1. No category

DEEPER KASAUTI (MH-CET) - 2015
R
Date : 12th April, 2015
Model Answers with Solutions
-
Answers are given in brief with reasoning.
No reasoning is given, where it is not neceessary.
These model solutions are offered exclusively to DEEPER students.
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without permission. Any person who does any unauthorised act in relation to this publication,
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means may liable for the damages, phease not it carefully.
PART - A - PHYSICS
Q.1
Ans. (D)
1
mw2 A2
2
T. E. =
It is same every where.
Q.2
Ans. (D)
e=
dj
1
=
wb/sec
dt
60
 Induced emf e =
Q.3
Ans. (D)
m C  Q = mgh
gh 9.8500
=
= 1.160C
C
4.2103
 ΔQ =
Q.4
1
V
60
Ans. (C)
potential


1
radius
V1 r2
16 2r1
= 
=
V2 r1
V2 r1
 V2 = 8V
Q.5
Ans. (C)
(1)
Q.6
Ans. (A)
mv2
2000 × 10 × 10
Fe =
=
=104 N
r
20
Q.7
Ans. (C)
3.2 × 1.6 × 10-19
K.E. = e Vs =
ev = 3.2 eV
1.6 × 10-19
Q.8
Ans. (A)
V = n
=
V 1200
=
= 4m
n
300
list between node and antinode =

Q.9

4
 4
=
= 1 m
4
4
Ans. (C)
A1 = 10 ; A 2 

52  (5 3)2  25  75  100  10
A1 10 1
=
= =1:1
A 2 10 1
Q.10 Ans. (A)
poission's Ratio
longiludinal stress
=
latual strain
longitudinal strain
=
lateral strain
longitudinal strain/Y
=
lateral strain × Y
possion's Ratio
10
 2  1012
= 100
0.25
= 2  3  1011
= 8  1011 N/m2
(2)
Q.11
Ans. (A)
M2 = 3M1, R2 = 3R1
2GM
R
Ve =

V1
m1 R 2
=
=
V2
m2 R1
=
m1
3R 1

3m2
R1
= 1
 V1 = V2  Ve' = ee
Q.12 Ans. (A)
λ=

h
h
=
m1v1
m2 v 2
V1 m 2
=
V2 m1
Q.13 Ans. (B)
B1 =
B2 =
μ 0 2m
4μ 0 M
=
3
4π  d 
πd 3
 
2
μ0 m
2μ 0 M
=
4π  d 3
πd3
 
2
 B = B12 + B22 
 20.
μ 0M
πd 3
 2 5.
μ 0M
πd 3
μ 0M
16  4
πd 3
(3)
Q.14 Ans. (D)
1
1
1
1
30
1
=
+
+ =
=
C 12 24 8 120 4
C = 4f
V=

q
q
q
+
+ = 60
12 24 8
30q
= 60
120
q
 60
4
q = 240 f
 Vs =
q
240
=
= 10V ; q = CsV = 5  10 = 50C
C
24
Q.15 Ans. (D)
has sharp breaking domat high voltage.
Q.16 Ans. (B)
V = 0.75 V
Ig = 15 mA = 15  10-3 A
G=
V
0.75
750
=
=
= 50Ω
-3
Ig 15  10
15
S=
Ig
×G
I - Ig
=
15  103
 50
25  15  103
15  103  50
15  50
= 3
=
10 (25000  15) 24985
 0.03 
Q.17 Ans. (A)
n20 = 2n1
n20 = n1 + (20-1)  4
3n1 = n1 + 19  4
2n1 = 19  4
n1 = 19  2
n1 = 38 Hz
(4)
Q.18 Ans. (D)
p = tan-1(g)
= tan-1 (1.5)
= 570
Q.19 Ans. (D)
λ=
0.693
T
T
λ
0.693

0.693
6.93  104
T = 103 = 1000 min
Q.20 Ans. (D)
 R 2 L2 
+ 
I = M 
4
12 

R =
D
2
 D 2 L2 
I = M 
+ 
 16 12 
Q.21 Ans. (C)
E = nhr
=
nhc
λ
 pt =
p=
E
t
nhc
λ
n
pλ
6  108  6.63  103
=
=
t
hc
6.63  103  3  108
= 2  1015 photons/s
(5)
Q.22 Ans. (A)
a sin = n
a sin 30 = 5.460  10-7  1
a
= 5.460  10-7
2
a = 10.92  10-7
a = 1.092  10-6 m
a = 1.092  10-4 cm
Q.23 Ans. (B)
y = A sin wt
Hence
A
2π
= A sin t = A sin
t
2
T
1 π
 sin -1   = t
2 2
π
π
 t
2
6
t =
1
second
3
Q.24 Ans. (D)
Third avertone has frequency 7n which means total length of pipe l 
7

4
7
l    3 full loops + one half loop which would make four nodes and four antinodes
4
Q.25 Ans. (C)
 12.26  0
λ= 
A
 V 
(12.26)2
V =
λ2
(12.26) 2
=
(0.12) 2
= 10.44 kV
(6)
Q.26 Ans. (A)
E = I(R + r)
0.5 (2 + r) = 0.25(5 + r)
4 + 2r = 5 + r
r = 1
 E = 0.5 (2 + 1)
=
1
3
2
= 1.5 V
Q.27 Ans. (B)
Breaking force = Breaking stresix Area of Cross section
= Weight of wire
6
10  A = A l  pq
106
l 
pq
106
=
 34 m
3  103  9.8
Q.28 Ans. (B)
I1 = 2A,
B1 = 4  10-3 wb/m2
B2 = 8  10-3 wb/m2, Q1 = Q2
 C 
I = 
Q
 nAB 
 C 
 C 
 I1 = 
Q
,
I
=
 1

 Q2
2
nAB
nAB

1 

2 
I1
B2 8×10-3 2
 =
=
=
I2
B1 4×10-3 1
 I2 =
I1
2
= = 1A
2
2
Q.29 Ans. (C)
A = 1m2
q = 8.85 μC
12
1 q2
f=

2 0
2 A 2 0
(7)
1
8.85  1012
  2
2 1 8.85  1012
= 0.5 N
Q.30 Ans. (A)
 50 - 30  20k
=
t
 t 
R1 = 
R1
 Q - 30 
= k

2
 t 
R2=
20k
R
20
t
 1

R1  Q  30  Q  30
k
2  t 
2 
20
Q  30
2Q - 60 = 20
2Q = 20 + 60
2Q = 80
Q = 400C
Q.31 Ans. (A)
Angle between plane of the coil and the magnetic field = 300
 Angle between A + B = 600

 = nABcosQ = nAB  cos60 = nAB 
1 = 20  10  10-4  0.5  cos60
for 2  Q = 00
 2 = nAB
1 
nAB  -1 
Q - Q2
2 
q= 1
=
R
R
 1
20  10  104  0.5   
 2   104 C

50
= 100  10-6 C = 100C
(8)
1
2
Q.32 Ans. (C)
J = nIABsinQ
= 100  10  25  10-4  1.2  sin 300
= 1.5 Nm
Q.33 Ans. (A)
T=

2πr
Vc
2π(2R)
Vc
= 4 sec
Q.34 Ans. (A)
R
8400 84000
=
=
= 21000 J/kmole K
γ-1
0.4
4
Cv =
21000 10500
=
= 750J/kgK
28
14
 Cp= Cv = 14  750 = 1050 J/kgK
 Cv =
Q.35 Ans. (C)
W = T  dA
= T  2  4 (r22 - r12)
= 28  10-3  2  4  3.14  (16  10-4 - 4  10-4)
= 28  8  3.14  12  10-3  10-4
= 8440  10-7
= 8.44  10-4 J
Q.36 Ans. (B)
Q.37 Ans. (B)
n
=
1
2π
k
m
1 225π 2
2π 0.01
= 75 Hz
 50 Hz
 No resonance
(9)
Q.38 Ans. (C)
path diff = 27.5  10-6m = 55

27.5  106
55
= 5  10-7 m
= 5000 AU
Q.39 Ans. (C)
Q.40 Ans. (B)
1
2
Rot KE 2 IW
=
RollKE 7 mv2
10
1 2 2 2
 mr w
2
5
=
7
mr 2 w 2
10
1 10
 
5 7

2
7
Q.41 Ans. (D)
2
2
g2  R   R  1
  
 
g1  r   2 R  4
% charge =
4 1
 100 = 75%
4
Q.42 Ans. (B)
r = 1 (0.8 + 0.4) = 0.16
Q = 100 kcal
Qr = rQ = 0.16  100
= 16 kcal
(10)
Q.43 Ans. (C)
L = 4m,
l1 = 150 cm
l2 = 100 cm
R = 10
 l 1 l2 

r = R 
 l1 
 150  100 

 150 
10 = 10 
=
10  50
100
= 5 
Q.44 Ans. (B)
By caw of constation of angular momentum
I1W1 = I2W2
 MR 2

MR 2
w1 = 
+ mR 2  w 2
2
 2

M
M

n1 = 
+ m  n2
2
 2

50
 50

× 48 = 
+ 5 n2
2
 2

25  48 = (25 + 5)  n2
n2 =
25  48
= 40 rpm
30
Q.45 Ans. (C)
w = 720  10-9 m
μg
3
3 9 λw
=
 = =
μw
2
4 8 λg
 w  8 720  8  109
 g 

9
9
= 640  10-9
= 640 nm
(11)
Q.46 Ans. (D)
T=
0.693
,  = 1.388 per year
λ
=
0.693
1.386
=
1
2
= 0.5 year
Q.47 Ans. (C)
E1 = W0 + K.E.1 , E1 = E
W0 = constant
E2 = W0 + K.E.2 , E2 = 2E
E2 - E1 = K.E.2 - K.E.1
2E - E = KE2 - K.E.1
E = KE2 - K.E.1
i.e. E1 = KE2 - K.E.1
Wot K.E.1 = KE2 - K.E1
K.E.2 = W0 + K.E.1 + K.E.1
K.E2 = W0 + 2K.E.1
K.E.2 = 2 K.E.1
Q.48 Ans. (B)
Vi = 20V,  = 50, IB = 2mA, RL = 4k
V0 = IB RL
= 50  2  10-3  4  103
= 400 V
Av =
=
Vo
Vi
400
20
= 20
Q.49 Ans. (A)
1 q1q 2
9  109  2  2  10-6  10-6
F=
=T=
4π 0 r2
22
= 9  10-3 N
(12)
n=
1 T
2l M
1
9  10 3


2 2
0.001
3  103
=
4
= 0.75  10-3 Hz
= 0.75 mHz
Q.50 Ans. (D)
E =
F
= Eq
= 234  0.5
= 117 N
W = U = F. S.
= 117  2  10-2
= 234  10-2 J
= 2340 mJ
(13)
PART - B - CHEMISTRY
MH-CET-2015 Solution Version-II Answer and Solution Key Chemistry
Q.51 Ans. (B)
Veal/videal knowledge base
Q.52 Ans. (A)
I as size ‘! down the group it is basecily “! I is weakest base
Q.53 Ans. (B)
Amylase
Q.54 Ans. (B)
0.098 atm
Q.55 Ans. (B)
knowledge base
Q.56 Ans. (B)
knowledge base
Q.57 Ans. (C)
knowledge base
+10KJ
Q.58 Ans. (D)
knowledge base question
Q.59 Ans. (D)
4
Q.60 Ans. (A)
Zero order reaction
Q.61 Ans. (B)
lnA knowledge base theory question.
Q.62 Ans. (C)
He+O2 information
Q.63 Ans. (C)
F
Q.64 Ans. (B)
knowledge base question
Q.65 Ans. (A)
A2B3 (x = 2 y = 3)
(14)
Q.66 Ans. (B)
CO informative
Q.67 Ans. (B)
H2O + CO2
Q.68 Ans. (C)
Informative
Q.69 Ans. (B)
(n-2) F1-14 , (n-1)d0-1 ns2 f block element last two orbitals are not completely filled.
Q.70 Ans. (B)
It form BaSO4 while ppt with BaCl2 solution.
Q.71 Ans. (B)
5[Pt(NH3)6]Cl4  4Cl(-) + [Pt(NH3)6]4+ hence 5 ions
Q.72 Ans. (D)
(Ethyl cyanide) compound B is ethyl cynocide is incorrect statement
Na/C2H5OH
Red P
CH3-OH and I
2
CH3-I
(A)
Acoholic
KCN
CH3 CN
(B)
hyd
CH3 CN2NH2
roly
(C )
H+ sis
CH3 COOH
(D)
Q.73 Ans. (C)
Terhiary butyl halide will undergo faster SN' becouse it form Terhiary butyl carbonium ion tnen it
gives terhiary butanol.
Q.74 Ans. (B)
butab-2-ol
CH3 - CH2-CH-CH3
|
OH
(X)
Conc
H2SO4
CH3-CH=CH-CH3
(Y)
Q.75 Ans. (B)
OH
Br
OH
Br
3Br2
+3HBr
water
CH3
CH3
Br
Tru Bromo Derivative
(15)
1) O3
2) H2O/Zn
2CH3-C-H
||
O
(C2H4O) + H2O2
Q.76 Ans. (C)
O
O
||
||
CH3-CH2 - CH2-C-CH3 Pentan – 2-one has - C - CH3
grouping hence Iodoform testnis +ve for Pentan -2- one and not for pentan -3- one.
Q.77 Ans. (B)
p-Ethylb benzaldehyde on oxidation gives 1-4 benzene dicarboxylic acid hence compound x is
p-Ethylbenzaldehyde
Q.78 Ans. (B)
Information base
Q.79 Ans. (C)
Knowledge base question.
Q.80 Ans. (B)
knowledge base question.
Q.81 Ans. (A)
Sucrose knowledge base
Q.82 Ans. (A)
Informative
Q.83 Ans. (A)
CH2 - CH - CH2
|
|
Zn dust CH3 - CH = CH2 + ZnBr2
Br Br
Q.84 Ans. (D)
As branching ‘! in alkane it’s B. P.
hence D
Q.85 Ans. (C)
CH2-CH3
+ CH3-CH2-Cl
AlCl3
Ethyl Benzene m.f.(C8H10)
Q.86 Ans. (D)
0, 0, knowledge base if structure is know.
Q.87 Ans. (D)
CH3-C
Cl
Cl
Cl
Cl
+ 6 Ag + Cl
Cl
C - CH 3
CH3-C  C-CH 3
(16)
But-2-yne + 6 AgCl 
Q.88 Ans. (D)
knowledge base
Q.89 Ans. (C)
Nacl
Q.90 Ans. (D)
0.3 M Bacl2 it gives more ions.
Q.91 Ans. (A)
knowledge base.
Q.92 Ans. (B)
Anode knowledge base
Q.93 Ans. (B)
CH(OH)2.CHCO3 knowledge base
Q.94 Ans. (C)
Group-15 knowledge base
Q.95 Ans. (A)
knowledge base aliphatic primary amine
Q.96 Ans. (D)
Vitamin B is water soluable
Q.97 Ans. (A)
Glucose and fructose form soluable
Q.98 Ans. (A)
Chloroprene is monomer of natural rubber
Q.99 Ans. (D)
Barbitaric acid derivative is sleep inducing drug
Q.100 Ans. (C)
Sucrose is not an artificial sweetener
(17)
PART - C - BIOLOGY
MH-CET-2015 Solution Version-22 Answer and Solution Key
Q 101 – D Phytophthora palmivora it is used as microbial herbicide not biopesticide.
Q 102 – A Mg++
Q 103 – A Pusa gaurav this variety of mustard is resistant to aphids
Q 104 – A Leaching.
Q 105 – D Atlass-66
Q 106 – B Cleaving DNA segment with endonuclease and rejoining them with ligase.
Q 107 – B boolworms
Q 108 – C Coli
Q 109 – B Vit-D and vit –E these are fat soluble vitamins
Q 110 – C Phosphorous
Q 111 – D 2
Q 112 – C C4 plants agranal chloroplast is found in c4 plants
Q 113 – C eutrophication
Q 114 – C CH3 , CH3 group in chlorophyll –a is replaced with CHO group in chlorophyll –b
Q 115 – A Double fertilization.
Q 116 – A proton gradient
Q 117 – B Enoclase
Q 118 – B a-3 b-4 c-1 d-2
1.
Oxaloagcetate - 3). 4 carbon compund
2.
Phosphoglyceraldehyde -4). 3 carbon compund
3.
Icocitarate -1). 6 carbon compund
4.
α-ketoglutarate -2) 5 carbon compound
5.
Q 119 – C all hybride plants are tall when a tall plant was grown in nutrient deficient soil its
genotype remain same.
Q 120 – D infinity
Q 121 – B maternal sporophytic tissue in ovule
Q 122 – B -8 f1 progency shows genotype AaBbCc it will produce 8 different type of gametes.
Q 123 – C A polypeptide of 24 amino acids will be formed when 25th codon (VAV ) is mutated to VAA
then After forming polypeptide of 24 amino acids protein synthesis will be stopped.
Q 124 – C fumaric acid
Q 125 – D Succinic acid → fumaric acid
Q 126 – A open during night and close during the day
Q 127 – A Azospirillum
Q 128 – C a-3, b-5, c-1, d-2
a) monohybride cross–3) TtXTt
b) Test-Cross -5) Tt-Xtt
c) Alleles -1) T and t
d) homozygous tall–2) TT
Q 129 – D Bryophylllum and kalanchoe In these plants vegetative propogation occurs by leaf.
Q 130 – A regulatory gene
Q 131 – D PS II - P700
Q 132 – D 6PGAL and 120 ATP
Q 133 – B 264g
For synthesis of 1809 of glucose. 6 CO2 molecules are required
Molecular wt. of C-is-12 12×6=72
Molecular wt. of O-is-16
O2 = 16×2=32, 32×6=192, (72+192=264)
Q 134 – A extra chromosomal material
Q 135 – A 5’AAU3’ condon can be read only in 5’-3’ direction so far Anticondon 3’ UUA5’
condon is 5’AAU3’ direction so far anticodon 3’ UUA5’condon is 5’AAU3’
Q 136 – A denaturation, annealing synthesis
Q 137 – D-lichens.
Q 138 – C-100-125 rpm
Q 139 – B- segregation
Q 140 – B-Filiform apparatus
Q 141 – D-N-formylmethionine in eujcarytp AA, t-RNA complex has amino acid methionine and in
Prokaryote AA1 t-RNA complex has amino acid M-Formyl methionine
Q 142 – C 58 and 15 Root is part of stock so chromosome in root are 58. While egg cell is part of
scion and is haploid so it contain-15 chromosome.
Q 143 – A -light red
Q 144 – A- 2 celled pollen grain
Q 145 – D-Rryy and RRyy
Q 146 – B-198
In 34 nm DNA strand has 10pairs of spiral so 200 nucleotides are present. In between two
Successive nucleotides, a single phosphodiester bond is present one nucleotide of each
end of DNA strand do not form linkage DNSA is double
200-2=198
Q 147 – C-Oosphere
Q 148 – D-UUA, CUU, leucine these two codon codes for leucine amino acid.
Q 149 – C-self sterility
Q 150 – C-Endolysin
Q 151 – A-Rare species
Rare species are those whose members are few, live in small geographical areas or in
unusual Environment.
Q 152 – C-Lateral walls of thalamencephalon
Thalamencephalon(Diencephalam) lateral walls are interconnected by Habencular
commisure.
Q 153 – C-Blue Print
The complete sequence of human genume is described as blue print.
Q 154 – B-Oenothera lamarkiana Hugo de varies (1901) a Dutch botanist proposed “mutation
theory” by studying the plant evening primerose [Oenothera lamarkiana]
Q 155 – B-50%
When y sperm fertilizes egg , then resulting zygote (XY) develops into male child(boy).
Possibility of fertilization of egg by either X-sperm or Y-sperm is always 50%. It is
independent of Number of previous children.
Q 156 – C-Replication
DNA amplification means to obtain more copies of DNA to in vitro replication by PCR
technique
Q 157 – B-Ureotetic—Tadpole larva of frog
Tadpole larva frog is ammonotetic.
Q 158 – D-Small Pox
Q 159 – D
Epoch
—
Animal life
A)Palaeocene
—
Rise of first prionate
B)Eocene
—
Diversification of placental mammals
C)Oligocene
—
Rise of first monkeys and apes
D)Miocenex
—
first man like open formed
Q 160 – B-interferon’s
Interferon’s (glycoproteins) make the surrounding cells resistant to viral infection.
Q 161 – B-Rh+ve husband and Rh-ve wife
Rh antigen induces a strong immunogenic response when introduce into RH-ve
individual. It results in haemolytic disease of new born.
Q 162 – B-Patients has antibodies ‘b’
Blood group
Genotype
AB
A
or
Antigen
Antibodies
Can donate blood
to
A&B
—
A, B, AB, O
A
b
A,O
Q 163 – Common name
—
Scientific name
a)sea anemone
—
Adamsia pallicata
b)Hermit crab
—
Euparagus prideauxi
c)Sucker fish
—
Echeneis
d)Pilot fish
—
Remora
Q164 – B inhibits cellular respiration in plants by reacting with cytochrome oxidase enzyme system.
Q165 – A Hormone
Ca ++, cyclic Amp & IP3 acts as a secondary messenger while hormone acts as a primary
messenger.
Q166) C Secondary oocyte
Spermatogenia, zygote and oogomia are diploid.
Q167) C Papaver somniferum
Heroin[ smack]/diacety/morphine is extracted from the latex of POPPY Plant.[Papaver
Somniferum]
Q168) B) Marrino ram,cross breeding
Blkaneri ewes X Marrino ram -> Hisardale
It is a eross breeding i.e superior male of one breed mated with superior breed of another
breed.
Q169) B) cck
Cholecystokinin [ckk] acts on pancreatia acini (exocrine port) and stimulate to secrete
pancreatic juice.
Q170) method
–
mode of action
a)Condom
–
prevents deposition of semen in the vagina.
b) Copper T
–
prevents fertilization/ implantation.
c)tubectomy
–
block gamete transport & prevent pregnancy.
d)spermicide
–
Immobilize and kill sperm.
Q171.) D)
The pulmonary trunk and systemic aorta are connected by ligamentum arteriosum that
represents remnants of ductus qryeriosus of foetus.
Q172.) B) systemic circulation pathway – LV -> RA
Left ventricle -> Aorta -> tissues -> veins -> Right auricle
Q173.) C) Iodopyracet.
In tubular secretion toxic,unfiltrated substances are actively secreted from blood into
nephron. eg. Iodopyracet
Q174.)D) Syngamy ( Fertilization)
Syngamy lead to mixing of maternal and paternal chromosomes which is a basis of variations
that leads to evolution.
Q 175.) B) Atria secrete ANF, B.P. decreases Atrial Natriuretic factor is secreted by Atria as blood
pressure increases. ANF acts as a vasodilator which results into decrease in blood pressure.
Q 176.) D) Apis indica And Apis meclifera.
Q 177.) B)Semicircular canals and otolith organ.
In internal ear, above the cochlea, vestibular apparatus is present, which consists of three
semicircular canals and otolith organ formed of the sacculus and utriculus.
Q 178.)C) Megameres of morula.
Zygote -> morula
micromeres -> trophoectoderm.
Megameres -> inner cell mass -> gastrnla i.e three germ layer
Q179.) A) Frontal lobe
In frontal lobe general motor area and association area is present.
Q 180.) D) Endocrine function of kidney.
Artificial kidney i.e. dialysis is treatment works on the principal of diffusion and helps in
osmoregulation and ultrafiltration.But fails to perform endocrine function of kidney i.e to secrete
hormones like calcitrial and erythropoietin.
Q181.) D) lenteotropin (prolactin) marintains cerpus lutenm during pregnancy. It secretes
progesterone. Increased level of progensterone suppress the menstrual cycle i.e no egg formation.
so no fertilization pregnancy.
Q182.) C Sensory
Trigeminal nerve ( Dentist nerve)/ IVth mixed type. It gives three branches asa) Opthalmic
b) Maxillary
c) Mandubular
-
Sensory
Sensory
mixed
Q183.) A) Certicosteroid
Pituitary of mature foetus secretes ACTH which causes release of certicosteriod hormones
from foetal adrenal cortex which diffuse across the placenta and accumulate in mothers blood till
they cause a decrease in progesterone production and increase in production of prostaglandins.
Q184) D) fructose
Seminal fluid contain fructose, fibrinogen and prostaglandins. Fructose present in seminal
vesicles secretion is not present elsewhere the body. Hence it provides a proof for forensic test in
case of rape.
Q185.)C) Minamata is a disease produced by mercury poisoning leads to crippling and fatal
disease(Japan 1950).
Q186.)D) Sterilization techniques.
More safest as compared to given option.
Q187.)D) Progesterone.
Decrease in progesterone level, causes shedding off endometrial lining, leads to
menstruation.
Q188.)A)
Human Ancester
Cramial capacity
1.
Homo habilis -
650-800cc
2.
Homo erectus -
940 cc/850-1200cc
3.
Neanderthal man -
1450cc
4.
Cro-magnon man -
1600cc
Q189.)D) Addison’s disease
Q190) B Origin of life on earth.
Q191.)A)23
Linkage group = haploid no. of chromosomes=number of pairs of chromosomes.
Q192.)B) Natural selection
Biston betularia and Biston carbonaria
Q193.)A) Opsonisation
Meaning/ definition
Q194.)B)
Column I
Column II
a)TPA -
Reverse blood clot
b)TGF-B -
Promotes new blood vessel formation
c)α-1 antitrypsin -
To treat emphysema
d)Blood Clotting Factor VIII-
To treat haemophilia
Q195.)
Normal Vision man
Y
×
Normal Vision Woman whose father was colorblind
×
Y
Y
Y
Normal
Female
Child
Carrier
Female
Child
Y
Normal
Male
Child
Colorblind
Male Child
50% Normal
Vision Male
50%
colorblind
Vision Male
Among Male
Children Only
Q196.)C) Sporozoite
Sporozoite infects liver cells.
Q197.)A) Chotong, assel, Brahma.
Q198.)B) Antithrombins.
Antithrombins are anticoagulants which prevents the clotting of blood inside blood vesssals.
Q199.)D) G Group.
Downs syndrome -> (21st thrisomy ) 46+1=47. Down’s syndrome is due to an extra
chromosome number 21(from Group G).
Q200.)D) Glomerulus
Glomerulus is a tuft of blood capillaries present in Bawman’s capsule. and acts as a filtering
units. As being blood capillaries truly it is a part of blood circulatory system.