All India Aakash Test Series for NEET - 2021...Test-4 (Code-A)_(Answers) All India Aakash Test Series for NEET-2021 Aakash Educational Services Limited - Regd. Office: Aakash Tower,

Test-4 (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 : 06/12/2020

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

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

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

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

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

TEST - 4 (Code-A)

All India Aakash Test Series for NEET-2021 Test-4 (Code-A)_(Hints & Solutions)


[PHYSICS]

1. Answer (4) Momentum is a vector quantity, Its value depends

on magnitude as well as direction. 2. Answer (1)

Hint : 2hkm

λ =

Sol. : k1 = k, m1 = m

1 2hkm

λ = = λ

2 2 2 , 2mk k m= =

2 22 22

h hm kmK

λ = = = λ× ×

3. Answer (1) Hint : de-Broglie wavelength

2 2h hKm qVm

λ = =

Sol. : Mass of proton = m Charge on proton = q Mass of α-particle = 4 m Charge of α-particle = 2q

2 2 2 4

h hqVm q mV ′

=× ×

8VV ′ =

4. Answer (3) Hint and Sol. : Tightly bound electron will emit

form 0 kinetic energy and loosely bound with maximum energy 2.6 eV. Because radiations are monochromatic, so energy of radiations are

E = 2.6 + 4.2 = 6.8 eV 5. Answer (2)

Hint : Energy of photon hcE =λ

Sol. : Power given by photons

–181.5 10hc nt

× = ×λ

–18 –7

–34 81.5 10 6.6 10 Photons5

6.6 10 3 10 secnt

× × ×= =

× × ×

6. Answer (3) Hint : Radius of path in magnetic field,

mv PrqB qB

= =

Sol. : pr rα=

2

pP PqB qB

α=

Pα = 2Pp

p

12

p

p

hPp

h pP

α α

α

λ= = =

λ

7. Answer (3) Hint and Sol. : Photoelectric current does not

depend upon frequency of incident light. Beyond threshold wavelength, photoelectric current becomes zero.

8. Answer (3) Hint : Photoelectric current ∝ Intensity of light Sol. : In quantum physics

Intensity, 2

n nIAt rlt

= =π

1 1I ir r

∝ ⇒ ∝

9. Answer (4)

Hint : 3 B

hk Tm

λ =

Sol. : 2 3 300 2H B

hk

λ =× ×

He 3 400 4B

hk

λ =× ×

283

H

He

λ=

λ

10. Answer (2) Hint and Sol. : Energy of incident radiation = 4.2 + 2.0 = 6.2 eV

Wavelength of radiation, λ = 124006.2

= 2000 Å (ultraviolet)

HINTS & SOLUTIONS

Test-4 (Code-A)_(Hints & Solutions) All India Aakash Test Series for NEET-2021


11. Answer (2) Hint : For shortest wavelength for any series

n = ∞

Sol. : ( )2 2

1 1 1–42BRR

= =

λ ∞

For Brackett series

( )2

1 1 1–164RR

= =

λ′ ∞

4 4B

λ′= ⇒ λ′ = λ

λ

12. Answer (1)

Hint and Sol. : As En = 2

213.6 z

n− , so on

increasing value of n difference of energy between consecutive energy level decreases.

13 Answer (4) Hint : Energy of nth level in hydrogen atom

En = – 213.6 eVn

Sol. : Energy given to hydrogen atom = 12.75 eV Energy of electron in excited state = – 13.6 + 12.75

= – 0.85 eV

= – 0.85 = 213.6n

−

n2 = 16 n = 4 Number of spectral lines in emission

( – 1) 4 3 62 2

n nN ×= = =

14. Answer (4) Hint and Sol. : Lyman series transition takes place

to energy level n = 1, so there will be maximum difference of energy

15. Answer (4) Hint : Moment of linear momentum is

2nhmvr =

π

Sol. : For second orbit n = 2

mvr = hπ

16. Answer (1) Hint : Apply energy conservation E31 = E32 + E31

Sol. : 31 32 21

hc hc hc= +

λ λ λ

1 1 140 60x

= +

X = 120 nm 17. Answer (2)

Hint : Centripetal acceleration 2n

cn

var

=

Sol. : Speed of electron in nth orbit, 137nC zv

n=

radius of nth orbit 2

0.53nnrz

=

2 3

4n

c cn

v za ar n

= ⇒ ∝

81

He

H

aa

=

18. Answer (3) Hint : For excitation (λ ≤ λ0), where λ0 is threshold

wavelength.

Sol. : ( )12400 eV

ÅgE =

λ

λ0 = 12400 4960 Å

2.5=

So λ = 4000 Å may be detected 19. Delete

Hint and Sol. : Input of AND gate is A and B, so =Y AB

20. Answer (3)

Hint : 0and 180o ii

VAV

∆= φ = φ + °

∆

Sol. : ∆V0 = 150 × 2 = 300 V ∆V0 = 2 × 150 = 300 V Φ0 = 10° + 180° = 190° V = 300sin(15t + 190°) ׄ 21. Answer (4)

Hint : It is combination of two OR gate and one AND gate.

Sol. : Output of upper OR gate = W + X Output of lower OR gate = W + Y Net output F = (W + X) (W + Y) = W + WY + XW + XY (Since WW = W) F = W(1 + Y) + XW + XY Since (1 + Y) = 1



F = W ⋅ 1 + XW + XY = W (1 + X) + XY = W + XY 22. Answer (2) Hint : 1 Bq = 1 decay/sec

Sol. : 180.6931 102 86400

dN Ndt

= λ = ××

= 4.0 × 1012 Bq

23. Answer (2) Hint : and Sol. : On emission of α particle mass

number decrease by 4 and atomic number by 2, and on emission of β atomic number increase by 1. So 3α and 4β particle will emit.

24. Answer (4) Hint and Sol. : In β-decay, β -particle of all energy

from 0 to a maximum value emit from sample of same element.

25. Answer (1) Hint and Sol. : Reaction (i) is of fusion and (iv) is

of fission. So energy will release in (i) and (iv). 26. Answer (4) Hint : Apply law of conservation of momentum.

Sol. : 3

1 1 1

1 2 2

18

m V d Rm V d R

= = =

As i.e. where v1 and v2 are volumes and d is the density of nucleus.

0 = m1v1 – m2v2

1 2

2 1

81

v mv m

= =

27. Answer (1) Hint : In pair production energy of γ-ray photon

convert into mass. Sol. : hν → –1β° + +1β° Rest Mass of photon = 0 Mass of electron-positron pair = 2 × 9.1 × 10–31 ∆E = ∆mC2 = 18.2 × 10–31 kg × 9 × 1016 Joule = 16.38 × 10–14 J = 1.02 MeV 28. Answer (2) Hint : Electric field in depletion layer is form n-side

to p-side, so when an electron enter from n to P side it will decelerate

Sol. : Electric field in depletion layer

E = –60.3

1 10Vd

=×

= 3 × 105 NC

Retardation of electron

= –19 5

16–31

1.6 10 3 10 5.3 109.1 10

× × ×= ×

×

v2 = 25 × 1010 – 10.6 × 1010 v = 3.8 × 105 m/s 29. Answer (3) Hint : Germanium diode conduct at 0.3 V, both Ge

and Si diodes are in parallel, so potential drop across both the diodes will remain 0.3 V

Sol. : Potential drop across load = 12 – 0.3 = 11.7 V

Current, i = 11.7 2.34 A5

VR

= =

30. Answer (4) Hint and Sol. : In full wave rectifier ∴ Ripple frequency = 2 × frequency of A.C. 31. Answer (1) Hint and Sol. : In common emitter amplifier phase

difference between input and output signal is π 32. Answer (1) Hint : It is a balanced Wheatstone bridge.

Sol : Reff = 1 21 2

6 12 46 12

R RR R

×= = Ω

+ +

33. Answer (3) Hint and Sol. : The dominant mechanism in

forward bias is due to diffusion of majority carrier and in reverse bias it is due to drift of minority carriers.

34. Answer (2)

Hint : In a semiconductor, 2i e hn n n=

Sol. : e h e hn n n n′ ′=

1.5 × 1016 × 1.5 × 1016 = ne′ × 4.5 × 1022

32

221.5 1.5 10

4.5 10en × ×′ =

×

95 10en′ = × per m3

35. Answer (1)

Hint : 13

0R R A=

Sol. : 13

Be Be

Ge Ge

R AR A

=



139

2 GeRR A

=

AGe = 8 × 9 = 72

36. Answer (2)

Hint : Using momentum and energy conservation

we get – 4AK QAα

=

Sol. : 4 216 5.5 5.4 MeV220

AK QAα− = = × =

Kinetic energy 2 1

2Pkm m

= ∝

kN : kα = 1 1:

216 4

∴ kα = 216 5.5220

× = 5.4 MeV

37. Answer (1)

Hint : h and eE mamv

λ = =

Sol. : λ = hmv

2( )d h adt m atλ

= −

2d hdt matλ

= −

2d hdt eEtλ

= −

38. Answer (1)

Hint : Activity R = λN = loge2NT

Sol. : 1 1 22 2 1

R N TR N T

=

1 1 2 22 2 1 2

2 10 45 1

T N R NT N R N

= × = × =

This is satisfied in option (1)

39. Answer (2)

Hint and Sol. : A Zener diode is used in reverse bias to regulate voltage across load.

40. Answer (4)

Hint and Sol. : = ⋅ ⋅( )Y A B C

41. Answer (3) Hint : It is combination of NOR gate, NAND gate

and Not gate Sol. :

A B C D Y

0 0 1 0 1

1 0 0 1 0

0 1 0 1 0

1 1 0 1 0

It is truth table of NOR gate. 42. Answer (2)

Hint and Sol. : 0 10 VDCVV = =π π

43. Answer (4) Hint and Sol. :

Current through 1 kΩ, i = ( )325 – 12

13 mA1 10

=×

Current in 2 kΩ, 312 6 mA

2 10=

×

Current through Zener, i2 = i – i1 i2 = 13 – 6 = 7 mA 44. Answer (1) Hint : Use Kirchhoff’s voltage law Sol. : Use K.V.L. in Base-emitter closed circuit

– 8.6 × 103 iB – VBE + 5 = 0

35 – 0.7 0.5 mA

8.6 10Bi = =

×

Collector current, ic = βIB

= 100 × 0.5 = 50 mA Use K.V.L. in collector emitter circuit ic × 100 – VCE + 10 = 0 VCE = 10 – ic × 100 = 10 – 50 × 10–3 × 100



VCE = 5 V 45. Answer (3)

Hint and Sol. : On increasing impurity level width of depletion layer decrease.

[CHEMISTRY] 46. Answer (2) Hint : Electron withdrawing group present at p-

position to the benzaldehyde makes carbonyl carbon electron deficient.

Sol. : In p-nitrobenzaldehyde carbonyl carbon is electron deficient due to strong – R effect of – NO2 group which assists the nucleophile to attack at carbonyl carbon most easily.

47. Answer (4) Hint : Electron withdrawing group increases the

acidic strength of carboxylic acid. Sol. : – F has strong – I effect hence FCH2COOH

is strongest acid among the given options. The acidity order is FCH2COOH > C6H5COOH > C6H5CH2COOH > CH3COOH.

48. Answer (1) Hint : Acid chloride on hydrogenation over Pd and

BaSO4 forms benzaldehyde. Benzaldehyde undergoes Cannizzaro reaction in

presence of 50% NaOH solution. Sol. :

49. Answer (4)

Hint : Compounds containing – group or – group form iodoform by I2/NaOH.

Sol. : Acetaldehyde and isopropyl alcohol both will form iodoform hence cannot be distinguished by I2/NaOH.

50. Answer (2) Hint : Benzyl bromide is obtained on reaction of

benzyl alcohol with HBr. Sol. :

51. Answer (2) Hint : Electron donating group present at para

position to –NH2 group will increase the basic nature of amine.

Sol. : +R effect of –OCH3 increases electron density on nitrogen of NH2 more than hyperconjugation effect of –CH3 group.

52. Answer (1) Hint : Nylon-6,6 is prepared by the condensation

polymerisation of hexamethylenediamine with adipic acid

Sol. :

Nylon-6,6 is a polyamide 53. Answer (1)

Hint : Orlon is

Sol. :



54. Answer (2) Hint : The polymers made by addition or

condensation polymerisation from two different monomers are called copolymers.

Sol. :

55. Answer (4) Hint : Cellulose is a natural biodegradable

polymer. Sol. : PHBV and Nylon-2-nylon-6 are synthetic

biodegradable polymers. 56. Answer (3) Hint : Bakelite is made of phenol and

formaldehyde. 57. Answer (4) Hint : Thermosetting polymers are cross linked

polymers. 58. Answer (2) Hint : The compounds having enolisable proton

will show keto-enol tautomerism.

Sol. :

59. Answer (2) Hint : Brompheniramine (Dimetapp) is an

antihistamine. 60. Answer (4)

Hint : Sucralose is 600 times sweeter than cane sugar (sucrose).

61. Answer (4) Hint : Furacine, soframycin and bithional are used

as antiseptics. 62. Answer (1) Hint : Glucose also has cyclic structure. Sol. : NaHSO3 addition does not take place to

glucose because of its cyclic structure. 63. Answer (1) Hint : Disaccharide and monosaccharide

containing free anomeric –OH will show mutarotation.

Sol. :

In sucrose free anomeric – OH is absent 64. Answer (3) Hint : The amino acid which does not contain

chiral centre is optically inactive. Sol. : Glycine does not contain chiral centre.

65. Answer (4) Hint : Sugar moiety in RNA is β-D-ribose. 66. Answer (3) Hint : Vitamin B-complex are water soluble. Sol. : Thiamine (Vitamin B1) riboflavin (Vitamin B2)

and Vitamin-C are water soluble. 67. Answer (3)

Hint :

β–D–(–)–Fructofuranose



68. Answer (4) Hint : Only primary amines both aliphatic and

aromatic give carbylamine reaction. Sol. : 69. Answer (2) Hint : Nitrobenzene is reduced to hydroazobenzene in

presence of Zn/NaOH.

Sol. : Zn/NaOH2PhNO PhNHNHPh→

70. Answer (1) Hint : Diazonium salt undergoes coupling reaction

with phenol in faintly alkaline medium. Sol. :

71. Answer (2) Hint : Hinsberg’s reagent is benzenesulphonyl

chloride. Sol. : Aniline on reaction with Br2/H2O forms a

white precipitate of 2,4,6-tribronoaniline but benzyl amine gives no such characteristic reaction with Br2/H2O

72. Answer (2) Hint : The given reaction is Hofmann degradation

reaction in which amide is converted to primary amine.

Sol. :

73. Answer (3) Hint : Aldehyde on reaction with excess alcohol in

presence of dry HCl gas form acetal.

Sol. :

(Acetal) 74. Answer (2) Hint : – I effect of chlorine increases electrophilic

character of carbonyl carbon of ester. Sol. : Two chlorine atoms in (iii) has highest – I

effect among the given compounds which enhances electrophilicity of carbonyl carbon hence ester hydrolysis in alkaline medium is facilitated.

75. Answer (2) Hint : The given reaction is known as Hell Volhard

Zelinsky reaction. Sol. :

76. Answer (3) Hint : Alkyl cyanide on acidic hydrolysis produces

carboxylic acid.

Sol. : 3H O3 2 3 2CH CH CN CH CH COOH+

→

3H O3 2 2 3 2CH CH CONH CH CH COOH+

→

3H O3 2 3 2 2CH CH NC CH CH NH HCOOH+

→ +

77. Answer (3) Hint : – NHCOCH3 group is o/p directing group in

electrophilic aromatic substitution reaction. Sol. :

78. Answer (4) Hint : Benzenediazonium chloride is colourless

crystalline compound and it is water soluble. 79. Answer (1) Hint : Arenediazonium chloride on reaction with

fluoroboric acid forms arene diazonium fluoroborate.

Sol. : 80. Answer (1)

Hint : β-keto acid decarboxylates most easily on heating.

Sol. :



81. Answer (2) Hint : Phorone is 2,6-dimethylhepta-2,5-dien-4-one

Sol. : 3CH3COCH3 HCl→ (CH3)2C = CHCOCH

= C(CH3)3 + 2H2O 82. Answer (4) Hint : α-hydrogens in carbonyl compounds are

enolisable hydrogens.

Sol. :

The above compound has 3-α-hydrogens. 83. Answer (3) Hint : Gabriel phthalimide synthesis is used in the

preparation of primary amine. Sol. :

84. Answer (2) Hint : Amide on reaction with LiAlH4 forms primary

amine.

Sol. :

85. Answer (3)

Hint : is semicarbazide

Sol. :

86. Answer (3) Hint : 2° alcohol is oxidised to ketone in presence

of PCC. (PCC is pyridinium chlorochromate)

Sol. :

87. Answer (3) Hint : Reagent used in etard reaction is CrO2Cl2.

Sol. :

88. Answer (3) Hint : Intermolecular hydrogen bonding increases

the intermolecular association hence boiling point increases.

Sol. : Acetic acid is polar and is associated extensively by hydrogen bonding.

89. Answer (1) Hint : Alkyl cadmium is an alkylating reagent.

Sol. :

90. Answer (1) Hint : Zn – Hg/C ⋅ HCl reduces butanone to

n-butane.

Sol. : (Clemmensen reduction)



[BIOLOGY]91. Answer (2) Hint: Man-made ecosystem is called

anthropogenic ecosystem. Sol.: Forest, lakes and estuaries are natural

ecosystem. Crop fields are agroecosystem which are largest anthropogenic ecosystem.

92. Answer (1) Hint: Aquarium is an anthropogenic aquatic

ecosystem. Sol.: Anthropogenic ecosystem (aquarium) does

not possess self-regulatory mechanism. It has simple food chain and little biodiversity. Crop fields show high productivity

93. Answer (2) Hint: Edaphic factor is related to soil. Sol.: Edaphic factor is concerned with the

composition of soil including its chemical and physical properties.

94. Answer (3) Hint: In some organisms, one stage of their life

cycle show herbivorous food habit while the other stage show carnivorous.

Sol.: Tadpoles are herbivorous and the adult frogs are carnivorous. Therefore, they are placed at different trophic levels

95. Answer (4) Hint: The biotic component of an ecosystem has

both autotrophs and heterotrophs. Sol.: Many arboreal animals occupy the top layer

of the forest ecosystem. 96. Answer (4) Hint: Productivity in an ecosystem is the rate of

biomass production. Sol.: Due to several biotic and abiotic factors, the

productivity of desert and deep sea is very less. 97. Answer (2) Hint: Water -soluble substances go down the soil

by percolating water. Sol.: Leaching is the process in which water-

soluble substances present in detritus go down into the soil and get precipitated as unavailable salts.

98. Answer (1) Hint: Decomposition of detritus is enzymatic

process. Sol.: An optimum temperature is required for the

activity of enzymes which is around 25°C or little more. Above 45°, the enzymatic activity becomes very slow or enzymes become inactive.

99. Answer (1) Hint: When in a food chain, energy flows from one

trophic level to the next, maximum energy is lost in the form of heat or other activities.

Sol.: According to second law of thermodynamics, no transfer of energy occurs unless it is accompanied by degradation or dissipation.

100. Answer (1) Hint: At each transfer in a food chain, only 10% of

the total energy is actually available to the next trophic level.

Sol.: Energy in 10 ants = 10 calories Energy gained by the anteater

10=10× 1 calory100

= .

101. Answer (3) Hint: Standing crop is measured as the mass of

living organisms. Sol.: Standing crop is the amount of living material

present in different trophic levels at a given time. 102. Answer (4) Hint: Biomass is the amount of living matter at any

particular trophic level at a given time. Sol.: For a tree ecosystem, the biomass of a tree

is always more as compared to the organisms at other trophic levels.

103. Answer (2) Hint: Ecological pyramids do not include

insectivorous plants. Sol.: Ecological pyramids assume food chain only.

Saprophytes are also not given a place in the ecological pyramid.

104. Answer (1) Hint: Hydrarch succession starts in aquatic

habitat. Sol.: Pioneer community is first biotic community

that develops in a bare area. Phytoplanktons and zooplanktons are first biotic community in aquatic habitat.



105. Answer (3) Hint: Primary succession starts at barren area,

never having vegetation of any type. Sol.: Sand dunes are barren areas. In these areas

primary succession starts. 106. Answer (1) Hint: The maximum cost of ecosystem services

accounts for soil formation. Sol.: Soil formation accounts for about 50% of the

total cost of ecosystem services. 107. Answer (2) Hint: Rice is a single species. Sol.: Different strains of rice show genetic

diversity. 108. Answer (2) Sol.: According to estimate made by Robert May,

the global species diversity is about 7 million. 109. Answer (3) Hint: Sector B includes algae Sol.: Sector A–Angiosperms Sector C–Fungi 110. Answer (4) Hint: Number of species of birds is more than that

of reptiles and mammals. Sol.: Number of species of different groups of

animals in Amazonia rain forest are as follows: Mammals – 427 Birds – 1300 Reptiles – 378 Amphibians – 427 Fishes – 3000 111. Answer (3) Hin: Tropics are less seasonal and more constant. Sol.: Tropics have warm temperature and high

humidity. Due to receiving more solar energy, the tropical communities are more productive and can support wide range of species.

112. Answer (1) Hint: On a logarithmic scale, the species area

relationship shows a straight line. Sol.: For a very large area, the slope of the line

w.r.t. species area relationship on a logarithmic scale is much steeper.

113. Answer (3) Hint: Increased diversity contribute to higher

productivity. Sol.: Stable community does not show much

variation in productivity from time to time and it is more resistant to the invasions of alien species.

114. Answer (3) Hint: According to Rivet popper hypothesis, rivets

are considered as species. Sol.: By taking the example of an airplane

(considered as ecosystem), Paul Ehrlich gave Rivet popper hypothesis to explain the ecosystem health.

115. Answer (2) Sol.: During the long period of life on earth, there

were five episodes of mass extinction of species. 116. Answer (1) Hint: Eichhornia is an aquatic plant. Sol.: Eichhornia was introduced by Europeans in

India. It clogges water bodies and resulting in death of several aquatic plants and animals.

117. Answer (3) Hint: Humans derive many direct economical

benefits from nature which are categorised as narrowly utilitarian.

Sol.: Food, firefood, drugs and industrial products are under narrowly utilitarian of ecosystem services.

118. Answer (3) Hint: One of the animal belongs to the national

park situated in Sikkim is snow leopard. Sol.: Rhino is found in Kaziranga National Park. 119. Answer (1) Hint: To control the pollution, our Government has

passed Environment Act. Sol.: Our Government has passed Environment

(Protection) Act in 1986 to control the environmental pollution.

120. Answer (4) Hint: Secondary pollutants are formed by the

reactions of primary pollutants. Sol.: Ozone is secondary pollutant which is formed

when heat and sunlight cause chemical reactions between oxides of nitrogen and hydrocarbons.



121. Answer (2) Hint: Aerosols cause ozone depletion. Sol.: Aerosols are vapour chemicals in the form of

fluorocarbons, chlorofluorocarbons, oxides of sulphur and nitrogen.

122. Answer (1) Hint: Electrode wires are negatively charged. Sol.: The electrons from negatively charged

electrode wires attach to dust particles giving them a net negative charge.

123. Answer (4) Hint: The calcium in limestone combines

chemically with sulphur. Sol.: Calcium in CaCO3 combines with sulphur to

produce CaSO4 which is separately collected. 124. Answer (3) Hint: One of the methods to reduce noise pollution

is planting trees under green muffler scheme. Sol.: Noise is measured by a sound meter and is

expressed in a unit called decibel (dB). 125. Answer (1) Hint: Lead causes pollution. Sol.: Use of unleaded petrol is one of the

measures taken by the Government to reduce the pollution.

126. Answer (1) Hint: Acid rain is a cocktail of H2SO4 and HNO3. Sol.: Oxides of sulphur and nitrogen combine with

moisture to form H2SO4 and HNO3 respectively. 127. Answer (2) Hint: Decomposition of organic wastes cause drop

in DO of water. Sol.: O2 consumption by organisms and

decomposition of organic matter are the factors which determine the amount of dissolved O2 in water.

128. Answer (2) Hint: Eichhornia, an aquatic plant is called ‘Terror

of Bengal’ Sol.: Algal bloom is excessive growth of

phytoplankton. Biological magnification of DDT in birds causes thinning of egg shell.

129. Answer (4) Hint: Minamata and black foot diseases are due to

mercury and arsenic respectively. Sol.: Itai-itai disease is caused by the consumption

of food and water contaminated with cadmium. 130. Answer (3) Hint: In a young lake, the water is cold and clear. Sol.: As the age of lake increase, lake grows

shallower and warmer. 131. Answer (3) Hint: Town of Arcata is an example of initiating

waste water treatment. Sol.: The citizens group called Friends Of the

Arcata Marsh (FOAM) are responsible for the upkeep and safeguarding of the marshes by integrated waste water treatment.

132. Answer (1) Hint: Human excreta can be recycled into a

resource. Sol.: EcoSan toilets are sustainable system for

handling human excreta. The recycled human excreta is a natural fertiliser which reduces the need of chemical fertilisers.

133. Answer (4) Hint: Due to deforestation, water holding capacity

of soil decreases. Sol.: Due to lack of vegetation, water does not stay

on the ground and run away in the form of flood. Since, water does not stay on ground, it does not percolate down the layer of soil and therefore not added to the ground water.

134. Answer (2) Hint: UV rays act on CFCs releasing Cl atoms. Sol.: The Cl atom released from CFC by the act of

UV rays catalyse the conversion of O3 into O2. 135. Answer (2) Sol.: Montreal protocol was signed to control the

emission of ODS. 136. Answer (1) Hint : Plasmid has its own ‘ori’ Sol. : Plasmid is an autonomously replicating

closed circular extrachromosomal DNA without any vital gene.



137. Answer (1) Hint : Two strands are mirror images with opposite

orientation. Sol. : The palindrome in DNA is a sequence of

base pairs that reads same on the two strands when orientation of reading is kept the same.

Eg : 5′ – GAATTC – 3′

3′ – CTTAAG – 5′ 138. Answer (3) Hint : In EcoRI, E is derived from the genus. Sol. : In Eco RI, E is derived from the genus,

Escherichia, co from the name of the species coli, roman number ‘I’ indicates the order of isolation from that strain of bacteria.

139. Answer (2) Hint : Enzyme forming phosphodiester bonds. Sol. : Action of restriction endonucleases

produces sticky ends. Lyases catalyse the breakdown of substrate into two parts without the use of water.

140. Answer (3) Hint : Thickest DNA band closest to loading wells. Sol. : Undigested DNA sample is in lane 1. 141. Answer (4) Hint : Agarose is a matrix which is a natural

polymer. Sol. : Agarose is extracted from sea weeds. The

separated bands of DNA are cut out from agarose gel and extracted from gel piece. This is called elution. Separated DNA fragments can be visualized after staining the DNA with ethidium bromide. While isolating DNA, the precipitated DNA can be separated out on glass rod, a step called spooling

142. Answer (2) Hint : Like plasmids, they are autonomously

replicating. Sol. : Bacteriophages because of their high copy

number per cell, have very high copy numbers of their genome within the bacterial cells.

143. Answer (3) Hint : β-galactosidase Sol. : Recombinants produce white colonies due

to insertional inactivation of β-galactosidase while

non-recombinants produce blue colonies. Ampicillin resistant gene is present in both pUC8 and pBR322.

Identification of recombinants requires single step i.e. plating onto agar medium containing ampicillin and X-gal.

pUC8 is 2.7 kb in size while pBR322 is 4.3 kb in size.

144. Answer (4) Hint : Pvu II recognises sequence within ‘rop’ of

pBR322 Sol. : Rop codes for the proteins involved in the

replication of the plasmid. When alien DNA is inserted in the recognition sequence of Pvu II, this causes insertional inactivation of rop. Therefore, there will be unavailability of proteins for replication of plasmid.

145. Answer (4) Hint : Transform normal cells to cancerous cells. Sol. : Retroviruses have been disarmed and used

to deliver genes into animal cells. 146. Answer (3) Hint : Identify a second messenger Sol. : Calcium chloride treatment is used to make

competent cells take up DNA. 147. Answer (1) Hint : Gene gun method. Sol. : Plant cells are bombarded with high velocity

microparticles of gold or tungsten coated with DNA in a method called gene gun/biolistics.

BAC (Bacterial artificial chromosome), and Agrobacterium tumefaciens are the vectors used in vector mediated transfer of DNA.

148. Answer (2) Hint : Chitin is present in fungi. Sol. : Isolation of genetic material can be achieved

by treating the cells with enzymes such as lysozyme (bacteria), cellulose (plant cells) and chitinase (fungus).

149. Answer (3) Hint : C gene with flanking sequence is produced

on digestion with Hind III Sol. :



150. Answer (4) Hint : Transferred DNA Sol. : β-galactosidase gene and antibiotic

resistance gene are features of pUC8 plasmid. Ti plasmid has its own ‘origin of replication’ [ori] gene. Ti plasmid has been disarmed by replacement of transferred DNA (T-DNA) with our gene of interest.

151. Answer (4) Hint : Streptomyces albus

Sol. :

152. Answer (3) Hint : It does not digest nucleic acids. Sol. : RNA can be removed by Ribonuclease,

whereas proteins can be removed by treatment with proteases. Chilled ethanol precipitates purified DNA. Calcium chloride is used to make competent cells.

153. Answer (1) Hint : Primer extension is the last step of PCR. Sol. : Two sets of primers are added to the

reaction mixture of PCR. These primers are small chemically synthesized oligonucleotides that are complementary to the region of template DNA.

154. Answer (3) Hint : It cuts internal bonds. Sol. : EcoRI cleaves internal phosphodiester

bonds. Its an endonuclease that cleaves specific palindromic sequence whereas exonucleases remove nucleotides from the terminal ends of a DNA molecule.

155. Answer (2) Hint : Its a dephosphorylating enzyme. Sol.: Alkaline phosphatase is used for the removal

of single phosphate groups from 5′ ends of linear vectors to prevent recircularization during cloning or to dephosphorylate DNA.

156. Answer (4) Hint : 1 billion copies are made at the end of 30

PCR cycles. Sol. : In PCR, Number of fragments amplified = 2(n)

n = 8

∴ 2(8) = 256 157. Answer (3) Hint : Culture of cells in log phase leads to high

yield. Sol. : In continuous culture system, the used

medium is drained out from one side while fresh medium is added from other to maintain cells in their physiologically most active log/exponential phase and produces a larger biomass leading to higher yields of desired product.

158. Answer (2) Hint : Biosynthetic stage is upstream processing. Sol. : Biosynthetic stage includes the process of

fermentation. Step of cutting and extracting DNA from the gel piece after gel electrophoresis is called elution.

159. Answer (3) Hint : Their names differ based on this difference. Sol. : Both simple stirred tank and sparged stirred

tank bioreactor have a stirrer that facilitates even mixing and oxygen availability throughout the bioreactor, a foam control system, sample port and pH control system. Sparged stirred tank bioreactor has a system through which sterile air bubbles are sparged to increase surface area for oxygen transfer.

160. Answer (3) Hint : Microneedles are used. Sol. : Biolistic/gene gun is suitable for plants. Heat

shock method of transformation requires a brief 42°C temperature treatment that enable bacteria to take up foreign DNA.

161. Answer (2) Hint : Blue white selection is done on X-gal

substrate. Sol. : One can see bright orange coloured bands

of DNA in a EtBr stained gel exposed to UV rays. Gram staining is done for identifying bacterial culture.



162. Answer (3) Hint : Marine flora. Sol. : Agarose is a commonly used matrix

extracted from sea weeds. Restriction endonucleases are extracted from bacteria.

163. Answer (3) Hint : Selectable markers of pBR322. Sol. : pBR322 has two selectable marker genes

ampR and tetR. The normal E.coli cells do not carry resistance against any of these antibiotics.

LacZ gene is present in pUC8 plasmid of E.coli. 164. Answer (2) Hint : This property prevents early exhaustion of

fertility of soil. Sol. : Genetically modified plants have increased

efficiency of mineral usage that prevents early exhaustion of soil fertility.

165. Answer (3) Hint : Name of bacterium. Soil. : Bt toxin is produced by bacteria called

Bacillus thuringiensis (Bt in short). 166. Answer (3) Hint : Crystallised toxin Sol. : Bt toxin protein exists as an inactive protoxin

but once an insect ingest the inactive toxin, it is converted into an active from of toxin due to alkaline pH of insect gut.

167. Answer (2) Hint : It is also controlled by cryIIAb gene encoded

protein. Sol. : Protein encoded by the genes cryIAc and

cryIIAb control the cotton bollworms and that of cryIAb controls corn borer.

168. Answer (3) Hint : Its colour is provided by the presence of

precursor of vitamin A. Sol. : Golden rice is a biofortified crop that

combats vitamin A and iron deficiency as it has the gene for beta carotene encoded in it.

169. Answer (2) Hint : A gene subtraction technique Sol. : RNA interference (RNAi) prevents

infestation of root of tobacco plants by nematode M. incognita.

RAPD : Random amplification of polymorphic DNA.

RFLP : Restriction fragment length polymorphism.

SDS-PAGE : Sodium dodecyl sulphate and polyacrylamide gel.

170. Answer (3) Hint : Sense and anti-sense RNA both are formed

in host cell. Sol. : RNAi involves silencing of specific RNA due

to formation of dsRNA molecule formed by binding of complementary RNA (anti-sense RNA) molecule to original mRNA, thereby preventing translation of original mRNA.

171. Answer (4) Hint : Longest extra stretch of polypeptide. Sol. : C peptide chain is not present in the mature

insulin and is removed during maturation into insulin.

172. Answer (2) Hint : Proper folding is an essential step. Sol. : Chain A and B were produced separately,

extracted and combined by creating disulphide bonds to form human insulin as the main challenge was getting it assembled into a mature form.

173. Answer (4) Hint : This enzyme is crucial for the immune

system to function. Sol. : ERT is enzyme replacement therapy that is

used to treat ADA deficiency. Alkaline phosphatase removes 5′ phosphate

group from DNA molecule. 174. Answer (1) Hint : Milk contained human alpha-lactalbumin. Sol. : Rosie was the first transgenic cow. 175. Answer (3) Hint : These molecules are employed in

autoradiography. Sol. : Probes are allowed to hybridise to its

complementary DNA in a clone of cells followed by detection using autoradiography.

176. Answer (3) Hint : Patent controversy. Sol. : There are an estimated 200000 varieties of

rice in India alone whereas 27 documented varieties of basmati rice are grown in India.



177. Answer (2) Hint : It should be novel. Sol. : Patents are supposed to satisfy three criteria

of ‘novelty’, non obviousness and utility. 178. Answer (3) Hint : Safety of GMO’s. GEAC : Genetic Engineering Approval

Committee RCGM : Research Committee on Genetic

manipulation. CDRI : Central Drug Research Institute,

Lucknow, UP.

179. Answer (4)

Hint : A traditional method of diagnosis. Sol.: Serum and urine analysis are traditional

methods of diagnosis in which early detection is not possible.

180. Answer (4)

Hint : Resistance to insects. Sol. : Bt toxin provide resistance to insects without

the need for insecticides in effect created a bio-pesticides. Golden rice is a biofortified crop. Roundup ready crops are herbicide tolerant crops.

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