All India Aakash Test Series for NEET - 2021 · All India Aakash Test Series for NEET-2021 Test-2 (Code-A)_(Hints & Solutions) Aakash Educational Services Limited - Regd. Office:

Test-2 (Code-A)_(Answers) All India Aakash Test Series for NEET-2021

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 1/16

All India Aakash Test Series for NEET - 2021

Test Date : 20/09/2020

ANSWERS

1. (1) 2. (4) 3. (2) 4. (1) 5. (1) 6. (2) 7. (2) 8. (4) 9. (1) 10. (2) 11. (2) 12. (1) 13. (4) 14. (3) 15. (4) 16. (3) 17. (4) 18. (3) 19. (2) 20. (4) 21. (3) 22. (2) 23. (1) 24. (1) 25. (3) 26. (3) 27. (3) 28. (2) 29. (1) 30. (3) 31. (1) 32. (3) 33. (3) 34. (1) 35. (1) 36. (3)

37. (2) 38. (1) 39. (1) 40. (3) 41. (2) 42. (4) 43. (3) 44. (2) 45. (4) 46. (3) 47. (3) 48. (2) 49. (3) 50. (1) 51. (3) 52. (1) 53. (4) 54. (4) 55. (3) 56. (4) 57. (4) 58. (4) 59. (4) 60. (1) 61. (3) 62. (1) 63. (4) 64. (4) 65. (2) 66. (4) 67. (2) 68. (3) 69. (4) 70. (4) 71. (2) 72. (2)

73. (1) 74. (4) 75. (4) 76. (4) 77. (1) 78. (4) 79. (1) 80. (3) 81. (2) 82. (4) 83. (1) 84. (3) 85. (4) 86. (3) 87. (4) 88. (4) 89. (4) 90. (3) 91. (1) 92. (4) 93. (1) 94. (2) 95. (3) 96. (1) 97. (4) 98. (2) 99. (4) 100. (2) 101. (2) 102. (4) 103. (1) 104. (3) 105. (3) 106. (4) 107. (4) 108. (3)

109. (1) 110. (3) 111. (1) 112. (2) 113. (1) 114. (2) 115. (4) 116. (3) 117. (2) 118. (4) 119. (1) 120. (3) 121. (3) 122. (4) 123. (1) 124. (3) 125. (3) 126. (4) 127. (3) 128. (3) 129. (2) 130. (1) 131. (1) 132. (Deleted) 133. (3) 134. (1) 135. (2) 136. (4) 137. (2) 138. (1) 139. (3) 140. (2) 141. (4) 142. (4) 143. (3) 144. (3)

145. (3) 146. (3) 147. (4) 148. (2) 149. (3) 150. (4) 151. (3) 152. (2) 153. (2) 154. (2) 155. (1) 156. (4) 157. (2) 158. (3) 159. (3) 160. (2) 161. (1) 162. (3) 163. (2) 164. (3) 165. (2) 166. (2) 167. (3) 168. (1) 169. (2) 170. (4) 171. (4) 172. (4) 173. (4) 174. (2) 175. (2) 176. (3) 177. (3) 178. (2) 179. (4) 180. (3)

TEST - 2 (Code-A)

All India Aakash Test Series for NEET-2021 Test-2 (Code-A)_(Hints & Solutions)


[PHYSICS] 1. Answer (1)

Hint: mvrqB

Sol. 212

eV mv

122

eVvm

122

m eVrqB m

12

22

mVreB

.

2. Answer (4)

Hint: . F i l B

Sol. . sin F i lB

F will be maximum when = 90°. 3. Answer (2)

Hint: M B

Sol. M = r2I

= MBsin60°

2 3

2

r IB

23

2

r IB .

4. Answer (1) Hint & Sol.: Moving charge can produce both

electric and magnetic field 5. Answer (1)

Hint & sol.: 0

2

I

Br

1B

r

6. Answer (2) Hint & sol.:

20

32 2 22

InrB

r x

20

32 2 22 8

InrB

r r

2

032 27

InrB

r

054

InBr

7. Answer (2) Hint: At neutral point magnetic field due to

conductor, balance the horizontal magnetic field of earth.

Sol.:

02

IBr

02

H

IB

r

772 10 10 2 10

r

r = 10 m. 8. Answer (4) Hint: Magnetic field of the centre of the circular

loop having current I and number of turn n will be 0 I2

nr

Sol.:

HINTS & SOLUTIONS

Test-2 (Code-A)_(Hints & Solutions) All India Aakash Test Series for NEET-2021


B = B1 – B2

0 1 1 0 2 2

1 22 2

I n I nr r

0 02 2

10 24 15 182 25 10 2 15 10

= 240 × 20 – 9000

= 4800 – 9000

= 4200 T

9. Answer (1)

Hint : Current sensitivity = NBAk

Sol.:

A

B

Current senstivityCurrent senstivity

A

B

NN

200100

(current sensitivity)A = 2(Current sensitivity)B 10. Answer (2)

Hint :

G G

sG

R IRI I

Sol.: 60020

GR

= 30

3

330 20 1010 20 10

SR

60010000 20

SR

Rs 0.06 11. Answer (2) Hint: To not enter in the region x > 3a deflection of

particle when it will reach at x = 3a, should be greater than 90°.

Sol.:

Rsin d Rsin90 2a

2mv aqB

2aqBvm

max2aqBV

m

12. Answer (1) Hint : magnetic field due to semi-circular arc at its

center is 02 I

r.

Magnetic field due to semi-infinite wire at distance

d is 04

Id

.

Sol.:

B = B1 + B2 + B3

0 0 0ˆ ˆ ˆ4 4 4

I I I

k k kr r r

0 ˆ4

I

kr

.

13. Answer (4) Hint: Magnetic field due to current carrying long

wire at a distance B is 02

Ir

.

Sol.: 1 2

2 1

B rB r

2

93

BB

2 3

BB

14. Answer (3) Hint: Magnetic field at the centre of circular wire is

02 I

R and magnetic moment of the coil is IR2 .

Sol.: 0

22

IB RM I R

032

R



15. Answer (4) Hint and Sol. : When charged particle moves in

uniform magnetic field the kinetic energy always remains constant because magnetic field does no work.

16. Answer (3) Hint: Use Ampere’s circuital law. Sol.: For r < a

2

1 2 3

I aI xa

1 9

II

0 1.

B dl I

01 2

3 9

IaB

01 6

IBa

For r > a I2 = I 0.

B dl I

2 0322aB I

02 3

I

Ba

0

1

02

6

3

IB a

IBa

12

17. Answer (4) Hint: magnetic field inside the current carrying

toroid 02

NIBr

.

Sol.: 7

24 10 3000 10

2 25 10B

B = 24 mT.

18. Answer (3)

Sol.: 2

qBf

m

19

271.6 10 1.4

2 3.14 1.67 10

= 2.2 × 107 s–1 19. Answer (2) Hint: Use Ampere circuital law

Sol.: 0 enc.B dI I

enc 4 1I = 3 A

0. 3

B dI .

20. Answer (4) Hint: Work done by magnetic field be zero. Sol.: By work energy theorem

net W K

22 21 6 82E BW W m

Q x0 E0 = 50m

00

50mxqE

00

50xE

.

21. Answer (3) Hint and sol.: The direction of magnetic field lines

of force produced by passing a direct current in conductor is given by Right hand palm rule.

22. Answer (2) Hint: Magnetic field due to bar magnetic at its axial

position is 03

24

Md

and on its equatorial position

034

Md

.

Sol.:



|B1| = |B3| = 2|B2|

|B2| = 102

= 5 T

|B2| = 10 T 23. Answer (1) Hint: B = 0(H + I) Sol.:

Intensity of magnetisation Magnetic moment Volume

66

30 10

I

6

510 2 105

I

7 550004 10 2 104

B

B = 5 × 10–4 + 8 × 10–2 = 0.2517 T. 24. Answer (1)

Hint: 2H

ITMB

Sol.: 2H

ITMB

0 2H

ITMB

2 IT

MB

cos

HBB

cos2H

ITMB

122 cos

H

ITMB

120 cosT T

2 20 cos T T .

25. Answer (3) Hint & sol.: r = (1 + m)

= (1 + 0.725) = 1.725. 26. Answer (3) Hint and sol.: A magnet attract ferromagnetic

substance strongly and paramagnetic substance weakly. But it repel diamagnetic substance.

27. Answer (3) Hint: Resultant magnetic moment will be equal to

vector sum of magnetic moment of all the magnets.

Sol.:

Net ˆ ˆ ˆ ˆ2 cos 60 3 cos 60 2 sin60 M Mi M i M i M j

ˆ3 sin 60 M j

net3ˆ ˆ ˆsin602

MM M M i i M j

net3 3ˆ ˆ2 2

M MM i j

net| | 9 32

MM

= 3 M .

28. Answer (2) Hint and sol. When all the molecule in a magnet arrange

themselves in the direction of magnetic field the condition is called saturation.

29. Answer (1) Hint: Use work energy theorem. Sol.: Wnet = k –U = k –(Uf – Ui) = kf – ki –(–MB cos)= kf kf = 80 J.



30. Answer (3) Hint: In stable equilibrium position angle between

net magnetic moment and magnetic field will be zero.

Sol.:

3tan MM

= 60. 31. Answer (1)

Hint : At a position where angle of dip is . The horizontal component of earth’s magnetic field BH = Bcos and vertical component of earth’s magnetic field BV = Bsin.

Sol. :

sin30cos60

v A

H B

B BB B

= 1 : 1.

32. Answer (3)

Hint : Rate of heat dissipation dHdt

= I2R = 2E

R.

Sol.: E = Bvl

2 2 2dH B v

dt R

2dH vdt

2 14dH dHdt dt

33. Answer (3)

Hint: ddt

.

Sol.: = BAcos

= Br2

2d B rdt

2 drB rdt

4 Ba

34. Answer (1)

Hint: . B A

Sol.: 4 4ˆ ˆ ˆ20 9 10 .10 10 i k k

= –90 × 10–8 = –900 ×10–9 || = 900 nWb

35. Answer (1)

Hint: . B A .

Sol.: Magnetic field produced by coil having current will be perpendicular to area vector of other coil = BAcos90°

= 0. Change in flux in other coil will be zero therefore

no current will flow in other coil. 36. Answer (3)

Hint: sPdIM

dt

Sol.: 10( 0)2000 mI

t

Im = 200 × 10–3 Im = 0.2 A. 37. Answer (2) Hint: Heat developed in R1 will be equal to energy

stored in the inductor

Sol.: 212

H U LI

2

EIR

2

22

12

LEHR

.

38. Answer (1)

Hint: . effv B l

Sol.: leff = 2lcos45



= 2l 2Blv By right hand palm rule P will be at higher

potential. 39. Answer (1) Hint and sol.: By Lenz’s law induce current

produced in ring oppose its cause. 40. Answer (3) Hint and Sol.: The mutual inductance of the pair

of coil depends upon position and orientation of the coil.

41. Answer (2) Hint.: Rotational emf. induced in a rod of length l

inside the uniform perpendicular magnetic field B

rotating about one end is 2

2B l

.

Sol. :

22(i)

2O Cl B

V V

= 2l2B

2

(ii)2O Al BV V

From (i) and (ii)

2

222

A C

l BV V l B

23

2

l B .

42. Answer (4) Hint : Use KVL Sol. : Case–I

a bdIV V L IRdt

12 = L + 3R …(i) Case–II

a bdIV V L IRdt

3 2 (ii) L R 15 = 5R R = 3

3 = – L + 6 L = 3 H. 43. Answer (3) Hint: Voltage across the inductor will be equal to

voltage across the 10 resistor. Sol.: At t = 0 current in the inductor will be zero but

rate of change of current will be maximum

315

I

1 A5

I

dIL I Rdt

1 105

dILdt

2 A/s5

dIdt

44. Answer (2)

Hint: ddt

Sol.: For 0 < t < 2T , d

dtis constant and negative,

ddt

which will be positive constant.

For 2T < t < T d

dt is constant and positive,

ddt

which will be negative constant.

45. Answer (4)

Hint: andd idt R

Sol.: 24 12 7t t

8 12d tdt

I will be zero, when ddt will be zero

Hence 8 12 0d tdt

12 1.5 s8

t



[CHEMISTRY]

46. Answer (3) Hint.: Chemisorption is specific in nature Sol.: Physical adsorption is not specific in nature. 47. Answer (3) Hint: Adsorption is exothermic in nature Sol. Higher is the temperature, less will be the

adsorption 48. Answer (2) Hint: Gas of highest critical temperature will be

adsorbed most easily Sol. CO2(g) has highest critical temperature

among the given gases. So, it is most easily adsorbed.

49. Answer (3) Hint: When physical state of catalyst is different

from physical state of reactants and products then it is an example of heterogeneous catalysis.

Sol.: 4NH3(g) + 5O2(g) Pt(s) 4NO(g) + 6 H2O(g) is an example of heterogeneous catalysis

50. Answer (1)

Hint: Sucrose Invertase Glucose + fructose 51. Answer (3) Hint: Paint is an example of sol Sol.: Paints contain solid as dispersed phase and

liquid as dispersion medium 52. Answer (1) Hint: For soaps the CMC value is 10–4 to 10–3 mol

L–1 53. Answer (4) Hint: CdS and charcoal are negatively charged sol Sol.: Al2O3 xH2O is a positively charged sol 54. Answer (4) Hint: The emulsifying agent forms an interfacial

film between suspended particles and the medium.

Sol.: The principal emulsifying agents for oil in water emulsion are proteins, gums, natural and synthetic soaps etc.

55. Answer (3) Hint: mmol of electrolyte required for coagulation

of 1000 mL of sol is called coagulation value. Sol. For 20 mL, mmol of NaCl required = 1 Hence, for 1000 ml, mmol of NaCl required = 50

56. Answer (4) Hint: Lyophilic sols are more stable than lyophobic

sols Sol. Smaller is the size and lesser the viscosity,

faster is the Brownian movement 57. Answer (4) Hint: Boiling, persistant dialysis and mixing of two

oppositely charged sols resulted in coagulation Sol. Peptization is the process of converting a

precipitate into colloidal sol by shaking it with dispersion medium in the presence of a small amount of electrolyte.

58. Answer (4) Hint.: Magnetite is Fe3O4 59. Answer (4) Hint: Removal of unwanted material from the ore

is known as concentration of an ore Sol.: Calcination is not a method of concentration. 60. Answer (1) Hint: Roasting is carried out for sulphides ores to

convert into oxide Sol.: 2ZnS + 3O2 2ZnO + 2SO2 61. Answer (3)

Hint: Reaction in which metal oxide on reaction with metal sulphide forms metal is self reduction

62. Answer (1) Hint: Vapour phase refining is used for refining of

Nickel.

Sol: Ni + 4CO 330 – 350K Ni(CO)4

Ni(CO)4 450– 470K Ni + 4CO

63. Answer (4) Hint: Silver and gold is concentrated by leaching

using aqueous NaCN solution. Sol.:

64. Answer (4) Hint: Zone refining method is very useful for

producing semiconductor and other metals of very high purity



65. Answer (2) Hint: Depressant, selectively prevents ZnS from

coming to the froth but allows PbS to come with the froth.

Sol.: NaCN is used to convert ZnS into water soluble complex Na2[Zn(CN)4]

66. Answer (4) Hint: Bauxite ore contains impurities of SiO2, Iron

oxides and titanium oxide (TiO2) 67. Answer (2) Hint: Ellingham diagram is based only on the

thermodynamics concepts. Sol.: Ellingham diagram does not say about the

kinetics of the reduction processes. 68. Answer (3) Hint: In lower temperature range (500 K – 800 K) in

blast furnace, Fe2O3 is mainly reduced by CO. Sol.: At 900–1500 K (higher temperature range in

the blast furnace) reactions are C + CO2 2CO FeO + CO Fe + CO2 69. Answer (4) Hint: Cu2S : copper glance 70. Answer (4) Hint: NaNO3 is called as chile saltpetre Sol.: KNO3 is known as Indian saltpetre 71. Answer (2) Hint: AgCl(s) is white compound Sol.:

Zn(OH)2(s) is white compound Fe2O3 · xH2O(s) is brown compound [Cu(NH3)4]2+ is deep blue compound 72. Answer (2) Hint: NO is neutral oxide N2O3, N2O4 and N2O5 are acidic in nature Sol.: N2O3 is blue coloured solid which is acidic in

nature 73. Answer (1)

Hint: Hypophosphorous acid is also known as phosphinic acid Sol.: Phosphinic acid is H3PO2

74. Answer (4) Hint: In solid state, PCl5 exists as an ionic solid [PCl4]+ [PCl6]– Sol.: The hybridization of P in [PCl4]+ is sp3

75. Answer (4) Hint: Species with central atom having 1 lone pair

and 4 bond pair of electrons acquire see saw shape

Sol.: SF6(g) has octahedral shape. 76. Answer (4) Hint: Generally larger is the bond length, smaller

is the bond enthalpy Sol.: I2 has least bond enthalpy among the

halogen molecules. 77. Answer (1) Hint: Weakest acid has highest pKa value. Sol.: Order of pKa; HF > HCl > HBr > HI 78. Answer (4) Hint: 3Cu + 8HNO3(dilute) 3Cu(NO3)2 + 2NO + 4H2O 79. Answer (1) Hint: Ionic character of metal halides decreases

as the size of halide increases Sol.: Ionic Character MF > MCl > MBr > MI (Where M is a monovalent metal) 80. Answer (3) Hint: Compound containing hydrogen on reaction

with Cl2, forms HCl. Sol. 2 Ca(OH)2 + 2 Cl2 Ca(OCl)2 + CaCl2

+ 2H2O 81. Answer (2) Hint: Halous acid of only chlorine exist. Sol.: HOBrO is least likely to exist 82. Answer (4) Hint: 6XeF4 + 12H2O 4Xe + 2XeO3 + 24HF + 3O2

83. Answer (1) Hint: Molecule with 1 lone pair and 6 bond pairs of

electrons around central atom has distorted octahedral shape.

Sol.

XeF6 :



84. Answer (3) Hint: XeF2 and XeF4 are fluorinating reagent. Sol.: 143K

4 2 2 6 2XeF + O F XeF + O

2Xe:F = 1: 202 6573 K, 60 – 70 barXe(g) + 3F (g) XeF (s)

85. Answer (4) Hint: Depending upon the relative excess amount

of NH3 and Cl2, different products are formed. Sol.: NH3 + 3Cl2 (excess) NCl3 + 3HCl 8 NH3 (excess) + 3Cl2 6NH4Cl + N2 86. Answer (3) Hint: Catalyst used in Deacon’s process is CuCl2 Sol. Catalyst used in contact process is V2O5

Catalyst used in Haber’s process is Fe2O3 with K2O and Al2O3. Catalyst used in ostwald’s process is platinized asbestos.

87. Answer (4) Hint: Due to presence of H-bonding, Melting point

of NH3 is exceptionally high Sol.: Order of melting point of NH3 > SbH3 > AsH3 > PH3

88. Answer (4) Hint: The bond enthalpy of N2 is high Sol.: 3 22 NaN 2 Na + 3N (pure)

At high temperature, with some metals it predominantly form ionic nitrides e.g.

2 36Li + N 2Li N

2 3 23Mg N Mg N

89. Answer (4) Hint: H2SO4 gives charring action with

carbohydrates Sol.: Ca3P2 + 6H2O 2PH3(g) + 3Ca(OH)2

(A)

2PH3(g) + 3CuSO4 Cu3P2 + 3H2SO4 (B) (C)

PH3 is a colorless gas with rotten fish smell. It is slightly soluble in water. It explodes in contact of trace amount of HNO3.

90. Answer (3) Hint: In all oxoacids of phosphorous one P = O

bond is present for every phosphorous atom. Sol.: Formula Structure

Phosphonic Acid H3PO3

Orthophosphoric Acid H3PO4

Hypophosphoric acid H4P2O6

Pyrophosphoric acid H4P2O7

[BIOLOGY]91. Answer (1) Hint : Recessive traits are expressed in

homozygous condition only

Sol. : Terminal flower position, constricted pod

– Recessive traits

Yellow seeds, Violet flower

– Dominant traits

92. Answer (4) Hint : Yellow pod is recessive trait of pea plants. Sol. : To produce recessive feature in F1

generation, both the parents must have at least a recessive allele.

Hence YY × yy

Yy all green pods Will not have any progeny with yellow pods. 93. Answer (1) Hint : Formula to calculate number of different

types of gametes produced is = 2n. Sol. : Number of heterozygous locus ‘n’ = 3. Hence 2n = 23 = 8 types of gametes. 94. Answer (2) Hint : Flower colour in snapdragon plant is an

example of incomplete dominance.



Sol. : In the given cross

F2 phenotypic ratio is 1 : 2 : 1 95. Answer (3) Hint : To have child with O blood group both the

parent must have one IO allele. Sol. :

96. Answer (1) Hint : A modified allele can produce normal or less

efficient enzyme. Sol. : If modified allele is produced due to silent

mutation it will have same phenotype and it can be a dominant allele.

97. Answer (4) Sol. : Punnett square is the graphical

representation of possible genotypes of offsprings of a genetic cross given by R. C. Punnett.

98. Answer (2) Hint : In Mendelian dihybrid cross 8 offsprings

have genotype heterozygous for single trait.

Sol. : Total number of offsprings heterozygous for both the traits = 4.

99. Answer (4) Hint : In test cross an individual with recessive

phenotype is crossed with individual with dominant phenotype.

Sol. : In 4 O’ clock plant, flower colour show incomplete dominance. So for this plant, if plant is showing red coloured flowers it means plant is homozygous dominant for red colour.

100. Answer (2) Sol. : Term linkage was coined by T. H. Morgan. 101. Answer (2) Hint : Pleiotropic gene controls multiple

phenotypes. Sol. : In pea starch synthesis gene shows

incomplete dominance. 102. Answer (4) Hint : Kernel colour in wheat show polygenic

inheritance. Sol. : Intermediate red kerneled wheat = AaBb White kerneled wheat = aabb AaBb × aabb

ab

AB AaBb Intermediate red wheat

Ab Aabb Light red wheat

aB aaBb Light red wheat

ab aabb White kernel wheat

Phenotypic ratio = 1 : 2 : 1 Genotypic ratio = 1 : 1 : 1 : 1 Thus only statement (ii) is correct 103. Answer (1) Sol. : Chromosomal theory of inheritance was

given by Sutton and Boveri. 104. Answer (3) Hint : An experimental material must be easily

grown and have short life span. Sol. : Both Drosophila and garden pea have short

life span. But only Drosophila can be grown in laboratory

and have distinguishable opposite sexes. 105. Answer (3) Hint : Recombination frequency (RF) is directly

proportional to distance between genes.



Sol. : Recombination or crossing over occurs between homologous chromosomes. Higher RF value indicates that genes are distantly situated on a chromosome not on separate chromosomes.

106. Answer (4) Sol. : In moth, females produce two types of

gametes i.e. heterogametic. 107. Answer (4) Hint : Infra red radiations due to high wavelength

have very low energy. Sol. : Being low energy radiations, infra red

radiations are not capable to directly damage the DNA.

108. Answer (3) Hint : Male cannot be carrier for X-linked recessive

disorders. Sol. : For haemophilia male can either be

diseased or normal

109. Answer (1) Sol. : Myotonic dystrophy is an autosomal

dominant disorders. 110. Answer (3) Hint : Colour blindness is X-linked recessive

disorder. Sol. : If a female is suffering from Colour blindness

then she must have genotype 44 + XCXC. This female can pass Colour blindness gene to her daughters as well as sons. Female ‘X’ must have received Colour blindness gene from both of her parents.

111. Answer (1) Hint : In sickle cell anaemia HbS peptide have

valine in place of glutamic acid. Sol. : m-RNA HbA peptide – 5 – GAG – 3 HbS peptide – 5 – GUG – 3 112. Answer (2) Hint : Phenylketonuria gene controls multiple

phenotypic expressions. Sol. : Gene responsible for phenylketonuria is a

pleiotropic gene. 113. Answer (1) Sol. : Four phenotypes A, B, AB and O are

possible for blood group in human population. 114. Answer (2) Sol. : Turner female have 44 + XO genetic

complement.

115. Answer (4) Hint : Cytidine is a nucleoside of RNA molecule. Sol. : A cytidine contains a ribose sugar a

pyrimidine nitrogenous base (6 membered) and N-glycosidic bond.

But since it’s a nucleoside it lacks phosphoester bond.

116. Answer (3) Sol. : Guanine is a purine base, rest three are

pyrimidine nitrogenous bases. 117. Answer (2) Hint : Watson and Crick were the first to propose

the idea of base complimentarity in DNA. Sol. : Watson and Crick were the first to propose

the base pairing between two strands of DNA. 118. Answer (4) Sol. : DNA packaging in prokaryotes involves non-

histone basic protein polyamines. 119. Answer (1) Hint : Reverse transcriptase catalyse the

synthesis of DNA over RNA. Sol. : Reverse transcriptase is found in retro

viruses and is RNA dependent DNA polymerase. It follows principle of complementarity. Viruses have very few enzymes such as reverse transcriptase, neuraminidase etc.

120. Answer (3) Sol. : Hershey and Chase proved that DNA act as

genetic material. 121. Answer (3) Hint : RNA transmits genetic information from

DNA to protein. Sol. : RNA is better material for transmission of

genetic information than DNA. 122. Answer (4) Hint : DNA polymerase has both exonuclease and

polymerase activity. Sol. : DNA polymerase is capable to synthesise

DNA as well as remove RNA primer. Joining of Okazaki fragments is a function of ligase enzyme.

123. Answer (1) Hint : m-RNA have bases sequences

complimentary to the non-coding DNA strand. Sol. : m-RNA transcribed from the given DNA

strand will be.

5th codon is stop codon i.e. UGA, hence only four

amino acids will be coded by this m-RNA.



124. Answer (3) Hint : Translation occurs inside cytoplasm. Sol. : Charging of t-RNA a step of translation occur

inside cytoplasm. 125. Answer (3) Hint : Most of the rRNA are synthesised by RNA

polymerase I. Sol. : RNA polymerase I : 5.8 S, 18 S, 28 S rRNA RNA polymerase III : 5 S rRNA RNA polymerase II does not catalyse synthesis of

any rRNA. 126. Answer (4) Hint : VNTR is polymorphic DNA. Sol. : VNTR are/have Size 0.1 to 20kb Found in eukaryotes Non expressed part of DNA 127. Answer (3) Sol. : Semiconservative mode of DNA replication

was experimentally verified by Meselson and Stahl using heavy nitrogen i.e. N15.

128. Answer (3) Hint : Snurp catalyse splicing process. Sol. : Splicing i.e. removal of introns is not found

in prokaryotes. 129. Answer (2) Hint : is the initiation factor of transcription. Sol. : (2) is holoenzyme. Only 2 is

considered as core enzyme. 130. Answer (1) Hint : There are no t-RNAs for stop codons. Sol. : UAG is a stop codon for which there is no t-

RNA. 131. Answer (1) Hint : -galactosidase hydrolyses lactose into

glucose and galactose. Sol. : The gene lac z codes for enzymes

-galactosidase. 132. Deleted 133. Answer (3) Hint : Repeated sequences are non-coding

sequences. Sol. : A very large portion of human genome

contain non coding sequences or repetitive DNA.

134. Answer (1) Sol. : Chromosome 1 has the most genes i.e.

2968. 135. Answer (2) Hint : UTRs are found towards 5 and 3 end. Sol. : UTRs are found toward 5 end before start

codon and 3 end after stop codon in m-RNA. 136. Answer (4) Hint : Identify a bacterial disease. Sol. : Gonorrhoea is caused by bacteria Neisseria

gonorrhoeae. Genital herpes – Herpes simplex virus Genital warts – Human papilloma virus Hepatitis B – Hepatitis B virus 137. Answer (2) Hint : It is called fertile period. Sol. : During periodic abstinence, couples abstain

from coitus from day 10-17 of menstrual cycle, when ovulation is expected.

138. Answer (1) Hint : Identify a copper releasing IUD. Sol. : Cu ions present in copper releasing IUDs

suppress sperm motility and fertilizing capacity of sperms. Progestasert and LNG-20 are hormone releasing IUDs.

139. Answer (3) Hint : Identify the hormone released by growing

follicles. Sol. : Progestogens prevent ovulation and

implantation and alter the quality of cervical mucus which does not let sperms meet the ovum. Estrogen alone is not effective to carry out these functions.

140. Answer (2) Hint : Identify a technique. Sol. : Infertile childless couples could be assisted

to have children through special techniques known as assisted reproductive technologies (ART). A popular method is test tube baby programme.

MTP – Medical Termination of Pregnancy RCH – Reproductive and Child Health Care STI – Sexually Transmitted Infection 141. Answer (4) Hint : Amniotic fluid contains foetal cells. Sol. : In amniocentesis, amniotic fluid around the

developing fetus is taken to analyse fetal cells to check for genetic disorders.



142. Answer (4) Hint : Not a goal of RCH. Sol. : Unprotected sexual co-habitation will lead to

more pregnancies, hence it encourages population growth.

143. Answer (3) Hint : Natural methods involve lactational

amenorrhea. Sol. : Natural methods of contraception work on

the principle of avoiding chances of meeting between ova and sperms. IUDs are one of the most widely used contraceptives in India.

144. Answer (3) Hint : Ovum donor is required. Sol. : Gamete Intra Fallopian Transfer involves

transfer of an ovum collected from a donor into the fallopian tube of another female who cannot produce one, but can provide suitable environment for fertilisation and further development.

145. Answer (3) Hint : MTP Sol. : Complications of reproductive tract

infections, if not treated early include pelvic inflammatory disease, abortions, still births, ectopic pregnancies, infertility or even cancer of reproductive tract. Haemophilia is a genetic disease and is not related with RTIs.

146. Answer (3) Hint : Viral STIs are generally incurable. Sol. : Hepatitis B, HIV and genital herpes are

incurable. Family planning programmes in India were launched at national level to attain total reproductive health as social goal.

147. Answer (4) Hint : Non-steroidal pill. Sol. : ‘Saheli’ the new oral contraceptive for

females contains a non-steroidal preparation – centchroman. It was developed by CDRI, Lucknow. It’s a once a week pill with very few side effects and high contraceptive value.

148. Answer (2) Hint : Chemical contraceptives include foams,

jellies etc. Sol. : Femidoms, Nirodh and chemical

contraceptives can be placed by the user itself. IUDs and implants are inserted by doctors or expert nurses. Injections are also administered by healthcare providers.

149. Answer (3) Hint : First trimester. Sol. : MTP is considered relatively safe during the

first trimester i.e. upto 12 weeks of pregnancy, second trimester abortions are much more riskier.

150. Answer (4) Hint : Zygote is single celled. Sol. : The zygote or early embryos formed after in-

vitro fertilisation are transferred into fallopian tube (ZIFT) and embryos with more than 8 blastomeres into the uterus (IUT), to complete its further development.

151. Answer (3) Hint : Vasectomy and tubectomy Sol. : Surgical methods also called sterilisation are

generally considered as a terminal methods to prevent any more pregnancies. These techniques are highly effective but their reversibility is very poor.

152. Answer (2) Hint : 2 percent. Sol. : According to 2011 census report, the

population growth rate was less than 2 percent i.e., 20/1000/year, a rate at which our population could increase rapidly.

153. Answer (2) Hint : Reducing atmosphere. Sol. : The early atmosphere was reducing in

nature containing CH4, NH3, H2, etc. 154. Answer (2) Hint : Swan neck flask experiment. Sol. : Louis Pasteur disproved the theory of

spontaneous generation which stated that life arose from decaying and rotting matter like straw, mud etc.

155. Answer (1) Hint : Selection by nature. Sol. : According to Darwin, fitness is the end result

of the ability to adapt and get selected by nature. 156. Answer (4) Hint : Hypothesis on chemical origin of first life. Sol. : Experimental evidence of chemical evolution

was given by Urey and Miller. This hypothesis was given by Oparin and Haldane who proposed that first form of life could have come from pre-existing non-living organic molecules.



157. Answer (2) Hint : Biogenesis. Sol. : Water vapour condensed to form water

bodies on earth. Thus, first forms of life were probably unicellular and originated in water environment only.

158. Answer (3) Hint : They evolved oxygen. Sol. : First mammals were like shrews. Plants

invaded land first and released oxygen which was followed by arrival of animals.

159. Answer (3) Hint : Reproductive fitness. Sol. : The pre-existing advantageous mutations

when selected by nature lead to appearance of new phenotypes. This leads to speciation. Variations occurring due to various reasons result in changed frequency of genes and alleles in future generation.

160. Answer (2) Hint : Its height was 20 feet. Sol. : Dinosaurs were land reptiles.

Tyrannosaurus rex was about 20 feet in height and had huge fearsome dagger like teeth.

161. Answer (1) Hint : Ichthyosaurs. Sol. : Most dinosaurs probably evolved into birds.

Coelacanth which was earlier thought to be extinct was the ancestor of modern day frogs and salamanders.

162. Answer (3) Hint : Convergent evolution. Sol. : The given example shows convergent

evolution between placental mammals and Australian marsupials. Spotted cuscus (Australian marsupial) and Lemur (placental mammal) show convergent evolution.

163. Answer (2) Hint : Abrupt evolutionary change. Sol. : Hugo de Vries worked on evening primrose

and brought forth the idea of mutation – large differences arising suddenly in a population. According to him, mutations cause evolution.

Concept of branching descent was given by Darwin. Genetic drift involves a sudden changes.

164. Answer (3) Hint : Directional selection. Sol. : Selection of one type of moth (melanised)

against non-melanised moth due to camouflage represents selection of one type of species while the other is eliminated. Thus it represents directional selection

165. Answer (2) Hint : Modification of axillary branch. Sol. : Divergent evolution is based on homology.

Thorns of Bougainvillea and tendrils of Cucurbita are modification of axillary branches.

Tendrils of pea (Pisum) and spines of Cactus are modified leaves.

166. Answer (2) Hint : Homology indicates common ancestry. Sol. : When the same structure develops along

different direction due to adaptations to different needs, it results in divergent evolution and these structures are homologous.

167. Answer (3) Hint : Common ancestor of cabbage and broccoli. Sol. : Artificial selection of different parts of wild

mustard was done to make broccoli, kohlrabi, cauliflower etc.

Biston carbonaria was selected due to camouflage during industrial melanism, Darwin’s finches exemplify adaptation according to needs.

168. Answer (1) Hint : Sauropsids evolved into thecodonts. Sol. : Crocodiles and birds evolved from

thecodonts. Turtles, Lizards, Snakes and Tuataras evolved

from sauropsids. 169. Answer (2) Hint : Variations are small. Sol. : According to Darwin, evolution is not a

directionless process. It is a gradual process based on chance events in nature. According to Hugo de Vries, mutations cause evolution which are sudden and directionless.

170. Answer (4) Hint : Adaptive convergence. Sol. : Flying phalanger shows adaptive

convergence with flying squirrel. Wombat, Bandicoot and sugar glider exhibit adaptive radiation.



171. Answer (4)

Hint : Adaptive radiation.

Sol. : Darwin’s finches evolved due to adaptive radiation on Galapagos island. Others are examples of evolution by anthropogenic action.

172. Answer (4)

Hint : Heterozygotes are 2 pq.

Sol. : A population in Hardy Weinberg equilibrium is represented by binomial expansion of (p + q)2 i.e. p2 + 2pq + q2 where p2 represents homozygous dominant, 2pq represents heterozygous and q2 represents homozygous recessive individuals.

173. Answer (4)

Hint : Embryology.

Sol. : Ernst von Baer disapproved the theory of Ernst Haeckel which stated that certain features during embryonic stage are common to all vertebrates, that may be absent in adult. Reproductive fitness was a concept given by Darwin.

174. Answer (2)

Hint : Connecting link between ape and man.

Sol. : Australopithecines probably lived in East-African grasslands and evidence suggests that they ate fruits. Hominids were first human like beings. Chimpanzees were apes.

175. Answer (2)

Hint : Hominids

Sol. : Cranial capacities.

Homo erectus – 900 cc

Homo habilis – 650-800 cc

Neanderthal man – 1400 cc

Cro magnon man – 1650 cc

176. Answer (3) Hint : Raisen district Sol. : Cave paintings by pre historic humans can

be seen at Bhimbetka rock shelter in Raisen district of Madhya Pradesh.

177. Answer (3) Hint : Bryophytes lack vessels. Sol. : Bryophytes evolved from chlorophyte

ancestors during Paleozoic era. Tracheophyte ancestors gave rise to Zosterophyllum in Silurian period.

178. Answer (2) Hint : Convergent evolution. Sol. : Marsupials in Australia have eutherian look

alikes with superficially similar adaptation. Eutherians complete their embryonic development in uterus whereas marsupials are born in under develop form and complete their development in an external pouch. Superficial similar adaptations are due to convergent evolution. Others are examples of divergent evolution.

179. Answer (4) Hint : Unit of evolution. Sol. : Micro evolution is the evolution on smallest

scale that occurs due to change in allele frequencies in a population over generation. Genetic variations refer to differences among individuals in the composition of their genes or other DNA sequences.

180. Answer (3) Hint : Chance event. Sol. : Genetic drift operates mostly on small

populations. It is a chance event that can cause allele frequencies to fluctuate unpredictably from one generation to next.

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