Regd. Office: Aakash Tower, 8, Pusa Road, New Delhi-110005, Ph.

011-47623456

Aakash Rank Booster Test Series for NEET-2020

MM : 720 Test - 10 Time : 3.00 Hrs.

Answer Key
1. (2) 37. (2) 73. (2) 109. (3) 145. (2)
2. (1) 38. (4) 74. (4) 110. (3) 146. (3)
3. (2) 39. (3) 75. (4) 111. (1) 147. (2)
4. (2) 40. (2) 76. (2) 112. (3) 148. (2)
5. (3) 41. (2) 77. (4) 113. (2) 149. (3)
6. (2) 42. (3) 78. (1) 114. (1) 150. (2)
7. (3) 43. (4) 79. (3) 115. (1) 151. (1)
8. (4) 44. (1) 80. (3) 116. (3) 152. (3)
9. (3) 45. (4) 81. (2) 117. (4) 153. (3)
10. (4) 46. (1) 82. (1) 118. (3) 154. (3)
11. (1) 47. (4) 83. (4) 119. (2) 155. (4)
12. (3) 48. (2) 84. (1) 120. (2) 156. (3)
13. (4) 49. (3) 85. (4) 121. (3) 157. (1)
14. (1) 50. (1) 86. (2) 122. (1) 158. (2)
15. (2) 51. (1) 87. (4) 123. (1) 159. (2)
16. (3) 52. (3) 88. (3) 124. (1) 160. (1)
17. (3) 53. (2) 89. (1) 125. (4) 161. (2)
18. (4) 54. (4) 90. (1) 126. (1) 162. (1)
19. (3) 55. (1) 91. (3) 127. (3) 163. (2)
20. (4) 56. (4) 92. (4) 128. (2) 164. (1)
21. (1) 57. (1) 93. (3) 129. (1) 165. (2)
22. (3) 58. (2) 94. (4) 130. (2) 166. (3)
23. (1) 59. (3) 95. (2) 131. (2) 167. (3)
24. (2) 60. (3) 96. (4) 132. (1) 168. (2)
25. (3) 61. (1) 97. (2) 133. (1) 169. (2)
26. (2) 62. (4) 98. (2) 134. (2) 170. (4)
27. (4) 63. (4) 99. (1) 135. (4) 171. (2)
28. (2) 64. (2) 100. (3) 136. (3) 172. (4)
29. (1) 65. (1) 101. (1) 137. (4) 173. (4)
30. (1) 66. (1) 102. (4) 138. (2) 174. (1)
31. (3) 67. (1) 103. (1) 139. (3) 175. (2)
32. (4) 68. (1) 104. (1) 140. (1) 176. (2)
33. (1) 69. (3) 105. (1) 141. (2) 177. (1)
34. (3) 70. (3) 106. (1) 142. (2) 178. (1)
35. (4) 71. (4) 107. (2) 143. (3) 179. (4)
36. (4) 72. (2) 108. (1) 144. (3) 180. (3)
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Test-10_(Hints & Solutions) Aak ash Rank Boo ster Test Serie s fo r NEET -2020

Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005, Ph.011-47623456

Aakash Rank Booster Test Series for NEET-2020

MM : 720 Test - 10 Time : 3:00 Hrs.

Hints and Solutions

PHYSICS
1. Answer (2) 4. Answer (2)
Reading = Main scale reading + least count × V.S
reading
= 3.9 cm + (0.01 × 3)
= 3.93 cm.
5. Answer (3)
dv
 v 3
dt
v
dv v
 3
   t
v u
u

   1
v
t
Angle between P and P  Q is  2
= t 
2v u 0

90   90  32 1 1
2 2
 = t
2v 2u 2
= 58°
2. Answer (1) 1 1
   2t
   v2 u2
v  u  a t
 1 1  2u 2t

 v  2 jˆ  2 iˆ  jˆ v2

u2
 
 2iˆ  4 jˆ m/s
v
u
.
3. Answer (2) 1  2u 2t
M = K d x Dy L z 6. Answer (2)
M R = u cos× T
 100
M 1 2
R´= u cos× T + aT
 d   D   L  2
 x  100   y   100   z   100 
 d   D   L  1 g  2u sin  
2

% M = (mx + ny + lz)% = u cos × T +  

24  g 

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Aakash Rank Boo ster T est Series for NEET -2020 Test-10 _(Hin ts & Solutio ns)

g 4u 2 sin2  12. Answer (3)

=R+ 
8 g2
u 2 sin2 
=R 
2g
R = R + H
7. Answer (3)
4mg  mg
a1   3g  a 
2
m I  4  m 
4mg  mg 3  2
a2   g
5m 5 ma2
 4  2ma2
4mg  mg 3 2
a3   g
2m 2
I  2  2  22
8. Answer (4)
= 16 kg m 2
When F < N then f = F
13. Answer (4)
F  N then f = N = constant
9. Answer (3)
T = ma2 = m42f 2
T 500 100 F – f = ma …(i)
f2  2
 2

ml 4  5  1 4  4 2
2 a
5 f .R  mR 2.
f  rps . 5 R

2
10. Answer (4) f  ma
5
W   Fdx
5
ma  f (ii)
4 2
 0 (40  0.5 x )dx
5f 7f
4 F f  
 x2  2 2
  40 x  
 4  0 2
f  F.
= [160 + 4] 7
= 164 J. 14. Answer (1)
11. Answer (1) GMm
mR 2 
R2
2
 2  4 
R 3    G  Rs3  s
T
   3 

4  2R 3 3 R 3
s   2 3
.
4
T 2G Rs3 GT Rs
3
v sin 
tan   15. Answer (2)
ev cos 
tan  L
tan ´ T  2
e g
 tan   At depth g decreases i.e T will increase. Hence
´  tan 1  
 e  stop watch Runs slow.

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Test-10_(Hints & S olutions) Aak ash Rank Boo ster Test Serie s for NEET -2020

16. Answer (3) 24. Answer (2)

FL 40  103
l  I 2
 10 2 W
AY 4  1
2
l1 A2  3  9 I   10 2 
    .
l2 A1  1  1 L  10log   = 10log10  12 
 10 
 I0   
17. Answer (3)
L = 100 dB
From continuity equation and Bernoulli’s
25. Answer (3)
Principle.
qin,max = 2(2R) = 4R
1
P  Pv 2 constant …(i) qin 4R
2 Flux   .
0 0
and A1v 1 = A2v2 …(ii)
26. Answer (2)
Where area is less, pressure be less.
18. Answer (4) q2
 40  106 J
For Reynold number more than 3000, flow 2  2C 
becomes turbulent.
q2
19. Answer (3)  160  106
C
 2 < 1  volume of cube increases more than q 2 = 10 × 10–6 × 160 × 10 –6
the corresponding decrease in density of liquid.  q = 40 C
Hence the fraction of cube will decrease with
increase in temperature of the system. q
 qC1 
2
20. Answer (4)
27. Answer (4)
For isothermal compression U = 0
Surface charge density is inversely proportional
For adiabatic compression U = +ve to radius of curvature of surface.
For isobaric compression U = –ve 28. Answer (2)
21. Answer (1) E  l 
r    1 R   1  1 R
For rigid container V = constant  n is fixed V   l2 
1
Hence   is independent of temperature  75  5 
2nd 2   1 4    1 4
 60  4 
22. Answer (3) r=1
T1 T 29. Answer (1)
T   t1  t2  2
2 2 2Req
Req  1 
m 2l1 2l 2 m 2  Req
   
K v0 v 0 4K Req2 + 2Req = 2 + Req + 2R eq
3 m l  l  Req2 – Req – 2 = 0
 2 1 2
2 K v0 1  1  8
Req   2
23. Answer (1) 2

5v 5v 10
i= = 5 A.
For open pipe, f5    4 th overtone 2
2L 2L
30. Answer (1)
9v 9v
For closed pipe, f9   4th overtone In one cycle proton will cross the gap two times.
4L 4L Hence in one cycle energy gained by proton is
5v E = 2e × 10 4 V
f5 2L 10
Now,   12  106 eV
f9 9v 9 Now n =  600 .
4L 2  104 eV

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Aakash Rank Boo ster T est Series for NEET -2020 Test-10 _(Hin ts & Solutio ns)

31. Answer (3) 3

Magnetic length is about 0.84 times of the n
2
geometric length.
2n = 3
32. Answer (4)
dc ammeter reads average value of current. 40. Answer (2)
33. Answer (1) h
  
Since V and B are either parallel or antiparallel. 2meV
Hence emf. Induced in the rod will be zero hence
6.6  1034 Js
VP~ VQ = 0. 
34. Answer (3) 2  1.67  10 27  1.6  10 19
d d 0.286
VP    40t  5   40 V  Å.
dt dt V
V N 41. Answer (2)
Now S  S
VP NP Absorption transitions = (n - 1)
500 n–1=4
VS  40   200 V .
100  n = 5.
35. Answer (4) Now number of emission transitions
C 3  108 3 between n = 5 to n = 1
v    108
 r r 82 4 n   n  1 54
   10 .
= 7.5 × 107 m/s. 2 2
36. Answer (4) 42. Answer (3)
f 20 N = N 0e -t
fl ,eq    10 cm
2 2  1
  
1 2 1 2 1 N  N0 e 
Now    
f fl ,eq fm 10 10 N0
N  0.37 N0
10 e
f  cm .
3 Decayed nuclei = N0 – N = 0.63 N0
37. Answer (2) 43. Answer (4)
u = – 25 cm, v = – 75 cm, f = ? 1
1
1 1 1 0 n 1 H 01 e  
 
f v u 44. Answer (1)
1 1 1 2 I0
   Transconductance g m 
f 75 25 75 Vin
75 I0 A 25
f  gm  = I   5  10 3  1
2 Ii Ri Ri 5  103
100  2 45. Answer (4)
P  2.67 D .
75 When both A and B are low then no current flow
38. Answer (4) through resistor and hence Y = 0. Now if any of A
 and B or both (A and B) become high then
Resolving power  . current starts flowing through resistor and output

39. Answer (3) Y becomes high. Hence truth table becomes of
OR gate.
4D ´ 5D
y4  y3  A B Y
d 2d
 5  D 0 0 0
y 4  y 3´   4  
 2 d 0 1 1
D  8  5  D 1 0 1
n   1 1 1
d  2  d

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Test-10_(Hints & Solutions) Aak ash Rank Boo ster Test Serie s fo r NEET -2020

CHEMISTRY
46. Answer (1) 58. Answer (2)
More the number of Ions, more will be the
conductance.
47. Answer (4)
Rate constant depends on temperature and
catalyst both.
48. Answer (2)
59. Answer (3)
Zeta potential is the potential difference
between fixed charged layer and the diffused
layer having opposite charge. Methanamide.
49. Answer (3) 60. Answer (3)
1n
 mol 
Unit of rate constant (k)    time1
  
50. Answer (1)

º 0.0591 [A 2 ]
Ecell  Ecell  log10 2
2 [B ]
51. Answer (1) 61. Answer (1)
Milk is an emulsion of oil in water. Pure NaCl does not absorb moisture.
52. Answer (3) 62. Answer (4)
For CaF2 and NaCl, Z = 4. Hydrolith : CaH2
53. Answer (2) 63. Answer (4)

 = CRT = 0.1×0.0821×300 = 2.46 atm.

54. Answer (4)
For fcc lattice, number of nearest neighbours =
x = +6
12.
64. Answer (2)
55. Answer (1)
Maximum and minimum oxidation state of N is
2KI + HgI2  K2[HgI4] +5 and –3 respectively.
Due to complex formation, number of Ions in 65. Answer (1)
solution decreases. In [Al(H2O)6]3+, covalency of Al is 6.
56. Answer (4) 66. Answer (1)
More the charge on cation, more will be
covalent character.
67. Answer (1)

Silicones :

57. Answer (1) 68. Answer (1)

PAN is a secondary pollutant. N2O is a neutral oxide.

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Aakash Rank Boo ster T est Series for NEET -2020 Test-10 _(Hin ts & Solutio ns)

69. Answer (3) 80. Answer (3)

Intensive properties do not depend on mass of a 2a
TB  and Ti  .
system. Rb Rb
70. Answer (3) 81. Answer (2)
During adiabatic compression, internal energy 1
of system increases. H2O  H2  O2 .
18 kg 2 kg 2
71. Answer (4) 16 kg

82. Answer (1)

NH4 Cl  H2O  NH4OH  HCl
Weak Base Strong acid meq. of H2SO4 = meq. of metal
72. Answer (2) 0.9
100×1 =  1000
On adding inert gas during constant pressure, E
reaction goes in that direction where gaseous E=9
mole increases. 83. Answer (4)
73. Answer (2) Total spectral lines obtained = (4 – 1) = 3
ng = (n P)g – (nR)g = (2) – (0) = 2. =3+2+1=6
KP = KC (RT)ng = KC(RT)2 84. Answer (1)
74. Answer (4) Acetone gives positive Iodoform test.
85. Answer (4)
For neutral solution, [OH –] = K w  108
Bakelite and polyethene are not biodegradable
 pOH = –log(10–8) = 8. polymers.
75. Answer (4) 86. Answer (2)
dz2 is involved in sp 3d hybridized orbitals.
76. Answer (2)
118 : Un + un + Oct  Uuo. (DDT)
77. Answer (4) 87. Answer (4)
3°-Alcohol reacts fastest with Lucas reagent.


PCl5  
 [PCl4 ]  [PCl6 ]


(Solid) sp3 sp 3d 2 88. Answer (3)

Dettol is used as an antiseptic.
78. Answer (1)
89. Answer (1)
Species CO N2 O2 H2 No. of stereoisomers = 24 = 16
Bond order 3 2.5 2.5 1 90. Answer (1)
79. Answer (3) Benzylamine is most basic due to localized lone
First period contains only s-orbital. pair of electrons on N-atom.

BOTANY
91. Answer (3) The amount of DNA (genetic material) doubles
but the number of chromosome remains same
The orange colour of carrot roots is due to
in S-phase. Hence according to given question.
chromoplast.
Initial amount of DNA = 30 C
92. Answer (4)
and, number of chromosome in a diploid cell =
Telomeres are ends of the chromosome and 12
they seal their ends. Telomere also prevent Therefore in S-phase of cell cycle, amount of
shortening of chromosome (or chromosome DNA = 2(30 C) = 60 C
loss).
and, number of chromosomes will remain 12 in
93. Answer (3) the diploid cell.

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Test-10_(Hints & S olutions) Aak ash Rank Boo ster Test Serie s for NEET -2020

94. Answer (4) 108. Answer (1)

The chromosomes are thickest and shortest in Rhizobium is associated with legumes, while,
metaphase. Hence, this is the best stage to Frankia and Anabaena are associated with
study morphology of chromosome. The best non-legumes. Azotobacter is free-living.
stage to study shapes of chromosomes is 109. Answer (3)
anaphase
In C 3 pathway, 3 ATP are consumed to fix one
95. Answer (2) CO2. Hence, to fix three molecules of CO2; 9 (or
Growth is an intrinsic property of living organism 3 × 3) ATPs are required.
through which it increases mass and number of 110. Answer (3)
cells in their bodies. In 1905, Blackman gave law of limiting factor.
96. Answer (4) 111. Answer (1)
Spirogyra. -Ketoglutaric acid is raw material for amino
97. Answer (2) acid synthesis.
Phycomycetes. 112. Answer (3)
98. Answer (2) RQ of oxalic acid is 4.0 while it is 1.33 for malic
Chrysanthemum stunt is a disease caused by acid.
viroids. 113. Answer (2)
99. Answer (1) Cytokinins delay senescence (ageing) of leaves
Muliathi belongs to the family fabaceae. and other organs by controlling protein
synthesis and mobilisation of nutrients as
100. Answer (3) resources (Richmond lang effect).
Flower is epigynous in bittergourd. 114. Answer (1)
101. Answer (1) A–Auxin, B–Cytokinin
The gritty texture of guava is due to the 115. Answer (1)
presence of hard cells or sclereids in their pulp.
Embryogenesis does not occur in algae.
102. Answer (4)
116. Answer (3)
Root hairs facilitate absorption of water and
Chara produces morphologically distinct
minerals.
gametes (heterogametes).
103. Answer (1)
117. Answer (4)
Wolfia is the smallest angiosperm.
Tapetal cells nourish the developing pollen
104. Answer (1) grains.
Bryophytes are non-vascular terrestrial plants of 118. Answer (3)
moist habitat. Meiotic division to produce ‘n’ number of seeds
105. Answer (1) n
in an angiosperms = n 
The value of p i.e., pressure potential would be 4
more positive for a turgid cell while it will be 200
zero (or, negligible) for a flaccid cell.  200   250 .
4
106. Answer (1)
119. Answer (2)
When a cell is flaccid, its DPD = OP
Genotype of plant = AaBBccDdEe. Hence, it
107. Answer (2) has 3 heterozygous pairs (n = 3)
Nitrogen  Major constituent of protein, nucleic Hence, types of gametes = 2n  2 3 8
acid 120. Answer (2)
Calcium  Activates ATPase Transversion.
Magnesium  Binding of ribosome subunit 121. Answer (3)
Iron Formation of chlorophyll Turner’s syndrome  44 × XO (Monosomy).
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Aakash Rank Boo ster T est Series for NEET -2020 Test-10 _(Hin ts & Solutio ns)
122. Answer (1) Barnacles growing on the back of a whale
In the presence of enzyme aminoacyl -tRNA benefits in the shelter. Whale remains
synthetase, specific amino acid binds with ATP. unaffected. Hence, this shows commensalism.
Hence, amino acids get activated. 130. Answer (2)
123. Answer (1) Anthropogenic ecosystems have high
productivity.
Lac z codes for -galactosidase enzyme in lac
operon. 131. Answer (2)
124. Answer (1) Deep sea is least productive ecosystem.
132. Answer (1)
A–11 million tonnes
There are 14 biosphere reserves in India.
B–75 million tonnes
133. Answer (1)
125. Answer (4)
Quagga is recently become extinct from Africa.
Cowpea.
134. Answer (2)
126. Answer (1) Montreal protocol was signed at Montreal
In our stomach, lactic acid bacteria (LAB) play (Canada) in 1987 (effective in 1989) to control
beneficial role in checking diseases causing the emission of ozone-depleting substances
microbes. (ODS).
127. Answer (3) 135. Answer (4)
Statins are produced by Monascus purpureus Chipko movement  Chandi Prasad Bhatt
128. Answer (2) Appiko movement  Pandurang Hegde
Mortality. Wildlife protection award  Amrita Devi
129. Answer (1) Remedy for plastic waste  Ahmed Khan

ZOOLOGY
136. Answer (3) fructose participate in formation of glycosidic
bond between the two.
140. Answer (1)
Parietal cells in mucosa of stomach secrete HCl
which converts pepsinogen secreted by peptic
cells of gastric mucosa into its active form
pepsin (proteolytic enzyme). Goblet cells are
mucus secreting unicellular glands. Brunner’s
137. Answer (4) glands are found in submucosa of duodenum.
Smooth muscle fibres do not show any light and 141. Answer (2)
dark bands. That is why, they are called smooth
Micelles formed in intestinal lumen are
or unstriped or non-striated muscles.
absorbed by intestinal cells which are reformed
138. Answer (2)
into chylomicrons in the enterocytes.
It represents a molecule of palmitic acid which
142. Answer (2)
has 16 carbons including carboxyl carbon and
is a saturated fatty acid. Arachidonic acid has Residual volume is the volume of air remaining
20 carbon atoms including the carboxyl atom. in the lungs even after a forcible expiration. This
averages 1100 ml to1200 ml.
139. Answer (3)
143. Answer (3)
All reducing sugars possess a free aldehyde or
ketone group. All monosaccharides have this About 20-25% of CO2 is carried by haemoglobin
ability. Among disaccharides, sucrose (glucose as carbamino-hemoglobin, whereas 70% of it is
+ fructose) is non-reducing because both carried as bicarbonate. About 7% of CO2 is
aldehyde group of glucose and ketone group of carried in a dissolved state through plasma.

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Test-10_(Hints & S olutions) Aak ash Rank Boo ster Test Serie s for NEET -2020

144. Answer (3) 151. Answer (1)

RBCs are biconcave disc shaped cells. Fovea is a thinned-out portion of the retina
Monocytes often change into macrophages in where only the cones are densely packed.
connective tissues. Lymphocytes exist as B and Space between the lens and cornea is called
T-lymphocytes and are responsible for immune aqueous chamber and is filled with aqueous
responses of the body. humor. Macula lutea is a yellowish pigmented
spot with a central pit called the fovea.
145. Answer (2)
152. Answer (3)
Normally heart beats 72 times per minute, i.e., Thermoregulation – Hypothalamus
that many cardiac cycles are performed per
Long term memory – Hippocampus
minute.
Language comprehension – Cerebrum
Duration of 1 cardiac cycle
153. Answer (3)
60 seconds Insulin promotes glucose uptake by the cells.

72 beats Sexcorticoids help develop and maintain
= 0.8 second secondary sex characters. Calcitonin is
secreted by thyroid gland.
In case of raised heart rate say
154. Answer (3)
120 beats/min., duration of each cardiac cycle
reduces to 0.5 second. Reptilian : Poikilothermy, pulmonary
features circulation, oviparity
146 Answer (3)
Avian : Homeothermy, complete double
Blood Antigens Antibodies Donor’s features circulation, oviparity, pneumatic
Group on RBCs in Plasma Group bones
A A anti-B A, O
155. Answer (4)
Salpa is a urochordate, Urochordata is a
B B anti-A B, O subphylum of phylum Chordata.
AB A, B nil AB, A, B,O 156. Answer (3)
Labeo (Rohu), Clarias and Pterophyllum are
O nil anti-A, B O
fresh water fish.
147. Answer (2) Scoliodon and Pristis have placoid scales.
157. Answer (1)
Increase in aldosterone levels cause an
increase in sodium reabsorption in the distal Male cockroaches bear a pair of short, thread-
convoluted tubule of nephron. like anal styles which are absent in females. In
both sexes, the 10th segment bears a pair of
148. Answer (2) jointed filamentous structures called anal cerci.
GFR in a healthy individual is approximately 158. Answer (2)
125 ml/minute i.e., 180 litres per day Hormones like human chorionic gonadotropin
149. Answer (3) (hCG) and human placental lactogen (hPL) are
secreted by placenta. Relaxin is also secreted
Fibrous joint : between skull bones
by ovary in later phases of pregnancy.
Cartilaginous joint : between adjacent vertebrae, 159. Answer (2)
sternum-ribs, at pubic
Haploid cells – Ovum, sperm, secondary
symphysis
spermatocyte
150. Answer (2) 160. Answer (1)
Binding of Ca++ to troponin results in unmasking Earthworms, sponge, tapeworm and leech are
of myosin binding site on actin and leads to examples of monoecious animals that possess
cross bridge formation. A-bands retain their both male and female reproductive organs.
length whereas I-bands get reduced in length Roundworm is an example of a dioecious
during shortening of the sarcomere. animal.
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Aakash Rank Boo ster T est Series for NEET -2020 Test-10 _(Hin ts & Solutio ns)

161. Answer (2) 169. Answer (2)

The second meiotic division is unequal and The yellowish colostrum secreted by mother
results in the formation of a second polar body during the initial days of lactation has abundant
and a haploid ovum (ootid). Soon the haploid antibodies (IgA) to protect the infant. This is
nucleus of the sperm and that of the ovum fuse also an example of passive naturally acquired
together to from a diploid zygote. immunity.
162. Answer (1) 170. Answer (4)
In vasectomy, a small part of the vas deferens Dengue and chikungunya are viral diseases
is removed or tied up through a small incision whose pathogens are commonly spread by
on the scrotum. No sperm occurs in semen. Aedes mosquito (vector). Pathogen of
163. Answer (2) elephantiasis or filariasis is transmitted through
Non-medicated IUD : Lippes loop the bite of female Culex mosquito. Psoriasis is
an autoimmune disease.
Hormonal implant : Norplant
171. Answer (2)
Non-steroidal pill : Saheli
164. Answer (1) NACO is National AIDS Control Organisation.

p = 0.8 172. Answer (4)

q = 0.2 ( q + p = 1) Each antibody molecule has four peptide

chains, two small called light chains (L) and two
ww = q2 = (0.2)2 = 0.4 = 4 % longer called heavy chains (H). Hence, an
WW = p 2 = (0.8)2 = 0.64 = 64 % antibody is represented as H 2L 2.
Ww = 2qp = 2(0.2)(0.8) = 0.32 = 32% 173. Answer (4)
165. Answer (2) Motile sporozoites penetrate the salivary glands
Sweet potato (root modification) and potato of mosquito.
(stem modification) is an example of analogy 174. Answer (1)
since these two structures are not anatomically Continued inbreeding, especially close
similar structures though they perform similar
inbreeding can cause inbreeding depression. In
functions.
interspecific hybridisation male and female
166. Answer (3) animals of two different related species are
Homo habilis – 650 to 800 cc mated. MOET is a programme for herd
improvement.
Homo erectus – 900 cc
Cro Magnon man – 1650 cc 175. Answer (2)

Neanderthal man – 1400 - 1450 cc In a method known as biolistics or gene gun

plant cells are bombarded with high -velocity
167. Answer (3)
microparticles of gold or tungsten coated with
Papaver somniferum  morphine DNA. Elution is separation and extraction of
Acetylation of morphine gives heroin DNA fragments from the gel piece after gel
electrophoresis.
Cannabis sativa  marijuana, hashish, charas
and ganja. 176. Answer (2)
168. Answer (2) Ampicillin resistant gene (ampR) and
R
Asthma is difficulty in breathing causing Tetracycline resistant gene (tet ) are selectable
wheezing due to inflammation of bronchi and markers present in pBR322. ampR has site for
bronchioles. Allergic asthma is the most Bam HI whereas tetR has site for Pst I.
common type of asthma. Gout is inflammation 177. Answer (1)
of joints due to accumulation of uric acid
Sticky ends : Sal I, Hind III, Eco RI
crystals. Alzheimer’s disease is a
neurodegenerative disease. Flush ends : Eco RV, Pvu II and Sma I

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Test-10_(Hints & S olutions) Aak ash Rank Boo ster Test Serie s for NEET -2020
178. Answer (1) 180. Answer (3)
Being hydrophilic, DNA molecules cannot pass Golden rice is enriched with vitamin-A. This
through cell membrane.
method of breeding crops to increase their
179. Answer (4)
nutritional value is called biofortification. In ‘flavr
A nematode Meloidogyne incognita infects the
savr’ transgenic tomato, expression of gene
roots of tobacco plants causing a great
reduction in yield. RNAi technique is used to which produces polygalacturonase is blocked,
bioengineer resistant plants that prevents the thus over-ripening of tomato is prevented
nematode from feeding on the roots. increasing its shelf life.

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