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