 and t = 6 seconds
Now s(t) = 2t² + 3t
Average velocity

(ii) f(t) = 2t² + 3t
f'(t) = 4t + 3
f'(3) = 4(3) + 3 = 15
f'(6) = 4(3) + 3 = 15

Question 2.
A camera is accidentally knocked off an edge of a cliff 400 ft high. The camera falls a distance of s = 16t² in t seconds.
(i) How long does the camera fall before it hits the ground?
(ii) What is the average velocity with which the camera falls during the last 2 seconds?
(iii) What is the instantaneous velocity of the camera when it hits the ground?
Solution:
(i) The camera falls a distance of s = 16t² in t sec.
s = 400 ft
∴ 16t² =400
t² = \(\frac { 400 }{ 16 }\) = 25
t = 5 sec
∴ Camera falls for 5 sec before it hits the ground.

(ii) In 5 sec camera falls 400 ft (given)
∴ Average velocity in 2 sec
= \(\frac { s(5)-s(3) }{ 5-3 }\)
= \(\frac { 16(5^2)-16(3^2) }{ 2 }\)
= \(\frac { 400-144 }{ 2 }\)
= \(\frac { 256 }{ 2 }\)
= 128 ft/sec

(iii) f(t) = 16t²
f'(t) = 32t
f'(t) at t = 5 = 32(5)
= 160 ft/sec

Question 3.
A particle moves along a line according to the law s(t) = 2t³ – 9t² + 12t – 4, where t ≥ 0.
(i) At what times the particle changes direction?
(ii) Find the total distance travelled by the particle in the first 4 seconds.
(iii) Find the particle’s acceleration each time the velocity is zero.
Solution:
s (t) = 2t³ – 9t² + 12t – 4, t ≥ 0
velocity v = \(\frac { ds }{ dt }\) = 6t² – 18t + 12
When the particle changes its direction, v = 0
6t² – 18t + 12 = 0 (÷6)
t² – 3t + 2 = 0
(t – 2) (t – 1) = 0
t = 1, 2
∴ When time t = 1 sec and t = 2 sec, the particle changes its direction.

(ii) The distance travelled in the first 4 seconds is
|s(0) – s(1)| + |s(1) – s(2)| + |s(2) – s(3)| + |s(3) – s(4)|
Here, s(t) = 2t³ – 9t² + 12t – 4
s(0) = -4
s(1) = 1
s(2) = 0
s(3) = 5
and s(4) = 28
∴ Distance travelled in the first 4 seconds
= |-4 – 1| + |1 – 0| + |0 – 5| + |5 – 28|
= 5 + 1 + 5 + 23 = 34 m

(iii) s (t) = 2t³ – 9t² + 12t – 4
velocity v = \(\frac { ds }{ dt }\) = 6t² – 18t + 12
v = 0 ⇒ 6(t² – 3t + 2) = 0 ⇒ t = 1, 2
Acceleration = \(\frac { d^2s }{ dt^2 }\) = 12t – 18
at t = 1, Acceleration = 12(1) – 18 = -6m/sec²
at t = 2, Acceleration = 12 (2) – 18 = 6 m/sec²

Question 4.
If the volume of a cube of side length x is v = x³. Find the rate of change of the volume with respect to x when x = 5 units.
Solution:
volume of a cube v = x³
Rate of change \(\frac { dv }{ dx }\) = 3x²
When x = 5 units, \(\frac { dv }{ dx }\) = 3(5)² = 3(25) = 75 units.

Question 5.
If the mass m(x) (in kilograms) of a thin rod of length x (in metres) is given by, m(x) = \(\sqrt { 3x }\) then what is the rate of change of mass with respect to the length when it is x = 3 and x = 27 metres.
Solution:

Question 6.
A stone is dropped into a pond causing ripples in the from of concentric circles. The radius r of the outer ripple is increasing at a constant rate of 2 cm per second. When the radius is 5 cm find the rate of changing of the total area of the disturbed water?
Solution:
radius = r, Rate of changes of radius \(\frac { dr }{ dt }\) = 2 and
given r = 5 cm
Area of circle A = πr²
Differentiating w.r.t ‘t’,
\(\frac { dA }{ dt }\)= 2πr\(\frac { dr }{ dt }\)
= 2π (5) (2)
= 20 π
∴ Area of circle (ripple) is increasing at the rate of 20 π cm²/sec.

Question 7.
A beacon makes one revolution every 10 seconds. It is located on a ship which is anchored 5 km from a straight shore line. How fast is the beam moving along the shoreline when it makes an angle of 45° with the shore?
Solution:

Time for one revolution = 10 sec
Now, angular velocity \(\frac { dv }{ dt }\) = \(\frac { 2π }{ 10 }\) = \(\frac { π }{ 5 }\)
From the figure, tan 45° = \(\frac { AB }{ OA }\)
1 = \(\frac { x }{ 5 }\) ⇒ x = 5
Again, tan θ = \(\frac { x }{ 5 }\)
x = 5 tan θ
Differentiating w.r.t. ‘t’
\(\frac { dx }{ dt }\) = 5 sec² θ \(\frac { dθ }{ dt }\)
= 5 sec² (45°) (\(\frac { π }{ 5 }\))
= (√2)² π = 2π
∴ The beam is moving at the rate of 2π km/sec.

Question 8.
A conical water tank with a vertex down of 12 metres height has a radius of 5 metres at the top. If water flows into the tank at a rate of 10 cubic m/min, how fast is the depth of the water increases when the water is 8 metres deep?
Solution:

From the figure \(\frac { r}{ h }\) = \(\frac { 5 }{ 12 }\)
r = \(\frac { 5h }{ 12 }\)
given rate of change of volume \(\frac { dV }{ dt }\) = 10
When h = 8 to find \(\frac { dh }{ dt }\)
Volume of cone V = \(\frac { 1 }{ 3 }\) πr² h

The depth of the water increasing at the rate of \(\frac { 9 }{ 10π }\) m/min

Question 9.
A ladder 17 metre long is leaning against the wall. The base of the ladder is pulled away from the wall at a rate of 5 m/s. When the base of the ladder is 8 metres from the wall.
(i) How fast is the top of the ladder moving down the wall?
(ii) At what rate, the area of the triangle formed by the ladder, wall, and floor is changing?
Solution:

Let the height of the wall where the ladder touches are ‘y’ m.
The bottom of the ladder is at a distance of ‘x’ m from the wall.
Given x = 8, \(\frac { dx }{ dt }\) = 5
x² + y² = 17²
(Pythagoras Theorem)
y² = 17² – x² = 289 – 64 = 225
∴ y = 15
Differentiating w.r.t. ‘t’

(i) The top of the ladder is moving down the wall at the rate of \(\frac { 8 }{ 3 }\) m/sec
(ii) Area of triangle formed by the ladder, wall and the floor is A = \(\frac { 1 }{ 2 }\) xy
differentiating w.r.t. ‘t’

∴ Area of the triangle is increasing at the rate of 26.83 m²/sec.

Question 10.
A police jeep, approaching an orthogonal intersection from the northern direction, is chasing a speeding car that has turned and moving straight east. When the jeep is 0.6 km north of the intersection and the car is 0.8 km to the east. The police determine with a radar that the distance between them and the car is increasing at 20 km/hr. If the jeep is moving at 60 km/hr at the instant of measurement, what is the speed of the car?
Solution:

given x = 0.8, y = 0.6, \(\frac { dy }{ dt }\) = -60
and \(\frac { ds }{ dt }\) = 20
from the figure
S² = x² + y²,
S² = (0.8)² + (0.6)² = 0.64 + 0.36 = 1
S² = 1 ⇒ S = 1
S² = x² + y²,
Differentiating w.r.t. ‘t’

∴ Speed of the car is 70 km/hr.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 6 Applications of Vector Algebra Ex 6.10 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10

Choose the most suitable answer from the given four alternatives

Question 1.
If \(\overline { a }\) and \(\overline { b }\) are parallel vectors, then [\(\overline { a }\), \(\overline { c }\), \(\overline { b }\)] is equal to
(a) 2
(b) -1
(c) 1
(d) 0
Solution:
(d) 0
Hint:
Since \(\overline { a }\) and \(\overline { b }\) are parallel ⇒ \(\overline { a }\) = λ\(\overline { b }\)
[ \(\overline { a }, \overline { c }, \overline { b }\)] = [λ\(\overline { b }, \overline { c }, \overline { b }\) ]
= λ[ \(\overline { b }, \overline { c }, \overline { b }\) ]
= λ(0) = 0

Question 2.
If a vector \(\overline { α }\) lies in the plane of \(\overline { ß }\) and \(\overline { γ }\), then
(a) [ \(\overline { α }, \overline { ß }, \overline { γ }\) ] = 1
(b) [ \(\overline { α }, \overline { ß }, \overline { γ }\) ] = -1
(c) [ \(\overline { α }, \overline { ß }, \overline { γ }\) ] = 0
(d) [ \(\overline { α }, \overline { ß }, \overline { γ }\) ] = 2
Solution:
(c) [ \(\overline { α }, \overline { ß }, \overline { γ }\) ] = 0
Hint:
If \(\overline { α }\) lies in \(\overline { ß }\) & \(\overline { γ }\) plane
we have [ \(\overline { α }, \overline { ß }, \overline { γ }\) ] = 0

Question 3.
If \(\overline { a }\).\(\overline { b }\) = \(\overline { b }\).\(\overline { c }\) = \(\overline { c }\).\(\overline { a }\) = 0, then the value of [ \(\overline { a }, \overline { b }, \overline { c }\) ] is
(a) |\(\overline { a }\)| |\(\overline { b }\)| |\(\overline { c }\)|
(b) \(\frac { 1 }{ 3 }\)|\(\overline { a }\)| |\(\overline { b }\)| |\(\overline { c }\)|
(c) 1
(d) -1
Solution:
(a) |\(\overline { a }\)| |\(\overline { b }\)| |\(\overline { c }\)|
Hint:

Question 4.
If \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) are three unit vectors such that \(\overline { a }\) is perpendicular to \(\overline { b }\) and is parallel to \(\overline { c }\) then \(\overline { a }\) × (\(\overline { b }\) × \(\overline { c }\)) is equal to
(a) \(\overline { a }\)
(b) \(\overline { b }\)
(c) \(\overline { c }\)
(d) \(\overline { 0 }\)
Solution:
(b) \(\overline { b }\)
Hint:

Question 5.
If [ \(\overline { a }, \overline { b }, \overline { c }\) ] = 1 then the value of

(a) 1
(b) -1
(c) 2
(d) 3
Solution:
(a) 1
Hint:

Question 6.
The volume of the parallelepiped with its edges represented by the vectors \(\hat { i }\) + \(\hat { j }\), \(\hat { i }\) + 2\(\hat { j }\), \(\hat { i }\) + \(\hat { j }\) + π\(\hat { k }\) is
(a) \(\frac { π }{ 2 }\)
(b) \(\frac { π }{ 3 }\)
(c) π
(d) \(\frac { π }{ 4 }\)
Solution:
(c) π
Hint:
\(\left|\begin{array}{lll}
1 & 1 & 0 \\
1 & 2 & 0 \\
1 & 1 & \pi
\end{array}\right|\) = π\(\left|\begin{array}{ll}
1 & 1 \\
1 & 2
\end{array}\right|\)
= π (2 – 1) = π

Question 7.
If \(\overline { a }\) and \(\overline { b }\) are unit vectors such that [\(\overline { a }\), \(\overline { b }\), \(\overline { a }\) × \(\overline { b }\)] = \(\frac { 1 }{ 4 }\), then the angle between \(\overline { a }\) and \(\overline { b }\) is
(a) \(\frac { π }{ 6 }\)
(b) \(\frac { π }{ 4 }\)
(c) \(\frac { π }{ 3 }\)
(d) \(\frac { π }{ 2 }\)
Solution:
(a) \(\frac { π }{ 6 }\)
Hint:

Question 8.
If \(\overline { a }\) = \(\hat { i }\) + \(\hat { j }\) + \(\hat { k }\), \(\overline { b }\) = \(\hat { i }\) + \(\hat { j }\), \(\overline { c }\) = \(\hat { i }\) and (\(\overline { a }\) × \(\overline { b }\))\(\overline { c }\) – λ\(\overline { a }\) + µ\(\overline { b }\) then the value of λ + µ is
(a) 0
(b) 1
(c) 6
(d) 3
Solution:
(a) 0
Hint:
\(\overline { a }\).\(\overline { c }\) = 1 and \(\overline { b }\).\(\overline { c }\) = 1
(\(\overline { a }\) × \(\overline { b }\))\(\overline { c }\) = (\(\overline { c }\) × \(\overline { a }\))\(\overline { b }\) – (\(\overline { c }\) × \(\overline { b }\))\(\overline { a }\) = λ\(\overline { a }\) + µ\(\overline { b }\)
⇒ µ = c; a = 1λ = -(\(\overline { c }\).\(\overline { b }\)) = -1
µ + λ = 1 – 1 = 0

Question 9.
If \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) are non-coplanar, non-zero vectors
such that [\(\overline { a }\), \(\overline { b }\), \(\overline { c }\)] = 3, then {[\(\overline { a }\) × \(\overline { b }\), \(\overline { b }\) × \(\overline { c }\), \(\overline { c }\) × \(\overline { a }\)]²} is equal to
is equal to
(a) 81
(b) 9
(c) 27
(d) 18
Solution:
(a) 81
Hint:

= 3⁴ = 81

Question 10.
If \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) are three non-coplanar vectors such that \(\overline { a }\) × (\(\overline { b }\) × \(\overline { c }\)) = \(\frac { \overline{b}+\overline{c} }{ √2 }\) then the angle between \(\overline { a }\) and \(\overline { b }\) is
(a) \(\frac { π }{ 2 }\)
(b) \(\frac { 3π }{ 4 }\)
(c) \(\frac { π }{ 4 }\)
(d) π
Solution:
(b) \(\frac { 3π }{ 4 }\)
Hint:

Question 11.
If the volume of the parallelepiped with \(\overline { a }\) × \(\overline { b }\), \(\overline { b }\) × \(\overline { c }\), \(\overline { c }\) × \(\overline { a }\) as coterminous edges is 8 cubic units, then the volume of the parallelepiped with (\(\overline { a }\) × \(\overline { b }\)) × (\(\overline { b }\) × \(\overline { c }\)), (\(\overline { b }\) × \(\overline { c }\)) × (\(\overline { c }\) × \(\overline { a }\)) and (\(\overline { c }\) × \(\overline { a }\)) × (\(\overline { a }\) × \(\overline { b }\)) as coterminous edges is
(a) 8 cubic units
(b) 512 cubic units
(c) 64 cubic units
(d) 24 cubic units
Solution:
(c) 64 cubic units
Hint:
Given volume of the parallelepiped with

Question 12.
Consider the vectors \(\overline { a }\), \(\overline { b }\), \(\overline { c }\), \(\overline { d }\) such that (\(\overline { a }\) × \(\overline { b }\)) × (\(\overline { c }\) × \(\overline { d }\)) = \(\overline { 0 }\) Let P₁ and P₂ be the planes determined by the pairs of vectors \(\overline { a }\), \(\overline { b }\) and \(\overline { c }\), \(\overline { d }\) respectively. Then the angle between P₁ and P₂ is
(a) 0°
(b) 45°
(c) 60°
(d) 90°
Solution:
(a) 0°
Hint:
A vector perpendicular to the plane P₁ of a, b is \(\overline { a }\) × \(\overline { b }\),
A vector perpendicular to the plane P₂ of c and d is \(\overline { c }\) × \(\overline { d }\)
∴ (\(\overline { a }\) × \(\overline { b }\)) × (\(\overline { c }\) × \(\overline { d }\)) = 0
⇒ (\(\overline { a }\) × \(\overline { b }\)) || \(\overline { c }\) × \(\overline { d }\)
⇒ The angle between the planes is \(\overline { 0 }\)

Question 13.
If \(\overline { a }\) × (\(\overline { b }\) × \(\overline { c }\)) = (\(\overline { a }\) × \(\overline { b }\)) × \(\overline { c }\) where \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) are any three vectors such that \(\overline { b }\).\(\overline { c }\) ≠ 0 and \(\overline { a }\).\(\overline { b }\) ≠ 0, then \(\overline { a }\) and \(\overline { c }\) are
(a) perpendicular
(b) parallel
(c) inclined at angle \(\frac { π }{ 3 }\)
(d) inclined at an angle \(\frac { π }{ 6 }\)
Solution:
(b) parallel
Hint:

Question 14.
If \(\overline { a }\) = 2\(\hat { i }\) + 3\(\hat { j }\) – \(\hat { k }\), \(\overline { b }\) = \(\hat { i }\) + 2\(\hat { j }\) – 5\(\hat { k }\), \(\overline { c }\) = 3\(\hat { i }\) + 5\(\hat { j }\) – \(\hat { k }\) then \(\overline { a }\) vector perpendicular to a and lies in the plane containing \(\overline { b }\) and \(\overline { c }\) is
(a) -17\(\hat { i }\) + 21\(\hat { j }\) – 97\(\hat { k }\)
(b) 17\(\hat { i }\) + 21\(\hat { j }\) – 123\(\hat { k }\)
(c) -17\(\hat { i }\) – 21\(\hat { j }\) + 97\(\hat { k }\)
(d) -17\(\hat { i }\) – 21\(\hat { j }\) – 97\(\hat { k }\)
Solution:
(d) -17\(\hat { i }\) – 21\(\hat { j }\) – 97\(\hat { k }\)
Hint:
A vector ⊥r to \(\overline { a }\) and lies in the plane containing \(\overline { b }\) and \(\overline { c }\)

Question 15.
The angle between the lines \(\frac { x-2 }{ 3 }\) = \(\frac { y+1 }{ -2 }\), z = 2 and \(\frac { x-1 }{ 1 }\) = \(\frac { 2y+3 }{ 3 }\) = \(\frac { z+5 }{ 2 }\) is
(a) \(\frac { π }{ 6 }\)
(b) \(\frac { π }{ 4 }\)
(c) \(\frac { π }{ 3 }\)
(d) \(\frac { π }{ 2 }\)
Solution:
(d) \(\frac { π }{ 2 }\)
Hint:

Question 16.
If the line \(\frac { x-2 }{ 3 }\) = \(\frac { y-1 }{ -5 }\) = \(\frac { z+2 }{ 2 }\) lies in the plane x + 3y – αz + ß = 0 then (α + ß) is
(a) (-5, 5)
(b) (-6, 7)
(c) (5, -5)
(d) (6, -7)
Solution:
(b) (-6, 7)
Hint:
\(\frac { x-2 }{ 3 }\) = \(\frac { y-1 }{ 5 }\) = \(\frac { z+2 }{ 2 }\) = λ ⇒ (3λ + 2, -5λ + 1, 2λ – 2)
which lie in x + 3y – αz + ß = 0
(3λ + 2) + 3(-5λ + 1) – α(2λ – 2) + ß = 0
3λ + 2 – 15λ + 3 – 2αλ + 2α + ß = 0.
(-12λ – 2αλ) + 2α + ß + 5 = 0.
-12λ – 2αλ = 0
2αλ = -12λ
α = -6
2α+ ß +5 = 0
-12 + ß + 5 = 0
ß – 7 = 0
ß = 7
(α, ß) = (-6, 7)

Question 17.
The angle between the line \(\overline { r }\) = (\(\hat { i }\) + 2\(\hat { j }\) – 3\(\hat { k }\)) + t(2\(\hat { i }\) + \(\hat { j }\) – 2\(\hat { k }\)) and the plane \(\overline { r }\) (\(\hat { i }\) + \(\hat { j }\)) + 4 = 0 is
(a) 0°
(b) 30°
(c) 45°
(d) 90°
Solution:
(c) 45°
Hint:

Question 18.
The co-ordinates of the point where the line \(\overline { r }\) = (6(\(\hat { i }\) – \(\hat { j }\) – 3\(\hat { k }\)) + t(-\(\hat { i }\) + \(\hat { k }\) meets the plane \(\overline { r }\) ((\(\hat { i }\) + (\(\hat { j }\) – (\(\hat { k }\)) = 3 are
(a) (2, 1, 0)
(b) (7, -1, -7)
(c) (1, 2, -6)
(d) (5, -1, 1)
Solution:
(d) (5, -1, 1)
Hint:
Given \(\overline { r }\) = (6(\(\hat { i }\) – (\(\hat { j }\) – 3(\(\hat { k }\)) + t(-(\(\hat { i }\) + (\(\hat { k }\))
\(\frac { x-6 }{ -1 }\) = \(\frac { y+1 }{ 0 }\) = \(\frac { z+3 }{ 4 }\) = t ⇒ (-t + 6, -1, 4t – 3)
which meets x + y – z = 3
-t + 6 – 1 – 4t + 3 = 3
-5t + 5 = 0
5t = 5
t = 1
∴ Co-ordinate is (5, -1, 1)

Question 19.
Distance from the origin to the plane 3x – 6y + 2z + 7 = 0 is
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(b) 1
Hint:
(x₁, y₁, z₁) = (o, 0, o)
(a, b, c) = (3, -6, 2); d = 7.
d = \(\frac { ax_1+by_1+cz_1+d }{ \sqrt{a^2+b^2+c^2} }\) = \(\frac { 7 }{ \sqrt{9+36+4} }\) = \(\frac { 7 }{ 7 }\) = 1

Question 20.
The distance between the planes
x + 2y + 3z + 7 = 0 and 2x + 4y + 6z + 7 = 0 is
(a) \(\frac { √7 }{ 2√2 }\)
(b) \(\frac { 7 }{ 2 }\)
(c) \(\frac { √7 }{ 2 }\)
(d) \(\frac { 7 }{ 2√2 }\)
Solution:
(a) \(\frac { √7 }{ 2√2 }\)
Hint:
x + 2y + 3z+7 = 0
2x + 4y + 6z + 7 = 0
(÷ 2) x + 2y + 3z + \(\frac { 7 }{ 2 }\) = 0
(1) and (2) are parallel planes

Question 21.
If the direction cosines of a line are \(\frac { 1 }{ c }\), \(\frac { 1 }{ c }\), \(\frac { 1 }{ c }\)
(a) c = ±3
(b) c = ±√3
(c) c > 0
(d) 0 < c < 1
Solution:
(b) c = ±√3
Hint:
cos²α + cos²ß + cos²γ = 1
\(\frac { 1 }{ c^2 }\) + \(\frac { 1 }{ c^2 }\) + \(\frac { 1 }{ c^2 }\) = 1
\(\frac { 3 }{ c ^2}\) = 1
c² = 3
c = ±√3

Question 22.
The vector equation \(\overline { r }\) = (\(\hat { i }\) – 2\(\hat { j }\) – \(\hat { k }\)) + t(6\(\hat { i }\) – \(\hat { k }\)) represents a straight line passing through the points
(a) (0, 6, -1) and (1, -2, -1)
(b) (0, 6, -1) and (-1, -4, -2)
(c) (1, -2, -1) and (1, 4, -2)
(d) (1, -2, -1) and (0, -6, 1)
Solution:
(c) (1, -2, -1) and (1, 4, -2)
Hint:
Given vector equation is

Question 23.
If the distance of the point (1, 1, 1) from the origin is half of its distance from the plane x + y + z + k = Q, then the values of k are
(a) ±3
(b) ±6
(c) -3, 9
(d) 3, -9
Solution:
(d) 3, -9
Hint:

Question 24.
If the planes \(\overline { r }\) (2\(\hat { i }\) – λ\(\hat { j }\) + \(\hat { k }\)) = 3 and \(\overline { r }\) (4\(\hat { i }\) + \(\hat { j }\) – µ\(\hat { k }\)) = 5 are parallel, then the value of λ and µ are
(a) \(\frac { 1 }{ 2 }\), -2
(b) –\(\frac { 1 }{ 2 }\), 2
(c) –\(\frac { 1 }{ 2 }\), -2
(d) \(\frac { 1 }{ 2 }\), 2
Solution:
(c) –\(\frac { 1 }{ 2 }\), -2
Hint:

Question 25.
If the length of the perpendicular from the origin to the plane 2x + 3y + λz = 1, λ > 0 is \(\frac { 1 }{ 5 }\), then the value of λ is
(a) 2√3
(b) 3√2
(c) 0
(d) 1
Solution:
(a) 2√3
Hint:

5 = \(\sqrt { 4+9+λ^2 }\)
25 = 4 + 9 + λ²
25 = 13 + λ²
λ² = 12
λ = 2√3

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Botany Guide Pdf Chapter 10 Economically Useful Plants and Entrepreneurial Botany Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Botany Solutions Chapter 10 Economically Useful Plants and Entrepreneurial Botany

12th Bio Botany Guide Economically Useful Plants and Entrepreneurial Botany Text Book Back Questions and Answers

I. Choose the correct answer :

Question 1.
Consider the following statements and choose the right option.
i) Cereals are members of grass family
ii) Most of the food grains come from monocotyledon
a) (i) is correct and (ii) is wrong
b) Both (i) and (ii) are correct
c) (i) is wrong and (ii) is correct
d) Both (i) and (ii) are wrong
Answer:
b) Both (i) and (ii) are correct

Question 2.
Assertion: Vegetables are important part of healthy eating.
Reason : Vegetables are succulent structures of plants with pleasant aroma and flavours.
a) Assertion is correct, Reason is wrong
b) Assertion is wrong, Reason is correct
c) Both are correct and reason is the correct explanation for assertion.
d) Both are correct and reason is not the correct explanation for assertion.
Answer:
a) Assertion is correct, Reason is wrong

Question 3.
Groundnut is native of ……………..
a) Philippines
b) India
c) North America
d) Brazil
Answer:
d) Brazil

Question 4.
Statement A : Coffee contains caffeine
Statement B : Drinking coffee enhances cancer
a) A is correct, B is wrong
b) A and B – both are correct
c) A is wrong, B is correct
d) A and B – both are wrong
Answer:
a) A is correct, B is wrong

Question 5.
Tectona grandis is coming under family.
a) Lamiaceae
b) Fabaceae
c) Dipterocaipaceae
d) Ebenaceae
Answer:
a) Lamiaceae

Question 6.
Tamarindus indica is indigenous to ……………..
a) Tropical African region
b) South India, Sri Lanka
c) South America, Greece
d) India alone
Answer:
a) Tropical African region

Question 7.
New world species of cotton
a) Gossipium arboretum
b) G. herbaceum
c) Both a and b
d) G.barbadense
Answer:
d) G. barbadense

Question 8.
Assertion : Turmeric fights various kinds of cancer.
Reason : Curcumin is an anti-oxidant present in turmeric.
a) Assertion is correct, Reason is wrong
b) Assertion is wrong, Reason is correct
c) Both are correct
d) Both are wrong
Answer:
c) Both are correct

Question 9.
Find out the correctly matched pair.
a) Rubber – Shorea robusta
b) Dye – Lawsonia inermis
c) Timber – Cyperus papyrus
d) Pulp – Hevea brasiliensis
Answer:
b) Dye : Lawsonia inermis

Question 10.
Observe the following statements and pick out the right option from the following.
Statement I : Perfumes are manufactured from essential oils.
Statement II : Essential oils are formed at different parts of the plants.
a) Statement I is correct
b) Statement II is correct
c) Both statements are correct
d) Both statements are wrong
Answer:
c) Both statements are correct

Question 11.
Observe the following statements and pick out the right option from the following.
Statement I : The drug sources of Siddha include plants, animal parts, ores and minerals.
Statement II: Minerals are used for preparing drugs with long shelf-life.
a) Statement I is correct
b) Statement II is correct
c) Both statements are correct
d) Both statements are wrong
Answer:
Both statements are correct

Question 12.
The active principle trans-tetra hydro canabial is present in
a) Opium
b) Curcuma
c) Marijuana
d) Andrographis
Answer:
c) Marijuana

Question 13.
Which one of the following matches is correct?
a) Palmyra – Native of Brazil
b) Saccharun – Abundant in Kanyakumari
c) Stevecide – Natural sweetener
d) Palmyra sap – Fermented to give ethanol
Answer:
c) Stevecide – Natural sweetener

Question 14.
The only cereal that has originated and domesticated from the New world.
a) Oryza sativa
b) Triticum asetumn
c) Triticum duram
d) Zea mays
Answer:
d) Zea mays

Question 15.
Write the cosmetic uses of Aloe.
Answer:
Aloe gel are used as skin tonic. It has a cooling effect and moisturizing characteristics and hence used in preparation of creams, lotions, shampoos, shaving creams, after shave lotions and allied products. It is used in gerontological applications for rejuvenation of aging skin. Products prepared from aloe leaves have multiple properties such as emollient, antibacterial, antioxidant, antifungal and antiseptic. Aloe vera gel is used in skin care cosmetics.

Question 16.
What is pseudo cereal? Give an example.
Answer:

• These are foods that are prepared and eaten as whole grain. Eg. quinoa (தினை) is a seed from Chenopodium quinoa plant. It belongs Amaranthaceae family.
• It is gluten free, whole grain carbohydrate.
• It is a whole protein with a essential amino acids.
• Taken for 6000 years in Andes hills.

Question 17.
Discuss which wood is better for making furniture.
Answer:
Teak wood is the ideal type of wood for making household furnitures because, it is highly durable and shows great resistance against the attack of termites and fungi. Moreover it doesnot split or crack and is a carpenter friendly wood.

Question 18.
A person got irritation while applying chemical dye. What would be your suggestion for alternative?
Answer:

• Henna is the best alternative dye.
• It is in North Africa, South west Asia. It is in Gujarat, Madhya Pradesh and Rajesthan.
• Orange dye henna is from leaves and shoots of
  Lawsonia inermis.
• Principal colouring matter is ‘lacosone’
• It is harmless causing no skin irritation.
• It is u sed to dye skin, hair and finger nails.

Question 19.
Name the humors that are responsible for the health of human beings.
Answer:
Vatam, Pittam and Kapam.

Question 20.
Give definitions for organic farming?
Answer:

• Alternative agricultural system.
• Plants and crops are cultivated in natural ways, by using biological inputs.
• It helps to maintain soil fertility and ecological balance.
• It minimizes pollution, wastage.

Question 21.
Which is called the “King of Bitters”? Mention their medicinal importance.
Answer:
Andrographis paniculata is called as King of Bitters. Andrographis is a potent hepatoprotective agent and is widely used to treat liver disorders. Concoction of Andrographis paniculata and eight other herbs (Nilavembu Kudineer) is effectively used to treat malaria and dengue.

Question 22.
Differentiate Bio-medicines and botanical medicines.
Answer:
Bio-medicines: Medicinally useful molecules obtained from plants are marketed as drugs. These are called bio-medicines.
Botanical Medicines: Medicinal plants are marketed as powders or in other modified forms. They are called Botanical medicines.

Question 23.
Write the origin and area of cultivation of green gram and red gram.
Answer:
Origin and area of cultivation of Green Gram.

• Native of India
• Archaeological evidence is in Maharashtra.
• Cultivated in Madhya Pradesh, Karnataka and Tamil Nadu.

Origin and area of cultivation of Red Gram

• The only pulse native of South India.
• Grown in Maharashtra, Andhra Pradesh,
  Madhya Pradesh, Karnataka, Gujarat.

Question 24.
What are millets? What are its types? Give example for each type.
Answer:
Millet’s: Small seeds cultivated by ancient people of Africa, Asia. Gluten-free with the less glycemic index.

Finger Millet (Ragi) (Eleusine coracana)
Came to India from East Africa. It is rich in calcium.’

Uses:

• Staple food in South Indian hills.
• Made into porridge, gruel.
• Ragi malt is a nutrient drink.
• Source of fermented beverages.

Sorghum vulgare.
Native of Africa. Major millet of the world with calcium, iron
Uses:

• Feed to poultry, birds, pigs, cattle
• Alcoholic beverage source.

Fox tail Millet (Setariaitalica)
Oldest traditional millet of India. Domesticated in China about 6000 years.
Uses: Strengthens heart, eye sight, lactation.

Kodo Millet (Paspalum scrobiculatum)
From West Africa.
Uses:

• Flour for pudding
• Diuretic, cures constipation.
• Reduce obesity, blood sugar, blood pressure.

Question 25.
If a person drinks a cup of coffee daily it will help him for his health. Is this correct? If it is correct, list out the benefits.
Answer:
Benefits of Coffee:

• Stimulates central nervous system.
• Mild diuretic
• Enhances acetyl choline release in brain.
• Enhances efficiency.
• Lower fatty liver diseases, cirrhosis, cancer.
• Reduce the risk of type 2 diabetes,

Question 26.
Enumerate the uses of turmeric.
Answer:
Turmeric is one of the most important and ancient Indian spices and used traditionally over thousands of years for culinary, cosmetic, dyeing and for medicinal purposes. It is an important constituent of curry powders. Turmeric is used as a colouring agent in pharmacy, confectionery and food industry. Rice coloured with turmeric (yellow) is considered sacred and auspicious which is used in ceremonies. It is also used for dyeing leather, fibre, paper and toys.

Curcumin extracted from turmeric is responsible for the yellow colour. Curcumin is a very good anti-oxidant which may help fight various kinds of cancer. It has anti-inflammatory, anti- ‘ diabetic, anti-bacterial, anti-fungal and anti-viral activities. It stops platelets from clotting in arteries, which leads to heart attack.

Question 27.
What is TSM? How does it classify and what does it focus on?
Answer:
Traditional System of Medicines (TSM)
It is classified into

1. Institutionalized (documented) system
2. Non-institutionalized (oral) system

Institutionalized system:

• It includes Siddha, Ayurvedha
• It is practiced for 2000 years.
• Text with symptoms, diagnosis, drugs, preparation of drugs, dosage, diet regimen.

Non-Institutionalized system:

• Do not have any records
• Practiced by rural, tribal people of India.
• Knowledge is in oral form.

Focus of TSM:

• Healthy lifestyle
• A healthy diet for good health
• Disease reversal.

Siddha system

• Siddha is the most popular, widely practiced and culturally accepted systm in Tamil Nadu.
• Siddha is principally based on the pancabute philosophy
• This system specializes in using minerals for preparing drugs with a long shelf-life.
• This system uses about 800 herbs as source of drugs.
• Great stress is laid on disease prevention, health promotion, rejuvenation and cure.

Ayurveda system:

• Ayurveda supposed to have originated from Brahma.
• The core knowledge is documented by charaka, sushruta and vagbhata in compendiums written by them.
• This system uses more of herbs and few animal parts as drug sources.
• Plant sources include a good propertion of Himalayan plants.
• The Ayurvedic pharmacopoeia of India lists about 500 plants used as source of drugs.

Folk system of medicine

• Major tribal communities in Tamil Nadu who are known for their medicinal knowledge indued Irulas, Malayalis, Kurumbas, paliyans and kaanis.
• Folk system survive as oral traditions among innumerable rural and tribal communities of India.

Question 28.
Write the uses of nuts you have studied.
Answer:
Cashews are commonly used for garnishing sweets or curries, or ground into a paste that forms a base of sauces for curries or some sweets. Roasted and raw kernels are used as snacks.

Question 29.
Give an account of the role of Jasminum in perfuming.
Answer:
Role of Jasminum in perfuming:

• Used in India for worship, ceremonial purposes, incense, fumigants.
• For making perfumed hair oil, cosmetics and soaps.
• Essential oil for soothing relaxing, antidepressant qualities.
• Blends with other perfumes.
• Used in modern perfumery and cosmetics.
• Popular in air fresheners, antiperspirants, talcum powder, shampoo, and deodorants.

Rose:
The average oil yield is a little less than 0.5 g from lOOOg of flowers.
Uses:

• Rose oil is largely used in perfumes, scenting
  soaps, flovouring soft drinks, liqueurs, and certain types of tobacco, particularly snuff of chewing tobacco.
• In India, water is much used in eye lotions
  and eyewashes.
• Rosewater (panner) containing much of phenyl ethyl alcohol and other compounds in dissolved confectioneries syrups and soft drinks.
• In addition, it is sprinkled on guests as a ceremonial welcome.

Question 30.
Give an account of active principle and medicinal values of any two plants you have studied.
Answer:
A) Medical importance of Keezhanelli (Phyllanthus amarus):
Active principle: Phyllanthus is a major chemical component.
Medical Importance:

• Hepatoprotective.
• Used in Tamil Nadu for jaundice treatment.
• Effective against hepatitis B virus.

B) Nilavembu (Andrographis paniculata) (King of Bitters)
Active principle: Andrographolides.
Medicinal Importance:

• Potent hepatoprotective
• Treats liver disorders.
• A concoction of Andrographis + 8 herbs
  (Nilavembu Kudineer) treats malaria, dengue.

Question 31.
Write the economic importance of rice.
Answer:
Rice is the easily digestible calorie rich cereal food which is used as a staple food in Southern and North East India. Various rice products such as Flaked rice (Aval), Puffed rice / parched rice (Pori) are used as breakfast cereal or as snack food in different parts of India. Rice bran oil obtained from the rice bran is used in culinary and industrial purposes. Husks are used as fuel, and in the manufacture of packing material and fertilizer.

Question 32.
Which TSM is widely practiced and culturally accepted in Tamil Nadu? explain.
Answer:
Siddha system of Medicine:

• It is widely practiced and culturally accepted in Tamil Nadu.
• Based on text of 18siddhars.
• Knowledge is documented as Tamil poems.
• Based on Pancabuta philosophy.
• Vatam, Pittam, Kapam are 3 humors. They are responsible for the health.
• Drug sources are plant, animal parts, marine products, minerals.
• Minerals are used for preparing drugs with long self life.
• 800 herbs are source of drugs.
• Disease prevention, health promotion, rejuvenation and cure are important.

Question 33.
What are psychoactive drugs? Add a note of Marijuana and Opium.
Answer:
Phytochemicals or drugs from some of the plants alter an individual’s perceptions of mind by producing hallucination are known as psychoactive drugs.

1. Marijuana: Marijuana is obtained from Cannabis sativa. The active principle in Marijuana is trans – tetrahydrocannabinol (TCH). It is used as pain killer and reduce hypertension. It is also used in the treatment of Glaucoma, cancer radiotherapy and asthma, etc.
2. Opium: Opium is obtained from the exudates of the fruits of papaver somniferum (poppy plants). It is used to induce sleep and relieve pain. Opium yields morphine which is used as a strong analgesic in surgeries.

Question 34.
What are the King and Queen of Spices? Explain about them and their uses.
Answer:
Queen of Spices: Cardamom (Elettaria Cardamomum)

Origin and area of cultivation:

• Indigenous to Southern India and Sri Lanka.
• Main cash crop in the Western Ghats, North-Eastern India.

Uses:

• For flavouring confectionaries, Bakery products, beverages.
• Seeds are used in curry powder, pickles and cakes.
• Medicinally, a stimulant and carminative (a drug for flatulence)
• Chewed as mouth fresheners.

King of Spices: Black Pepper (Piper nigrum)

Origin and area of cultivation:

• Indigenous to western ghats.
• Black gold of India.
• Kerala, Karnataka and Tamil Nadu are top producers in India.
• Pungency is due to alkaloid piperine.
• 2 types (Black pepper, white pepper)

Uses:

• Flavouring sauce, soup, curry, and pickles
• Aromatic stimulant for salivary gastric secretions as a stomachic.
• Pepper enhances the absorption of medicines.

Question 35.
How will you prepare an organic pesticide for your home garden with the vegetables available from your kitchen?
Answer:
Preparation of Organic Pesticide:

Step 1: Mix 120 g of hot chilies with 110 g of garlic or onion. Chop them thoroughly.

Step 2: Blend the vegetables together manually or using an electric grinder until it forms a thick paste.

Step 3: Add the vegetable paste to 500 ml of warm water. Give the ingredients a stir to thoroughly mix them together.

Step 4: Pour the solution into a glass container and leave it undisturbed for 24 hours. If possible, keep the container in a sunny location. If not, at least keep the mixture in a warm place.

Step 5: Strain the mixture. Pom- the solution through a strainer, remove the vegetables and collect the vegetable-infused water and pour into another container. This filtrate is the pesticide. Either discard the vegetables or use it as a compost.

Step 6: Pour the pesticide into a squirt bottle. Make sure that the spray bottle has first been cleaned with warm water and soap to get rid it of any potential contaminants. Use a funnel to transfer the liquid into the squirt bottle and replace the nozzle.

Step 7: Spray your plants with the pesticide. Treat the infected plants every 4 to 5 days with the solution. After 3 or 4 treatments, the pest will be eliminated. If the area is thoroughly covered with the solution, this pesticide should keep bugs away for the rest of the season.

12th Bio Botany Guide Economically Useful Plants and Entrepreneurial Botany Additional Important Questions and Answers

I. Choose the correct answer :

Question 1.
Staple food of North India is ………………….
a) Sorghum
b) Millet
c) Paddy
d) Wheat
Answer:
d) Wheat

Question 2.
Folk system of medicine is popular in ………………
a) Nigeria
b) USA
c) India
d) UK
Answer:
c) India

Question 3.
State not growing black gram
a) Uttar Pradesh
b) Tamil Nadu
c) Chattisgarh
d) Karnataka
Answer:
b) Tamil Nadu

Question 4.
The very common rubber yielding plant of Tamilnadu is ……………………
a) Manihot esculenta
b) Ficus elastica
c) Hevea benthamiana
d) Hevea brasiliensis
Answer:
d) Hevea brasiliensis

Question 5.
Not a major cultivar mango in India
a) Alphonsa
b) Neelam
c) Malgova
d) Salem Mango
Answer:
d) Salem Mango

Question 6.
Toddy is from ……………… tree
a) Palmyra
b) Coconut
c) Mango
d) Sugar cane
Answer:
a) Palmyra

Question 7.
Chillies are a good source of:
a) Vitamin A, C and E
b) Vitamin K
c) Vitamin D
d) Vitamin B complex and Vitamin D
Answer:
a) Vitamin A, C and E

Question 8.
Gingeelly or sesame is originated in ……………………..
a) Asia
b) Africa
c) China
d) Europe
Answer:
b) Africa

Question 9.
Coffee is native of ……………….
a) Nigeria
b) Cuba
c) Ethiopia
d) Egypt
Answer:
c) Ethiopia

Question 10.
India is the largest producer of
a) Chilly
b) Tamarind
c) Turmeric
d) Pepper
Answer:
c) Turmeric

Question 11.
World’s largest turmeric market is in ………………. of Tamil Nadu
a) Coimbatore
b) Erode
c) Madurai
d) Nagercoil
Answer:
b) Erode

Question 12.
Asia contributes …………… % of latex in world production.
a) 80
b) 90
c) 70
d) 50
Answer:
b) 90

Question 13.
……………….. is the largest producer of latex
a) Kerala
b) Karnataka
c) Andhra
d) Delhi
Answer:
a) Kerala

Question 14.
…………….. is native of Sudan
a) Henna
b) Aloe
c) Jasmine
d) Turmeric
Answer:
b) Aloe

Question 15.
Thovalai of Tamil Nadu produces ………………..
a) Aloe
b) Tamarind
c) Turmeric
d) Jasmine
Answer:
d) Jasmine

Question 16.
Paste of this plant is used in bone fracture
a) Ocimum
b) Phyllanthus
c) Cissus
d) Acalypha
Answer:
c) Cissus

Question 17.
Find the Matching Pair
a) Ocimum – Antiseptic
b) Phyllanthus – Ringworm disease
c) Acalypha – Immune modulator
d) Aegle marmelos – Bone fracture
Answer:
a) Ocimum – Antiseptic

Question 18.
Capsaicin is in ……………..
a) Chilly
b) Pepper
c) tea
d) coffee
Answer:
a) Chilly

Question 19.
Veldt grape is the common name of …………….
a) Ocimum
b) I’hyllanthus
c) Acalypha
d) Cissus
Answer:
d) Cissus

Question 20.
Find the Mismatching Pair
a) Pappaver somniferum – Opium
b) Cannabis sativa – Marijuana
c) Phyllanthus amarus – Keezhanelli
d) Andrographis paniculata – Turmeric
Answer:
d) Andrographis paniculata – Turmeric

Question 21.
Match

a) A-4, B-3, C-l, D-2
b) A-l, B-2, C-3, D-4
c) A-4, B-3, C-2, D-l
d) A-2, B-l, C-4, D-3
Answer:
a) A-4, B-3, C-l, D-2

Question 22.
Vigna mungo is the botanical name of ………………..
a) Black gram
b) Red gram
c) Green gram
d) Brown gram
Answer:
a) Black gram

Question 23.
Match

a) A-4, B-3, C-2, D-l
b) A-l, B-2, C-3, D-4
c) A-2, B-l, C-4, D-3
d) A-3, B-l, C-2, D-4
Answer:
a) A-4, B-3, C-2, D-l

Question 24.
Which one of the following is an incorrect pair?
a) Turmeric – Erode
b) Cardamom – Queen of spices
c) Rubber – Kerala
d) Banana – National fruit of India
Answer:
d)Banana – National fruit of India

Question 25.
Assertion (A): Rice is the staple food for most of people in the world.
Reason (R): It is easily digestible and calorie-rich food.
a) (A) correct; (R) wrong
b) (A) wrong; (R) correct
c) (A) correct; (R) correct; but (R) does not explain (A)
d) (A) correct; (R) correct; (R) explains (A)
Answer:
d) (A) correct; (R) correct; (R) explains (A)

Question 26.
………………… is the largest consumer of coffee in India?
a) Tamil Nadu
b) Andhra
c) Kerala
d) Karnataka
Answer:
a) Tamil Nadu

Question 27.
………………… is the largest coffee producing estate in India
a) Kerala
b) Karnataka
c) Tamil Nadu
d) Andhra
Answer:
b) Karnataka

Question 28.
Curcumin is extracted from
a) Turmeric
b) Chilly
c) Cardamom
d) Tamarind
Answer:
a) Turmeric

Question 29.
Vilvum belongs to ……………………
a) Lamiaceae
b) Rutaceae
c) Vitaceae
d)Euphorbiaceae
Answer:
b) Rutaceae

Question 30.
Dr. Thyagarajan of university of Madras proved effect of Phyllanthus amarus against
a) Hepatitis-B
b) Cirrhosis
c) Cancer
d) Typhoid
Answer:
a) Hepatitis-B

Question 31.
Which one of the following is highly effective against jaundice?
a) Nilavembu
b) Opium poppy
c) Marijuana
d) Phyllanthus
Answer:
d) Phyllanthus

Question 32.
……………… Are gluten free with less Glycemic indess
a) pulses
b) gram
c) vegetables
d) millets
Answer:
d) millets

Question 33.
…………….. is native to tropical region of Africa.
a) Sugar cane
b) Palmyra
c) Peanut
d) Sesame
Answer:
b) Palmyra

Question 34.
Nuts contain ……………. Oil
a) 54%
b) 45%
c) 44%
d) 54%
Answer:
c) 44%

Question 35.
The medicinal plant commonly known as “King of Bitters” is ……………………
a) Nilavembu
b) Holy basil
c) Adathodai
d) Turmeric
Answer:
a) Nilavembu

Question 36.
Pungency of cayenne pepper is ……………….. Scoville Heat Units (SHU)
a) 30,000 to 50,000
b) 1,349,000
c) 2,200,000
d) 1,200,000
Answer:
a) 30,000 to 50,000

Question 37.
Foxtail millet is domesticated in China ……………….. years ago
a) 4000
b) 3000
c) 5000
d) 6000
Answer:
d) 6000

Question 38.
Setaria italica is the scientific name of ……………………
a) kodo millet
b) foxtail millet
c) sorghum
d) finger millet
Answer:
b) foxtail millet

Question 39.
Lady’s finger is not grown in abundance in ……………..
a) Tamil Nadu
b) Assam
c) Maharashtra
d) Gujarat
Answer:
a) Tamil Nadu

Question 40.
Which is the temperature region fruit?
a) Mango
b) Jack
c) Banana
d) plum
Answer:
c) Banana and d) plum

Question 41.
The following are the activities of entrepreneurship
a) Mushroom cultivation
b) Single cell protein production
c) Organi farming
d) Above all
Answer:
d) Above all

Question 42.
…………………. is a bio-pest repellent
a) Tamarind
b) Chilly
c) Sesame
d) Neem
Answer:
d) Neem

Question 43.
Indigenous to western ghats of India
a) Black pepper
b) Cardamom
c) Turmeric
d) Red pepper
Answer:
a) Black pepper

Question 44.
Endosperm of ………………… is a refreshing summer food
a) Coconut
b) Groundnut
c) Gingelly
d) Palmyra
Answer:
d) Palmyra

Question 45.
……………….. Enhances salivary and gastric secretions
a) Cardamom
b) Black pepper
c) Red pepper
d) Turmeric
Answer:
b) Black pepper

Question 46.
………………… is used in gerontological applications
a) Aloe
b) Turmeric
c) Jasmine
d) Phyllanthus
Answer:
a) Aloe

Question 47.
Lacosone (Colouring Matter) is in ……………………
a) Aloe
b) Jasminum
c) Henna
d) Turmeric
Answer:
c) Henna

Question 48.
Paper pulp is made from ………………
a) Eucalyphis
b) Casuarina
c) Neolamarkia
d) all the above
Answer:
d) all the above

Question 49.
Eco friendly packaging material is ……………………
a) cotton
b) latex
c) wood pulp
d) jute
Answer:
d) jute

Question 50
……………….. is a ingredient of Ponga I of Tamil Nadu
a) Green gram
b) Red gram
c) Black gram
d) Brown gram
Answer:
a) Green gram

II Two Marks

Question 1.
Name the 3 grass species of food plants?
Answer:
Rice, Wheat, Maize.

Question 2.
What are the nutrients provided by cereals?
Answer:
Carbohydrates, proteins, fibres, vitamins and minerals.

Question 3.
Classify cereals based on size? Give example.
Answer:

• Major Cereals. Eg. Rice, Wheat
• Minor Cereals. Eg. Millet’s, Sorghum

Question 4.
Comment on Maida?
Answer:

• Processed wheat flour is called Maida.
• It is used in making Parota, Naan and Bakery products.

Question 5.
Explain the rice products?
Answer:

• Flaked rice (Aval)
• Puffed rice (Pori) are used as breakfast cereal (or) snack food in India.

Question 6.
What are millet’s?
Answer:
A variety of very small seeds. These were originally cultivated by ancient people in Africa. It is gluten free, less glycemic index.

Question 7.
Enlist the uses of finger millet?
Answer:

• Rich in calcium
• Staple food of south hilly regions in India.
• Ragi is made into porridge and gruel.
• Ragi malt is a nutrient drink
• It is the source of fermented beverage.

Question 8.
How is Sorghum useful?
Answer:

• It is used to feed poultry, birds, pigs, cattle.
• Source of fermented alcoholic beverage.

Question 9.
Discuss the medicinal uses of Fox tail millet?
Answer:

• Strengthens heart
• Improves eye sight
• Thinai porridge is given to lactating mother.

Question 10.
Kodo Millet is medicinally useful – Discuss?
Answer:

• It is a good diuretic
• It cures constipation
• It reduces obesity, blood sugar, blood pressure.

Question 11.
Which is the only pulse native to southern India? Give it’s uses?
Answer:
Red gram (Pigeon pea) Cajanus cajan is the only pulse native to south India.
Uses:

• Major ingredient of Sambar
• Roasted, salted, unsalted seeds are snacks.
• Young pods are cooked and consumed.

Question 12.
Enlist the nutrients in vegetables?
Answer:
Potassium, fibre, folic acid, vitamin A, E, C

Question 13.
Molecular farming plants are different from natural medicinial plants. How?
Answer:

Question 14.
What are the major cultivating states of okra in Tamil Nadu?
Answer:
Coimbatore, Dharmapuri, Vellore.

Question 15.
Classify fruits based on the climatic region in which they grow?
Answer:

• Temperate Eg. Apple, Pear, Plum
• Tropical fruits Eg. Mango, Jack, Banana.

Question 16.
Which is the National fruit of India? Give its origin and area of cultivation?
Answer:
Mango (Mangifera Indica)
Origin and area of cultivation.

• Native of southern Asia, Burma and Eastern India.
• Mango producing states are Andhra Pradesh, Bihar, Gujarat and Karnataka.
• Salem, Krishnagiri, Dharmapuri are major mango producing districts of Tamil Nadu.

Question 17.
Name the Major Cultivars of Mango in India.
Answer:
Alphonsa, Banganapalli, Neelam, Malgova.

Question 18.
Which food is the source of antioxidants?
Answer:
Dry fruits with hard shell and edible kernel are nuts. They are the good source of health fat, fibre, protein, vitamin, mineral, antioxidants.

Question 19.
Name the plants ideal for the extraction of commercial sugar?
Answer:
Sugar Cane, Palmyra

Question 20.
What is sugar?
Answer:
It is the generic name for sweet tasting soluble carbohydrate. They are used in food, beverages.

Question 21.
Give the sources of sugar?
Answer:
Roots of Sugar beet, Stems of Sugar cane, Fruits of Apple, Palmyra sap.

Question 22.
How is cultivated Saccharum officinarum evolved?
Answer:
By repeated back crossing of Saccharum officinarum of new guinea with wild Saccharum Spontaneum.

Question 23.
Toddy-Comment?
Answer:
The sap from Palmyra inflorescence is fermented to get toddy.

Question 24.
Name the 2 kinds of oils?
Answer:

• Essential oil
• Vegetable, fatty oil

Question 25.
Define Essential oil?
Answer:
They evaporate or volatilize in contact with air. So, they are called volatile oils.

Question 26.
Give the sources of essential oils?
Answer:
Flowers of jasmine, fruits of orange and roots of ginger.

Question 27.
Comment on vegetable oil?
Answer:
These are non-volatile oils or fixed oils. They do no evaporate. Eg. Whole seeds or endosperm are the sources.

Question 28.
What are spices?
Answer:
Aromatic plant products. They are of sweet or bitter taste. They give flavour and improve the palatability of food.

Question 29.
Comment on condiments?
Answer:
Flavouring substances with sharp taste. They are added to food after cooking Eg. Curry leaves.

Question 30.
Dates of India – Discuss.
Answer:
Tamarindus is an Arabian word. It means dates of India (Tamar – Taste; Indus – India)

Question 31.
Write any two uses of THC.
Answer:
THC is used in treating Glaucoma a condition in which presšure develops in the eyes.

• THC is also used in reducing nausea of cancer patients under going radiation and chemotherapy.
• It is an effective pain reliever and reduces hypertension.

III. Three Marks

Question 1.
Suggest the 4 commercial cotton species?
Answer:

• G. hirsutum
• G. barbadense
• G. arboreum
• G. herbaceum

Question 2.
Give the uses of cotton?
Answer:
Manufacturing of textile, hosiery products, toys. UsedinFlospitals.

Question 3.
Name the 2 species of plants from which Jute is derived?
Answer:

• Corchorus capsularis (Indo – Burmese origin)
• Corchorus olitorius (African origin)

Question 4.
Define Vulcanization?
Answer:
Heating rubber with sulphur under pressure at 150°C. It overcomes defect in rubber articles. (Vulcan is the Roman god of fire)

Question 5.
Name the woods used for making paper pulp?
Answer:

• Wood of Melia azadirachta.
• Neolamarkia chienensis
• Cauarinaspe, Eucalyptus spe.

Question 6.
Dyeing is in use since that ancient times? Substantiate?
Answer:

• Authentic records of dyeing is in the tomb painting of ancient Egypt.
• Colouring of mummy cements (wrapping) included saffron and indigo.
• Found in rock paintings of India.

Question 7.
Give the significance of Henna?
Answer:

• Orange dye henna is from the leaves and shoots of Lawsonia intermis.
• The colouring matter Lacosone is harmless and causes no skin irritation.
• This dye is used for skin, hair and finger nails.
• It is used for colouring leather for tails of horses and in hair dyes.

Question 8.
What do the south Indian people traditionally for skin and hair care?
Answer:
People of South India use turmeric, green gram powder, henna, sigaikai and usilai for skin, hair care.

Question 9.
What does the word’perfume’mean?
Answer:

• ‘Perfume’ is a word derived from Latin.
• Per (through) and fumus (to smoke) means through smoke.
• Age old tradition of burning scented woods at religious ceremonies.

Question 10.
Give an account of NCB?
Answer:
The Narcotics Control Bureau is the drug law enforcement and intelligence agency of India. It is responsible for drug trafficking and abuse of illegal substances.

Question 11.
Can we create new business using plant resources?
Answer:
Entrepreneurial botany is the study of new business created using plant resources.

Question 12.
What is entrepreneurship?
Answer:
Developing ideas to create new ventures among young people.

Question 13.
What is the role of an Entrepreneur?
Answer:

• One who works to create a product or service that people will buy
• He builds an organization to support the sales.

Question 14.
Discuss about ‘Capsaicin’
Answer:

• It is an active component of chillies.
• It has pain relieving properties.
• It gives pungency or spicy taste to chillies.
• Pungency is measured in Scoville Heat Unit (SHU)
• Eg. Naga piper is the hottest chilly of India with 1,349,000 SHU.

Question 15.
Name some plants used in making paper pulp?
Answer:

• Melia azadirachta
• Neolamarkia chinensis
• Casuarina spe
• Eucalyptus spe

Question 16.
Give the uses of purified dissolving pulp?
Answer:
It helps to manufacture rayon, artificial silk, fabrics, transparent films (cellophane, cellulose, acetate films), plastic. Viscose process of making rayon is common.

Question 17.
Which the second geographical Indication tag after Mysore Malli? How?
Answer:

• Madurai Malli is the second GI Tag.
• It has thick petals, long stalk.
• Distinct fragrance is due to chemicals like jasmine, alpha terpineol.

Question 18.
Name the major tribal communities in Tamil Nadu known for their medicinal knowledge?
Answer:
Irulas, Malayalis, Kurumbas, Paliyans and Kaanis

Question 19.
Discuss the origin and area of cultivation of black gram?
Answer:

• Archeo botanical evidence show the presence of black gram 3500 years ago.
• India gives 80 % of global production.
• Black gram is grown in Uttar Pradesh, Chattisgarh and Karnataka.

Question 20.
Suggest the nutrients in fruits?
Answer:
Potassium, dietary fibre, folic acid, vitamins.

Question 21.
What is rayon?
Answer:

• Rayon is purified dis solving pulp is used as a basic material in the manufacture of rayon or artificial silk, fabrics, transport films (cellophane), cellulose, acetate films), plastics.
• This viscose process of making rayon is the most common process.

Question 22.
What is known as sustainable development of agriculture?
Answer:

• Use of biofertilizers is one of the important components of integrated organic farm management, as they are cost effective and renewable source of plant nutrients to supplement the chemical fertilizers for sustainable agriculture.

IV . Five Marks

Question 1.
What kind of cereal can be eaten as a whole grain? Discuss?
Answer:

• Pseudocereal can be eaten as wholegrain.
• These are botanical outliers from grasses.
• Eg. Seed from the Chenopodium quinoa (Family: Amaranthaceae)
• Gluten-free, whole grain carbohydrate, whole protein with all essential amino acids.
• Eaten for 6000 years in Andes hill region.

Question 2.
Suggest the attributes of cereals as food plants?
Answer:

• Adaptability and colonisation on every type of habitat.
• Ease of cultivation.
• Tillering property gives high yield per unit area.
• Compact dry grains are easily handled, transported, stored without spoilage.
• High-calorie value provides energy.

Question 3.
How will you prepare a Bio-pest repellent?
Answer:

• Neem tree leaves are plucked.
• Put chopped leaves in 50 litre container half-filled with water. Leave it for 3 days to brow.
• Strain the mixture and spray on plants.
• 100 ml of cooking oil is added to make the repellent stick to the plants.
• Soap water is added to break down the oil.
• A stewed leaf mixture can be composted around the base of the plant.

Question 4.
Tabulate the uses of common medicinal plants?
Answer:

Question 5.
Give a detailed account on the ‘National Fruit of India’?
Answer:
Mango (Mangifera indica) belongs to the family Anacardiaceae
Origin and area of cultivation.

• Native of southern Asia, Burma and Eastern India.
• Andhra, Bihar, Gujarat and Karnataka are mango producing states.
• Salem, Krishnagiri, Dharmapuri are mango producing districts of Tamil Nadu.
• Major cultivars of Mango are Alphonsa, Banganapalli, Neelam and Malgova.

Uses:

• Major Indian table fruit.
• Rich in beta carotenes.
• Used as dessert, canned, dried, preserves in Indian cuisine.
• Unripe mangoes are used in chutney, pickle, side dishes, eaten raw with salt, chilli.
• Pulp is made as jelly
• Aerated, non aerated soft drinks are prepared.

Question 6.
Enlist the uses of Sugar cane.
Answer:
Botanical Name : Saccharum officinarum of
Poaceae family
Uses:

• Raw material for white sugar
• Industries supported are
• Sugar mills producing refined sugar
• Distilleries producing liquor grade ethanol.
• Jaggery manufacturing unit.
• Refreshing drink can be extracted.
• Gives molasses. It is the raw material for ethyl alcohol.

Question 7.
Detail on the State Tree of Tamil Nadu.
Answer:
Botanical Name : Borassus flabellifer of Arecaceae family
Origin and area of cultivation

• Native of tropical Africa, Asia, New Guinea.
• All over Tamil Nadu especially in coastal districts.

Uses:

• Exudate from inflorescence gives palm sugar
• Sap of inflorescence is a healthy drink
• Processed sap gives palm sugar
• Fermented sap gives toddy
• Endosperm is a refreshing summer food
• Elongated embryo of germinated seeds is edible.

Question 8.
Enlist the uses of Chilly / Red pepper?
Answer:
Botanical Name : Capsicum annuum of
Solanaceae family
Uses:

• Capsicum annuum is less pungent
• Capsicum annuum includes large sweet bell peppers.
• Long fruit cultivars called ‘Cayenne Pepper’ are crushed, powdered and used as condiment.
• Sauce, Curry powder, pickle can be prepared.
• Capsaicin has pain relieving property. Good source of vitamins A, C, E.

Question 9.
Which plant contributes to 90% of world production by Asia? Detail the uses?
Answer:
Rubber (Hevea brasiliensis) of Euphorbiaceae.
Origin and Cultivation

• Native of Brazil
• Kerala is the largest Indian producers

Uses:

• Tyre, automobile parts consume 70% of rubber.
• To manufacture footwear, wire, cable insulation, rain coats, household, hospital goods, shock absorbers, belts, sports goods, erasers, adhesives, rubber band
• Hard rubber is used in electrical and radio engineering
• Latex makes gloves, balloons, condoms.
• Foamed latex used for the manufacture of cushion, pillow, life belts.

Question 10.
Which system of medicine originated from Brahma? Explain?
Answer:

• Ayurveda system of Medicine
• Core knowledge is documented in compendium of Charaka, Sushruta and Vagbhata.
• It is based on 3 humor principles Vatha, Pitha, Kapha.
• Herbs, few animal parts are drug sources.
• Himalayan plants are plant sources.
• Ayurvedic pharmacopoeia of India list 500 plant sources.

Question 11.
Which system of medicine survives as oral tradition? Explain?
Answer:
Folk system of Medicine:

• It is a oral tradition in rural, tribal communities.
• Document of plants used by ethnic communities was launched by Ministry of Environment and Forest, Government of India.
• The document is All India Co-ordinated Research Project on Ethnobiology.
• 8000 species of medical plants are documented.
• Major tribal communities with medicinal knowledge are Irulas, Malayalis, Kurumbas, Paliyans and Kaanis.

Question 12.
Jute Industry occupies an important place in the national economy of India. Explain?
Answer:

• One of the largest exported fibre of India.
• Used for safe packaging of natural, renewable, bio degradable, Eco-friendly products.
• Used in bagging, wrapping textile.
• 75% is used to prepare sack, bag.
• Manufacture of blanket, rag, curtain
• Used in textiles recently.

Question 13.
Organic farming is considered as the movement towards the philosophy of Back to Nature. Explain
Answer:

• Organic farming is an alternative agricultural system in which plants / crops are cultivated in natural ways by using biological inputs to maintain soil fertility and ecological balance thereby minimizing pollution and wastage
• Indians were organic farmers by default until the green revolution came in to practice.
• Use of bio-fertilizers is one of the important components of integrated organic farm management, as they are cost effective and renewable source of plant nutrients to supplement the chemical fertilizers for sustainable agriculture.
• Several microorganisms and their association with crop plants are being exploited in the production of bio-fertilizers.
• Organic farming is thus considered as the movement directed towards the philosophy of Back to Nature.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Botany Guide Pdf Chapter 3 Chromosomal Basis of Inheritance Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Botany Solutions Chapter 3 Chromosomal Basis of Inheritance

12th Bio Botany Guide Chromosomal Basis of Inheritance Text Book Back Questions and Answers

I. Choose the correct answer

Question 1.
An allohexaploidy contains
a) Six different genomes
b) Six copies of three different genomes
c) Two copies of three different genomes
d) Six copies of one genome
Answer:
c) Two copies of three different genomes

Question 2.
The A and B genes are 10 cM apart on a chromosome. If an AB/ab heterozygote is test crossed to ab/ab, how many of each progeny class would you expect out of 100 total progeny?
a) 25 AB, 25 ab, 25 Ab, 25 aB
b) 10 AB,10 ab
c) 45 AB, 45 ab
d) 45 AB, 45 ab, 5 Ab, 5aB
Answer:
c) 45 AB, 45 ab

Question 3.
Match list I with list II

Question 4.
Which of the following sentences are correct?
1. The offspring exhibit only parental combinations due to incomplete linkage
2. The linked genes exhibit some crossing over in complete linkage
3. The separation of two linked genes are possible in incomplete linkage
4. Crossing over is absent in complete linkage
a) 1 and 2
b) 2 and 3
c) 3 and 4
d) 1 and 4
Answer:
c) 3 and 4

Question 5.
Accurate mapping of genes can be done by three point test cross because increases
a) Possibility of single cross over
b) Possibility of double cross over
c) Possibility of multiple cross over
d) Possibility of recombination frequency
Answer:
b) Possibility of double cross over

Question 6.
Due to incomplete linkage in maize, the ratio of parental and recombinants are
a) 50:50
b) 7:1:1:7
c) 96.4: 3.6
d) 1:7:7:1
Answer:
b) 7:1:1:7

Question 7.
Genes G S L H are located on same chromosome. The recombination percentage is between L and G isT5%, S and L is 50%, H and S are 20%. The correct order of genes is
a) GHSL
b) SHGL
c) SGHL
d) HSLG
Answer:
c) SGHL

Question 8.
The point mutation sequence for transition, transition, transversion and transversion in DNA are
a) A to T, T to A, C to G and G to C
b) A to G, C to T, C to G and T to A
c) C to G, A to G, T to A and G to A
d) G to C, A to T, T to A and C to G
Answer:
b) A to G, C to T, C to G and T to A

Question 9.
If haploid number in a cell is 18. The double monosomic and trisomic number will be
a) 35 and 37
b) 34 and 38
c) 37 and 35
d) 17 and 19
Answer:
a) 35 and 37

Question 10.
Changing the codon AGC to AGA represents
a) missense mutation
b) nonsense mutation
c) frameshift mutation
d) deletion mutation
Answer:
a) missense mutation

Question 11.
Assertion (A): Gamma rays are generally use to induce mutation in wheat varieties.
Reason (R): Because they carry lower energy to non-ionize electrons from atom
a) A is correct. R is correct explanation of A
b) A is correct. R is not correct explanation of A
c) A is correct. R is wrong explanation of A
d) A and R is wrong
Answer:
c) A is correct. R is wrong explanation of A

Question 12.
How many map units separate two alleles A and B if the recombination frequency is 0.09?
a) 900 cM
b) 90 cM
c) 9 cM
d) 0.9 cM
Answer:
c) 9 cM

Question 13.
When two different genes came from same parent they tend to remain together.
i) What is the name of this phenomenon?
ii) Draw the cross with suitable example.
iii) Write the observed phenotypic ratio.
Answer:
i) The name of this phenomenon is known as Linkage.
This is reported in Sweet pea Lathyrus odoratus by Willium Bateson & Reginald C Punnet in 1906.

Genes for Purple colour and Long pollen grain were found close together in the same homologous pair of chromosomes – They do not assort independently and this condition is known as linkage.

Question 14.
If you cross dominant genotype PV/PV male Drosophila with double recessive female and obtain FI hybrid. Now you cross FI male with double recessive female,
i) What type of linkage is seen?
ii) Draw the cross with correct genotype.
iii) What is the possible genotype in F₂ generation?

Question 15.

i) What is the name of this test cross?
ii) How will you construct gene mapping from the above given data?
iii) Find out the correct order of genes.
Answer:
i) It is three point test cross – It refers to analysing the inheritance, patterns of three alleles by crossing a triple recessive herterozygote with a triple recessive homozygote.
ii) The relative distance between the three alleles & the order in which they are located can be determined with the help of frequency of recombination between them.

All the loci are linked because all the RF values are considerable less then 50%. In AC loci show highest RF value, they must be farthest apart. There fore the B locus must lie between them. The order of genes should be abc. A genetic map can be drawn.

A final point note that two small map distances. 19.9 m.u and 21.75 m.u is add up to 41.95 m.u which is greater then 40.16 m.u the stance calculated for 1 and g. We must identify the two least number of progenius (totalling 8) in relation to recombination of AC . These two least progenius are double cross over. The two least progenies not only counted once should have count each of them twice because each represents a double recombinant progeny. Hence, We can correct the value adding the number 114 + 125 + 116 + 128 + 5 + 14 + 4 = 500 of the total 1200 this number exactly 41.65% which is identical with the same of two component values.
The test cross parental combination can be rewritten as follows.

Gene order showing double recombinant.

Question 16.
What is the difference between missense and nonsense mutation?
Answer:
Mis-sense Mutation:
The mutation where the codon for one amino acid is changed into a codon for another amino acid is called Missense or non-synonymous mutations.

Non-sense Mutation:
The mutations where the codon for one amino acid is changed into a termination or stop codon is called Nonsense mutation.

Question 17.

From the above figure identify the type of mutation and explain it.
Answer:

• It is a change in the arrangement of gene loci,
• Here the duplicated segment is located immediately aftear the normal segment but the gene sepuence order will be reversed – (Paracentric inversion)

Question 18.
Write the salient features of Sutton and Boveri concept.
Answer:
Salient features of the chromosomal theory of inheritance:

1. Somatic cells of organisms are derived from the zygote by repeated cell division (mitosis). These consist of two identical sets of chromosomes. One set is received from female parent (maternal) and the other from male parent (paternal). These two chromosomes constitute the homologous pair.
2. Chromosomes retain their structural uniqueness and individuality throughout the life cycle of an organism.
3. Each chromosome carries specific determiners or Mendelian factors which are now termed as genes.
4. The behaviour of chromosomes during the gamete formation (meiosis) provides evidence to the fact that genes or factors are located on chromosomes.

Question 19.
Explain the mechanism of crossing over.
Answer:
Crossing Over – it is a very significant biological process
It is a precise one with several stages

i) Synapsis:
During zygotene – of prophase. I of meiosis I the homologous chromosomes come and align side by side known as – bivalents.
This pairing – is known as synapsis or syndesis.
Types of synapsis

ii) Tetrad Formation:
Each homologous chromosome of – a bivalent begin to form two identical sister chromatids – held together by a centromere.
Each bivalent has 4 chromatids – (tetrad stage),

iii) Cross Over:
At pachytene stage cross over occur. The points of contact at one or more points between non-sister chromatids is called Chiasmata.
Crossing over is exchange of corresponding segments occur, in the chiasma region.

Synaptonemal Complex (SC)
The highly organised structure of filaments called SC – facilitate chiasma formation.

SC formation & chiasma formation – is absent in Drosophila

Terminalisation:
After crossing over, chiasma starts to moving towards the terminal end of chromatids is known as terminalisation. Complete separation of homologous chromosomes occurs after terminalization.

Question 20.
Write the steps involved in molecular mechanism of DNA recombination with diagram.
Answer:
Proposed by Robin Holliday in 1964

Steps:

• Homologous DNA molecules are paired side by side with their duplicated copies of DMAs
• One strand of both DNAs cut in one place by the enzyme endonuclease.
• The cut strands cross and join the homologous strands – Holliday junction.
• Holliday junction migrates away from the original site, a process called branch migration, as a result heteroduplex region is formed.
• DNA strands may cut along through the vertical (V) line or horizontal (H) line.
• The vertical cut will result in heteroduplexes with recombinants.
• The horizontal cut will result in heteroduplex with non recombinants.

Question 21.
How is Nicotiana exhibit self¬incompatibility? Explain its mechanism.
Answer:
In Nicotiana self sterility or self-incompatibility is due to multiple alleles.
The pollen from a plant is unable to germinate on its own stigma – and no fertilization.
The gene for self incompatibility can be – ‘S’ which has allelic series S₁, S₂, S₃, S₄ & S₅.
Cross-fertilizing tobacco – were not always homozygous as S₁S₁ or S₂S₂, but heterozygous
Crosses between different S₁S₂ plants, pollen tube did not develop normally.
But effective – development observed when cross was made with other than S₁S₂ Eg. S₃S₄.

Question 22.
How is sex determined in monoecious plants. Write the genes involved in it.
Answer:
Zeamays (maize) – monoecious plant
Made & Female flowers are present on the same plant.

• Terminal inflorescence – arise from tassel bear staminate flowers
• Lateral inflorescence – arise from ear or cob bear pistillate flowers.
• Unisexvality in maize – occurs through selective abortion of ear florets and pistils in tassel florets.
• The allele for barren plant (ba)- when homozygous makes the stalk staminate (eliminating silk and ears)
• The allele for tassel seed (ts) – transforms tassel into a pistillate structure (no pollen produced)
• Most of these mutations are shown to be defects in Gibberellins biosynthesis.
• Gibbercilins play an important role in the suppression of stamens in florets on the ears.

Question 23.
What is gene mapping? Write its uses.
Answer:
The diagrammatic representation of the position of genes and related distances between the adjacent genes is called genetic mapping. It is directly proportional to the frequency of recombination between them. It is also called a linkage map.
Uses of genetic mapping:

• It is used to determine gene order, identify the locus of a gene and calculate the distances between genes.
• They are useful in predicting the results of dihybrid and trihybrid crosses.
• It allows the geneticists to understand the overall genetic complexity of the particular organism.

Question 24.
Draw the diagram of different types of aneuploidy.
Answer:

Question 25.
Mention the name of man-made cereal. How it is formed?
Answer:

1. Tetraploidy: Crosses between diploid wheat and rye.
2. Hexaploidy: Crosses between tetraploid wheat Triticum durum (macaroni wheat) and rye.
3. Octoploidy: Crosses between hexaploid wheat T. aestivum (bread wheat) and rye. Hexaploidy Triticale hybrid plants

12th Bio Botany Guide Chromosomal Basis of Inheritance Additional Important Questions and Answers

I. Fill in the blanks

1. The scientists who independently rediscovered mendelian works were
Answer:
De Vries, Correns & Tschermak

2. The worm-shaped cells formed during cell division are called in the earlier period as
Answer:
Chromosomes

3. Who postulated that the chromosomes of a cell are responsible for transferring heredity
Answer:
Wilhelm Roux (1883)

4. ……………………… was the first to find out physical mutagen in Drosophila
Answer:
Muller (1927)

5. ………………………used X-rays for the first time to induce mutation in the fruit fly
Answer:
H.J. Muller

6. Induced mutations are planted was reported for the first time by
L.J. Stadler

7. Chemical mutagenesis was first reported by
Answer:
Auerback (1944)

8. Double nullisomy is
Answer:
2n-2-2

9. Trisomis were first reported by Blakeslee in
Datura Stramonium

10. All possible tetrasomics are available in ……………………… plant
Answer:
Wheat

11. The kind of Aneuploid are usually lethal are
Answer:
Nullisomy

12. The alkaloid used to induce polyploidy is
Answer:
Colchicine

13. Raphano brassicas the sterile hybrid of Radish & Cabbage was produced by
Answer:
G.D. Karpechenko (1927)

14. The cross between hexaploid wheat Triticum aestivum and rye produced is a
Answer:
Octoploidy

15. Colchicine is extracted from the root and corms of
Answer:
Colchicum autumnale

16. Who first reported duplication in drosophila
Answer:
Bridges (1919)

17. In which types of cells chromosomal aberration is commonly found?
Answer:
Cancer cells

18. Recombination frequencies are the same for
Answer:
CIS and trans heterozygotes

19. The map distance between gene A and B is 3 units between B & C is 10 units and between C & A is 7 units – the order of genes in a linkage map constructed on the about would perhaps be
Answer:
B-A-C

20. The percentage of crossing over will be more if
Answer:
Linked genes are located apart from each other

21. A point mutation that changes an amino acid coding codon into a stop codon, prematurely terminating synthesis of the encoded protein ………………………
Answer:
Nonsense mutation

22. Single base change in DNA is known as ………………………
Answer:
Point mutation

23. Genetic change in a non-sex cell is known as
Answer:
Somatic mulalion

24. A duplicated DNA sequence next to the original sequence is known as
Answer:
Tandem duplication

25. A missing sequence of DNA or part of a chromosome
Answer:
Deletion mutation

26. Mutation that alters the genes reading frame is known as
Answer:
Frame shift mutation

27. A single base change mutation that alters and amino acid ………………………
Answer:
Missense

28. A substance that changes, adds, or deletes a DNA base
Answer:
Multagen

29. The mutation that introduces a section of aminoacids not normally found is known as ………………………
Answer:
Frame shift mutation

30. A mutation that changes an adenine to guanine is an example of a ………………………
Answer:
transition

31. A point mutation that has no obvious effect at all on the phenotype is called a ……………………… mutation
Answer:
silent

32. A point mutation that changes a codon specifying an amino acid into a stop codon is called a ………………………
Answer:
Non sense mutation

33. Changing the codon AGC to AGA represents ……………………… of a point mutation
Answer:
missense

34. A point multation that alters a codon so that the encoded aminoacid is substituted with another is called a ………………………mutation
Answer:
missense

35. A ………………………mutation occurs during the DNA replication that precedes meiosis. while a ……………………… mutation occurs during the DNA replication that preceeds mitosis.
Answer:
germline, somatic

36. A mutation that introduction of section of aminoacids not normally found is ………………………
Answer:
Frame shift mutation.

37. A point mutation altering a purine to pyrimidine or vice versa is ………………………
Answer:
transversion

38. A spontaneous mutation usually originates as an error in ………………………
Answer:
DNA replication

39. The codon for leucine is CUC. How many different aminoacids could possibly result from a single base substitution
Answer:
7

40. How may map units separate two alleles if the recombination frequency is o.o7?
Answer:
7cM

41. In a population of 1000 individuals 360 belong to genotype AA. 480 to Aa and the remaining 160 to aa – Based on the data, the frequency of allela A in the population is
Answer:
0.7

II. Find out the incorrect statement

Question 43.
Which one of the following is incorrect regarding chromosomal behaviour during cell division?
a) The alleles of a genotype are found in the some locus of a homologous chromosome
b) In the S phase of meiotic interphase each chromosome replicates forming two copies of each allele, one on each chromatid.
c) The Homologus chromosomes segregate in metaphase I, thereby separating two different alleles.
d) In anaphase II of meiosis separation of sister chromatid of homologous chromosomes takes place.
Answer:
c) The Homologus chromosomes segregate in metaphase I, thereby separating two different alleles.

Question 44.
a) 94% of all flowering plants are sexually monomorphic
b) When three or more allelic forms of a gense occupy the same locus in given pair of homologous chromosome they are known as Multiple alleles
c) The mutation that result in the change of one
codon of an aminoacid changed into codon of another amino acid is known as Frame shift mutation.
d) Muller (1928) first time used x-rays to induce mutation in Drosophila
Answer:
c) The mutation that result in the change of one codon of an aminoacid changed into codon of another amino acid is known as Frame shift mutation.

Question 45.
Which of the following statement is not correct of two genes that show 50% recombination frequency?
a) The genes may be on different chromosomes
b) The genes are tightly linked
c) The genes show independent assortment
d) If the genes are present on the same chromosome, they undergo more than one crossover in every meiosis.
Answer:
d) If the genes are present on the same chromosome, they undergo more than one crossover in every meiosis.

Question 46.
a) Selfing of monosomic plants produce nullisomics.
b) In a true diploid both the monoploid and haploid chromosome number are same.
c) An Auto triploids can be produced artificially by crossing between haploid and a diploid.
d) An increase in the number of chromosome
sets has been an important factor in the origin of new plant species.
Answer:
c) An Auto triploids can be produced artificially by crossing between haploid and a diploid.

III. Match the following

a) (i) C (ii) D (iii) B (iv) A
b) (i) B (ii) C (iii) D (iv) A
c) (i) C (ii) B (iii) A (iv) D
d) (i) D (ii) C (iii) A (iv) B
Answer
d) (i) D (ii) C (iii) A (iv) B

IV. Choose the correct statement

Question 48.
When red eyed female Drosophila is crossed with white eyed male, the FI offsprings would be
a) Females are with white eye and males are with red eye.
b) Males are with red eye and females are with yellow eye.
c) Both males and females are with red eye
d) Both males and females are with white eye.
Answer:
c) Both males and females are with red eye

V. Find the Odd man out with reference to Allopolyploidy

Question 49.
a) All organisms which possess two or more basic sets of chromosomes derived from two
different specie’s.
b) They have four or six copies of its own genome – induced by doubling of the diploid species.
c) They can be developed by inter-specific crosses and fertility is restored by chromosome doubling with colchicine treatment.
d) They are formed between closely related species only..
Answer:
b) They have four or six copies of its own genome – induced by doubling of the diploid species.

VI. Find the Odd man out with reference to Altotriploidy

Question 50.
a) Thev can be produced artificially bv crossing between autotetraploid & diploid.
b) They are highly fertile due to large number of gametes.
c) Cultivated triploid bananas are seedless having larger fruits than diploid.
d) Common doob cross is a natural autotriploid.
Answer:
b) They are highly fertile due to large number of gametes.

VII. Find the Odd man out regarding crossing over

Question 51.
a) It occur in germinal cells during gametogenesis.
b) Take place during Pachytene state of prophase I of meiosis..
c) It is directly proportional to the frequency of recombination between them.
d) It has universal occurrence has great significance.
Answer:
c) It is directly proportional to the frequency of recombination between them.

VIII. Choose the wrongly matched pair

Question 52.

Answer:
d

Question 53.

Answer:
d

Question 54.

Answer:
c

IX. Choose the incorrect statement with reference to Deletion

Question 55.
a) Deletions occur due to chemicals, drugs & radiation.
b) On the basis of location of breakage on chromosome it is divided in to Terminal deletion & inter calary deletion
c) Larger deletions have evolutionary significance.
d) Deletions are recorded in Drosophila & Maize
Answer:
c) Larger deletions have evolutionary significance.

Question 56.
How can we reverse the sterility of FI hybrid?
a) Genetic Engineering
b) Protoplasmic fusion
c) Induced Mutation
d) Induced chromosomal aberration
Answer:
d) Induced chromosomal aberration

Question 57.
If haploid number in a cell is 23. The double monosomic and pentasomy number will be
a) 44 and 49
b) 17 and 34
c) 47 and 46
d) 45 and 48
Answer:
a) 44 and 49

Question 58.
Genes located close together on the same chromosome and inherited together represented as
a) linked genes
b) unlinked gene
c) syntenic genes
d) trans genes
Answer:
a) linked genes

XI. Assertion (A) & Reason (R)

Question 59.
Assertion (A): Arabidopsis plant chromosomes have more repeats of TTT nucleotide sequences in the telomeres.
Reason (R) : Restriction endonuclease enzyme is used in the formation of nucleotide sequence (Telomeres) mui
a) (A) is incorrect, (R) is correct
b) (A) is correct, (R) is the correct explanation (A)
c) (A) is correct, (R) is the incorrect explanation (A)
d) (A) and (R) are wrong.
Answer:
b) (A) is correct, (R) is the correct explanation (A)

Question 60.
Assertion (A) : Linkage and crossing over are two processes that have opposite effects. Reason (R) : Linkage keeps particular genes together but crossing over mixes them.
Answer:
a) If both the Assertion (A) & Reason (R) are true and the reason is a correct explanation of the Assertion. .

Question 61.
Assertion (A) : Increase in temperature increases the rate of mutation.
Reason (R) : While rise in temperature hydrolyses DNA by the restriction endonuclease which degrade Nucleotides.
Answer:
c) Assertion (A) is true but Reason (R) is false

XII. Two Marks

Question 1.
Define chromosome theory of inheritance.
Answer:
It states the Mendelian factors (genes) have specific locus on chromosomes & they carry information from one generation to the next generation.

Question 2.
State the number of chromosomes of the given organism.
Answer:
1) Ophioglossum 2) Arabiodopsis 3) Sugarcane 4) Rice 5) Potato 6) Maize
Answer:
1) -1262;
2) -10;
3) 80;
4) 24;
5) 48;
6) 20

Question 3.
What are Fossil Genes?
Answer:

• Some junk DNA is made up of pseudogenes, once working but have lost their ability to make proteins.
• They are fossilized parts act as evidence for evolution.

Question 4.
State the works of T.H. Morgan
Answer:

• His works on Drosophila melanogaster – Sex linkage – helped to confirm chromosome theory of heredity.
• He received Nobel prize in Physiology of medicine in 1933 fot it.
• He coined the term crossing over.

Question 5.
What are co-mutagens ?
Answer:
Compounds which are not having own mutagenic properties – but enhance the effects of known mutagens.
Eg. Ascorbic acid – increase the damage caused by hydrogen peroxide.
Caffeine – increase the toxicity of methotrexate.

Question 6.
Differentiate between Euploidy & Aneuploidy
Answer:
Eupoidy :

1. Ploidy involving entire sets of chromosomes is known as euploidy
2. Triploidy (3x); Tetraploidy (4x); Poly ploidy ( ∞n)

Aneuploidy :

1. Here the diploid number is altered either by addition or deletion of one or more chromosomes
2. Trisomy; Tetrasomv; Monosomy; Nullisomy (2n+1)(2n+2)(2n-1)(2n-2)

Question 7.
Distinguish between Monoploidy & Haploidy
Answer:
Monoploidy
In Monoploidy the chromosome number is referred as x .
Eg.
Hexaploidy wheat
(2n) = 6 x = 72
haplaid = (n) 36
Monoploidy = x = 12

Haploidy :
Half the number of somatic chromosomes is referred as gametic chromosome number called haploidy (n) Human of haploid = 23 (n) Wheat of haploid = 36 (n)

Question 8.
Independent assortment & Linkage are alternatives of each other – Discuss
Answer:

Question 9.
How does the strength and weakness of linkage depend on linked genes?
Answer:

• The strength of linkage increases as the distance between linked genes decreases.
• The linkage becomes weaker with the increase in the distance between genes.

Question 10.
Distinguish between crossing over & Reciprocal Translocation.
Crossing over

1. It is legitimate & natural
2. Occurs between nonsister chromatids of homologous chromosomes
3. Occurs between nonsister chromatids of homologous chromosomes

Reciprocal Translocation :

1. It is illegitimate & chromosomal abnormality
2. Occurs between non sister chromatids of non homologous chromosomes
3. Also play major rome in formation of species

Question 11.
Distinguish between tetrad & bivalent Tetrad:
Answer:

• During Synapsis homologous chromosomes come together side by side resulting in bivalents
• As the stage during which each bivalent has 4 chromatids & the stage is known as tetrad stage.

Question 12.
Define Recombination.
Answer:
In this segments of DNA one broken and recombined to produce new combination of alleles – known as Recombination.

Question 13.
What is RF (Recombination Frequency)
Answer:
The frequency with which recombination occur is a certain condition

Question 14.
A diploid organism is heterozygous for 4 loci. How many types of gametes cars be produced.
Answer:
The formula 2n is applied – Organism hetr oizy gous f or 4 loci = n = 4.
So 2n = 24 = 2 x 2 x 2 x 2 = 16.
The organism produces 16 types of gametes.

Question 15.
Notes on Colchicine.
Answer:

• Alkaloid, extracted from – root and corms of colchicum autumnale
• In low concentration to the growing lips it induce polyploidy
• It does not affect the source plant due to the presence of Anticolchicine

Question 16.
Write down the significance of ploidy.
Answer:

• Polyploids – More vigorous & more adaptive
• Ornamental flowers – (Autotetraploids) larger flowers – longer flowering duration
• Increase in fresh weight (due to more water content)
• Aneuploids – help to determine the phenotypic effects (loss or gain of different chromosomes
• Allopolyploids of angiosperms play a role in an evolution of plants.

Question 17.
Distinguish between Mendelian disorder & Chromosomal disorder.
Answer:
Mendelian disorder:
Occur due to mutation of single gene & follow the well known Mendelian pattern of inheritance.
Eg. Sickle cell anaemia

Chromosomal disorder :
Chromosomal disorders are produced due to alteration in the number of chromosomes.
Eg. Down syndrome

Question 18.

Answer:
a) Which type of crossing over is mentioned in the
above diagram? I h M I
Single cross over
b) Mention the percentage of Recombination Frequency (RF)
RF = 2/4 x 100 = 50%

Question 19.
This is a type of Numerical chromosomal abnormality find it out give a note on it.

Answer:

• This numerical chromosomal abnormality is known as double monosomy (2n-l-l)
• From a diploid set of chromosome if one chromosomes is lost, the condition is known as monosomy (2n-l)
• If another chromosome is also lost it is known as double monosomy (2n-l-l)

Question 20.
Bring out the difference between Linkage & Crossingover in inheritance
Answer:
Linkage

1. It is the tendency of genes in a chromosome to stay close together
2. It involves same chromosome of homologous chromosome
3. It reduces new gene combinations
4. Not very significant in evolution

Crossing over :

1. It leads to separation of linked gene
2. It involves exchange of segments between nonsister chromatids of homologous chromosome.
3. It increases – variability by forming new combinations → lead to formation of new organism
4. play important role in evolution

Question 21.
What is chiasmata?
Answer:

• The non – sister chromatids of homologous pair make a contact at one or more points.
• These points of contact between non-sister chromatids of homologous chromosomes are called chiasmata.

Question 22.
What is multiple alleles?
Answer:
When any of the three or more allelic forms of a gene occupy the same locus in a given pair of homologous chromosomes, they are said be called multiple alleles.

Question 23.
What is monomorphic?
Answer:

• About 94% of all flowering plants have only one type of individual, which produces flowers with male organs (the stamens) and female organs (the carpels).
• Such plants are termed as sexually monomorphic.

Question 24.
What is Dimorphic?
Answer:
Some 6% of flowering plants which have two separate sexes are called dimorphic.

XIII. Three Marks

Question 1.
Differentiate tetrasomy from tetraploidy
Answer:

Question 2.
Give a tabulation comparing the behaviour of gene & Chromosome
Answer:

Question 3.
The important aspects about the chromosome behaviour during cell devision.
Answer:

• Alleles of a genotype – found in the same locus of a homologous chromosome (A/a)
• In ‘S’ – Phase of meiotic interphase – the replication of chromosome occur – (two copies of each allele (AA/aa) one on each chromatid
• Anaphase II of meiosis, separation of sister chromatids of homologous chromosomes. So each daughter cell (gamete) carries only a single allele of a character (A), (A), (a) and (a)

Question 4.
Write the differences between coupling and Repulsion
Answer:
Coupling

1. The two dominant alleles or recessive alleles called repulsion or trans configuration
2. It tend to inherit together into same gametes

Repulsion :

1. If dominant or recessive alleles are present on occur in the same homologous chromosomes two different but homologous chromosomes.
2. If they inherit apart in to different game es are

Question 5.
Define synapsis.
What are the types of Synapsis
Answer:

• During cygotene stage of prophase I of meiosis I – homologous chromosomes are aligned side by side resulting in a pair called (bivalents).
• This pairing phenomenon is called synapsis or syndesis.
  Based on the starting poiring of pairing there are 3 types of synapsis

  Procentric	Proterminal	Random
  Starts from middle	Starts from the telomeres	Starts from any where

Question 6.
Distinguish between sharbati sonora & Castor Aruna.
Answer:
Sharbati sonora :

1. Multant variety of Wheat – approved in 1967
2. Sonora 64 (Mexican variety subjected to gamma rays to produce sharbati sonora
3. Developed by Dr. M.D. Swaminathan
4. Early maturing & high protein content high kneading quality

Castor Aruna :

1. Mutant castor variety
2. Seeds treated with thermal neutrons
3. Early maturing – (120 days instead of 270 dyas) & High yielding.

Question 7.
How do increase in temperature cause mutation?
Answer:
Rise temperature breaks the hydrogen bonds between two DNA nucleotides – & affects the process of replication & transcription.

Question 8.
Distinguish between the impact of ionizing & non ionizing radiation in causing mutation. Ionizing radiation Non Ionizing radiation
Answer:
Ionizing radiation :

• Short wave length and carry enough higher energy to ionize electrons from atoms. They breaks the chromosome & chromatids. Ex. x-rays – gamma rays, alfa rays, beta rays & cosmic rays.

Non Ionizing radiation :

• Longer wave lengths and carry lower energy so they have lover penetrating power used to treat unicellular microbes – spores pollengrains – which have nuclei – near surface membrance. Eg. UV rays

Question 9.
What is significance of ploidy?
Answer:

• Many polyploids are more vigorous and more adaptable than diploids.
• Many ornamental plants are autotetraploids and have large flower and longer flowering duration than diploids.
• Auto polyploids usually have increase in fresh weight due to more water content.
• Aneuploids are useful to determine the phenotypic effects of loss or gain of different chromosome.
• Many angiosperms are allopolyploids and they play a role in an evolution of plants.

Question 10.
What is chemical mutagens? Give an example?
Answer:
Chemical which include mutation are called mutagens.
Example:
Nitrous oxide alters the nitrogen bases of DNA and disturb the replication and transcription that leads to the formation of incomplete and defective polypeptide during translation.

Question 11.
What is cis configuration (or) coupling?
Answer:
The two dominant alleles or recessive alleles occur in the same homologus chromosomes, tend to inherit together into same gamete are called coupling (or) cis configuration

XIV. Five Marks

Question 1.
Whose works supported the chromosomal theory of heredity? Explain.
Answer:

• T.H. Morgan works on fruit fly supported the chromosomal theory of inheritance.
• The alleles for red or white eye colour are present on the X – chromosome but there is no counter part for this gene on the Y chromosome.
• The genes for yellow body colour and miniature wings are also carried on the X – chromosome.
• By understanding the sex linked inheritance of these characters it is proved that genes are located on the chromosomes.
• Thus T.H. Morgan’s works on Drophila came as a support to the chromosomal theory of inheritance.

Question 2.
Write down the steps in the Holliday’s hybrid DNA model.
Answer:

• Homologous DNA molearles are paired side by side with their duplicated copies of DMAs.
• One strand of both DNAs cut in one place by the enzyme endonuclease.
• The cut strands cross & join the homologous strands forming the Holliday junction
• Holliday junction – migrates away from the original site, by branch migration – as a result
  heteroduplex region is formed.
• DNA strands may cut along the vertical (V) or horizontal (H) line.
• The vertical cut will result in heteroduplexes with recombinants & the Horizontal with non recombinants.

Question 3.
Explain sex determination is Silene latifolia (Melandrium album)
Answer:

• C.E Allen (1917) discovered sex determination in plants.
• Complex precess determined by
  1. genes
  2. environment
  3. hormones

Sex determination silene latifolia – is controlled by 3 distinct regions in a sex chromosome

• Y – Chromosome – determines maleners
• X – Chromosome – specify femaleness
• X & Y – show different segments (I, II, III, IV, & V)

Question 4.
How do Hawaii explain the sex determination in Papaya
Answer:

• Carica papaya 2n = 36
  
• The sex chromosomes look like autosomes
• Developed from autosomes
• Y- chromosome carries the genes for male organ
• X- chromosomes bear the gene for female organ development.

Question 5.
Explain sex determination in Sphaerocarpos donnelli. It is also known as Bottle liverwort (Bryophyta)
Answer:

• gametophyte – haploid with 8 chromosome (n).
• The sporophyte – diploid & heterogametic
• Male sfemale gameto phyte – seven autosomes are similar.
• In female 8th chromosome is X – Larger than the seven autosomes.
• In male 8th chromosome is Y – Smaller than the autosomes.
• In sporophyte – contain XY – combinations produces two types of meiospores
• Meiospore with X – produce – female gemetophyte

Question 6.
What are the various types of crossing over.
Answer:

Question 7.
The two loci A/a and D/d are so tightly linked that no recombination is ever observed. If AA dd is crossed to aa DD what phonotypes will be seen in the F2 and in what proportions.
Answer:
Genotypes of the parents are Ad/ Ad x aD x aD – If genes are so tightly linked, the only possible types of gametes produced by parents are
Ad and aD respectively (parental or nonrecombinant gametes)
FI will be all Ad / aD
only types of gametes from each FI can be Ad (50%)oraD(50%)
F2 frequencies can be calculated from these F2 will be Ad/ Ad (1/4i) Ad/ aD (1/2) aD / aD (1/4)

Question 8.
Classify maj or types of mutations.
Answer:

Question 9.
Define point mutation & explain it’s types
Answer:
Definition:
Mutation affecting single base or base pair of DNA
Types:

• Indel mutation : (Base pair insertions or . addition. Addition or deletions of nucleotide
  pairs.
• Substitution : one base pair is replaced by another

Types – (Two)

1. (Purine replaced by Purine)
2. Pyrimidine replaced by Pyrimidine
3. Transversion purine replaced by pyrimidin or pyridine replaced

Synonymous or silent mutations:
Here change in one codon for an amino acid into another codon for that same amino acid

Missense or Non synonymous mutations
Here the codon for one amino acid is changed in to -a termination or stop codon.

Frameshift mutations.
Additions or deletions of a single base pair of DNA, – changed the reading frame for translation – so there is complete loss of normal protein structure & function.

Question 10.
Explain how translocation chromosomal aberration is different from crossing over?
Answer:

Question 11.
Explain structural changes in chromosome with reference to changed to changes in the number of gene loci
Answer:

• There are 2 types
  1. Deletion
  2. Duplication

Deletion or Deficiency :

• (loss of a portion of chromosome)
• 2 types
  1. Terminal deletion (break in any one end
  2. Intercalary deletion (two breaks & reunion of terminal parts leaving the middle.

• > Unpaired loops some times formed known as deficiency loops (during meiotic prophase)
• > Larger deletions may have lethal effect Duplication or Repeat
• > Same order of genes repeated more than once in the same chromosome.
  Eg. Drosophila

Duplication
3 types

1. Tandem duplication
2. Reverse tandem
3. Displaced duplication

i) Tandem duplication
Duplicated segment is located immediately after the normal segment in the same order.

ii) Reverse tandem
Duplicated segment, immediately after the normal segment but gene sequence order will be reversed.

ii) Reverse tandem
Duplicated segment away from the normal segment.
Duplication play a maj or role in evolution.

Question 12.
Explain translocation.
Definition : The transfer of a segment of chromosome to – a non homologous chromosome is called translocation.
Types-3
i) Simple translocation:
Single break in only one chromosome the broken segment gets attached to one end of a non homologous chromosome – (Very rare occurrence)

ii) Shift translocation.
Broken segment of one chromosome gets inserted interstitially in a non homologous chromosome.

iii) Reciprocal translocation.
Mutual exchange of chromosomal segments between two non homologous chromosomes – Illegitimate crossing over)

a) Homozygous translocation.

• Both the chromosomes of two pairs are involved.
• Two homologous of each translocated chromosomes are identical.

b) Heterozygous translocation.
Only me of the chromosome from each pair of two homologous are involved others remain normal.

Question 13.
Consider two hypothetical recessive auto¬somal genes a and b, where a heterozygote is testcrossed to a double homozygous mutant. Predict the phenotypic ratios under the following conditions:
a) a and b are located on separate autosomes.
b) a and b are linked on the same autosome but are so far apart that a crossover occurs between them.
c) a and b are linked on the same autosome but are so close together that a crossover almost never occurs.
Answer:
a) The problem involves an understanding of linkage, crossing over & independent assortment.
F₂geno & phenotypic ratio =

b) When crossing over occurs the result in same as the question (a)
F₂ geno & phenotypic ratio =

c) When a & b linked with out crossing on the heterozygolic parent can be AB / ab – (cis) or Ab / ab – (tr ans)
However there won’t be any recombinant gametes because no. crossing over occur.
It will produce Ab & aB (50 % of each)
The progeny of test cross will be Ab/ ab&aB/ab

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Botany Guide Pdf Chapter 2 Classical Genetics Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Botany Solutions Chapter 2 Classical Genetics

12th Bio Botany Guide Classical Genetics Text Book Back Questions and Answers

I. Choose the correct answer from the given option

Question 1.
Extra nuclear inheritance is a consequence of presence of genes in
a) Mitrochondria and chloroplasts
b) Endoplasmic reticulum and mitrochondria
c) Ribosomes and chloroplast
d) Lysososmes and ribosomes
Answer:
a) Mitrochondria and chloroplasts

Question 2.
In order to find out the different types of gametes produced by a pea plant having the genotype AaBb, it should be crossed to a plant with the genotype
a) aaBB
b) AaBB
c) AABB
d) aabb
Answer:
d) aabb

Question 3.
How many different kinds of gametes will be produced by a plant having the genotype AABbCC?
a) Three
b) Four
c) Nine
d) Two
Answer:
b) Four

Question 4.
Which one of the following is an example of polygenic inheritance?
a) Flower colour in Mirabilis jalapa
b) production of male honey bee
c) Pod shape in garden pea
d) Skin colour in humans
Answer:
d) Skin colour in humans

Question 5.
In Mendel’s experiments with garden pea round seed shape (RR) was dominant over wrinkled seeds (rr), Yellow cotyledon on (YY) was dominant over green cotyledon (yy). What are the expected phenotypes in the F₂ generation of the cross RRYY x rryy?
a) Only round seeds with green cotyledons
b) Only wrinkled seeds with yellow cotyledons
c) Only wrinkled seeds with green cotyledons
d) Round seeds with yellow cotyledons and wrinkled seeds with yellow cotyledons
Answer:
d) Round seeds with yellow cotyledons and wrinkled seeds with yellow cotyledons

Question 6.
Test cross involves
a) Crossing between two genotypes with a recessive trait
b) Crossing between two F₁ hybrids
c) Crossing the F₁ hybrid with a double recessive genotype
d) Crossing between two genotypes with dominant trait
Answer:
c) Crossing the Fx hybrid with a double recessive genotype

Question 7.
In pea plants, yellow seeds are dominant to green. If a heterozygous yellow seed plant is crossed with a green seeded plant, what ratio of yellow and green seeded plants would you expect in FI generation?
a) 9:1
b) 1:3
c) 3:1
d) 50:50
Answer:
d) 50 : 50

Question 8.
The genotype of a plant showing the dominant phenotype can be determined by
a) Back cross
b) Test cross
c) Dihybrid cross
d) Pedigree analysis
Answer:
b) Test cross

Question 9.
Select the correct statement from the ones given below with respect to dihybrid cross
a) Tightly linked genes on the same chromosomes show very few combinations
b) Tightly linked genes on the same chromosomes show higher combinations
c) Genes far apart on the same chromosomes show very few recombinations
d) Genes loosely linked on the same chromosomes show similar recombinations as the tightly linked ones
Answer:
a) Tightly linked genes on the same chromosomes show very few combinations

Question 10.
Which Mendelian idea is depicted by a cross in which the Fx generation resembles both the parents
a) Incomplete dominance
b) Law of dominance
c) Inheritance of one gene
d) Codominance
Answer:
d) Codominance

Question 11.
Fruit colour in squash is an example of
a) Recessive epistasis
b) Dominant epistasis
c) Complementary genes
d) Inhibitory genes
Answer:
b) Dominant epistasis

Question 12.
In his classic experiments on Pea plants, Mendel did not use
a) Flowering position
b) seed colour
c) pod length
d) Seed shape
Answer:
c) pod length

Question 13.
The epistatic effect, in which the dihybrid cross 9:3:3:1 between AaBb Aabb is modified as
a) Dominance of one allele on another allele of both loci
b) Interaction between two alleles of different loci
c) Dominance of one allele to another allele of same loci
d) Interaction between two alleles of some loci
Answer:
b) Interaction between two alleles of different loci

Question 14.
In a test cross involving FI dihybrid flies, more parental type offspring were produced than the recombination type offspring. This indicates
a) The two genes are located on two different chromosomes
b) Chromosomes failed to separate during meiosis
c) The two genes are linked and present on the same chromosome
d) Both of the characters are controlled by more than one gene
Answer:
c) The two genes are linked and present on the same chromosome

Question 15.
The genes controlling the seven pea characters studied by Mendel are known to be located on how many different chromosomes?
a) Seven
b) Six
c) Five
d) Four
Answer:
a) Seven

Question 16.
Which of the following explains how progeny can possess the combinations of traits that none of the parents possessed?
a) law of segregation
b) Chromosome theory
c) Law of independent assortment
d) Polygenic inheritance
Answer:
c) Law of independent assortment

Question 17.
“Gametes are never hybrid” This is a statement of
a) Law of dominance
b) Law of independent assortment
c) law of segregation
d) Law of random fertilization
Answer:
c) law of segregation

Question 18.
Gene which suppresses other genes activity but does not lie on the same locus is called as
a) Epistatic
b) Supplement only
c) Hypostatic
d) Codominant
Answer:
a) Epistatic

Question 19.
Pure tall plants are crossed with the pure dwarf plants. In the FI generation, all plants were tall. These tall plants of the F1 generation were selfed and the ratio of tall to dwarf plants obtained was 3 : 1. This is called
a) Dominance
b) Inheritance
c) Codominance
d) Heredity
Answer:
a) Dominance

Question 20.
The dominant epistatis ratio is
a) 9:3:3:1
b) 12:3:1
c) 9:3:4
d) 9:6:1
Answer:
b) 12:3:1

Question 21.
Select the period for Mendel’s hybridiza tion experiments
a) 1856 -1863
b) 1850 -1870
c) 1857 – 1869
d) 1870 – 1877
Answer:
a) 1856 -1863

Question 22.
Among the following characters which one was not considered by Mendel in his experimentation pea ?
a) Stem – Tall or dwarf
b) Trichomal glandular or non – glandular
c) Seed – Green or yellow
d) Pod – Inflated or constricted
Answer:
b) Trichomalgalandular or non – glandular

Question 23.
Name the seven contrasting traits of Mendel.
Answer:
Plant Height, Seed Shape, Cotyledon colour, Flower colour, Pod colour, Pod form, Flower position

Question 24.
What is meant by true-breeding or pure breeding lines/strain?
Answer:

• True breeding lines (pure breeding strains) means it has undergone continuous self-pollination having specific phenotype trait inheritance from parent to offspring.
• Mating within pure breeding lines produces offsprings having, specific parental traits that are the same in inheritance and expression for many generations.
• Parents are homozygous for every trait.

Question 25.
Give the names of the scientists who rediscovered Mendelism.
Answer:
Mendel’s experiments were rediscovered by three biologists, Hugo de Vries of Holland, Car Correns of Germany and Erich von Tschermak of Austria.

Question 26.
what is back cross?
Answer:

• back cross is a cross off Fi offsprings with either one of the parental genotypes.
• The recessive back cross helps to identify the heterozygosity of the hybrid.
• It involves the cross between the fi offspring with either of the parents dominant.

Question 27.
Define Genetics.
Answer:
“Genetics” is the branch of biological science which deals with the mechanism of transmission of characters from parents to offsprings. The term Genetics was introduced by W. Bateson in 1906.

Question 28.
What are multiple alleles?
Answer:

• Alleles are alternative form of gene and they are responsible for differences in the phenotypic expression of a given trait. A gene for which atleast two alleles exist is to be polymorphic, so a particular gene may exist in three or more allelic forms known as multiple alleles
• eg) ABO of human blood is controlled by three alleles

Question 29.
What are the reasons for Mendels successes in his breeding experiment? Pisum sativum a wise choice, because
Answer:
Mendel was successful because:

1. He applied mathematics and statistical methods to biology and laws of probability to his breeding experiments.
2. He followed scientific methods and kept accurate and detailed records that include quantitative data of the outcome of his crosses.
3. His experiments were carefully planned and he used large samples.
4. The pairs of contrasting characters which were controlled by a factor (genes) were present on separate chromosomes.
5. The parents selected by Mendel were pure breed lines and the purity was tested by self crossing the progeny for many generations.

Question 30.
Explain the law of dominance in monohybrid cross.
Answer:
Law of dominance
In cross of parents that are pure for contrasting traits only one form of the trait will appear in the next generation. They have hybrid or dominant trait in the phenotype.
eg) Monohybrid cross

Regarding F₁ generation the recessive allele is not lost, and it remain hidden or masked. But it reappears in the next generation.

Question 31.
Differentiate incomplete dominance and co-dominance.
Answer:
Incomplete Dominance:

1. In incomplete dominance, neither of the allele is not completely dominant to another allele rather combine and produce new trait
2. New phenotype is formed due to character blending (not alleles)
3. Example : Pink flowers of Mirabilis Jalapa

Co-dominance:

1. In co-dominance, both the alleles in a heterozygote are dominant and the traits are equally expressed (joint expression)
2. No formation of new phenotype rather both dominant traits are expressed, conjointly
3. Example: Red and white flowers of camellia

Question 32.
What is meant by cytoplasmic inheritance?
Answer:

• DNA is a universal genetic material.
• Genes located in nuclear chromosomes follow Mendelian inheritance.
• Certain traits are governed by the chloroplast (or) mitochondrial genes which is known as extranuclear inheritance.
• It is a kind of Non – Mendelian inheritance.
• The cytoplasmic organelles chloroplast and mitochondrion act as inheritance vectors so-called cytoplasmic inheritance.
• It is based on self – replicating extrachromosomal unit called plasminogen in the cytoplasmic Organelles, Chloroplast, and mitochondria.

Question 33.
Describe dominant epistasis with an example.
Answer:
Epistasis can be defined as a gene interaction whereby one gene interferes with the phenotypic expression of another non-allelic gene. The gene or locus which suppresses or masks the action of a gene at another locus is called the epistatic gene. The gene or locus where expressions are suppressed by an epistatic gene is called gene hypostatic.
Dominant epistasis A dominant epistasis suppresses the expression of a non-allelic gene, (dominant (or) recessive)
The F₂ phenotypic ratio is 12:3:1

Question 34.
Explain polygenic inheritance with an example
Answer:

• Polygenic inheritance, also known as quantitative inheritance, refers to a single inherited phenotypic trait that is combined bv two or more different: genes.
  (or)
• Several genes combine to affect a single trait. A group of genes that together determine (or) contribute a characteristic of an organism is called polygenic Inheritance
  (or)
• Polyinheritance occur when one characteristic is controlled bv two or more genes.
  Eg. Human skin colour & eye colour and weight.
• H.Nilsson -Ehle (1909), a Swedish geneticist discovered a polygenic inheritance in wheat (kernel colour). Kernel colour is controlled by two genes each with two alleles, one with red kernel colour was dominant to white. He crossed the pure breeding wheat varieties dark red and a white.
• Dark red genotypes R₁R₁R₂R₂ crosed unit r₁ r₁ r₂ r₂. In the F₁ generation medium red were obtained with genotype R₂ r₁ R, r₂. So the intensity of the red colour is determined by the number of R genes in the F₂ generation
• Four R genes: A dark red kernel colour is obtained.
• Three R genes: Medium – dark red kernel colour is obtained.
• Two R genes: Medium-red kernel colour is obtained.
• One R gene: Light red kernel colour is obtained.
• Absence of R gene:Results in White kernel colour is obtained.
  

The data produces a bell shaped curve which demonstrate continuous variation in wheat kernel from dark red to white in F₂ when the number F₁ were self crossed five different phenotypic classes appeared in F₂ in into ratio of 1:4:6:4:1
The phenotype ratio is Dark red :1 Medium dark red :4 Medium red : 6
light red : 4 white : 1
Hence the total ratio is 63 red : 1 white in F2 generation
1:6:15 :20:15:6:1 in generation

He found that In F₂ generation plants have Kernel’s with range of colour variation. This is due to the fact that the genes are segregating and recombination takes place.

Question 35.
Differentiate continuous variation with discontinuous variation .
Answer:
Variation is the difference between individual with in a species. This can be caused by inherited or environmental factors. It can be continuous and discontinuous. Height, and weight of the human being are best examples of continuous variation. Human blood group, gender identity and eye colour are best example of discontinuous variation

Question 36.
Explain with an example how single genes affect multiple traits and alleles the phenotype of an organism.
Answer:

• There are several patterns responsible for the inheritance of traits, gene causes one trait. But in some cases one gene is responsible for multiple traits. Sometimes two or more gene are required to produce one trait.
• It is otherwise called pleiotropy. It means, where one gene will code and control the phenotype or expression of several different and unrelated traits.
• Eg. Phenylketenuria disease.
• A gene that produces multiple or effect is called a Pleitropic gene. Multiple effects of a single gene is know as pleiotropy. A Pleitropic gene is a single gene that controls more that one trait.
• Eg. Human genetic disorder are often pleitropic ie, unusual tall height, thin finger and toes, dislocation of the lens of the eye, heart in the aorta (heart function)
• Eg : Pisum sativum plant with purple brown seeds and dark spot on the axis of the leaves were crossed with a variety of a peas having white flowers light coloured seed and no spot on the axils of the leaves, the three traits for peas colour, seed colour and a leaf axil spot all were inherited together as a single exist. This is due to the pattern of inheritances controlled by a single gene with dominant and recessive alleles,
• eg .Sickle cell anemia
• eg .Marfan syndrome
• A human genetic disorder called marfan syndrome is caused by a mutation in one gene, yet it affects many aspects of growth and development inducing height, vision and heart function. This is an example of pleiotropy or one gene affecting multiple characteristics.

• Gene also interact in pattern such a partial dominance or co-dominance, the trait is expressed a mix between two gene , Those are possibilities for one gene. Most trait are influenced by many genes. There are many different way for these gene to influence how trait is expressed.

Question 37.
Bring out the inheritance of chloroplast gene with an example.
Answer:

• It is found in 4 O’clock plant (Mirabilis jalapa)
• There are dark green leaved plants and pale green leaved plants.
• When the pollen of dark green leaved plant (male) is transferred to the stigma of pale green leaved plant (female) the pollen of pale green leaved plant is transferred to the stigma of dark green leaved plant, the F₁ generation of both the crosses is identical as per mendelian inheritance.
• In the reciprocal cross the F₁ plant differs from each other.
• The F₁ plant reveals the character of the plant.
• The inheritance is due to the chloroplast gene found in the ovum of the female plant which contributes the cytoplasm during fertilization.
• The male gamete contribute only the nucleus.

12th Bio Botany Guide Classical Genetics Additional Important Questions and Answers

I. Match the following

Question 1.

Answer:
(a) Tall – (iv) dwarf
(b) Purple – (i) white
(c) arial – (iii) terminal
(d) Round – (ii) wrinkled

Question 2.

Answer:
(a) Dominant epistasis – (ii) 12:3:1
(b) Duplicate genes – (iii) 15 :1
(c) Recessive epistasis – (iv) 9:3:4
(d) Complementary gene – (i) 9:7

Question 3.

Answer:
(a) Genetics – (ii) W. Batson
(b) Mendel – (iii) Father of Geetics
(c) lethal gene – (i) E. Baeur
(d) H. Nillsson Ehle – (iv) Kernel colour

Question 4.

Answer:
(a) Polygenic inherence – (iv) wheat kernel colour
(b) 4 O’clock pea plant – (iii) Mirabilis jalapa
(c) Garden pea plant – (i) Pisum sativm
(d) H. NillssanEhle – (ii) genetic material

II. Choose the correct statement

Question 5.
a. HbA and Hbs alleles of normal and single-cell hemoglobin are multiple alleles
b. HbA and Hbs alleles of normal and single-cell hemoglobin are dominant recessive allele
c. HbA and HbA alleles of normal and single cell heamoglobin are codominant allele
d. HbA and Hb & alleles of normal and single-cell hemoglobin are recessive alleles
Answer:
c) HbA and HbA alleles of normal and single cell heamoglobin are codominant allele

Question 6.
a. When alleles of the two contrasting characters are present together, one of the character ex-press and the other remains hidden. There is the law of purity of gametes.
b. When alleles of the contrasting characters are present together, one of the character express and the other remains hidden . There is a law of dominance.
c. When alleles of the contrasting characters are present together with one of the character express and the other remain hidden This is law of segregation
d. When allele of two contrasting character are present together, one of the character express and remain hidden. This is law of independent assortment.
Answer:
b) When alleles of the contrasting characters are present together, one of the character express and the other remains hidden. This is the law of dominance

Question 7.
a. Monohybrid ratio is 9:3:3:1
b. The crossing of FI to any one of the parent is called test cross
c. The phenotypic ratio of a monohybrid cross is 1:2:1
d. A cross in which parents differ in a single pair of contrasting character is called a dihybrid cross
Answer:
c) The phenotypic ratio of a monohybrid cross is 1:2:1

Question 8.
a. The hybrid progeny in the first generation is called F2
b. The major reasons for the success of Mendelian experiment was the true-breeding of Garden Pea plant
c. X and Y are examples of alleles.
d. A pedigree chart shows the genotypes of any parent.
Answer:
b) The major reason for the success of mendelian experiment was true-breeding of Garden Pea plant

III. Choose the correct pair

Question 9.
a. Discontinuous variation – qualitative inheritance
b. Continuous variation – qualitative inheritance
c. Duplicate gene – 13: 3
d. Recessive epilate – 9:7
Answer:
a) Discontinuous variation – qualitative inheritance

Question 10.
a. Monohybrid – 9:3:3:1
b. Dihybrid – 1: 2: 1
c. recessive epistasis — 9: 3 : 4
d. extra chromosomal
inheritance — Mendelian inheritance
Answer:
c) recessive epistasis – 9 : 3 : 4

Question 11.
a. Emasculation – removal of anther
b. Tt – homozygous
c. genetic constitution – phenotype
d. mono hybrid cross – law of independent
assortment
Answer:
a) Emasculation – removal of anther

Question 12.
a. polygenic trait – Traits that are controlled by multiple gene
b. Multiple alleles – A gene that is controlled by one allele
c. Pleiotropy – one gene cannot affects multiple characters
d. Phenotype – genetic makeup of an organism.
Answer:
a) polygenic trait – Traits that are controlled by multiple gene

IV. Choose the incorrect statement

Question 13.
a. A pedigree charts are shown which genes are co-dominant
b. A true-breeding is a kind of breeding where the parents would produce offspring that would carry the same phenotype
c. In polygenic inheritance, traits are determined by interaction of single gene
d. The interactions between separate genes, in which one masks the effect of another is called epistasis.
Answer:
c) In polygenic inheritance traits are determined by interaction of single gene

Question 14.
a. The outward appearance resulting from an individual’s genotype for a particular characteristic is called phenotype
b. The recessive allele of the same gene represented by lower case letter.
c. Blood group is a human characteristic that shown discrete variation
d. The name given to different form of the same gene is gametes
Answer:
d) The name given to different form of the same gene is gametes

Question 15.
a. An allele is a viable DNA, coding that occupies a given locus on a chromosome
b. An allele is an alternative form of gene
c. An organism which has two different alleles of the gene is called homozygous
d. A person with one ‘A’ blood type and one ‘B’ blood type allele would have a blood type of “AB” ”
Answer:
c) An organism which has two different alleles of the gene is called homozygous

Question 16.
a. A pleiotropic gene is a single gene that more than one trait
b. A single gene affects multiple traits and alter the phenotype of the organism called as pleiotropy
c. Marfans syndrome is an example of pleiotropy.
d. one (or) single gene that cannot affect multiple traits are called pleiotropy.
Answer:
d) one (or) single gene that cannot affect multiple traits are called pleiotropy.

V. Choose the Incorrect Pair

Question 17.
a. Genotype – Genetic makeup of organism
b. recessive – A trait that is hidden
c. probability – The chance that an event will take place
d. Independent assortment – Mendel’s first law
Answer:
d) Independent assortment – Mendel’s first law

Question 18.
a. Dominant Allele – RR
b. Recessive allele – rr
c. Heterozygous – Tt
d. Homozygous recessive – TT
Answer:
d) Homozygous recessive – TT

Question 19.
a. Intra-locus interaction – allelic interactions
b. Inter-locus interaction – non-allelic interactions
c. Epistatic – allelic interactions
d. Polygenic interaction – non-allelic interaction
Ans:
c) Epistatic – allelic interactions

Question 20.
a. Complementary gene – 9:7
b. Co -dominance -1:2:1
c. Dominant epistatics – 9:3:4
d. Inhibitor gene -13:3
Answer:
c) Dominant epistatics – 9:3:4

VI. Choose the Odd one out

Question 21.
a) Mirabilis jalapa
b) Snapdragon
c) ABO Blood system
d) Epistasis
Explanation: a,b,c are F₂ phenotypic ratio is 1:2:1
Answer:
d) Epistasis

Question 22.
a. DNA
b. mitochondrial inheritance
c. Chloroplast inheritance
d. Atavism
Explanation : a,b,c are used as genetic material.
Answer:
d) Atavism

Question 23.
a. Monohybrid cross
b. checkerboard
c. genotype
d. phenotype
Answer:
b) checkerboard

Question 24.
a) co-dominance
b) Duplicate gene
c) inhibitor gene
d) supplementary gene
Explanation : b,c and d are intergenic or non¬allele interaction
Answer:
b) Duplicate gene

VII. Assertion and Reason

Question 25.
A : Polygenic inheritance
R : Several genes combine to affect a single trait
(a) A is correct (b) R is false
(c) R is the correct explanation of A
(d) R only correct
Answer:
c) R is the correct explanation of A

Question 26.
A : Atavism is a modification of biological structure whereby an ancestral trait reappears after having been lost through evolutionary changes in the previous generation
R : Reemergence of sexual reproduction in the flowering plant Hieracium pilosella is the best example for Atavism in plants
(a) A is correct R is the correct explanation of A
(b) A only true
(c) R only True (d) A false & R is true
Answer:
(a) A is correct R is the correct explana-tion of A

Question 27.
A: The physical expression of an individual gene called phenotype
R: Phenotype is physical observable charactertics of an organism
a) A & R True
b) A & R False
c) A is correct
d) R is correct
Answer:
(a) A & R True

Question 28.
A : Interaction between two alleles of the same loci is the effect of epistasis
R : The epistasis is the kind of intergenic and allelic interaction.
(a) A is correct R is false
(b) R alone correct
(c) R & A are true
(d) R is the correct explanation of A
Answer:
(a) A is correct R is false

VIII. Choose the best answer

Question 29.
If you do dihybrid cross in Pisum sativum on the traits of pod shape and plant height, Will you get 9:3:3:1 ratio in F₂ ?
a. Yes, because they are independently assorting genes.
b. No, they are linked genes.
c. Yes, because thev are situated on different chromosomes
d. No, we can not do experiments on these two traits.
Answer:
a) Yes, because they are independently assorting genes.

Question 30.
A single characteristic is controlled by a number of genes is called
a. Inheritance
b. Epistasis
c. Polygenic inheritance
d. Co-dominance
Answer:
c) Polygenic inheritance

Question 31.
An allele is
a. a homozygous genotype
b. a heterozygous genotype
c. another word for gene
d. several possible form of gene
Answer:
c) another word for gene

Question 32.
Continuous variation is due to
a. effect of polygenes
b. effect of environment
c. effect of polygenes and environment
d. effect of one or two genes.
Answer:
c) effect of polygenes and environment

Question 33.
A variation in a characteristic in which individuals show two or a few traits with large differences between them.
a. dominant
b. continuous variation
c. discontinuous variation
d. recessive
Answer:
c)discontinuous variation

Question 34.
A trait that masks the expression of another trait when both versions of the gene are present in an individual
a. variation
b. recessive
c. co-dominance
d. dominant
Answer:
d) dominant

Question 35.
Which one of the following is not a correct pair regarding genes of pea plant,
a. Seed shape – Chromosome number 6
b. Pod colour – Chromosome number 5
c. Flower position – Chromosome number 4
d. Seed colour – Chromosome number 1
Answer:
a) Seed shape – Chromosome number 6

Question 36.
The study of heredity behaviour of several genes by Gregor Mendel.
a. Molecular genetics
b. Population genetics
c. Quantitative genetics
d. Transmission genetics
Answer:
d) Transmission genetics

Question 37.
Transmission of characters from parents to offsprings
a. variation
b. dominance
c. heredity
d. growth
Answer:
c) heredity

Question 38.
Species that shows a difference in the characteristics of the same natural population is called
a. heredity
b. variation
c. recessive
d. co -dominace
Answer:
b) variation

Question 39.
Qualitative inheritance is otherwise called
a. co – dominance
b. continuous variation
c. discontinuous variation
d. heredity
Answer:
c) discontinuous variation

Question 40.
“Experiments on plant hybrids” is a
a. book
b. research paper
c. journal
d. Magazine
Answer:
b) research paper

Question 41.
Mendels theory of inheritance is based on
a. Particulate theory
b. mass
c. hybridization
d. variation theory
Answer:
a) Particulate theory

Question 42.
Removal of the anther is called
a. Atavism
b. Epistasis
c. Hybridization
d. Emasculation
Answer:
d) Emasculation

Question 43.
Botanical name of garden pea is
a. Solanum tuberosum
Question b. Coccus nucitera
c. Pisum sativum
d. pea
Answer:
c) Pisum sativum

44.
Mendel’s experiments were rediscovered by
a. Hugo de vries & carl correns
b. E. Baur
c. H. Nilsson
d. T.H.Morgan
Answer:
a) Hugo de vries & carl correns

Question 45.
If a homozygous red flowered plant is crossed with a homozygous white flower plant then the off-spring will be_
a. All red flowered
b. Half white flowered
c. Half red flowered
d. All white flowered
Answer:
c) Half red flowered

Question 46.
…………….. is he best example for chloroplast inheritance
a. Mirabilis jalapa
b. Sorgum vulgare
c. Triticum vulgare
d. Musa paradisiaca
Answer:
a) Mirabilis jalapa

Question 47.
Among the pea plant cell which one has the ability to convert a precursor molecule into an active inform
a. Le:le
b. GA₁
c. Le
d. le
Answer:
b) GA₁

Question 48.
Gene interaction concept was introduced and explained by
a. W. Bateson
b. Morgan
c. E. Baur
d. Nilsson
Answer:
a) W. Bateson

Question 49.
An allele which has the potential to cause the death of an organism is called
a. Genetic interaction
b. lethal alleles/lethal gene
c. Atavism
d. Autism
Answer:
b) lethal alleles/lethal genes

Question 50.
The gene whose expression is interfered by non- alletic gene and prevents from exhibiting its character is known as
a. hypostatic
b. epistatic
c. metastatic
d. hipostatic
Answer:
a) hypostatic

Question 51.
Height and skin colour in human are controlled by
a. two pair of genes
b. three pair of genes
c. five pair of genes
d. a pair of genes
Answer:
b) three pair of genes

Question 52.
The genotypic ratio of monohybrid cross is
a. 3:1
b. 1:2:1
c. 3:1:1
d. 9:3:3:1
Answer:
b) 1:2:1

Question 53.
Which of the following statements are true regarding law of segregation
a. alleles separate with each other during gametogenesis
b. The segregation of factors is due to the segregation of chromosomes during meiosis
c. Law of segregation is called as law of purity of gametes
d. all of the above
Answer:
d) all of the above

Question 54.
The crossing of Fj to anyone of the parents is called
a. test cross
b. back cross
c. FI cross
d. all of these
Answer:
b) back cross

Question 55.
The character that is express in to the F₂ is called
a. recessive character
b. co-dominant character
c. dominant character
d. none of these
Answer:
c) dominant character

Question 56.
The recessive character will express in
a. F₁
b. F₂
c. both a & b
d. F₃ only
Answer:
b) F₂

Question 57.
Which of the following pair is not correct
a. KK=dominant
b. hybrid = heterogeneous
c. heterozygous = Kk
d. homozygous = Rr
Answer:
a) KK=dominat

Question 58.
What is the phenotype of wheat kernal colour for the genotype: R₁ R₁ r₂ r₂ ?
a. Dark red
b. Medium dark red
c. Medium red
d. Light red
Answer:
c) Medium red

Question 59.
Mendel worked at the rules of inheritance and arrived at the correct mechanism. But
a. without any knowledge of cellular mechanism
b. knowledge of cellular mechanism
c. heredity mechanism
d. growth mechanism
Answer:
a) without any knowledge of cellular mechanism

Question 60.
is crossing an individual of unknown one pair of a genes is called genetic genotype with a homogeneous recessive.
a. back cross
b. test cross
c. monohybrid cross
d. dihybrid cross
Answer:
b) test cross

Question 61.
……………….is the expression of a single character by the interaction of more than interaction or interaction of genes.
a. factor hypothesis/ Bateson factor hypo¬thesis
b. alternative hypothesis
c. nell hypothesis
d. All of the above
Answer:
d) All of the above

IX. One Mark Question

1. The genetic constitution of the individual is called
Answer:
Genotype

2. The observable characteristics of an organism are called
Answer:
Phenotype

3. Who is father of genetics?
Answer:
Gregor Johann Mendel

4. Name the Mendel’s published work.
Answer:
Experiments on plant Hybrids.

5. Name the publication of Mendel research work
Answer:
1899

6. What is the year of published work Mendel’s Research paper?
Answer:
The proceedings of the Brunn Society & Natural History.

7. What is an allele?
Answer:
It is another word for a Gene.

8. Individuals show a range of traits with small difference between them.
Answer:
Continuous variation

9. When an individual show two or a few traits with large differences between them. This type of variation is called.
Answer:
discontinuous variation

10. Human height is the good example of ………….. variation.
Answer:
Continuous variation

11. Human skin colour is the good example of …………….. variation.
Answer:
Continuous variation

12. Mention any two examples of continuous variation.
Answer:
a. Human height
b. Human skin colour

13. Mention any two examples of discontinuous variation.
Answer:
Style length of Primula & Height of the garden pea.

14. A trait that makes the expression of another trait when both version of the gene are present in the individual called
Answer:
Dominant.

15. What is F₁?
Answer:
It is the first filial generation in a cross; the offspring of the parental generation.

16. The letter ‘P’ denoted in genetics is
Answer:
The parental generation in a cross

17. A variation in an inherited characteristics is
Answer:
Trait

18. One pair genes can completely makes the expression of another pair of genes known as
Answer:
Epistasis

19. Who discovered incomplete dominance?
Answer:
Correns. (Germany)

20. Crosses between F1 offsprings with either of the two parents (hybrids) are known as
Answer:
Back cross

21. Diploid organisms that have two different allele at a specific gene locus are said to be
Answer:
Heterozygous

22. TT referred as…………….
Answer:
Homogenous dominant variety.

23. ‘tt’ referred as ……………
Answer:
Homozygous recessive character.

24. ‘Tt’ denotes for …………….
Answer:
Heterogeneous hybrid variety.

25. The superiority of hybrid over either of its parents in one or more traits known as
Answer:
Hybrid vigour or Heterosis

26. The site or position of a particular gene on a chromosome is
Answer:
locus

27. An allele which has the potential to cause the death of an organism is called ……………….
Answer:
Lethal genes

28. A single gene affects multiple traits are called ……………..
Answer:
Pleiotropy

29. A single gene affects multiple traits and alter the phenotype of the organism is
Answer:
Pleiotropy

30. Several genes combine to affect a single trait of an organism.
This kind of inheritance is ……………
Answer:
Polygenic inheritance.

31. Who demonstrated first experiment on polygenic inheritance.
Answer:
Swedish Geneticist H. Nilsson – Ehle (1909)

32. Which plant to use to identify the polygenic inheritance?
Answer:
Wheat – Kernel colour (dark red & white variety)

33. List any two intragenic or allele interaction.
Answer:

1. Incomplete Dominance
2. Co-dominance

34. List any two intergenic or non-allele interaction
Answer:

1. Dominant Epistasis
2. Recessive Epistasis

35. Corren has used plant for studied incomplete dominance.
Answer:
Mirabilis jalapa (4′ O clock plant)

36. Mention the botanical name of 4′ O clock plant.
Answer:
Mirabilis jalapa.

37. Duplicate genes with cumulative effect of non-alleleic interaction is derived in
Answer:
Fruit shape in Summer squash.

38. What is the FI phenotypic ratio of inhibitor genes in the intergenic interaction?
Answer:
13:3

39. When the heterozygote exhibits a mixture of phenotypic character of both homozygous called as
Answer:
Co-Dominance.

40. Name the two gene interaction.
Answer:

1. Intralocus interaction (allelic interaction)
2. Interlocus interaction (non-allelic interaction)

41. A chart shows which genes are co-dominant. This is known as
Answer:
A pedigree charts.

42. Each character is controlled by distinct units called factor, which occur in pairs. If the pairs are heterozygous, one wiil always dominant other. This is known as
Answer:
First law of inheritance or Law of Dominance.

43. The second law of inheritance otherwise called as
Answer:
Law of Segregation.

44. Give the name of the scientists who re-discovered Mendelism
Answer:

1. Hugo Devries
2. Carl Correns
3. Erich Von Tschermak.

45. is the prerequisite for Hybridization technique.
Answer:
Emasculation.

46. Transmission of genes that occur outside the nucleus is called………………
Answer:
Cytoplasmic Inheritance or Extra Nuclear

47. Cytoplasmic inheritance are found in
Answer:
Mitochondria & Chloroplast

48. The interaction between separate gene in which one makes the effect of another
Answer:
Epistasis

49. The acquisition of traits or conditions controlled by self replicating substances within the cytoplasm. This is a type of
Answer:
Cytoplasmic Inheritance.

50. The hybrid progeny in the first generation is called as
Answer:
F₁

51. The innate tendency of offspring to resemble their parents is called
Answer:
Heredity

52. The tendency of offspring to differ from parents is called
Answer:
Variation

53. Multiple allelic inheritances is otherwise called as
Answer:
Co – dominance

54. What is the use of pedigree analysis in genetics?
Answer:
It helps in genetic counselling.

55. Who proposed the genetic theory of inheritance?
Answer:
T.H.Morgan

56. Give one good example for Atavism in plants.
Answer:
Reemergence of sexual reproduction in Hieracium pilosella.

57. In pea plant, yellow seeds are dominant to green. If a heterozygous yellow seeded plant is crossed with a green seeded plant. What ratio of yellow and green seeded plants would you expect in FI generation?
Answer:
50:50 (or) 1:1

58. Some genes have allele that prevents survival when homozygous or heterozygous. What is the kind of allele?
Answer:
Lethal alleles

59. Recessive alleles of two different genes may give the same phenotype; This kind of genes also called
Answer:
Complementary gene.

60. A gene is a functional unit of DNA which codes for a
Answer:
Polypetide chain

61. Allele are the alternative form of the
Answer:
gene

62. discovered incomplete dominance.
Answer:
Correns

63. Human blood group is an example of variation.
Answer:
Discontinuous

X. Two marks

Question 1.
Write short note on Genotype.
Answer:
It is the genetic makeup of an organism responsible for a particular trait.

Question 2.
What is phenotype?
Answer:
It is the outward appearance or observable physical attributes of that trait.

Question 3.
Briefly explain monohybrid inheritance.
Answer:
Monohybrid inheritance looks at the inheritance of a single trait (a characteristics such as eye color, round or wrinkled seed type) coded by a single gene locus on a chromosome

Question 4.
Define Mendel’s first law.
Answer:
Mendel’s first law is ‘The law of segregation’. Segregation means separation. The two alleles are separated from each other during meiosis, so each gamete produced is haploids that is contain one allele of each gene.

Question 5.
What is epistasis?
Answer:
It is a term which describes how genes interact to affect a phenotype whereby an allele at one locus prevents an allele at another locus from manifesting its effect.
(or)
One gene is effectively interfering with or masking the effects of another gene.

Question 6.
What is epistatic?
The gene that suppresses or masks the . phenotypic expression of a gene at another locus is known as epistatic.

Question 7.
Write a note on hypostatic?
Answer:
In epistatsis, the gene whose expression is interfered by non- allelic genes and prevents from exhibiting its character is known as hypostatic.

Question 8.
Describe one of the reason that made the garden pea an excellent choice of Mendel system for studying inheritance.
Answer:
It is easily available self pollinated crop.

Question 9.
What is continuous variation with examples?
Answer:
A variation in a characteristics in which individuals show a range of traits with small difference between them. Eg: Human height and skin colour.

Question 10.
Write a note on discontinuous variation with suitable examples.
Answer:
Discontinuous is a variation in characteristic in which individuals show two or a few traits with large differences between them. (Eg) Height or Length of a plant.

Question 11.
What is hybridization?
Answer:
The process of mating two individuals that differ, with the goal of achieving a certain characteristics in their offspring.

Question 12.
Briefly explain ‘F₂‘.
Answer:
The second filial generation produced when Fi individuals are self-crossed or fertilized with each other.

Question 13.
Write a short note on Punnett square or checkerboard?
Answer:
A sort of cross multiplication matrix used in the prediction of the outcome of a genetic cross, in which male and female gametes and their frequencies are arranged along the edges.

Question 14.
List out the ‘R’ gene on responsible for polygenic inheritance in wheat (kernel colour)
Answer:
Four R genes are produced dark red kernel color. Three R genes are produced medium dark red kernel colour. Two R genes are produced medium red kernel colour. One R gene is produced medium red kernel colour and absence of R genes in results in white kernel colour.

Question 15.
Explain the role of genes in the formation of purple colour in the flowers of pisum sativum.
Answer:

• It was called Pea Gene A which encodes a protein that functions as a transcription factor which is responsible for the production of anthocyanin pigment.
• So the flowers are purple. Pea plants with white flowers do not have anthocyanin, even though they have the gene that encodes the enzyme involved in anthocyanin synthesis.

Question 16.
Write a note on Mendel’s Law of Dominance.
Answer:
It states that a dominant allele expresses itself in a monohybrid cross and suppresses the expression of recessive allele. However this recessive allele for a character is not lost and remain that hidden or masked in the progenies of F:l generation and reappear in the next generation.

Question 17.
What are multiple alleles?
Answer:
Alleles are alternative form of a gene. A gene for which at least two alleles exist is said to be polymorphic. Instances in which a particular gene may exist in three or more allelic forms are known as multiple allele conditions.

Question 18.
Briefly explain Mendelian Genetics.
Answer:
The set of theories prepared by Gregor Mendel, which attempt to explain the inheritance pattern of genetic characteristics based on simple breeding experiments involving single gene on chromosome pairs.

Question 19.
Write a note on Gene interaction.
Answer:
A single phenotype is controlled by more than one set of genes, each of which has two or more alleles. This phenomenon is called gene interaction.

Question 20.
Explain the three kinds of plants that have recersive lethal gene in Antirrhinum sp.
Answer:

1. Green plants with chlorophyll (CC)
2. Yellowish green plants with carotenoids are referred to as pale green, golden or a urea plants (Cc)
3. White plants without any chlorophyll, (cc)
4. The genotype of the homozygous green plants is CC. The genotype of the homozy¬gous white plant is cc.

Question 21.
Write a note on incomplete dominance.
Answer:
It refers to genetic situation in which one allele does not completely dominate another allele, and therefore results in a new phenotype.
(or)
It is a form of intermediate inheritance in which one allele for a specific trait is not completely expressed over its paired allele. This results in third phenotype in which the expressed physical traits is a combination of the phenotypes of both alleles.

Question 22.
A diagram that shows the possible outcomes of breeding between two individuals.
Answer:
Punnett Square or Checkerboard

Question 23.
Write a note on Punnet Square.
Answer:
It is a square type of a diagram that shows the possible outcomes of breeding between two individuals.

Question 24.
What do you mean by genetics ?
Answer:
Genetics is the study of how living things receive common traits from previous generation.

Question 25.
What are genes ?
Answer:
Genes are functional unit of inheritance. It is the basic unit of heredity (biological information) which transmits biochemical, anatomical and behavioural traits from parents to off springs.

Question 26.
What is population Genetics ?
Answer:
It deals with heredity in groups of individuals for trait which is determined by a few genes.
(or)
Population genetics is the study of genetic variation with in population, and the examination and moddling of changes in the frequencies of gene and allele in populations over space and time.

Question 27.
Define Molecular genetics .
Answer:
It is the field of that biology that studies the structure and function of genes at a molecular level
(or)
Study of structure and function of genes at molecular level
(or)
A branch of genetics that deals with structure and function of genes at molecular leve

Question 28.
Define Mutation.
Answer:
A permanent, heritable change in the nucleotide sequence in a genes or a chromosome , the process in which such a change occurs in a gene or in chromosome.

Question 29.
What do you mean by genetic transmission ?
Answer:
Genetic transmission is the transfer of genetic information (From parent to offspring), almost synonymous with heredity, or from one location in a cell to another .

Question 30.
Define Transmission Genetics :
Answer:
The study of the mechanisms involved in the passage of gene from one generation to the next.

Question 31.
What are polygenes ?
Answer:
A gene where individual effect on a phenotype is too small to be observed but which can act together with others to produce observable variation.
(or)
Characters are determined by two or more gene pairs, and they have additive or cumulative effect. Such genes are called polygenes or multiple factors or cumulative gene. Eg. Human skin colour.

Question 32.
Define Polygene ?
Answer:
Inheritance of phenotype is determined by the combined effects of many genes with environmental factor . These gene are called as polygene

Question 33.
Mendel was successful, why?
Answer:

• He applied mathematical method of law of probability to his breeding experiments
• He used pairs of contrasting characters in their experiment.

Question 34.
Write a note on self fertization.
Answer:
> (2 Marks) Fertilisation in a plant or animal by the fusion of male and female gametes produced by the same individual
(or)
> (3 Marks) Fertilisation that occurs when male and female gamete produced by the organism unite self fertilisation occur in many protozoans and invertebrate animal. It result from self pollination in plants. Seeds fertilization allows an isolated individual organism to reproduce but restricts the genetic diversity of a community.

Question 35.
What is cross fertilisation ?
Answer:
The fertilisation of an organism by the fusion of an egg from one individual with a sperm or male gamete from a different individual’s is opposite to the self.
(or)
Cross fertilisation is a term used in the field of biological reproduction describing the fertilisation of an occurs from one individual with spermatozoa of another. It is also called allogamy.
(or)
The fusion of male and female gamete (sex cells) from different individual of the same species.
It is mostly occur in dieocious plant and in animal species which they are separate male and female individual.

Question 36.
Does pure breeding means homozygous?
Answer:
If they are pure breeding that mean they are homozygous . So A group of identical individual that always produce offspring and same phenotype when intercrossed

Question 37.
What is the relationship between pure breeding and true breeding ?
Answer:
True breeding means that the parents with also pass down a specific phenotypic trait to their offspring. True breeding organism will have a pure genotype (genetic expression of a trait) and they will produce a certain phenotype. True breed is sometime also called pure breed.

Question 38.
Write a short note on Anthocyanin pigment.
Answer:

• Anthocyanin are naturally occurring pigment of red, purple and blue.
• Anthocyanin pigments are more stable at low PH (Acidic condition) which gives a red pigment. Measurable higher the PH value of anthocyanin will provide of colour fading of the colour blue or purple.

Question 39.
What is the mean ‘progeny’?
Answer:
The word progeny is the progeny of the Latin verb “progignere” meaning “to beget” . In biology, offspring are the young born of living organism, produced either by a single organism or in the case of sexual reproduction, true organism. Collective offspring may be known as a brood or progeny. It is also called as offspring of animals or plants or the children and other descendants.

Question 40.
Point out the mechanism of Trihybird cross.
Answer:
A cross between homozygous parent that differ in three gene pairs is called to trihybrid cross. A self fertilising trihybird plants forms 8 different gemeter and 64 different zygote. So these combination of three pair crosses operating together.

Phenotypic ratio -27:9:9:9:3:3:3:1 F₂
Mendel laws of segregation and independent assortment are also applicable to three pairs of contrasting traits ie. Trihybrid cross

Question 41.
What is back cross ?
Answer:
The cross between the F₁ offspring with either of the two parents. The parent may be dominant or recessive
(or)
When F₁ individuals are crossed with one of the true parenst from which they were derived, then such cross is called back cross
Explanation

• When TT is crossed with tt we get Tt as F₁ generation
• TT x tt = Tt
• when Tt (F₁ ) is crossed with either TT or tt (parent) it is called a back cross .

Question 42.
What are the classification of gene interactions?
Answer:
Interactions take place between the alleles o the same gene.
alleles at the same locus is called intragenic or intralocus gene interactions.

• Incomplete dominance
• co dominance
• multiple alleles
• pleiotropic genes.

Question 43.
Inheritance of chloroplast and mitochondria characters are non-mendelian inheritance
pattern why?
Answer:
The chloroplast arid mitochondrial genes show special pattern of inheritance known as Extra chromosomal inheritance.

Chromosomal inheritance:
The other aspects are

• They have vegetative segregation involving cytoplasmic plasmagenes. .
• It has uniparental inheritance (only from female parent)
• Both have reduced rate of recombinations.

Question 44.
What is hybrids?
Answer:
Mendels non-true breeding plants ae heterozygous called as hybrids.

Question 45.
What is Dihybrid cross?
Answer:

• It is a genetic cross which involves individuals differing in two characters.
• Dihvbrid inheritance is the inheritance of two separate genes each with two alleles.

XI. Three marks

Question 1.
Explain Bateson’s factor hypothesis ?
Answer:
Mendelian experiments prove that a single gene controls one character. But in the post mendelion findings, various exception have been noticed, in which different types of interaction are possible between the genes. FTence the expression of a single character by the interaction of more one pair of genes is called genic interaction or interaction of genes. According to this hypothesis some character are produced by the interaction of two or more pairs factor (gene).

Question 2.
What is the human ABO phenotype blood type based on?
Answer:
It is the major human blood group system. The ABO type of a person depends on they presence of absence of two gene, A and B. These gene determine the configuration of the red blood cell surface. A person who has two A gene or an A and O gene has bloodcells of type A. There are four main group of blood A,B,AB and O. The phenotype ratio is given below.
Blood group inheritance phenotype only

Question 3.
Explain the Genetic inheritance of pattern of human blood system ?
Answer:
An individually ABO type results from the inheritance of 193 alleles is A,B,0 from each parent . The possible out comes are given below

Both A and B alleles are dominant over O. As a results individual who have an AO gene type will- have an A phenotype. People who are type O have OO genotype. In other words, they inherited a recessive ‘O’ allele from both parents . The A and B alleles are co-dominant. Therefore, if an A is inherited from one parent and a B from the other the phenotype will be AB.

Question 4.
In blood type co-dominance or incomplete dominance ?
Answer:
It is closely related to incomplete dominance is co-dominance is which both alleles are simultaneously expressed in the heterozygote. In both co-dominonce and incomplete dominance both alleles for a trait are dominate in co-dominance a hetrozygous individual express both simultaneously with out any blending. People who are to type O have OO genotype. In other words they inherited a recessive O allele from both parents. The A and B alleles are co-dominant. Therefc )re is an A is inherited from one parent and a B from other the phenotype will be AB

Question 5.
In sickle cell co-dominant or incomplete dominance ?
Answer:
sickle cell anemia is a disease, in which the haemoglobin protein is produced incorrectly and the red bloodcells have a sickle shape. A person that is homozygous recessive for the sickle cells traits wills have red blood cells that all have the incorrect haemoglobin.

Question 6.
Write a note on co-dominance ?
Answer:
Co-dominance occurs when the phenotype of both parents are simultaneously expressed in the same offspring . An example of co¬dominance occurs in the human ABO blood group

Question 7.
Across between Bbcc and Bbcc. What is the probability of Bbcc?

Answer:
Solution
Probability of Bbcc = (Probability Bb) . (Probability Cc)

Question 8.
Write a note on Homologous chromosome or homologous.
Answer:
Morphologically, physiologically and genetically similar chromosome present is a diploid cell are called homologous or homologous chromosomes. In each pair of homologous chromosomes, one chromosome maternal and the other is paternal.

Question 9.
Write a note Emasculation.
Answer:
Removal of stamen well before another is called emasculation . It is done in bud condition to prevent self -pollination.

Question 10.
What is Punnett square or checker board?
Answer:
Punnett square is a graphical representation to calculate the probability of all possible genotypes of offsprings in a genetic cross. It was developed by Reginald C.Punnett.

Question 11.
Distinguish between homozygous and heterozygous
Answer:
homozygous :

1. Organism having identical alleles for a character are homozygous.
2. It is pure or true breeding
3. They form only one type of gametes
4. (eg) Tall (TT) dwarf (tt)

heterozygous :

1. Organism having dismillar alleles for a character are heterozygous.
2. It is hybrid
3. They form more then one type of gametes.
4. es (Tt)

Question 12.
Differentiate dominant from recessive character.
Answer:

Question 13.
Differentiate between Phenotype and Genotype
Answer:
Phenotype

1. It is the physical appearance of and organism
2. It can be directly seen
3. phenotype can be determined from genotype, (eg) Tt =Tall

Genotype

1. It is the genetic constitution of an organism
2. It is determined by inheritance pattern
3. Genotype can not be determined from phenotype (eg) Tall can either Tt (or) TT

Question 14.
Listant/Point out/Enlist the several traits in pea selected by mendel
Answer:

Question 15.
Draw the flow chart for heterozygous tall X homozygous dwarf pisum sativum plants If heterozygous tall test cross
Answer:

Question 16.
Distinguish between monohybrid cross and dihybrid cross
Answer:
Monohybrid cross :

1. The cross between to pure parents differing in a single pair of contrasting character is called Monohybrid cross
2. Phenotypic ratio is 3:1
3. Genotypic ratio is 1:2:1
4.  The law of segregation is explained by this method

Dihybrid cross :

1. The cross between two pure parents differing in two pairs of contrasting character is called dihybrid cross
2. phenotype ratio is 9:3:3:1
3. genotype ratio is 1:2:2:4:1:2:1:2:1
4. The law a independent is explained by this cross.

Question 17.
Distinguish between Test cross and Back cross
Answer:
Test Cross

1. The cross between F1 hybrid and its recessive parent is called test cross
2. A test cross is always a back cross
3. Test cross determines the genetic constitution of an organism
4. Test cross produces both dominant and recessive character is equal proportion

Back Cross :

1. The cross between F1 hybrid and any one of its parents (either dominant or recessive) is called back cross.
2. A back cross is not always a test cross
3. Back cross helps in improving and obtaining desirable character
4. Back cross helps in improving and obtaining desirable character

Question 18.
What is genetic testing?
Answer:
Genetic testing is analysing an individuals genetic material to determine predisposition to a particular health condition or to confirm a diagnosis of genetic disease

Question 19.
What are genetic disorder ?
Answer:
Genetic disorders are nothing but malfunctioning of genes due to some changes in their arrangement brought by mutation. Often these disorders characterized by absence or inactive protein products.

Question 20.
Write a short note on ‘Mutation’?
Answer:
Sudden heritable change in DNA or chromosome is called mutation. There are agents which cause mutation called Mutagens. Due to mutations many abnormalities will appear in new generations which may be useful or harmful.

Question 21.
Co-dominance is an example of intragenic gene interaction. How?
Answer:

• The phenomenon in which two alleles are both expressed in the heterzygous individual is known as codominance
  Example:
• Red and white flowers of camellia, inheritance of sickle cell haemoglobin.
• ABO blood group system in human beings.
• In humanbeings, IA and IB alleles of I gene are codominant which follows mendels law of segregation.
• The co-dominance was demonstrated in plants with the help of electrophoresis or chromatography for protein or flavonoid substance.

Question 23.
What is the different between sex linked and sex influenced diseases ?
Answer:
In sex linked diseases the defected gene are present on the sex chromosomes attached to them whereas in sex influenced diseases defective gene are present on the other chromosome but affects the sex chromosomes.

Question 24.
What is Genome ?
Answer:
A complete set off gene is an organism is called genome .

Question 25.
What are lethal gene or lethal allele ?
Answer:
Lethal allele are alleles that cause the death of the organism that carries them. They are unusually a result of mutation is gene that are essential to growth or development. Lethal allele may be recessive, dominant or conditional depending on the genes or genes involved.

Question 26.
What do you mean by inheritance of sickle cell anemia in man.
Answer:
The diseases sickle cell anemia is causes by a gene (Hbs) which is lethal in homozygous condition. But has a slight denotable effect is the heterozygous conditions, producing sickle cell trait. The homozygous for this gene (Hbs/HbS) generally die of fatal anemia. The hetrozygotes or carriers for Hbs. (ie) HbA/HbS) show signs of mild anemia as their RBC become sickle – shaped in oxygen deficiency. A marriage between two carriers, therefore results in carrier and normal offspring in the ratio 2:1

Question 27.
What is cytoplasmic male sterility ?
Answer:
Plants that fail to produce functional pollengrains are said to be male-sterile. If the traits conditioning the sterility is not inherited according to mendelion rules, but is instead maternally transmitted, it is referred to as cytoplasmic male sterility(cms). So in this male-sterility is inherited maternally.
The gene for cytoplasmic male sterility is found in the mitochondrial DNA
(or)
When plants fails to produce functional pollengrain, they are called male sterile mole. Male sterility may be conditioned by either nuclear or Cytoplasmic genes. If the sterility trait is inherited is a non -Mandelian fashion, it is designated as cytoplasmic male sterility (CMS). Cytoplasmic gene are most often maternally transmitted in plants.

Question 28.
Briefly explain ‘Atavism’ with suitable examples.
Answer:
Atavism derives via French from Latin atavius, meaning “ancestor”. Avus in Latin means ‘grand father’; and its is believed that the ‘at’ is related to atta a word for “Daddy”. Atavism is a term rooted in evolutionary study referring to instances when an organism possesses trait closer to a more remote ancestor, rather than its own parents. It is modification of a biological structure whereby an ancestral traits re appears after having been lost through evolutionary changes is the previous generations.

(eg) Re-emergence of sexual reproduction in the flowering plant Hieracium pilosella is the best example for Atavism in plants

Question 29.
How to do test for homozygosity of a trait in plant.
Answer:

• To identify whether an organism exhibiting a dominant trait is homozygous or hesterozy- gous for a specific allele a scientist can perform a test cross.
• The organism in question can be crossed with an organism that is homozygous for the recessive trait – and the offspring’s of the test cross are examined.

XII. Five Marks

Question 1.

Difference between Pleiotropy and polygenic inheritance with suitable examples.
Answer:
Pleiotropy is when one gene affect multiple characters eg. Marfan syndrome and polygene inheritance is when one traits is controlled by multiple genes (eg), skin colour (or) skin pigmentation

Question 2.
Co-dominance and incomplete dominance are not the same? why?
Answer:

• In co-dominance neither allele is dominant over the other, so both will be expressed equally in the heterozygote.
• In incomplete dominance, there is an intermediate heterozygote. Such as pink flower when the parent phenotypes are red. and white.

Question 3.
Difference between Monohybrid cross and Reciprocal cross.
Answer:
Monohybrid

1.  It is one sided or both sided
2.  It is used to study the inheritance of single pair of alleles.
3. It cannot distinguish between nuclear and Cytoplasmic (or) sex linked and autosomal traits

Reciprocal cross

1. It is both sided cross in which female of one type is crossed with male of the second type and vice versa
2. It may study inheritance of one, two or more traits
3. It can distinguish between nuclear and cytoplasmic inheritance as well as sea linked l j and autosomal inheritance.

Question 4.
Difference between Monohybrid and Dihybrid cross
Answer:
Monohybrid :

1. Mono refer to single and hybrid means mixed breed
2. It is used to study the inheritance of single pair of alleles.
3. Genotype ratio is 1:2:1
4. Phenotypic ratio is 3:1
5. One pair of contrasting character are involved

Dihybrid

1. Di refers to two or double and hybrid means breed.
2. It is used to study the inheritance of two different alleles.
3. Genotype is ratio is 1:2:1:2:4:2:1:2:1
4. Phenotypic ratio is 9:3:3:1
5. Two pair of contrasting character are involved.

Question 5.
Explain Monohybird cross.
Answer:
A monohybird is a genetic cross which occurs between two individuals, focusing on the inheritance of one trait at one time. Monohybrid cross is also known as single trait cross. Two homozygous parent are selected for this cross.

Each homozygous parent in the P generation produces only one kind of gamete.
The heterozygous F] offspring produces two kinds of gamete
The heterozygous Fi offspring produces two kinds of gamete

Question 6.
Explain Dihybrid cross.
Answer:
A dihybrid cross is a genetic cross that occurs between two individuals, focusing on the inheritance of two independent traits at one time. It is also known as two trait cross.
Two parents considered for this cross have two independent traits (eg: pea colour and pea shapes of plants). Thus a dihybrid cross involves two pairs of genes. The following figure explains the process of dihybrid crossing.

F₁ phenotype: All round yellow cotyledon
Fi genotype : All RrYy
RrYy x RrYy (Fj generation selfied)
Ry Ry rY rY x Ry Ry ry ry (Haploid gametes)

How to do a Dihybrid Cross

• Analyze the data!
• Make a tally of all possible phenotypes.

In a dihybrid cross, traits are considered as not linked, and they have an equal probability of sharing up in offspring. Each pair of alleles segregates independently of the gametes. Offspring is predicted and assessed for two trait inheritance. The phenotypic ratio of the offspring generation is 9:3:3:1 in a dihybrid cross.

Question 7.
Briefly explain Trihybrid cross.
Answer:
A trihybrid cross is between two individuals that are homozygous for three different traits. (Eg: Pea shape, colour and pea shape)
(or)
A cross between homozygous parents that differ in three gene pairs, (ie: producing trihybrid) is called trihybrid cross. A seed fertilizing trihybrid plant forms 8 different gametes and 64 different zygotes. So a combination of three single pair crosses operating together. The three contrasting characters of a trihybrid crosses are

• F₂ Phenotypic ratio – 27:9:9:9:3:3:3:1
• 27 – round, green, smooth pod
• 9 – round, green, constructed pod
• 9 – round, yellow, smooth pod
• 9 – wrinkled, green, smooth pod
• 3 – round, yellow, constructed pod
• 3 – wrinkled, green, constructed pod
• 3 – wrinkled, yellow, smooth pod
• 1- wrinkled, yellow, constructed pod

Question 8.
What traits are determined by multiple alleles?
Answer:
A trait controlled by one gene but multiple allele is blood type. There are four phenotypes A, B, AB, O. Type A and B are co-dominant and ‘O’ is recessive to A and B. None are dominant. Some traits are controlled by a single gene with two alleles. Mendelian heredity had only two alternative expression or alleles. However many genes can change in several different ways or changes. Those changes give rise to several alternative states which are called multiple alleles.
(or)
Blood type is an example of a common multiple allele trait. There are three different alleles for blood type A, B & O. A is dominant to O, B is also dominant to O. A and B are both co-dominant.

Question 9.
What is a gene?
Answer:
A gene is a segment of DNA that spells out the gentic code for a particular trait. A trait is a physical characteristics.

Question 10.
What is Incomplete dominance with example.
Answer:

• Carl Correns’s (1905) experimented in 40′ clock plant, Mirabilis jalapn.
• When the pure breeding homozygous red (R¹R¹) parent is crossed with homozygous white (R¹R¹)
• The phenotype of the F₁ hybrid is heterozygous pink (R²R²)
• The F1 heterozygous phenotype differs from both the parental homozygous phenotype.
• This cross did not exhibit the character of the dominant parent but has an intermediate colour pink.
• The phenotypic and genotypic ratios were found to be same as 1:2:1 (1 red : 2 pink : 1 white). Genotypic ratio is 1 R¹R¹: 2 R¹R²: 1 R²R² in F₂ interbreed.
• In the F2 generation, R¹ and R² genes segregate and recombine to produce red, pink and white in the ratio of 1:2:1.
• R¹ allele codes for an enzyme responsible for the formation of red pigment and R² allele codes for defective enzyme. R¹ and R² genotypes produce only enough red pigments to make the flower pink.
• Mendel’s particulate inheritance takes place in this cross which is confirmed by the reappearance of original phenoty in F₂
  
  

Question 11.
Briefly explain about lethal gene.
Answer:
Allele that cause an organism to die are called lethal alleles or lethal genes. Lethal genes are usually a result of mutations in genes that are essential to the growth or development. Lethal gene can cause death of an organism prenatally or anytime after birth. Lethal genes are first discovered by Lucien cuenot in the study of coat colour in mice.

Question 12.
Explain epistatsis and its two types.
Answer:
Epistasis is a type of polygenic interaction where one gene controls the phenotype of another gene for a trait. Both genes have an influence on the physical appearance of the traits, but the one that shows epistasis masks the effect of the other. Eg: albinism.

Dominant epistatsis: It happens when the dominant allele of one gene masks the expression of all allele of another gene.

Recessive Epistasis:
Recessive epistasis is when the recessive allele of one gene in a homozygous slate masks the phenotypic expression of the dominant allele of another gene.
(eg) Mice,
In Mice, body Colour is determined by a gene A. A is hypostatic to an allele C of another gene, which mean that C marks the expression of A. C is the presence of a gives cinnamon mice, While C in the prsence of A gives agouti mice.

Recessive Epistasis :

In dominant epistasis, the majority of the individuals are affected. There is a 12:3:1 ration.

Genes that show recessive epistasis can only mask a phenotype if two alleles are present The ratio is 9:3:4

Question 13.
Briefly explain duplicate recessive gene in Intergenic interaction (or) complementary gene interaction
Answer:
If both gene loci have homozygous recessive alleles and both of them produce identical phenotype the F2 ratio 9:3:3:1 would be 9:7. The genotype aaBB, aaBb, AAbb, Aabb and aabb produce same phenotype. Both dominant alleles when are present together only than they can complement each other. This is known as complementary gene.
Complementary Genes (9:7)
Ex: In Lathyrusodoratus ,Bateson and punnet crossed two varieties(CCpp x ccPP),each with white flowers.

• Eg: complete dominance at both gene pairs, but either recessive homozygote is epistatic to the effect of the other gene.
• In sweet pea flower colour.
• Gene pair A = purple dominant over white
• Gene pair B = colour dominant over white
• Interaction = Homozygous recessive of either gene A or B produce white

Question 14.
Explain duplicate gene with cumulative effect (9:6:1)
Answer:

• Certain phenotype traits despond on the dominant alleles of two gene loci. When dominant is present it will share its phenotype. The ratio will be 9:6;1 Eg: Fruit shape in summer squash.
• Complete dominance at both gene pair, interaction between, both dominance to give new phenotype.
• Gene pair ‘A’ sphere shape dominant over long.
• Gene pair ‘B’ sphere shape dominant over long.
• So interaction at ‘AB’ when present together form disc shaped fruit.
• Finally disc shaped fruit 9/16 Sphere shaped fruit 6/16
  

Duplicate genes with cumulative effect (9:6:1) :

Question 15.
What are Duplicate dominant gene (15:1) or duplicate gene?
Answer:

• If a dominant allele of both gene low produces the same phenotype without cumulative effect i.e., independently the ratio will be 15:1
• Eg : seed capsule of shephered’s purse complete dominance at both gene pair, but either gene when dominant, epistatic to the other.
• Gene pair ‘A’=Triangular shape dominant over ovoid
  Gene pair ‘B’=Triangular shape dominant over ovoid (double recessive)

Duplicate dominant genes (15:1):
15/16 = Triangular
1/16= Ovoid (top shaped)
AABB x aabb
Triangular ovoid
AaBb x AaBb
Triangular 

Question 16.
Explain dominant and recessive interaction (or) inhibitor gene (13:3)
Answer:
Sometimes the dominant alleles of one gene locus (A) in homozygous and heterozygous (AA, Aa) condition and homozygous recessive alleles bb of another locus (B) produces the same phenotype. The F2 ratio will become 13:3. The genotype AABB, AaBB, AAbb, Aabb and aabb produce one type of phenotype and genotype aaBb, aaBB, will produce another type of phenotype.

• Eg: Feather colour of Fowl
• Complete dominance at both gene pair, but are gene when dominant epistatic to the other and the second gene when homozygous recessive epistatic to the first.
• Gene ‘A’ colour inhibition is dominant to colour appearance.
• Gene ‘B’ colour in dominant to white.

Interaction:

• Dominant colour inhibitors prevents colour even when colour is present, colour gene, when homozygous recessive prevents colour when dominant inhibitor is present.

Dominant and recessive interaction (13:3):

13/16 = white
3/16 = coloured

Question 17.
Male sterility found in pearl maize (sorgum Vulgare) is the best example for mitochondria cytoplasmic inheritance.
Answer: Male sterility found in pearl maize (sorgum Vulgare) is the best example for mitochondria cytoplasmic inheritance. so it is called cytoplasmic male sterility.
In this, male sterility is inherited maternally.
The gene for cytoplasmic male sterility is found in the mitochondrial DNA.

In this plant there are two types, one with normal cytoplasm (N) which is male fertile and the other one with aberrant cytoplasm (s) which is male sterile.
These types also exhibit reciprocal differences as found in Mirabilis jalapa

Recently it has been discovered that cytoplasmic genetic male sterility is common in many plant species.
This sterility maintained by the influence of both nuclear and cytoplasmic genes.
There are commonly two types of cytoplasm N (Normal) and S (Sterile)
The genes for these are found in mitochondrian.
There are also restores of fertility (Rf) genes. Even though these genes are nuclear genes, they are distinct from genetic male sterility genes of other plants.
Because the Rf genes do not have any expression of their own, unless the sterile cytoplasm is present.
Rf genes are required to restore fertility in S cytoplasm which is responsible for sterility.
So the combination of N cytoplasm with rfrf and s cytoplasm with RfRf products plants with fertile pollens, while S cytoplasm with rfrf produces only male sterile plants.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 6 Applications of Vector Algebra Ex 6.9 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.9

Question 1.
Find the equation of the plane passing through the line of intersection of the planes \(\overline { r }\) = (2\(\hat { i }\) – 7\(\hat { j }\) + 4\(\hat { k }\)) = 3 and 3x – 5y + 4z + 11 = 0 and the point (- 2, 1, 3).
Solution:
Given planes are
\(\vec{r} \cdot(2 \hat{i}-7 \hat{j}+4 \hat{k})\) = 3
2x – 7y + 4z – 3 = 0 and 3x – 5y + 4z + 11 = 0
Equation of a plane which passes through the line of intersection of the planes
(2x – 7y + 4z – 3) + λ(3x – 5y + 4z + 11) = 0 …………… (1)
This passes through the point (-2, 1, 3).
(1) ⇒ (-4 – 7 + 12 – 3) + λ(-6 – 5 + 12 + 11) = 0
-2 + λ(12) = 0 ⇒ 12λ = 2
λ = \(\frac{2}{12}\) ⇒ λ = \(\frac{1}{6}\)
The required equation is
(1) ⇒ (2x – 7y + 4z – 3) + \(\frac{1}{6}\) (3x – 5y + 4z + 11) = 0
12x – 42y + 24z – 18 + 3x – 5y + 4z + 11 = 0
15x – 47y + 28z – 7 = 0

Question 2.
Find the equation of the plane passing through the line of intersection of the planes x + 2y + 3z = 2 and x – y + z = 3 and at a distance \(\frac { 2 }{ √3 }\) from the point (3, 1, -1).
Solution:
Required equation of the plane
(x + 2y + 3z – 2) + λ(x – y + z – 3) = 0 ………. (1)
(1 + λ)x + (2 – λ)y + (3 + λ)z + (-2 – 3λ) = 0 ………(2)
Distance from (2) to the point (3, 1, -1) is \(\frac { 2 }{ √3 }\)

putting
λ = \(\frac { -7 }{ 2 }\) in (1)
The required equation
(x + 2y + 3z – 2) – \(\frac { 7 }{ 2 }\) (x – y + z – 3) = 0
2x + 4y + 6z – 4 – 7x + 7y – 7z + 21 = 0
-5x + 11y – z + 17 = 0
5x – 11y + z – 17 = 0

Question 3.
Find the angle between the line
\(\overline { r }\) = (2\(\hat { i }\) – \(\hat { j }\) + \(\hat { k }\)) + (\(\hat { i }\) + 2\(\hat { j }\) – 2\(\hat { k }\)) and the plane \(\overline { r }\) (6\(\hat { i }\) + 3\(\hat { j }\) + 2\(\hat { k }\)) = 8
Solution:

Question 4.
Find the angle between the planes \(\overline { r }\) (\(\hat { i }\) + \(\hat { j }\) – 2\(\hat { k }\)) = 3 and 2x – 2y + z = 2.
Solution:

Question 5.
Find the equation of the plane which passes through the point (3, 4, -1) and is parallel to the plane 2x – 3y + 5z + 7 = 0. Also, find the distance between the two planes.
Solution:
The required equation parallel to the plane
2x – 3y + 5z + 7 = 0 ………….. (1)
2x – 3y + 5z + λ = 0 ………….. (2)
This passes through (3, 4, -1)
(2) ⇒ 2(3) – 3(4) + 5(-1) + λ = 0
6 – 12 – 5 + 1 = 0
λ = 11
(2) ⇒ The required equation is 2x – 3y + 5z + 11 =0 …………… (3)
∴ Now, distance between the above parallel lines (1) and (3)
a = 2, b = -3, c = 5, d₁ = 7, d₂ = 11

Question 6.
Find the length of the perpendicular from the point (1, -2, 3) to the plane x – y + z = 5.
Solution:

Question 7.
Find the point of intersection of the line with the plane (x – 1) = \(\frac { y }{ 2 }\) = z + 1 with the plane 2x – y – 2z = 2. Also, the angle between the line and the plane.
Solution:
Any point on the line x – 1 = \(\frac{y}{2}\) = z + 1 is
x – 1 = \(\frac{y}{2}\) = z + 1 = λ,(say)
(λ + 1, 2λ, λ – 1)
This passes through the plane 2x – y + 2z = 2
2(λ + 1) – 2λ + 2(λ – 1) = 2
2λ + 2 – 2λ + 2λ – 2 = 2
λ = 1
∴ The required point of intersection is (2, 2, 0)

Question 8.
Find the co-ordinates of the foot of the perpendicular and length of the perpendicular from the point (4, 3, 2) to the plane x + 2y + 3z = 2.
Solution:
Let us take the point P(4, 3, 2) and Q(x₁, y₁, z₁)
⇒ (x₁ – 4, y₁ – 3, z₁ – 2)
Plane x + 2y + 3z = 2 ………. (1)
Cartesian equation of PQ, \(\frac { x_1-4 }{ 1 }\) = \(\frac { y_1-3 }{ 2 }\) = \(\frac { z_3-2 }{ 3 }\) = λ
(λ + 4, 2λ + 3, 3λ + 2) lies in (1)
(λ + 4) + 2(2λ + 3) + 3(3λ + 2) – 2 = 0
λ + 4 + 4λ + 6 + 9λ + 6 – 2 = 0
14λ + 14 = 0
14λ = -14 .
λ = -1
Co-ordinates of the foot of the ⊥r is (3, 1, -1).
Distance PQ = \(\sqrt{(4-3)^2 + (3-1)^2 + (2+1)^2}\)
= \(\sqrt{1^2 + 2^2 + 3^2}\) = \(\sqrt{1 + 4 + 9}\)
= \(\sqrt{14}\) units.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Bio Botany Guide Pdf Chapter 1 Asexual and Sexual Reproduction in Plants Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Bio Botany Solutions Chapter 1 Asexual and Sexual Reproduction in Plants

12th Bio Botany Guide Asexual and Sexual Reproduction in Plants Text Book Back Questions and Answers

I. Choose the correct answer

Question 1.
Choose the correct statement from the following
a) Gametes are involved in asexual reproduction
b) Bacteria reproduce asexually by budding
c) Conidia formation is a method of sexual reproduction
d) Yeast reproduce by budding
Answer:
d) Yeast reproduce by budding

Question 2.
An eminent Indian embryologist is
a) S.R.Kashyap
b) P.Maheswari
c) M.S. Swaminathan
d) K.C.Mehta
Answer:
b) P.Maheswari

Question 3.
Identify the correctly matched pair
a) Tuber – Allium cepa
b) Sucker – Pistia
c) Rhizome – Musa
d) Stolon – Zingiber
Answer:
c) Rhizome – Musa

Question 4.
Pollen tube was discovered by
a) J.G.Kolreuter
b) G.B.Amici
c) E.Strasburger
d) E.Hanning
Answer:
b) G.B.Amici

Question 5.
Size of pollen grain in Myosotis
a) 10 micrometer
b) 20 micrometer
c) 200 micrometer
d) 2000 micrometer
Answer:
a) 10 micrometer

Question 6.
First cell of male gametophyte in angiosperm is
a) Microspore
b) megaspore
c) Nucleus
d) Primary Endosperm Nucleus
Answer:
a) Microspore

Question 7.
Match the following

Answer:
a) I—iv; ll—i; III—ii; I’V—iii
b) 1—iii; J1—iv; III—i; V—ii
c) I—iii; I1—iv; III—ii, IV—i
d) I—iii; II—i; III—iv; IV—ii
Answer:
b) I—iii;II—iv;III—i;1 V—ii

Question 8.
Arrange the layers of anther wail from locus to periphery
a) Epidermis,middle layers, tapetum, endothecium
b) Tapetum, middle layers, epidermis, endothecium
c) Endothecium, epidermis, middle layers, tapetum
d)Tapetum, middle layers endothecium epidermis
Answer:
d) Tapetum, middle layers endothecium epidermis

Question 9.
Identify the incorrect pair
a) sporopollenin – exine of pollen grain
b) tapetum – nutritive tissue for developing microspores
c) Nucellus – nutritive tissue for developing embryo
d) obturator – directs the pollen tube into micropyle
Answer:
c) Nucellus – nutritive tissue for developing embryo

Question 10.
Assertion : Sporopollenin preserves pollen in fossil deposits.
Reason : Sporopollenin is resistant to physical and biological decomposition.
a) Assertion is true; reason is false
b) Assertion is false; reason is true
c) Both Assertion and reason are not true
d) Both Assertion and reason are true
Answer:
d) Both Assertion and reason are true

Question 11.
Choose the correct statement(s) about tenuinucellate ovule
a) Sporogenous cell is hypodermal
b) Ovules have fairly large nucellus
c) sporogenous cell is epidermal
d) ovules have single layer of nucellus tissue
Answer:
a) Sporogenous cell is hypodermal and d)ovules have single layer of nucellus tissue

Question 12.
Which of the following represent megagametophyte
a) Ovule
b)Embryo sac
c) Nucellus
d)Endosperm
Answer:
b) Embryo sac

Question 13.
In Haplopappus gracilis, number of chromosomes in cells of nucellus is 4. What will be the chromosome number in Primary endosperm cell?
a) 8
b) 12
c) 6
d) 2
Answer:
C) 6 (3n)

Question 14.
Transmitting tissue is found in
a) Micropylar region of ovule
b) Pollen tube wall
c) Stylar region of gynoecium
d) Integument
Answer:
c) Stylar region of gynoecium

Question 15.
The scar left by funiculus in the seed is
a) tegmen
b) radicle
c) epicotyl
d) hilum
Answer:
d) hilum

Question 16.
A Plant called X possesses small flower with reduced perianth and versatile anther. The probable agent for pollination would be
a) water
b) air
c) butterflies
d) beetles
Answer:
b) air

Question 17.
Consider the following statement(s)
i) In Protandrous flowers pistil matures earlier
ii) In Protogynous flowers pistil matures earlier
iii) Herkogamy is noticed in unisexual flowers
iv) Distyly is present in Primula
a) i and ii are correct
b) ii and iv are correct
c) ii and iii are correct
d) i and iv are correct
Answer:
b) ii and iv are correct

Question 18.
Coelorhiza is found in
a) Paddy
b)Bean
c) Pea
d) Tridax
Answer:
a) Paddy

Question 19.
Parthenocarpic fruits lack
a) Endocarp
b) Epicarp
c) Mesocarp
d) seed
Answer:
d) seed

Question 20.
In the majority of plants, pollen is liberated at
a) 1 celled stage
b) 2 celled stage
c) 3 celled stage
d) 4 celled stage
Answer:
b) 2 celled stage

Question 21.
What is reproduction?
Answer:
Reproduction is the biological process of producing young ones of their own kind. It is a vital process for the existence of a species and it also brings suitable changes through variation in the offsprings for their survival on ear

Question 22.
Mention the contribution of Hofmeister towards Embryology.
Answer:

• He worked on flowering plant embryology.
• Discovered alternation of generation in plants.
• He described the structure of pollen tetrad.

Question 23.
List out two sub-aerial stem modifications with example.
Answer:
Subaerial stem modifications.
The stem is partly aerial and partly underground.

a) Runner. (Ex. oxalis, Centella Asiatica)

• It is running horizontally on the soil surface.
• Nodes have axillary buds, scale leaves, and adventitious roots.
• Runner arises from the axillary bud.
• Mother plant produces many runners in all directions.
• They break off and grow into individual plants.

b) Sucker. (Ex. Musa (banana), chrysanthemum)
Grows horizontally for a distance under the soil. Then it emerges obliquely upwards.

c) Stolon (Ex. Strawberry, Vallisneria)
Develop from underground stems.
They grow horizontally outwards.

d) Offset (condensed runners)
Unlike runners, they produce tilt of leaves above and duster of roots below Ex. Pistia, Eichhornia.

Question 24.
What is layering?
Answer:

• It is an artificial method of vegetative propagation.
• The stem of the parent plant is allowed to develop roots while still intact.
• The root develops. The rooted part is cut. It is planted to grow as a new plant.
• Ex. Ixora, Jasminum.

Question 25.
What are clones?
Answer:
Individuals developed by asexual reproduction are morphologically and genetically identical. Such individuals are called clones.

Question 26.
A detached leaf of Bryophyllum produces new plants. How?
Answer:

• Bryophyllum undergoes vegetative reproduction in the leaf.
• The succulent leaf is notched in its margin.
• Adventitious buds develop at these notches. They are called epiphyllous buds.
• These buds develop a root system. When the leaf decays, they become independent plants.

Question 27.
Differentiate Grafting and Layering.
Answer:
Grafting:

1. In grafting, two different plants (stock & scion) are used to develop new plants.
2. The new plant will support to possess the characters of both the parents or new variation can be noticed.

Layering:

1. In layering, only one plant is used to develop a new plant.
2. Variation cannot be expected. The new individual is exactly similar to the parent plant.

Question 28.
“Tissue culture is the best method for propagating rare and endangered plant species” Discuss.
Answer:
Micropropagation.
The growth of plant tissue in special culture medium under suitable controlled conditions is known as “tissue culture”.
it is the regeneration of a whole plant from a single cell or tissue.

Advantages.

• Rare, Endangered plants are propagated.
• In a short duration, plants with desirable characteristics can be multiplied.
• Produce Genetically identical plants.
• Done in any season.
• Plants without viable seeds (or) difficult to germinate can be propagated.
• Meristem culture produces disease-free plants.
• Cells can be genetically modified or transformed.

Question 29.
Distinguish mound layering and air layering.
Answer:
Mound Layering:
In mound layering, a lower flexible branch with leaves is bent to the ground and a part of the stem is buried in the soil and the tip of the branch is exposed above the soil. After the roots emerge from the buried stem, a cut is made in the parent plant so that the buried plant grows into a new plant.

Question 30.
Explain the conventional methods adopted in the vegetative propagation of higher plants.
Answer:
Conventional methods of vegetative propagation.
a) Cutting (Ex. Hibiscus)

• Plant parts like stem, leaf are cut from the parent plant.
• Cut part is placed in suitable medium,
• It produces root and grows into a new plant.

b) Grafting (Ex. Citrus, Mango)

• Two different plants are joined.
• They grow as one plant.
• Plant in soil is called stock.
• Plant used for grafting is the scion.
• It is of 5 types.

i) Bud grafting – scion is placed in the incision of stock.
ii) Approach grafting – Cut surfaces of stock scion are tied together.
iii) Crown Grafting – Wedge-shaped scion is inserted into the cleft of stock.
iv) Tongue grafting – Stock and scion are cut obliquely scion is fit into stock and bound with tape.
v) Wedge grafting – Twig of the scion is inserted into slot in the stock.

c) Layering
Stem of parent plant is allowed to develop roots while still intact. The root develops. The rooted part is cut and planted to grow as a new plant.

I) Mound Layering

• Flexible branch is buried in soil.
• Roots emerge from buried stem. It grows into a new plant.

ii) Air Layering

• Nodal region is girdled.
• Hormones are applied.
• Rooting is promoted.
• This area is covered by moist soil.
• Roots emerge in 2-4 months.
• Roots branches are removed from parent. They are grown separately.

Question 31.
Highlight the milestones from the history of plant embryology.
Answer:

1. 1682 – Nehemiah Grew mentioned stamens as the male organ of a flower.
2. 1694 – R.J. Camerarius described the structure of a flower, anther, pollen, and ovule
3. 1761 – J.G. Kolreuter gave a detailed account of the importance of insects in pollination.
4. 1824 – G.B. Amici discovered the pollen tube.
5. 1848 – Hofmeister described the structure of pollen tetrad.
6. 1870 – Hanstein described the development of embryos in Capsella and Alisma.
7. 1878 – E. Strasburger reported polyembryony.
8. 1884 – E. Strasburger discovered the process of Syngamy.
9. 1899 – S.G. Nawaschin and L. Guignard independently discovered Double fertilization.
10. 1904- E. Hanning initiated embryo culture.
11. 1950 – D.A. Johansen proposed classification for embryo development.
12. 1964 – S. Guha and S.C. Maheswari raised haploids from Datura pollen grains
13. 1991 – E.S. Coen and E.M. Meyerowitz proposed the ABC model to describe the genetics of initiation and development of floral parts
14. 2015 – K.V. Krishnamurthy summarized the molecular aspects of pre and post-fertilization reproductive development in flowering plants.

Question 32.
Discuss the importance of Modern methods in reproduction of plants.
Answer:
The genetic ability of a plant cell to produce the entire plant under suitable condition is said to be totipotency.

• This characteristic feature of a cell is utilized in horticulture, forestry and industries to propagate plants.
• The mature phloem parenchyma cells removed from the carrot were placed in a suitable medium under controlled conditions.
• It stimulate to start dividing again to produce a new carrot plant.

Importance of modern methods of reproduction in plants.

• Rapid Multiplication of desired plants in short duration.
• Genetically identical plants are produced.
• Tissue culture can be done at any season
• Plants without viable seeds (or) difficult to germinate can be propagated.
• Rare, Endangered plants are propagated.
• Meristem culture produces disease-free plants.
• Cells are genetically modified or transformed.

Question 33.
What is Cantharophily?
Answer:

• It is the cross-pollination of flowers by beetles. They feed on pollen or juicy tissues of their flower.
• The plants using this mode of pollination
• Er. Nymphaea species of plants – Rhinoceros beetle.
• Giant Water lily – Scarab beetle
• Illicium plant – Diptera files.

Question 34.
List any two strategy adopted by bisexual flowers to prevent self-pollination.
Answer:
1) Dichogamy
Anthers and stigmas mature at different times.

•  Protandry – Stamens mature earlier.
• Protogyny – Stigmas mature earlier.

2) Herkogamy

• Self pollination is impossible by the arrangement of stamens and stigmas.
• Ex: In Hibiscus, stigmas project above the stamens.
• In some plants, when the pollen grain of a flower reaches the stigma of the same.
• It is unable to germinate or prevented to germinate on its own stigma.
• It is a genetic mechanism.
  Example: Abutilon, passiflora.

Question 35.
What is the endothelium?
Answer:
In the Asteraceae species, the inner layer of the integument gets specialized for nourishing the embryosac and this is called the integumentary tapetum or endothelium.

Question 36.
“The endosperm of angiosperm is different from gymnosperm”. Do you agree? Justify your answer.
Answer:
Endosperm of Angiosperms

1. Triploid Endosperm
2. Endosperm is formed by triple fusion.
3. Endosperm surrounds the embryo.

Endosperm of Gymnosperm

1. Haploid endosperm.
2. The endosperm is formed before fertilisation.
3. Gymnosperms (Ex; pine) produce embryos. It provides nutrition as starch. with many cotyledons. Primary Endosperm is used as food.

Question 37.
Define the term Diplospory.
Answer:
Diplospory is a condition where a diploid embryosac is formed from megaspore mother cells without a regular meiotic division.
E.g: Eupatorium.

Question 38.
What is polyembryony? How can it be commercially exploited?
Answer:
Polyembryony

• The occurrence of more than one embryo in a seed is called poly embroyony.

Practical Applications.

• Seedlings from nucellar tissue of citrus are better clones for orchards.
• Embryos from polyembryonic are virus-free.

Question 39.
Why does the zygote divide only after the division of the Primary endosperm cell?
Answer:
The primary endosperm nuclear (PEN) divides prior to zygotic division and form endosperm. Endosperm acts as a nutritive tissue and nourishes the developing embryo.

Question 40.
What is Mellitophily?
Answer:
Pollination by honeybee is called mellitophily (Latin word mellitus= honey or sweet), Among the insects the bees are the main flower visitors and dominant pollinators.

Question 41.
“Endothecium is associated with dehiscence of anther” Justify the statement.
Answer:
The inner tangential wall develops bands (sometimes radial walls also) of cellulose (sometimes also slightly lignified). The cells are hygroscopic. The cells along the junction of the two sporangia of an anther lobe lack these thickenings. This region is called stomium. This region along with the hygroscopic nature of endothecium helps in the dehiscence of anther at maturity.

Question 42.
List out the functions of tapetum.
Tapetum is the innermost layers of anther wall.
Answer:

• Supplies nutrition to developing microspores.
• Contributes sporopollenin through ubisch bodies. They play role in pollen wall formation.
• Pollenkitt material is contributed by tapetal cells. It is layer transferred to pollen surface.
• Exine proteins for rejection reaction are derived from tapetal cells.

Question 43.
Write a short note on Pollen kitt.
Answer:
Pollenkitt is contributed by the tapetum and coloured yellow or orange and is chiefly made of carotenoids or flavonoids. It is an oily layer forming a thick viscous coating over pollen surface. It attracts insects and protects damage from UV radiation.

Question 44.
Distinguish tenuinucellate and crassinucellate ovules.
Answer:
Tenuinucellate ovule

1. The sporogenous cell is hypodermal
2. It has single layer of nuclear tissue.
3. It has very small nucellus

Crassinucellate ovule

1. These ovules have sub-hypodermal sporogenous cell
2. They have large nucellus.
3. Many layers of cells are seen.

Question 45.
‘Pollination in Gymnosperms is different from Angiosperms’ – Give reasons.
Answer:
In gymnosperms, the ovules are exposed and the pollens are deposited directly on it. Hence the pollution is direct in a gymnosperm. Whereas in angiosperms it is said to be indirect, as the pollens are deposited on stigma or the pistil.

Question 46.
Write a short note on Heterostyly.
Answer:

1. Heterostyly is a mechanism to promote cross-pollination.
2. Different forms of flowers with different lengths of stamen and style.
3. Pollination takes place between organs of same length.

a) Distyly. (Ex. Primula)

• Thrum-eyed flowers have short styles. Anthers of pin have short stamen.
• Anthers of thrum-eyed flowers and stigma of the pin are of the same height (both are long). This helps in effective pollination.

b) Tristyly (Ex. Lythrum)
3 kinds of flowers are there, with respect to the length of style and stamens. Flower of one type can’t pollinate their own type. They pollinate the other 2 types.

Question 47.
Enumerate the characteristic features of Entomophilous flowers.
Answer:
The characteristic features of entomophilous flowers are as follows:

1. Flowers are generally large or if small they are aggregated in dense inflorescence. Example: Asteraceae flowers.
2. Flowers are brightly coloured. The adjacent parts of the flowers may also be brightly coloured to attract insects. For example in Poinsettia and Bougainvillea, the bracts become coloured.
3. Flowers are scented and produce nectar.
4. Flowers in which there is no secretion of nectar, the pollen is either consumed as food or used in building up of its hive by the honeybees. Pollen and nectar are floral rewards for visitors.
5. Flowers pollinated by flies and beetles produce foul odour to attract pollinators.
6. In some flowers, juicy cells are present which are pierced and the contents are sucked by the insects.

Question 48.
Discuss the steps involved in
Microsporogenesis.
Microsporogenesis.
Answer:
Formation of haploid microspores from diploid microspore mother cell by meiosis.

• The primary sporogeneous cells undergo mitosis to form sporogenous tissue. ‘
• Sporogenous tissue functions as microspore mother cells.
• Microspore mother cell divides meiotically to form a tetrad (4 haploid microspores)
• Microspores get separated. They remain free in the anther locule. They develop into pollen grains.
• Microspores are held together by pollinium. Filament (or thread) like part form pollinium is called retinaculum. Through retinaculum pollinia are attached to clip like corpusculum. This structure is called Translator (Y shapled).
  
  

Question 49.
With a suitable diagram explain the structure of an ovule.
Structure of ovule (Megasporangium)
Answer:

• Ovule of ovule (Megasporangium)
• It has a stalk and a body.
• stalk (funiculus) is at the base of ovule. It attaches ovule to the placenta.
• Hilum is the junction (point of attachment) between ovule and funicle.
• In an inverted ovule, the funicle is fused to the body of ovule. Thus a ridge called raphe is formed.
• Body of ovule has central mass of reserve food called nucellus.
• Nucellus is covered by 2 layers, called integuments.
• Integument covers the nucellus completely except at the top. This forms a pore called micropyle.
• Ovule with single integument is called unitegmic.
• At the base of body, nucellus, integument and funicle meet. This is called chalaza.
• Sac like structure in nucellus towards micropylar end is called embryosac (or) female gametophyte. It is formed from functional megaspore of nucellus.
• The nutritive inner intergument layer is called integumentary tapetum or endothelium.
• Tenuinucellate type ovule has hypodermal sporogenous cell. It has single layer of nucellar tissue.
• Crassinucellate type, ovule has subhypodermal sporogenous cell.
• Group of cells between chalaza and embryosac is called hypostase.
• Thick walled cells above micropyle are called epistase.

Question 50.
Give a concise account on steps involved in the fertilization of an angiosperm plant.
Answer:
Steps in the fertilization of Angiosperms
1. Germination of pollen to form pollen tube in the stigma.

• Pollens fall on receptive stigma.
• Compatible pollen germinates to form a tube.
• This is helped by stigmatic fluid in wet stigma and pellicle in dry stigma.
• Compatibility is decided by recognition, rejection protein reaction, between pollen and stigma surface.
• Pollen undergoes hydration. Pollen wall proteins cire released.
• The entire content moves into pollen tube.
• Growth is at the cytoplasmic contents at the tip.
• The remaining part of pollen tube is occupied by a vacuole.
• It is cut off from tip by callose plug.
• The hemispherical, transparent pollen tip of pollen tube is called ‘cap block.
• The “cape block” disappears and the growth of the pollen tube stops.

2. Growth of pollen tube in the style.

• Hollow style glandular canal cells secrete mucilaginous substance. These secretions are nutrition for growing pollen tube. They control compatibility of style and pollen tube.
• In solid style the pollen tube grows through the intercellular space of transmitting tissue. Semisolid style is intermediate between solid and open type.

3. Entry of the pollen tube into the ovule.

• Propgamy – Pollen tube enters through the micropyle.
• Chalazogamy – Pollen tube enters through chalaza.
• Monogamy – Pollen tube enters through integument.

4. Entry of pollen tube into the embryo sac.

• Pollen tube enters embryosac at the micropylar end. It is guided by an obturator.
• Pollen tube enters into one of the synergids.

5. Double fertilization and Triple fusion.

• In Angiosperms, both the male gametes are involved in fertilization, it is called double fertilisation.
• One of the male gametes fuses with the egg nucleus (syngamy). Thus zygote is formed.
• The second gamete migrates to central cell. It fuses with polar nuclei (or) secondary nucleus. Thus primary Endosperm nucleus is formed. This involves the fusion of 3 nuclei so it is called Triple fusion.

Question 51.
What is endosperm. Explain the types. (OR) Write the three fusion of Antisper- mous plant fertilization.
The zygote divides into an endosperm.
The primary Endosperm Nucleus is the regulatory structure. It nourishes the developing embryo.
These types of endosperms are based on the mode of development.

a) Nuclear Endosperm. (Ex.Arachis)

• Primary Endosperm Nucleus undergoes mitosis.
• No cell wall formation.
• A free nuclear condition exists

b) Cellular Endosperm (Ex. Helianthus)

• Primary Endosperm Nucleus divides into 2 nuclei.
• It is followed by wall formation.

c) Helobial Endosperm. (Ex. Vallisneria)

• Primary Endosperm Nucleus moves towards the base of the embryo sac. It divides into 2 nuclei.
• Cell wall is formed. It divides large micropylar chamber into form the small chalazal chamber.
• Nucleus of micropylar chamber divides. The Chalazal chamber nucleus does not divide.

Question 52.
Differentiate the structure of Dicot and Monocot seed.
Answer:
Structure of Dicot seed

1. Two cotyledons
2. Two seeds may be seen
3. The seed coat has outer coat testa and inner tegmen.
4. In pea the cotyledons store the food. In castor the endosperm, stores reserve food.
5. Coleoptile (sheath of plumule) coleorhiza (sheath of radicle) are absent.

Structure of Monocot seed :

1. Only one cotyledon
2. Paddy is one-seeded.
3. Seed is enclosed by husk. The brown membranous seed coat closely adheres to grair
4. Scutellum supplies embryo with food from endosperm through epithelium
5. Coleoptile and coleorhiza are seen.

Question 53.
Give a detailed account of parthenocarpy. Add a note on its significance.
Answer:
In some plants, fruit-like structures may develop from the ovary without the act of fertilization. Such fruits are called parthenocarpic fruits. Invariably they will not have true seeds. Many commercial fruits are made seedless.
Examples: Banana, Grapes, and Papaya. Nitsch in 1963 classified the parthenocarpy into the following types:

1. Genetic Parthenocarpy: Parthenocarpy arises due to hybridization or mutation.
   Examples: Citrus, Cucurbita.
2. Environmental Parthenocarpy: Environmental conditions like frost, fog, low temperature, high temperature etc., induce Parthenocarpy. For example, low temperature for 3-19 hours induces parthenocarpy in Pear. Chemically
3. induced Parthenocarpy: Application of growth-promoting substances like Auxins and Gibberellins induces parthenocarpy.
4. Significance: The seedless fruits have great significance in horticulture.
   • Seedless fruits have great commercial importance.
   • Seedless fruits are useful for the preparation of jams, jellies, sauces, fruit drinks, etc.
   • A high proportion of edible parts is available in parthenocarpic fruits due to the absence of seeds.

12th Bio Botany Guide Asexual and Sexual Reproduction in Plants in Animals Additional Important Questions and Answers

I. Choose the correct answer

Question 1.
Match the following

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

Question 2.
Find the Matching pair
a) Rhizome – Zingiber
b) Corm – Solanum
c) Tuber – Lilium
d) Bulb – Tuber
Answer:
a) Rhizome – Zingiber.

Question 3.
Find the mismatching pair
a) Runner – Centella
b) Sucker – Chrysanthemum
c) Stolon – Fragaria
d) Offset – Bryophyllum
Answer:
d) Offset – Bryophyllum.

Question 4.
Epiphyllous buds are in
a) Chrysanthemum
b) Agave
c) Curcuma
d) Scilla
Answer:
d) Scilla

Question 5.
Eyes of potato are referred to
a) adventitious roots
b) axillary buds
c) terminal buds
d) intercalary buds
Answer:
b) axillary buds

Question 6.
The T-shaped incision is made in grafting.
a) Bud
b) Approach
c) Tongue
d) Crown
Answer:
a) Bud

Question 7.
Plants propagated economically by vegetative propagation
a) Solanum tuberosum
b) Ixora
c) Jasminum
d) Chrysanthemum
Answer:
a) Solanum tuberosum

Question 8.
Steward produced……………plant from phloem
a) Beetroot
b) Carrot
c) Solanum
d) Radish
Answer:
b) Carrot

Question 9.
Arrange from the periphery to centre in another wall.
a) Endothecium, Middle layer, tapetum
b) Tapetum, middle layer, Endothecium
c) Endothecium, tapetum, middle layer
d) Middle layer, endothecium, tapetum
Answer:
a) Endothecium, Middle layer, tapetum

Question 10.
Microspores are held together by pollinium in
a) Hibiscus
b) Calotropis
c) Ixora
d) Datura
Answer:
b) Calotropis

Question 11.
…………… cells are hygroscopic in anther wall.
a) Epidermis
b) Endothecium
c) Middle layer
d) tapetum
Answer:
b) Endothecium

Question 12.
Find the wrong statement
a) Invasive tapetum is peri plasmodial.
b) Amoeboid tapetum is associated with male sterility
c) Middle layer is ephemeral.
d) Epithelium is hygroscopic
Answer:
d) Epithelium is hygroscopic

Question 13.
Find the correct statement
a) Carrot grass causes allergy
b) Bee pollen is an artificial substance.
c) Palynology is the study of honey pollen.
d) Mellitopalynology is the study of pollen grain.
Answer:
a) Carrot grass causes allergy

Question 14.
Not a shape of pollen grain
a) Globose
b) Ellipsoid
c) crescent-shaped
d) Cubical
Answer:
d) Cubical

Question 15.
…………… protects pollen grain from UV
radiation.
a) Sporopollenin
b) pollenkitt
c) Exine
d) callose.
Answer:
b) pollenkitt

Question 16.
Not in exine of pollen grain
a) Cellulose
b) Sporopollenin
c) Pollenkitt
d) Callose
Answer:
d) Callose

Question 17.
………………….. % of angiosperm pollen is liberated in 2 cell stage
a) 50
b) 60
c) 40
d) 30
Answer:
b) 60