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Test - 7 (Code A) (Answers & Hints) All India Aakash Test Series for Medical-2017

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1. (2)

2. (1)

3. (2)

4. (3)

5. (2)

6. (1)

7. (4)

8. (2)

9. (1)

10. (3)

11. (1)

12. (3)

13. (4)

14. (2)

15. (2)

16. (3)

17. (4)

18. (4)

19. (3)

20. (4)

21. (2)

22. (3)

23. (2)

24. (4)

25. (1)

26. (1)

27. (4)

28. (1)

29. (1)

30. (1)

31. (4)

32. (3)

33. (4)

34. (1)

35. (3)

36. (3)

ANSWERS

TEST - 7 (Code-A)

All India Aakash Test Series for Medical-2017

Test Date : 12-02-2017

37. (2)

38. (2)

39. (4)

40. (3)

41. (1)

42. (2)

43. (2)

44. (1)

45. (3)

46. (3)

47. (3)

48. (2)

49. (3)

50. (3)

51. (2)

52. (1)

53. (1)

54. (4)

55. (2)

56. (2)

57. (4)

58. (4)

59. (1)

60. (3)

61. (1)

62. (4)

63. (1)

64. (4)

65. (2)

66. (4)

67. (4)

68. (4)

69. (4)

70. (3)

71. (3)

72. (2)

73. (2)

74. (4)

75. (3)

76. (1)

77. (2)

78. (1)

79. (3)

80. (3)

81. (4)

82. (1)

83. (2)

84. (2)

85. (1)

86. (3)

87. (4)

88. (4)

89. (1)

90. (3)

91. (4)

92. (3)

93. (2)

94. (1)

95. (4)

96. (1)

97. (4)

98. (2)

99. (2)

100. (3)

101. (1)

102. (4)

103. (2)

104. (3)

105. (4)

106. (2)

107. (2)

108. (3)

109. (3)

110. (4)

111. (2)

112. (3)

113. (4)

114. (3)

115. (2)

116. (2)

117. (2)

118. (4)

119. (3)

120. (4)

121. (3)

122. (2)

123. (3)

124. (2)

125. (3)

126. (1)

127. (3)

128. (3)

129. (4)

130. (2)

131. (2)

132. (4)

133. (2)

134. (4)

135. (2)

136. (1)

137. (2)

138. (3)

139. (4)

140. (3)

141. (1)

142. (2)

143. (3)

144. (2)

145. (3)

146. (1)

147. (4)

148. (1)

149. (2)

150. (4)

151. (4)

152. (3)

153. (3)

154. (1)

155. (3)

156. (4)

157. (4)

158. (4)

159. (2)

160. (3)

161. (4)

162. (1)

163. (2)

164. (3)

165. (3)

166. (3)

167. (1)

168. (2)

169. (3)

170. (1)

171. (3)

172. (1)

173. (4)

174. (3)

175. (2)

176. (3)

177. (1)

178. (3)

179. (2)

180. (4)

All India Aakash Test Series for Medical-2017 Test - 7 (Code A) (Answers & Hints)

2/12

Hints to Selected Questions

[ PHYSICS]

1. Answer (2)

d sin = n

sindn

For n to be max sin = max = 1

max

dn

⎡ ⎤ ⎢ ⎥⎣ ⎦

Number of maxima on both sides

2d⎛ ⎞

⎜ ⎟⎝ ⎠=

2 70007

2000

2. Answer (1)

(2 1)

2 2

d n Dy

d

2

(2 1)

d

n D

2 2 2

, , 3 5

d d d

D D D

3. Answer (2)

4. Answer (3)

1

4 1

4( )

Dx

d

2

3 2

3( )

Dx

d

1 2

4 3

4 3,

D Dx x

d d

1

2

3

4

5. Answer (2)

At x = 0 path difference is 3, hence third order

maxima will be obtained.

At x = path difference = 0. Hence, zero order

maxima is obtained. In between 1st and 2nd order


6. Answer (1)

Distance between two consecutive maxima = 2

0.14 142

0.14 20.02 m

14

8

103 101.5 10 Hz

0.02

c ⇒

7. Answer (4)

8. Answer (2)

n = 2d sin

2 sind

n

max

= 2d = 2 × 2.8 × 10–8

= 5.6 × 10–8 m

9. Answer (1)

20 01

cos2 2 4

I II

⎛ ⎞ ⎜ ⎟⎝ ⎠ 0

12.5%8

I

10. Answer (3)

Direction of light

ˆ ˆ ˆˆ 2 3n i j k

Angle with y axis

2 2

2cos

| | 1 2 3

ya

a

1 12 2cos cos

1 4 9 14

⎛ ⎞ ⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠


3/12

11. Answer (1)

20

2

21

mE c

v

c

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

2 4 2 6

2 0 0

2 2 2

21

m c m cE

v c v

c

2 42

0

2 2 2

m cE

c c v

...(i)

0

2

21

m vP

v

c

2 2

2 0

2

21

m vP

v

c

2 2 2

2 0

2 2

m v cP

c v

...(ii)

Solving the equation (i) – (ii)

2 4 2 2 222 0 0

2 2 2 2 2

m c m v cEP

c c v c v

2 222 2 20

2 2 2( )

m cEP c v

c c v

22 2 2

02

EP m c

c

2 2 2 2 4

0E c P m c

2 2 2 2 4

0E P c m c

For photon m0 = 0

E2 = P2c2

E = Pc

For electron m0 0 E Pc

12. Answer (3)

0 03

hceV

⎡ ⎤ ⎢ ⎥⎣ ⎦

0 03

2

hceV

⎡ ⎤ ⎢ ⎥⎣ ⎦

0 03

hceV

0 0

33 3

2

hceV

_______________________

0

31 2

2

hc ⎛ ⎞ ⎜ ⎟⎝ ⎠

0

12

2

hc ⎛ ⎞ ⎜ ⎟⎝ ⎠

0

22

hc hc

0 = 4

13. Answer (4)

0

= h0

= e × 4.14 × 10–15 × 1 × 1014

= 0.414 eV

14. Answer (2)

Intensity (aperture)2

Since aperture (Diameter) remains same

Intensity will remain same.

15. Answer (2)

v2

v1

B

A

1 2 1 2

2 2cm

mv mv v vv

m

� � � �

�

Velocity of A w.r.t. C frame

1 2

12

AC A C

v v

v v v v

⎛ ⎞ ⎜ ⎟⎝ ⎠

� �

� � � �

1 2

2

v v� �


4/12

2 2 2 2

1 2 1 2| | , | |2 2

� �

AC CM

v v v v

v v

| | | |AC CM

v v� �

| |AC

h

m v

�

2 2

1 2

2h

m v v

2 2

1 2

2 ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

h

mh h

m m

2 2

1 2

2

1 1

h

m h

m

1 2

eq 2 2

1 2

2

16. Answer (3)

1

T

(327 273) 600 1

(927 273) 1200 2

2

17. Answer (4)

2

P

KeE

r

2

2

keE

r

EP = 2E = 2(–13.6)eV = – 27.2 eV

Potential energy of electron in the ground state of

Li2+ ion is – 32 × 27.2 eV = – 244.8 eV.

18. Answer (4)

2

nhL

19. Answer (3)

Making potential energy zero increases the value of

total energy by

13.6 –(–13.6) = 27.2 eV

Now actual energy in second orbit = – 3.4 eV

Hence, new value is (–3.4 + 27.2 eV) = 23.8 eV

20. Answer (4)

X Y Product

y yN

x xN

Net rate of formation of y at any time t is

y

x x y y

dNN N

dt

Ny is maximum when 0

ydN

dt

xNx =

yNy

21. Answer (2)

Total required energy = BE of electron in He atom

= (24.6 + 13.6 × 22) eV

= (24.6 + 54.4) eV

= 79 eV

22. Answer (3)

23. Answer (2)

y x h

3 3 3 ( )

4 4 7

x x hy h y y ⇒ ⇒ ⇒

4

7

hx

24. Answer (4)

d/2

d/2 O

\\\\\\\\\\\\ \\\\ \\\\\\\\\\\\\\\\\\\


5/12

When 2

df , the rays from one mirror after

reflection will reach parallel to the other mirror.

Second mirror will refocus them at O. When object

is at 2F, image is formed at 2f.

25. Answer (1)

1.52 1.33 1.52 1.33

8 2v

or v = – 21.3 cm ; 1.33( 21.3)

2.331.52 (8)

M

26. Answer (1)

3 cm

1.5 1 1.5 13cm

3 3v

v

27. Answer (4)

0.5x

r

45°

45°

45°

Length of shadow = x + 0.5 m

x = 0.5 tan r

sin45º sin r

1 4sin sin

32 r r

3sin

4 2r

3tan

23r

28. Answer (1)

If the rays are to retrace the path, light ray must fall

normal on the mirror. Hence, I should be 20 cm from

mirror and 15 cm from lens.

1 1 1

15 20x

or1 1 1 1

15 20 60x

x = – 60 cm, i.e., 60 cm away from lens.

\\\\\\\\\\\\ \\\\ \\\\\\\\\\\\\\\\\\\

15 cml

5

20

29. Answer (1)

1.22 1mm

3mm d

or

6

7

3 105 m

1.22 5 10d

30. Answer (1)

1 1 1

v u f

or1 1 1

200 2000 100v

or v = 200 cm

Using v I

u o

2 150 m or 5 cm

2000 20I

31. Answer (4)

1 2

2 2

t t

t

or

1 2

2 1 1


6/12

32. Answer (3)

From case (i) we get flens

= 15 cm

Since lens

= 1.5 flens

= Rlens

= 15 cm

In case (ii) focal length of the combination = 40 cm

[Combination of lens + combination of (planoconcave)

liquid lens]

liquid lens

1 1 1

40 15 f

or

liquid lens

1 1 1

15 40f

fliquid lens

= – 24 cm

liquid lens

1 1 1 11 or

24f R

⎡ ⎤ ⎢ ⎥⎣ ⎦

= 11 or

15

⎡ ⎤ ⎢ ⎥⎣ ⎦

13

8

33. Answer (4)

i

rr

4

117

5

2

i

1

1

1 sin 5tan ,

12 sin

17

ii

r

17

3.4 1.845

34. Answer (1)

35. Answer (3)

36. Answer (3)

37. Answer (2)

I 4

3 2

12 V

122 A

4 2I

38. Answer (2)

39. Answer (4)

Voltage drop across R = 4 volt

So3

4

10 10R

= 400

40. Answer (3)

out

(10 )L C B

V R I k I

1mV10 k 100 1 V

1k

⎛ ⎞ ⎜ ⎟⎝ ⎠

41. Answer (1)

+

20 V

250

1 k

I

5 V

I1

I2

15 V

2

5 V 15 V20 mA, = 15 mA

250 1kI I

= I1 = 5 mA

42. Answer (2)

= 100, RL = 2 k, R

1 = 1 k

Voltage magnification

out

in

L

i

V R

V Rin

2

V

3

3

(100)(2 10 )

1 10

Vin = 0.01 V = 10 mV

43. Answer (2)

44. Answer (1)

NAND

45. Answer (3)


7/12

[ CHEMISTRY]

46. Answer (3)

Phenoxide ion is more resonance stabilised,

therefore, phenol is more acidic.

47. Answer (3)

Alkyl/aryl group of Grignard reagent has a tendency

to form alkane/arene by accepting an acidic hydrogen.

C H MgBr + C H OH6 5 2 5

C H + Mg6 6

Br

OC H2 5

(Benzene)

48. Answer (2)

The structures of possible alcohols having formula

C4H

9OH are as

OH

Butanol OH

Butan-2-ol

OH

2-methylpropan-1-ol

OH

2-methylpropan-2-ol

Among these, 2-methylpropan-2-ol cannot be

prepared by the reduction of carbonyl compounds.

49. Answer (3)

2

2

Cl

2 5 3 3

form Cl

C H OH [O] CH CHO CCl CHO

(X) (Y)

50. Answer (3)

CH C CH OH3 2

CH3

CH3

H+

–H O2

H C C CH3 2

CH3

CH3

neo-pentylalcohol

CH C CH CH3 2 3

CH3

1, 2-CH shift3

–H+

CH C CH CH3 3

CH3

2-methyl-2-butene(Major product)

51. Answer (2)

The oxidation product of X reacts with phenyl

hydrazine, thus it contains C O group. The

same product does not respond to silver mirror test,

thus it is a ketone because only aldehydes give

this test. Thus, the compound X must be 2º alcohol,

as only secondary alcohols give ketones on

oxidation and hence, X is (CH3)2 CHOH.

(CH ) CHOH3 2

[O]

K Cr O + H SO2 2 7 2 4

CH C CH3 3

O

C H NHNH6 5 2

CH C NHNC H3 6 5

CH3

Phenylhydrazone

52. Answer (1)

Polyether is represented as

X YCrown

Total number of carbon and oxygen

Number of oxygen atom

53. Answer (1)

ONa + CH CH Br3 2

O CH CH2 3

(Williamson's synthesis reaction)

54. Answer (4)

It is a case of alkoxy mercuration demercuration :

Addition of solvent (CH3OH) takes place by

Markownikoff's rule without rearrangement.

55. Answer (2)

56. Answer (2)

Cl

ONa

Cl ONa O

Intermediate

57. Answer (4)

–NO2 group(EWG) increases acid strength of phenol

but its para isomer is more acidic than meta isomer.


8/12

58. Answer (4)

Maximum pKa value means least acidic (maximum

basic)

–NO2 : (EWG) increases acidic strength of phenol

–CH3 : (ERG) decreases acidic strength of phenol

In (4), lone pair on N-atom in ring makes it basic

(least acidic).

59. Answer (1)

C

C

O

O

O

Conc.

C

C

O

O

OH OH

Phenolphthalein(Colourless)

NaOH

C

COO–

O

OH

(Pink colour)

+

OH

H SO2 4

60. Answer (3)

Phenol is not sufficiently acidic to liberate CO2 from

NaHCO3 solution.

61. Answer (1)

62. Answer (4)

R C R HCN

KCN

O

C

R

R

OH

CN

LiAlH4

C

R

R

OH

CH NH2 2

63. Answer (1)

HCl is strong inorganic acid, hence it does not react

with ethanal whereas with other reagents, ethanal

reacts in following manner :

CH3CHO + Cl

2 CCl

3.CHO Chloral

CH3CHO + PCl

5 CH

3CHCl

2

CH CHO + NaHSO3 3

CH CH3

OH

SO Na3

CH3CHO + HCl No reaction

64. Answer (4)

Crotonaldehyde is produced by Aldol condensation of

acetaldehyde.

CH C H + H C CHO3

O–

+

H

H

dil. NaOH Nucleophilic addition

H C C C CHO3

H H

HO H

H CCH CH CHO3

–H O2

Elimination

Aldol

65. Answer (2)

Ca

HCOO

HCOO

Dry distillationHCHO + CaCO

3

Formaldehyde

Calcium formate

66. Answer (4)

-keto acid undergo decarboxylation readily.

67. Answer (4)

Phenol Benzoic acid

(1) Neutral FeCl3 sol. Violet colour Buff. coloured

ppt.

(2) NaHCO3

No reaction Brisk

effervescence

68. Answer (4)

CH COOH

H C3

H C3

P / I4 2

C COOH

H C3

H C3

I

COOH

COOH

O

O

O–H O

2

P O /2 5

Na, D

ry e

ther

Wurt

z r

eaction

69. Answer (4)

I Aromatic

II Aromatic but less basic than I due to

–NO2 (EWG)

III Aliphatic


9/12

IV Aromatic but less basic than I and II

both due to adjacent carbonyl group

Thus, order is

IV < II < I < III

70. Answer (3)

I Aromatic amine

II –CH2NH

2, thus maximum basic

III Aromatic but –NO2 group is EWG

IV C

O

directly attached with N, hence

least basic due to delocalization of lone

pair on nitrogen.

71. Answer (3)

2Br , KOH3 2 3 2

(1º amine)

CH CONH CH NH

4LiAlH3 3 2 2

(1º amine)

CH CN CH CH NH

4LiAlH3 3 3

(2º amine)

CH NC CH NHCH

4LiAlH3 2 3 2 2

(1º amine)

CH CONH CH CH NH

72. Answer (2)

It is 1° amine.

73. Answer (2)

Fact.

74. Answer (4)

Ethers neither react with metallic sodium nor with 2,

4 DNP.

75. Answer (3)

Diazotization in cold yield diazonium salt which on

reaction with dimethylaniline gives azo dye.

76. Answer (1)

X is less reactive towards nucleophilic

substitution reaction.

77. Answer (2)

NH2

N Cl2

OH

(X)

(Y)

H O2

78. Answer (1)

CHO CH NNHC H6 5

CHOH C NNHC H6 5

(CHOH)3

(CHOH)3

CH OH2

CH OH2

Glucose Osazone (Glucosazone)

3C H NHNH6 5 2

2H O, –NH ,

–C H NH2 3

6 5 2

79. Answer (3)

Isoelectric point of amino acid = 1 2a a

pK pK

2

= 2.3 9.7

2

= 12

2 = 6.0

80. Answer (3)

CHO COONH4

(CHOH)4

(CHOH)4

CH OH2

CH OH2

+ 2[Ag(NH ) ]OH3 2

Tollen's reagent

+ 2AgSilver

mirror

Glucose

81. Answer (4)

Polytetrafluoroethylene (teflon) is used to make ‘non-

stick’ cookware.

82. Answer (1)

The monomer units of Nylon-6, 6 are hexamethylene

diamine and adipic acid.

nH N (CH ) NH + nHO C (CH ) C OH2 2 6 2 2 4

O O

Adipic acid

N (CH ) N C (CH ) C 2 6 2 4

H H O O

nNylon-6, 6

553 K

High pressure

83. Answer (2)

Fact.

84. Answer (2)

85. Answer (1)

Ziegler-Natta catalyst [(C2H

5)3Al + TiCl

4] is used in

linear polymerisation to prepare high density

polythene (HDPE).


10/12

91. Answer (4)

D-Grassland, E - Arctic and Alpine tundra.

92. Answer (3)

93. Answer (2)

94. Answer (1)

Temperature is most ecologically relevant

environmental factor.

95. Answer (4)

Eurythermal

96. Answer (1)

Mammals are regulator

97. Answer (4)

98. Answer (2)

99. Answer (2)

When resources are limited then growth is logistic.

100. Answer (3)

101. Answer (1)

In mutualism and protocooperation both species are

benefited.

102. Answer (4)

103. Answer (2)

Population density increase when (Natality +

Immigration) > (Mortality + Emigration)

104. Answer (3)

105. Answer (4)

106. Answer (2)

Aquarium is an example of man-made ecosystem.

107. Answer (2)

Principle of competitive exclusion explained by

Gause.

[ BIOLOGY

]

108. Answer (3)

109. Answer (3)

Zooplanktons are primary consumers

110. Answer (4)

Desert and deep ocean have lowest NPP

111. Answer (2)

10% law for energy transfer was given by

Lindemann.

112. Answer (3)

Cow and zooplanktons are primary consumers.

113. Answer (4)

Pyramid of biomass in aquatic ecosystem is

inverted.

114. Answer (3)

115. Answer (2)

India has more than 50,000 genetically different strain

of rice.

116. Answer (2)

117. Answer (2)

A – 2.4%, B – 8.1%

118. Answer (4)

119. Answer (3)

Extinction of Steller ’s sea cow is due to over

exploitation by humans.

120. Answer (4)

Alexander Von Humboldt.

121. Answer (3)

Nile perch introduced into lake victoria led to

extinction of more than 200 species chichlid fish.

122. Answer (2)

Aravali Hills are in Rajasthan.

123. Answer (3)

86. Answer (3)

OCOCH3

COOH

Aspirin

87. Answer (4)

88. Answer (4)

Both reduces Tollen’s reagent.

89. Answer (1)

90. Answer (3)


11/12

124. Answer (2)

Species diversity decrease as we move away from

the equator towards pole.

125. Answer (3)

126. Answer (1)

127. Answer (3)

128. Answer (3)

CFC – 14%

129. Answer (4)

130. Answer (2)

DDT, SO2 and CO

2 are primary pollutant.

131. Answer (2)

CNG is cheaper.

132. Answer (4)

133. Answer (2)

Net primary productivity.

134. Answer (4)

135. Answer (2)

136. Answer (1)

Heroin is diacetylmorphine, muscimal drug is obtained

from a fungus Amantia muscimal, LSD from Claviceps

purpura. Amphetamines are synthetic stimulants that

are analogous to adrenaline.

137. Answer (2)

Apoptosis means programmed cell death. The

metastatic cells get vascularised at new locations.

138. Answer (3)

It uses RNA-dependent DNA polymerase i.e., reverse

transcriptase enzyme, uses CD4 protein present on

macrophages and TH surface.

139. Answer (4)

Telomerase-inhibitor, c-onc are related to cancer,

histamines and serotonins to allergy and

autoantibodies to auto-immune disorders.

140. Answer (3)

Opiates are depressants but interact with membrane

bound receptors on cells of GIT and CNS, produce

euphoria and analgesia and induce sleep.

141. Answer (1)

Charas, atropine, crack, mescaline.

142. Answer (2)

Alcohol is not a hallucinogen.

143. Answer (3)

HIV shows high mutation rate.

144. Answer (2)

Gynecomastia is breast enlargement in males.

145. Answer (3)

Anaplastic cells are poorly differentiated cells.

146. Answer (1)

Withdrawal symptoms appear when the regularly used

drug is abruptly discontinued.

147. Answer (4)

Peptic ulcer means gastric ulcer.

148. Answer (1)

Brooders pneumonia is a fungal disease of hens.

149. Answer (2)

FSH-like hormones induce superovulation.

150. Answer (4)

Rest three are fresh water fishes.

151. Answer (4)

Apiary is the place where bee-hives are kept.

152. Answer (3)

Outcrossing is mating of unrelated animals of same

breed.

153. Answer (3)

Tharparkar and kankrej - Cow

Surti - Buffalo

Leghorn and cornish - Exotic poultry

154. Answer (1)

To reverse inbreeding depression, mating should be

carried out between unrelated superior animals of the

same breed.

155. Answer (3)

Composite culturing is rearing of non-competitive

species in a single pond for maximum outputs.

156. Answer (4)

Maintenance of beehives in all seasons is needed.

157. Answer (4)

Ri-plasmid hairy root inducing plasmid.

Ti-plasmid tumor inducing plasmid.


12/12

� � �

158. Answer (4)

Taq polymerase is a thermostable enzyme and

optimum temperature is 72°C.

159. Answer (2)

Ethidium bromide is used as a stain, cos sites are

seen in lambda phage vector and Hind II is a

restriction endonuclease.

160. Answer (3)

Blunt ends are also called flush ends.

161. Answer (4)

YAC can carry upto 1 MB size, BAC upto 300-350 kB,

cosmid upto 45 kB, phage vector upto 23 kB and

plasmid upto 10 kB size of gene of interest.

162. Answer (1)

Multiple cloning site is also called as polylinker.

163. Answer (2)

OKT-3 is a monoclonal antibody that is used as an

immunosuppressive drug.

164. Answer (3)

Rest three are the parts of downstream processing.

165. Answer (3)

Ca2+

ions increase binding ability.

166. Answer (3)

All of them are used in RDT.

Reverse transcriptase is used to synthesise the copy

DNA i.e., complementary DNA (cDNA) by using

mRNA as a template.

167. Answer (1)

The method is suitable for plant cells and does not

use electrical impulses. Moreover the desired gene

is coated with gold or tungsten particles.

168. Answer (2)

Ampicillin and tetracycline are antibiotics.

-galactosidase gene is the selectable marker of

plasmids like PUC8.

169. Answer (3)

Thurioside and sporeins are the endotoxins produced

by Bacillus thuringinesis bacterium.

170. Answer (1)

It does not involve PCR.

171. Answer (3)

RFLP is restriction fragment length polymorphism.

172. Answer (1)

Coryza is the bacterial infection of poultry.

173. Answer (4)

Kosa slik is another name of tassar silk.

Muga silk is obtained from Antherea assamensis.

174. Answer (3)

Tracy sheep’s milk contained -1-antitrypsin protein.

175. Answer (2)

Antisense technology is based on RNA interference

which was discovered by Craig Mellow and Andrew

Fire. RNA interference naturally occurs in eukaryotes.

176. Answer (3)

X-rays, P32

and probe are being used in

autoradiography from before.

177. Answer (1)

ADA intravenously injected comes under ERT i.e.,

enzyme replacement therapy which does not come

under gene therapy.

178. Answer (3)

Brazzein protein is produced in a west african berry.

There are about 2 lakh varieties of rice in India.

Bioremediation is used of microorganism metabolism

to remove pollutants.

179. Answer (2)

Flavr savr tomato is resistant to over-ripening (an

abiotic stress) and Bt-soyabean is resistant to

insects (a biotic stress).

180. Answer (4)

cry genes are found in B.thuriengiensis bacterium.

Test - 7 (Code B) (Answers & Hints) All India Aakash Test Series for Medical-2017

1/12

1. (1)

2. (3)

3. (4)

4. (4)

5. (3)

6. (1)

7. (2)

8. (4)

9. (4)

10. (1)

11. (1)

12. (3)

13. (2)

14. (1)

15. (2)

16. (3)

17. (3)

18. (3)

19. (2)

20. (3)

21. (3)

22. (2)

23. (4)

24. (1)

25. (4)

26. (2)

27. (1)

28. (2)

29. (2)

30. (1)

31. (4)

32. (4)

33. (2)

34. (1)

35. (1)

36. (1)

ANSWERS

TEST - 7 (Code-B)

All India Aakash Test Series for Medical-2017

Test Date : 12-02-2017

37. (3)

38. (4)

39. (2)

40. (3)

41. (4)

42. (1)

43. (4)

44. (3)

45. (4)

46. (1)

47. (3)

48. (4)

49. (2)

50. (1)

51. (3)

52. (2)

53. (4)

54. (3)

55. (2)

56. (1)

57. (1)

58. (3)

59. (4)

60. (3)

61. (1)

62. (2)

63. (4)

64. (4)

65. (1)

66. (1)

67. (2)

68. (2)

69. (4)

70. (2)

71. (4)

72. (2)

73. (3)

74. (2)

75. (3)

76. (1)

77. (3)

78. (2)

79. (2)

80. (4)

81. (2)

82. (2)

83. (3)

84. (3)

85. (4)

86. (1)

87. (1)

88. (4)

89. (1)

90. (1)

91. (4)

92. (2)

93. (4)

94. (2)

95. (4)

96. (4)

97. (2)

98. (1)

99. (1)

100. (3)

101. (1)

102. (4)

103. (1)

104. (4)

105. (1)

106. (2)

107. (1)

108. (2)

109. (4)

110. (4)

111. (4)

112. (1)

113. (2)

114. (1)

115. (4)

116. (2)

117. (1)

118. (1)

119. (4)

120. (4)

121. (2)

122. (1)

123. (4)

124. (2)

125. (3)

126. (1)

127. (4)

128. (4)

129. (2)

130. (3)

131. (2)

132. (3)

133. (4)

134. (1)

135. (2)

136. (2)

137. (4)

138. (1)

139. (3)

140. (1)

141. (2)

142. (1)

143. (2)

144. (3)

145. (1)

146. (3)

147. (1)

148. (2)

149. (3)

150. (1)

151. (1)

152. (1)

153. (4)

154. (3)

155. (2)

156. (1)

157. (4)

158. (4)

159. (2)

160. (2)

161. (1)

162. (3)

163. (1)

164. (1)

165. (2)

166. (2)

167. (4)

168. (3)

169. (4)

170. (3)

171. (1)

172. (4)

173. (1)

174. (4)

175. (3)

176. (1)

177. (2)

178. (1)

179. (4)

180. (3)

All India Aakash Test Series for Medical-2017 Test - 7 (Code B) (Answers & Hints)

2/12

Hints to Selected Questions

[ PHYSICS]

1. Answer (1)

2. Answer (3)

NAND

3. Answer (4)

4. Answer (4)

= 100, RL = 2 k, R

1 = 1 k

Voltage magnification

out

in

L

i

V R

V Rin

2

V

3

3

(100)(2 10 )

1 10

Vin = 0.01 V = 10 mV

5. Answer (3)

+

20 V

250

1 k

I

5 V

I1

I2

15 V

2

5 V 15 V20 mA, = 15 mA

250 1kI I

= I1 = 5 mA

6. Answer (1)

out

(10 )L C B

V R I k I

1mV10 k 100 1 V

1k

⎛ ⎞ ⎜ ⎟⎝ ⎠

7. Answer (2)

Voltage drop across R = 4 volt

So3

4

10 10R

= 400

8. Answer (4)

9. Answer (4)

I 4

3 2

12 V

122 A

4 2I

10. Answer (1)

11. Answer (1)

12. Answer (3)

13. Answer (2)

i

rr

4

117

5

2

i

1

1

1 sin 5tan ,

12 sin

17

ii

r

17

3.4 1.845

14. Answer (1)

From case (i) we get flens

= 15 cm

Since lens

= 1.5 flens

= Rlens

= 15 cm

In case (ii) focal length of the combination = 40 cm

[Combination of lens + combination of (planoconcave)

liquid lens]


3/12

liquid lens

1 1 1

40 15 f

or

liquid lens

1 1 1

15 40f

fliquid lens

= – 24 cm

liquid lens

1 1 1 11 or

24f R

⎡ ⎤ ⎢ ⎥⎣ ⎦

= 11 or

15

⎡ ⎤ ⎢ ⎥⎣ ⎦

13

8

15. Answer (2)

1 2

2 2

t t

t

or

1 2

2 1 1

16. Answer (3)

1 1 1

v u f

or1 1 1

200 2000 100v

or v = 200 cm

Using v I

u o

2 150 m or 5 cm

2000 20 I

17. Answer (3)

1.22 1mm

3mm d

or

6

7

3 105 m

1.22 5 10

d

18. Answer (3)

If the rays are to retrace the path, light ray must fall

normal on the mirror. Hence, I should be 20 cm from

mirror and 15 cm from lens.

1 1 1

15 20x

or1 1 1 1

15 20 60x

x = – 60 cm, i.e., 60 cm away from lens.

\\\\\\\\\\\\ \\\\ \\\\\\\\\\\\\\\\\\\

15 cml

5

20

19. Answer (2)

0.5x

r

45°

45°

45°

Length of shadow = x + 0.5 m

x = 0.5 tan r

sin45º sin r

1 4sin sin

32 r r

3sin

4 2r

3tan

23r

20. Answer (3)

3 cm

1.5 1 1.5 13cm

3 3v

v


4/12

21. Answer (3)

1.52 1.33 1.52 1.33

8 2v

or v = – 21.3 cm ; 1.33( 21.3)

2.331.52 (8)

M

22. Answer (2)

d/2

d/2 O

\\\\\\\\\\\\ \\\\ \\\\\\\\\\\\\\\\\\\

When 2

df , the rays from one mirror after

reflection will reach parallel to the other mirror.

Second mirror will refocus them at O. When object

is at 2F, image is formed at 2f.

23. Answer (4)

y x h

3 3 3 ( )

4 4 7

x x hy h y y ⇒ ⇒ ⇒

4

7

hx

24. Answer (1)

25. Answer (4)

Total required energy = BE of electron in He atom

= (24.6 + 13.6 × 22) eV

= (24.6 + 54.4) eV

= 79 eV

26. Answer (2)

X Y Product

y yN

x xN

Net rate of formation of y at any time t is

y

x x y y

dNN N

dt

Ny is maximum when 0

ydN

dt

xNx =

yNy

27. Answer (1)

Making potential energy zero increases the value of

total energy by

13.6 –(–13.6) = 27.2 eV

Now actual energy in second orbit = – 3.4 eV

Hence, new value is (–3.4 + 27.2 eV) = 23.8 eV

28. Answer (2)

2

nhL

29. Answer (2)

2

P

KeE

r

2

2

keE

r

EP = 2E = 2(–13.6)eV = – 27.2 eV

Potential energy of electron in the ground state of

Li2+ ion is – 32 × 27.2 eV = – 244.8 eV.

30. Answer (1)

1

T

(327 273) 600 1

(927 273) 1200 2

2

31. Answer (4)

v2

v1

B

A

1 2 1 2

2 2cm

mv mv v vv

m

� � � �

�

Velocity of A w.r.t. C frame


5/12

1 2

12

AC A C

v v

v v v v

⎛ ⎞ ⎜ ⎟⎝ ⎠

� �

� � � �

1 2

2

v v� �

2 2 2 2

1 2 1 2| | , | |2 2

� �

AC CM

v v v v

v v

| | | |AC CM

v v� �

| |AC

h

m v

�

2 2

1 2

2h

m v v

2 2

1 2

2 ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

h

mh h

m m

2 2

1 2

2

1 1

h

m h

m

1 2

eq 2 2

1 2

2

32. Answer (4)

Intensity (aperture)2

Since aperture (Diameter) remains same

Intensity will remain same.

33. Answer (2)

0

= h0

= e × 4.14 × 10–15 × 1 × 1014

= 0.414 eV

34. Answer (1)

0 03

hceV

⎡ ⎤ ⎢ ⎥⎣ ⎦

0 03

2

hceV

⎡ ⎤ ⎢ ⎥⎣ ⎦

0 03

hceV

0 0

33 3

2

hceV

_______________________

0

31 2

2

hc ⎛ ⎞ ⎜ ⎟⎝ ⎠

0

12

2

hc ⎛ ⎞ ⎜ ⎟⎝ ⎠

0

22

hc hc

0 = 4

35. Answer (1)

20

2

21

mE c

v

c

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

2 4 2 6

2 0 0

2 2 2

21

m c m cE

v c v

c

2 42

0

2 2 2

m cE

c c v

...(i)

0

2

21

m vP

v

c

2 2

2 0

2

21

m vP

v

c

2 2 2

2 0

2 2

m v cP

c v

...(ii)

Solving the equation (i) – (ii)

2 4 2 2 222 0 0

2 2 2 2 2

m c m v cEP

c c v c v

2 222 2 20

2 2 2( )

m cEP c v

c c v

22 2 2

02

EP m c

c


6/12

2 2 2 2 4

0E c P m c

2 2 2 2 4

0E P c m c

For photon m0 = 0

E2 = P2c2

E = Pc

For electron m0 0 E Pc

36. Answer (1)

Direction of light

ˆ ˆ ˆˆ 2 3n i j k

Angle with y axis

2 2

2cos

| | 1 2 3

ya

a

1 12 2cos cos

1 4 9 14

⎛ ⎞ ⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

37. Answer (3)

20 01

cos2 2 4

I II

⎛ ⎞ ⎜ ⎟⎝ ⎠ 0

12.5%8

I

38. Answer (4)

n = 2d sin

2 sind

n

max

= 2d = 2 × 2.8 × 10–8

= 5.6 × 10–8 m

39. Answer (2)

40. Answer (3)

Distance between two consecutive maxima = 2

0.14 142

0.14 20.02 m

14

8

103 101.5 10 Hz

0.02

c ⇒

41. Answer (4)

At x = 0 path difference is 3, hence third order


At x = path difference = 0. Hence, zero order

maxima is obtained. In between 1st and 2nd order


42. Answer (1)

1

4 1

4( )

Dx

d

2

3 2

3( )

Dx

d

1 2

4 3

4 3,

D Dx x

d d

1

2

3

4

43. Answer (4)

44. Answer (3)

(2 1)

2 2

d n Dy

d

2

(2 1)

d

n D

2 2 2

, , 3 5

d d d

D D D

45. Answer (4)

d sin = n

sindn

For n to be max sin = max = 1

max

dn

⎡ ⎤ ⎢ ⎥⎣ ⎦

Number of maxima on both sides

2d⎛ ⎞

⎜ ⎟⎝ ⎠=

2 70007

2000


7/12

[ CHEMISTRY]

46. Answer (1)

47. Answer (3)

48. Answer (4)

Both reduces Tollen’s reagent.

49. Answer (2)

50. Answer (1)

OCOCH3

COOH

Aspirin

51. Answer (3)

Ziegler-Natta catalyst [(C2H

5)3Al + TiCl

4] is used in

linear polymerisation to prepare high density

polythene (HDPE).

52. Answer (2)

53. Answer (4)

Fact.

54. Answer (3)

The monomer units of Nylon-6, 6 are hexamethylene

diamine and adipic acid.

nH N (CH ) NH + nHO C (CH ) C OH2 2 6 2 2 4

O O

Adipic acid

N (CH ) N C (CH ) C 2 6 2 4

H H O O

nNylon-6, 6

553 K

High pressure

55. Answer (2)

Polytetrafluoroethylene (teflon) is used to make ‘non-

stick’ cookware.

56. Answer (1)

CHO COONH4

(CHOH)4

(CHOH)4

CH OH2

CH OH2

+ 2[Ag(NH ) ]OH3 2

Tollen's reagent

+ 2AgSilver

mirror

Glucose

57. Answer (1)

Isoelectric point of amino acid = 1 2a a

pK pK

2

= 2.3 9.7

2

= 12

2 = 6.0

58. Answer (3)

CHO CH NNHC H6 5

CHOH C NNHC H6 5

(CHOH)3

(CHOH)3

CH OH2

CH OH2

Glucose Osazone (Glucosazone)

3C H NHNH6 5 2

2H O, –NH ,

–C H NH2 3

6 5 2

59. Answer (4)

NH2

N Cl2

OH

(X)

(Y)

H O2

60. Answer (3)

X is less reactive towards nucleophilic

substitution reaction.

61. Answer (1)

Diazotization in cold yield diazonium salt which on

reaction with dimethylaniline gives azo dye.

62. Answer (2)

Ethers neither react with metallic sodium nor with 2,

4 DNP.

63. Answer (4)

Fact.

64. Answer (4)

It is 1° amine.


8/12

65. Answer (1)

2Br , KOH3 2 3 2

(1º amine)

CH CONH CH NH

4LiAlH3 3 2 2

(1º amine)

CH CN CH CH NH

4LiAlH3 3 3

(2º amine)

CH NC CH NHCH

4LiAlH3 2 3 2 2

(1º amine)

CH CONH CH CH NH

66. Answer (1)

I Aromatic amine

II –CH2NH

2, thus maximum basic

III Aromatic but –NO2 group is EWG

IV C

O

directly attached with N, hence

least basic due to delocalization of lone

pair on nitrogen.

67. Answer (2)

I Aromatic

II Aromatic but less basic than I due to

–NO2 (EWG)

III Aliphatic

IV Aromatic but less basic than I and II

both due to adjacent carbonyl group

Thus, order is

IV < II < I < III

68. Answer (2)

CH COOH

H C3

H C3

P / I4 2

C COOH

H C3

H C3

I

COOH

COOH

O

O

O–H O

2

P O /2 5

Na, D

ry e

ther

Wurt

z r

eaction

69. Answer (4)

Phenol Benzoic acid

(1) Neutral FeCl3 sol. Violet colour Buff. coloured

ppt.

(2) NaHCO3

No reaction Brisk

effervescence

70. Answer (2)

-keto acid undergo decarboxylation readily.

71. Answer (4)

Ca

HCOO

HCOO

Dry distillationHCHO + CaCO

3

Formaldehyde

Calcium formate

72. Answer (2)

Crotonaldehyde is produced by Aldol condensation of

acetaldehyde.

CH C H + H C CHO3

O–

+

H

H

dil. NaOH Nucleophilic addition

H C C C CHO3

H H

HO H

H CCH CH CHO3

–H O2

Elimination

Aldol

73. Answer (3)

HCl is strong inorganic acid, hence it does not react

with ethanal whereas with other reagents, ethanal

reacts in following manner :

CH3CHO + Cl

2 CCl

3.CHO Chloral

CH3CHO + PCl

5 CH

3CHCl

2

CH CHO + NaHSO3 3

CH CH3

OH

SO Na3

CH3CHO + HCl No reaction

74. Answer (2)

R C R HCN

KCN

O

C

R

R

OH

CN

LiAlH4

C

R

R

OH

CH NH2 2

75. Answer (3)

76. Answer (1)

Phenol is not sufficiently acidic to liberate CO2 from

NaHCO3 solution.


9/12

77. Answer (3)

C

C

O

O

O

Conc.

C

C

O

O

OH OH

Phenolphthalein(Colourless)

NaOH

C

COO–

O

OH

(Pink colour)

+

OH

H SO2 4

78. Answer (2)

Maximum pKa value means least acidic (maximum

basic)

–NO2 : (EWG) increases acidic strength of phenol

–CH3 : (ERG) decreases acidic strength of phenol

In (4), lone pair on N-atom in ring makes it basic

(least acidic).

79. Answer (2)

–NO2 group(EWG) increases acid strength of phenol

but its para isomer is more acidic than meta isomer.

80. Answer (4)

Cl

ONa

Cl ONa O

Intermediate

81. Answer (2)

82. Answer (2)

It is a case of alkoxy mercuration demercuration :

Addition of solvent (CH3OH) takes place by

Markownikoff's rule without rearrangement.

83. Answer (3)

ONa + CH CH Br3 2

O CH CH2 3

(Williamson's synthesis reaction)

84. Answer (3)

Polyether is represented as

X YCrown

Total number of carbon and oxygen

Number of oxygen atom

85. Answer (4)

The oxidation product of X reacts with phenyl

hydrazine, thus it contains C O group. The

same product does not respond to silver mirror test,

thus it is a ketone because only aldehydes give

this test. Thus, the compound X must be 2º alcohol,

as only secondary alcohols give ketones on

oxidation and hence, X is (CH3)2 CHOH.

(CH ) CHOH3 2

[O]

K Cr O + H SO2 2 7 2 4

CH C CH3 3

O

C H NHNH6 5 2

CH C NHNC H3 6 5

CH3

Phenylhydrazone

86. Answer (1)

CH C CH OH3 2

CH3

CH3

H+

–H O2

H C C CH3 2

CH3

CH3

neo-pentylalcohol

CH C CH CH3 2 3

CH3

1, 2-CH shift3

–H+

CH C CH CH3 3

CH3

2-methyl-2-butene(Major product)

87. Answer (1)

2

2

Cl

2 5 3 3

form Cl

C H OH [O] CH CHO CCl CHO

(X) (Y)


10/12

91. Answer (4)

92. Answer (2)

93. Answer (4)

Net primary productivity.

94. Answer (2)

95. Answer (4)

CNG is cheaper.

96. Answer (4)

DDT, SO2 and CO

2 are primary pollutant.

97. Answer (2)

98. Answer (1)

CFC – 14%

99. Answer (1)

100. Answer (3)

101. Answer (1)

102. Answer (4)

Species diversity decrease as we move away from

the equator towards pole.

103. Answer (1)

104. Answer (4)

Aravali Hills are in Rajasthan.

105. Answer (1)

Nile perch introduced into lake victoria led to

extinction of more than 200 species cichlid fish.

106. Answer (2)

Alexander Von Humboldt.

[ BIOLOGY

]

107. Answer (1)

Extinction of Steller ’s sea cow is due to over

exploitation by humans.

108. Answer (2)

109. Answer (4)

A – 2.4%, B – 8.1%

110. Answer (4)

111. Answer (4)

India has more than 50,000 genetically different strain

of rice.

112. Answer (1)

113. Answer (2)

Pyramid of biomass in aquatic ecosystem is

inverted.

114. Answer (1)

Cow and zooplanktons are primary consumers.

115. Answer (4)

10% law for energy transfer was given by

Lindemann.

116. Answer (2)

Desert and deep ocean have lowest NPP

117. Answer (1)

Zooplanktons are primary consumers

118. Answer (1)

119. Answer (4)

Principle of competitive exclusion explained by

Gause.

88. Answer (4)

The structures of possible alcohols having formula

C4H

9OH are as

OH

Butanol OH

Butan-2-ol

OH

2-methylpropan-1-ol

OH

2-methylpropan-2-ol

Among these, 2-methylpropan-2-ol cannot be

prepared by the reduction of carbonyl compounds.

89. Answer (1)

Alkyl/aryl group of Grignard reagent has a tendency

to form alkane/arene by accepting an acidic

hydrogen.

C H MgBr + C H OH6 5 2 5

C H + Mg6 6

Br

OC H2 5

(Benzene)

90. Answer (1)

Phenoxide ion is more resonance stabilised,

therefore, phenol is more acidic.


11/12

120. Answer (4)

Aquarium is an example of man-made ecosystem.

121. Answer (2)

122. Answer (1)

123. Answer (4)

Population density increase when (Natality +

Immigration) > (Mortality + Emigration)

124. Answer (2)

125. Answer (3)

In mutualism and protocooperation both species are

benefited.

126. Answer (1)

127. Answer (4)

When resources are limited then growth is logistic.

128. Answer (4)

129. Answer (2)

130. Answer (3)

Mammals are regulator

131. Answer (2)

Eurythermal

132. Answer (3)

Temperature is most ecologically relevant

environmental factor.

133. Answer (4)

134. Answer (1)

135. Answer (2)

D-Grassland, E - Arctic and Alpine tundra.

136. Answer (2)

Cry genes are found in B.thuriengiensis bacterium.

137. Answer (4)

Flavr savr tomato is resistant to over-ripening (an

abiotic stress) and Bt-soyabean is resistant to

insects (a biotic stress).

138. Answer (1)

Brazzein protein is produced in a west african berry.

There are about 2 lakh varieties of rice in India.

Bioremediation is used of microorganism metabolism

to remove pollutants.

139. Answer (3)

ADA intravenously injected comes under ERT i.e.,

enzyme replacement therapy which does not come

under gene therapy.

140. Answer (1)

X-rays, P32

and probe are being used in

autoradiography from before.

141. Answer (2)

Antisense technology is based on RNA interference

which was discovered by Craig Mello and Andrew Fire.

RNA interference naturally occurs in eukaryotes.

142. Answer (1)

Tracy sheep’s milk contained -1-antitrypsin protein.

143. Answer (2)

Kosa slik is another name of tassar silk.

Muga silk is obtained from Antherea assamensis.

144. Answer (3)

Coryza is the bacterial infection of poultry.

145. Answer (1)

RFLP is restriction fragment length polymorphism.

146. Answer (3)

It does not involve PCR.

147. Answer (1)

Thurioside and sporeins are the endotoxins produced

by Bacillus thuringinesis bacterium.

148. Answer (2)

Ampicillin and tetracycline are antibiotics.

-galactosidase gene is the selectable marker of

plasmids like PUC8.

149. Answer (3)

The method is suitable for plant cells and does not

use electrical impulses. Moreover the desired gene

is coated with gold or tungsten particles.

150. Answer (1)

All of them are used in RDT.

Reverse transcriptase is used to synthesise the copy

DNA i.e., complementary DNA (cDNA) by using

mRNA as a template.

151. Answer (1)

Ca2+

ions increase binding ability.

152. Answer (1)

Rest three are the parts of downstream processing.

153. Answer (4)

OKT-3 is a monoclonal antibody that is used as an

immunosuppressive drug.


12/12

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154. Answer (3)

Multiple cloning site is also called as polylinker.

155. Answer (2)

YAC can carry upto 1 MB size, BAC upto 300-350 kB,

cosmid upto 45 kB, phage vector upto 23 kB and

plasmid upto 10 kB size of gene of interest.

156. Answer (1)

Blunt ends are also called flush ends.

157. Answer (4)

Ethidium bromide is used as a stain, cos sites are

seen in lambda phage vector and Hind II is a

restriction endonuclease.

158. Answer (4)

Taq polymerase is a thermostable enzyme and

optimum temperature is 72°C.

159. Answer (2)

Ri-plasmid hairy root inducing plasmid.

Ti-plasmid tumor inducing plasmid.

160. Answer (2)

Maintenance of beehives in all seasons is needed.

161. Answer (1)

Composite culturing is rearing of non-competitive

species in a single pond for maximum outputs.

162. Answer (3)

To reverse inbreeding depression, mating should be

carried out between unrelated superior animals of the

same breed.

163. Answer (1)

Tharparkar and kankrej - Cow

Surti - Buffalo

Leghorn and cornish - Exotic poultry

164. Answer (1)

Outcrossing is mating of unrelated animals of same

breed.

165. Answer (2)

Apiary is the place where beehives are kept.

166. Answer (2)

Rest three are fresh water fishes.

167. Answer (4)

FSH-like hormones induce superovulation.

168. Answer (3)

Brooders pneumonia is a fungal disease of hens.

169. Answer (4)

Peptic ulcer means gastric ulcer.

170. Answer (3)

Withdrawal symptoms appear when the regularly used

drug is abruptly discontinued.

171. Answer (1)

Anaplastic cells are poorly differentiated cells.

172. Answer (4)

Gynecomastia is breast enlargement in males.

173. Answer (1)

HIV shows high mutation rate.

174. Answer (4)

Alcohol is not a hallucinogen.

175. Answer (3)

Charas, atropine, crack, mescaline.

176. Answer (1)

Opiates are depressants but interact with membrane

bound receptors on cells of GIT and CNS, produce

euphoria and analgesia and induce sleep.

177. Answer (2)

Telomerase-inhibitor, c-onc are related to cancer,

histamines and serotonins to allergy and

autoantibodies to auto-immune disorders.

178. Answer (1)

It uses RNA-dependent DNA polymerase i.e., reverse

transcriptase enzyme, uses CD4 protein present on

macrophages and TH surface.

179. Answer (4)

Apoptosis means programmed cell death. The

metastatic cells get vascularised at new locations.

180. Answer (3)

Heroin is diacetylmorphine, muscimal drug is obtained

from a fungus Amantia muscimal, LSD from Claviceps

purpura. Amphetamines are synthetic stimulants that

are analogous to adrenaline.

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