Student's Solution Copy [Code - 27650]

NEET PATTERN TEST Brahmastra Major Test-06

13th NEET - Phase 12
KOTA

Date: 02-Apr-2025

Duration: 3 Hours

Max Marks: 720

Physics - Section A

1. 2.
Answer: 1 Answer: 3
Sol: Sol:

ne nh = (ni )
2
Displacement = Area of graph with sign

ne is concentration of electron, Displacement =

1 1
( × 10 × 5)+(10 × 5) +( × 5 × 30)
nh is concentration of holes and ni is the 2
1 1
2

concentration of electron pairs in the intrinsic +(

2
× 5 × 20)−
2
(5)(20)

semi-conductor. = 25 + 50 + 75 + 50 − 50

Here; = 150 m

nh = 10
21
, ne =?, ni = 10
19 Distance → Area of graph with positive value

10
21
× ne = 10
19
× 10
19 Distance = 25 + 50 + 75 + 50 + 50 = 250
Distance 250 5
10
38
17 −3 = =
ne = 21
= 10 m Displacement 150 3
10

3.
Answer: 1
Sol:
Here from the graph we see that both x,y are
at same temperature but they are varying
w.r.t time, so
dT dT
( ) > ( )
dt x dt y

Now from the Kirchoff's law, we know that

emissivity is equal to absorptivity, so the
body which emits more will also absorb
more.
Therefore, ex > ey and Ax > Ay
4. 7.
Answer: 2 Answer: 1
Sol: Sol:

λ =
2π
= 1m By Kepler's third law, T
2
∝ R
3

2π

λ 1 2 3 2 3
T T T2 R2
= m
or
1 1
2 2 ∴ 2
= 3
( ) = ( )
T T T1 R1
2 2

5. 3
2 2 3.5R
⇒ T = T ( )
Answer: 3 2 1 7 R

Sol: (∴ R2 = R + 2. 4 R and R1 = R + 6 R)

2
T
The Magnetic force is given by or T
2
=
1
⇒ T2 =
T1

2 8
2√ 2

F = Bil
24
⇒ T2 =
where l = displacement length 2√2

l = length AB ∴ T2 = 6√2h

8.
2
2
here l = √L + (2r)

2 2 Answer: 4
l = √L + 4r

Sol:
then, force is equal to
2 2
According to given pattern of wave, the
F = Bi√L + 4r
wavelength can be gives as:
6. 3λ
= 6
2

Answer: 3
λ =
12
= 4m .
Sol:
3

So, option (4) is correct.

From the formula of Bulk modulus
–ΔP
9.
β =
Answer: 4
ΔV
( )
V

ΔV ΔP
(0–1×10 )
5

−7
Sol:
∴ = − = − 11
= 8 × 10
V β
Given,
1.25×10

KAg ⇒ 11Kfe
dQ ΔT
=
dt L

( )
Keq A

Let the temp. at the intersection be T

100−T T−0
H ⇒ =
L L

K A K A
fe Ag.

Kfe (100 – T) = KAg(T – 0)

Kfe (100 – T) = 11Kfe(T – 0)
100 – T = 11 T
100
T ⇒ ⇒ 8 .3 °C
12
10. 12.
Answer: 3 Answer: 4
Sol: Sol:
Velocity of train A
2V0 = √v
2
0
+ vx
2

km 5
vA = 90 = 90 × = 25 m/s
18
4V02 =
hr 2 2
v + vx
0
Velocity of train B
Vx2 = 3V02
km 5
vB = 54 = 54 × = 15 m/s
hr 18 qE t
0
∴ √3v0 =
m
Velocity of train B w.r.t. train A
√3 mv0

∣
→
∣ ∣
→
∣ t=
= v B − v A qE
∣ ∣ ∣ ∣ 0

13.
= 15 − (−25) m/s = 40 m/s

length of train
Answer: 2
Time of crossing =
relative velocity
Sol:

At depth,
ℓ ′ h d
(8) = g = g(1 − )or g(1 − )
40 R R

g d
ℓ = 8 × 40 = 320 meter ⇒ = g(1 − )
n R

11. n−1
or d = R(
n
)
Answer: 2
Sol: 14.

The rise of a liquid in a capillary is given as, Answer: 3

2Tcosθ Sol:
h =
D
( ) ρg
2
Inet = I1 + I2 + 2√I1 I2 cosϕ ----(1)
Since, we know that height of capillary rise is
Here,
o 1
I1 = I2 = Io andcos 60 =
inversely proportional to radii of tube, i.e., 2

height ∝
1 in equation (1), resultant intensity
radius
1
Inet = Io + Io + (2Io × ) = 3Io
h1 D2 22 2
⇒ = =
h2 D1 66
⇒ Inet = 3Io

⇒ D1 : D2 = 3 : 1
15. 18.
Answer: 3 Answer: 2
Sol: Sol:
→ 2
r = 10t î + 15t ĵ + 7k̂
There is silence at point D, o it is a point of
destructive interference.
→
→ d r
v =
dt
= 10 î + 30 t̂ ĵ For Minima,
→ 1
→ d v Δy = (n − )λ
a = = 30 ĵ 2
dt

1
→ 2ℓ − ℓ = λ
→ 2
F = (m) a = (m)(30 ĵ )
λ = 2ℓ

= along positive y − axis

19.
16.
Answer: 3
Answer: 2 Sol:
Sol:
We have, me = qE
Kinetic energy of N moluecule of gas mg mg
3 q = =
E
E = NkT V/d
2

–15
1.96×10 ×10
Initially and finally
3
E1 = N1 kT1 =
2 (400/0.02)
3
E2 = N2 kT2 –14
2 1.96×10 –18
= 4
= 0. 98 × 10
20×10
But according to proble E1 = E2 and N2 =
2N1 = 9.8×10–19 ≈ 6e
3 3 T1
∴ N1 kT11 = (2N) kT2 ⇒ T2 =
2 2 2

Since the kinetic energy constant

3 3
N1 kT1 = N2 kT2
2 2

N1 T1 = N2 T2 ∴ NT = constant

From ideal gas equation of N molecule PV =

NkT
P 1 V1 = P 2 V2 ∴ P 1 = P 2

[As V1 = V2 and NT = constant]

17.
Answer: 2
Sol:
dϕ –Ldi
∣ ∣ ∣ ∣
e = ∣ ∣ = ∣ ∣
dt dt

–3

= 5 mH
–LΔI eΔt 5×10
e = ∣
∣
∣ ⇒ L = ∣
∣ ∣
∣ =
∣
Δt ΔI 1
20. 23.
Answer: 4 Answer: 2
Sol: Sol:

Let the acceleration of both blocks is a.

Now, Vp = +2 − 4 + VQ

VP − VQ = 2 V

24.
Answer: 2
For 4 kg block applying second law : Sol:
40 – T = 4a The velocity of sound can be mathematically
expressed as
γP
v = √
d

Fk = 0. 02 × 40 × 10 = 8N
where is γ specific heat ratio, P is the
For 40 kg block applying Newton’s second pressure of gas and d is the density of the
law medium.
T – 8 = 40 a According to this, the velocity of sound in a
Solving above equations
medium is inversely proportional to the
square of the density of that medium,
8 −2
a = ms 1
11 v ∝
√d

21.
So, the velocity will be lower for denser
Answer: 1 medium and higher for rarer medium.
Sol:
Dω = −DU
Dω = −DU = −2J
22.
Answer: 3
Sol:

2
2
Z = √R + (2πvL)

≃ 50 ohm
25. 27.
Answer: 2 Answer: 4
Sol: Sol:
The Einstein's equation for photoelectric
effect is given by,
KEmax = hf − hf0

where, KEmax = maximum kinetic

energy of photo electron ,

f = f requency of incident light

−
→ and f0 = threshold f requency.
vA = vȷ̂

−
→
Since maximum kinetic energy of photo
vB = −v ı̂ electrons is greater than zero so

Time to reach from A to B =

2πR

4
×
1

v
=
πR

2v
hf − hf0 > 0

28.
Displacement from A to B = R√ 2

Answer: 3
Now, Average velocity from A to B
Displacement
Sol:
R√2 2√2v
= = =
Time πR

2v
π
The electrical potential producd by the
nucleus at the position of the electron,
Instantaneous velocity at B is −v ı̂
q
9
V = 9 × 10 ×
According to question, r

–19
(+1.6×10 )
9
instantaneous velocity π = 9 × 10 × = 27. 2V
= 0.53×10
–10

average velocity x√2

v π 29.
=
2√2v
x√ 2
π
Answer: 2
π π

2√2
=
x√2
Sol:

⇒ x = 2 For prism use have,

Li + Le = LA + LD . . .(i)
26. if i = 15°, then e = 60°

Answer: 2 and D = 30°

From eq.(i),
Sol: 15° + 60° = A + 30°

T2 ⇒ A = 45°
P =
T1 –T2

265
5 =
T1 –265

T1 – 265 = 53

T1 = 318 K

T1 = 45°C
30. 34.
Answer: 3 Answer: 1
Sol: Sol:
2
ω x = V∘ = ω √ A
2
− x
2 We have,
μ2 μ1 μ2 –μ1
2 2 2 2
A − x = x ω − =
v u R

Here u = –25 cm , R = 20 cm,

2 2 2
2π
(3) − (2) = ( ) × 4
T
μ1 = 1. 0 and μ2 = 1. 5
2
2 16π
T =
5 Putting the values in (i),
4π 1.5 1.0 1.5–1.0
T = + =
√5 v 25 cm 20 cm

or,
1.5 1 1
31. v
=
40 cm
–
25 cm

Answer: 2 or, v =– 100 cm.

Sol: As v is negative, the image is formed to the

left of the separating surface at a distance of
100 cm from it.
35.
Answer: 2
Sol:
Energy to remote electron is very low
because electron is bounded to nucleus at We have
outside. →
√3
ˆ 1 z
B =( i + ĵ )30 sin[ω(t − )]
So Ee < En 2 2 c

→ → →
32. E = B × c and E = B0 c

Answer: 4 →
√3 1
here, E( (− ĵ ) + î )
2 2
Sol:
and E0 = 30c
Potential difference across C si 10 V. →
1 √3 z
E =( î − ĵ )30 csin[ω(t − )]
∴ q = CV = 6×10 = 60 μC 2 2 c

33.
36.
Answer: 2
Answer: 4
Sol:
Sol:
2 4
I = 2[5(0. 2) + 2(0. 4) Zero (No potential difference across
voltmeter).
37. 39.
Answer: 3 Answer: 4
Sol: Sol:

KE of rotation = 1
Iω
2
=
1
× (
2 2
mr ) ω
2
Wheatstone bridge is balancd.
2 2 5

P R
=
= 1

2
×
2

5
× 1 × (
3×3

100×100
) (50 × 50) Q S

40 60 4
Joule
9
⇒ = =
10 15 1
20

38. ⇒ VAB = VAD

Answer: 2 So,

Sol: 40I1 = 60I2

⇒ I1 = 1. 5 I2
asinθ = nλ where n=1
Heat produced in AB = I
2
Rt
θ = sin
−1
(
λ

a
) .......... (1) 1

=
According to question 2 2
(1. 5I2 ) × 40 × t = 90I t
2
–3
λ = 2 × 10 m
Heat produced in BC, = I
2
Rt
–3 1
a = 4 × 10 m .......... (2)
=
From equation (1) and (2) 2 2
(1. 5I2 ) × 10 × t = 22. 5I t
2

θ = sin–1(1/2) ⇒ θ = 30°.
Heat produced in AD = I
2
2
Rt

=
2 2
I × 60 × t = 60I t
2 2

Heat produced in DC = I
2
2
Rt

=
2 2
I × 15 × t = 15I t
2 2

40.
Answer: 3
Sol:
3×16 = 6 ×v2
v2 = 8 m/s

KE = × 6 × (8)2
41. 43.
Answer: 4 Answer: 2
Sol: Sol:

Fringe width :
Dλ
β =
d

β2 λ2 D2 d1
According to question, β1
=
λ1 D1 d2

d1
Here, D2 = 2D1 & d2 =
2

β2 λ2 (2D1 ) d1 λ2 44.
= = 4
β1 λ1 D1 d λ1
Answer: 2
1
( / )
2

β2
= 4
λ2
=
4×4×10
−7

= 2. 5
Sol:
−7
β1 λ1 6.4×10

Stress is on x-axis
−4
β2 = 2. 5β1 = 2. 5 × 10
∵ slope of B is more
42.
∴ B is more elastic
Answer: 1
45.
Sol:
Answer: 3
T.E. = K.E. + P.E.
Sol:
=
[v]=[M
0 1
L T
−1
] ; [a]=[M
0 1
L T
−2
]

[F ]=[M
1 1
L T
−2
] ; [W ]=[M
1 2
L T
−2
]

= 2500 J
 Chemistry - Section A

46. 49.
Answer: 3 Answer: 2
Sol: Sol:
1 2 200
PS = × 100 + 50 × =
3 3 3

100/3
1
XB = =
200/3 2

50.

3, N, N–trimethylpentan-3-amine Answer: 3

47. Sol:

a)
+ − − −
Answer: 3 NaCl → Na (10e ) + Cl (18e )

Sol: b) BeCl2 → Be
2+
(2e
−
) + 2 Cl
−
(18e
−
)

Given V1 = 8cm3
c)
2+ − − −
MgF → Mg (10e ) + 2F (10e )
2

V2 = 80 cm3
d)
2+ − 2− −
CaS → Ca (18e ) + S (18e )
n=5
51.
T = 27 + 273 = 300K
Answer: 2
V2
ΔST = 2 .303 nRlog
V1 Sol:

on substituting the above value,

&
we get,
80
ΔST = 2 .303 ×5 × 8 .314 × log
8
Have different alkyl grouping around

ΔST = 2 .303 ×5 × 8 .314 polyvalent finctional group

ΔST = 19 .14 ×5

−1 Metamers
ΔST = 95 .74 JK

48.
Answer: 4
Sol:
% of Atomic Atomic Simplest
Element
element mass ratio ratio
80 6.67
C 80 12 12
= 6.67
6.67
= 1

20 20
H 20 1 1
= 20
6.67
= 3

Empirical formula = CH3

52. 54.
Answer: 1 Answer: 1
Sol: Sol:
Part 1: Enthalpy OSMOTIC PRESSURE
Enthalpy is a thermodynamics quantity that
measure heat changes at constant pressure. π = iCRT
It is useful to define a new state function and [π ∝ i]
it is refer as (H).
For NaCl, i = 2 max.
Part 2: Calculation of bond enthalpy
The reaction is proceed in the following for urea, glucose and sucrose, i = 1
ways-
NaCl gives maximum ion hence it will show
P(s) +
3
Cl2 (g) → PCl3 (g) highest osmotic pressure.
2

55.
The above given reaction results in the
formula - Answer: 4
Δf H(PCl3 , g) =ΔHatomization (P, s) + 3× Sol:
Δf H(Cl, g)−(B. E(P − Cl) × 3)
l2 is solid and sublimes at ordinary
Substituting the value in above formula- temperature because of weak vanderwaal's
force between l2 molecules.
306 kJ / mol= 314 kJ / mol

+3 × 121 kJ / mol−(B. E(P − Cl) × 3) 56.

B. E(P − Cl) = 123. 66 kJ / mol

Answer: 3
Sol:
53.
According to Bronsted-Lowry theory, acid is a
Answer: 2
substance which donates an H+ ion or a
Sol: proton and forms its conjugate base and the
base is a substance which accepts an H+ ion
Third excites state n=4 or a proton and forms its conjugate acid.
n–ℓ–1=2 CO32− + H+ →HCO3−
4–ℓ–1=2 57.
ℓ =1 Answer: 1
= √ℓ (ℓ + 1)ℏ Sol:
k2 Ea T2 −T1
= √1 (1 + 1) ℏ = √2 ℏ log [ ] = [ ]
k1 2.303R T1 T2

k2/k1= 2; T2 = 308, T1 = 298K

Ea 10
∴ log 2 = ×
2.303×8.314 308×298

3
Ea = 52. 903 × 10 J

or Ea = 52. 903 kJ
58. 60.
Answer: 3 Answer: 4
Sol: Sol:
Cr2O3 + 2Al → Al2O3 + 2Cr MnS is pink and the rest of them are black.
Here we can see that, aluminium metal 61.
displaces chromium metal and therefore this
is metal displacement reaction. Answer: 1

59. Sol:

Answer: 2 Initial Rate = k [A] [B]2

Sol: Rate1 = k [0.60] [0.80]2 .....(i)

Geometrical isomers are the compounds Rate2 = k [At] [Bt]|2 .....(ii)

having same molecular formula but different Given reaction,
arrangements of atoms in space. There are
two necessary conditions for a compound to A(g) + 2B(g) C(g) + D(g)
possess geometrical isomerism.
0.60atm 0.80atm 0 0 t=0
(i) It must have restricted rotation of bond.
(0.6 – 0.2) (0.8–0.4) 0.2 0.2atm
It can be due to double bond or ring.
0.40atm 0.40 atm 0.2atm 0.2 atm t=
(ii) There must be two unlike atoms or
t
groups linked to each double bonded C-
atom. Put value of pressure of t = t in rate equation
(II)
e.g. XYC = CXY and YXC = CXZ
Rate2 = k [.40] [.40]2
Alkenes containing identical groups are
atoms on doubly bonded C-atom do not show k[0.4][0.4]
2

geometrical isomerism. Rate2

=
Rate1 2
k[0.6][0.8]

eg. AAC = CAA, AAC = BB, AAC = AB etc.

Rate2 2×1 1
= =
3×4 6
(i) Compound contain identical
Rate1

Br atoms on one doubly bonded carbon,

hence does not show geometrical isomerism.

(ii) Compound contains all

different groups/atoms on doubly bonded C-

atoms, hence it cannot show geometrical
isomerism.
Compound (iii) and (iv) follow all the
conditions of geometrical isomerism and are
geometrical isomers of each other.

(iii) (iv)
62. 65.
Answer: 3 Answer: 1
Sol: Sol:
For WAWB type of salt Conjugate acid of (II) is stabilized by charge
1 delocalization.
pH = [pKw +pKa -pKb ]
2
66.
pKw of water at 25° C is 14.
Answer: 4
pKa of acetic acid
Sol:
pKa = − log Ka
Acidic buffer solution: Acidic buffer are
pKa = - log 1.8×10 -5 -5
= - (log 1.8 + log 10 ) solution of a mixture of weak acid and salt of
its conjugate base of that acid with a strong
= - (0.255 - 5) = 4.745 base.
pKb of ammonium hydroxide a) HCN + NaCN

pKb = − log Kb Basic bufer solution: Basic buffer are solution

of a mixture of weak base and salt of its
pKb = - log 1.8×10-5 = - (log 1.8 + log 10- conjugate acid of that base with a strong
5
) acid.
= - (0.255 - 5) = 4.745 c) NH4 OH + (NH4 )
2
SO4

As we know d) HCOOH + NaOH is not a buffer

pH =
1
[pK +pK -pK ]
solution.
2 w a b
Since formic acid is not present in excess
amount.
= 1

2
[14 + 4. 745 − 4. 745] =7

So pH = 7 (Neutral solution)
63.
Answer: 4
Sol:

64.
Answer: 3
Sol:
(1) Nitrogen can not form pentahalide due to
the absence of vacant d- d-orbitals in
outermost orbitals
(2) EN=N>P
67. 70.
Answer: 3 Answer: 1
Sol: Sol:
Both Assertion and Reason are correct and
Reason is correct for the Assertion.
71.
Answer: 2
S = [Ne] 3s2 3p4 3d0
Sol:
In the given reaction
S* = 3s2 3p3 3d1
2AgCl(s) + H2 (g) (1 bar) → 2HCl (aq) +
2Ag
Silver is undergoing reductino Ag+ → Ag
O = [He] 2s2 2p4
Hence it will act as cathode in following cell.
Pt(s) | H2(g), 1bar | 1m HCl (aq) |1mAg+
(aq) | Ag(s).
72.
Answer: 1
Sol:

68.
Answer: 3
Sol:

73.
69.
Answer: 3
Answer: 2
Sol:
Sol:
Anhydrous aluminium chloride is hydrolysed
E = E –
o 0.0591
log
1
partly with the moisture in the atmosphere
2
to give HCl gas. This HCl combines with the
+2
[Zn ]

moisture in the air and appears white in

E = E
o
+
0.0591
log [Zn
+2
] colour.
2

Hydrolysis of AlCl3 ​ gives HCl which fumes in

the air.
o 0.0591
E = E + × (– 2)
2

E=Eo–0.0591 AlCl3 + 3H2 O → Al(OH)3 + 3 HCl

74. 76.
Answer: 4 Answer: 1
Sol: Sol:
Given that - Bond order of N2 = 3
∘
Λm (H2 SO4 )= x S cm
2
mol
−1
Bond order of N2+ = 2.5
∘
Λm (K2 SO4 )= y S cm
2
mol
−1
Bond order of N2- = 2.5
Therefore, the correct answer is (A)
° 2 −1
Λm (CH3 COOK) = z S cm mol
77.
o
Λm (H2 SO4 )= Answer: 3

2 Λm (H
o + o
)+Λm (SO
2−
). . . . (1)
Sol:
4

Mole at equilibrium
o o +
Λm (K2 SO4 )= 2 Λm (K ) 2
x
KC =
(a–x)V
o 2−
+ Λm (SO ). . . (2)
4

Addition of inert gas at constant V has no

o
Λm (CH3 COOK)= effect on reactions having Δn=0 or Δn ≠ 0.
o
Λm (CH3 COO
− o
)+Λm (K
+
). . . . (3) But addition of inert gas at constant P has
effect on reactions having Δn ≠ 0; and no
o
Λm (CH3 COOH)= effect if Δn=0. The given reaction has
Δn ≠ 0 and thus choice (c) is correct. Also,
o − o +
Λm (CH3 COO )+Λm (H ). . . . (4)
the effect may be shown as : on addition of
2 × Equation (4) = Equation(3) inert gas at constant P, volume increases. To
×2 + Equation(1) − Equation(2) have KC constant, x must increase.
o
2Λm (CH3 COOH)= 2z + x − y 78.

o
2z+x−y Answer: 3
Λm (CH3 COOH)=
2

x−y
Sol:
o
Λm (CH3 COOH)= z +
z
Methyl Cyanide on reacting with a Grignard’s
75. reagent produces a ketone

Answer: 2 CH3 – C ≡ N −−−−−→

+ CH3 MgI

Sol: methyl cyanide

+H2 O

−−−
−−→
–MgINH
2
79. 83.
Answer: 4 Answer: 1
Sol: Sol:
Structure of [Cu(NH3)4]SO4 is as follows: Tetrahedral complex rarely formed low spin
complex because Δt is not large enough to
facilitete back pairing
84.
Answer: 2
Sol:

Thus, we can see, this molecule contains

covalent bond between N and H atoms,
coordinate bond between N and Cu atom and
electrovalent bond between SO42– and
[Cu(NH3)4]2+ ion. Thus,
80.
Answer: 1
Sol:

85.
Answer: 2
Sol:
The Mond process, sometimes known as the
carbonyl process, This process converts
81. nickel oxides into nickel metal with very high
purity being attainable in just a single
Answer: 4 process.
Sol: 330–350K

Ni + 4 CO −−−−−→ Ni (CO)
4
Square planer complex do not show optical Impure Nickel

isomerism due to presence of plane of 450–470

−−−
− → Ni + 4 CO
symmetry. pure

82.
Answer: 3
Sol:
Valine is one of the 20- essential amino acid
for metabolism.
86. 89.
Answer: 2 Answer: 3
Sol: Sol:
In aqueous solution, more than 99% of i) H can't precipitate the salt of Zn from its
glucose molecules, exist in cyclic isomeric solution because H has less reducing
form. This conversion to a ring form arises property as compared to Zn.
from the open-chain form by a nucleophilic
addition reaction between the aldehyde ii) H can't precipitate the salt of Cu from its
group at C-1 and the hydroxyl group at C-4 solution because H has less reducing
or C-5, yielding a hemiacetal group -C(OH)- property/power as compared to Cu.
O-, at C-1. The reaction between C-1 and C- iii) H can precipitate the salt of Ag from its
5 creates a molecule with a six-membered solution because H has higher reducing
ring, called pyranose. The position of the property/power as compared to Ag.
hydroxyl on the C-1 determines the alpha or
the beta form of the pyranose. In general, β- iv) H can't precipitate the salt of Fe from its
D-glucopyranose may also be called as β​-D- solution because H has less reducing
glucose. property/power as compared to Fe.
90.
Answer: 4
Sol:

87. As Cu+2 ion test in excess of KCN means this

form complex in presence of excess KCN .
Answer: 3
This is due to the formation of the complex
Sol: ion [Cu (CN) ]
3−
.
4

Other than La3+ and Lu3+ all other 3−

lanthanides are paramagnetic in +3 state. CuSO4 + KCN → K3 [Cu (CN) ]

4

88.
Answer: 2
Sol:
Prussian blue is a dark blue complex.
Synthetic pigment by oxidation, of ferrous
ferrocyanide salts. It contains ferric
hexacynoferrate (II) in a cubic lattice crystal
structure.

2F e2 (SO4 ) + 3K4 [F e(CN ) ]

3 6

→ F e4 [F e(CN ) ] + 6K2 SO4

6 3
 Biology - Section A

91. 94.
Answer: 3 Answer: 1
Sol: Sol:
Skeletal system consists of a framework of Accessory pigments like carotenoids and
bones and a few cartilages. In human chlorophyll b enhance the efficiency of
beings, this system is made up of 206 bones photosynthesis by absorbing a broader
and a few cartilages. It is grouped into two spectrum of light wavelengths, which allows
principal divisions – the axial and the plants to capture more energy from sunlight
appendicular skeleton. compared to relying solely on chlorophyll a.
Axial skeleton comprises 80 bones If both Assertion & Reason are True & the
distributed along the main axis of the Reason is a correct explanation of the
body. The skull, vertebral column, Assertion.
sternum and ribs constitute axial skeleton.
95.
The bones of the limbs alongwith their
girdles constitute the appendicular Answer: 1
skeleton. Sol:
92. 12th NCERT Page No. 28
Answer: 4 96.
Sol: Answer: 4
Perianth is the collective name of the Sol:
nonessential floral organs. Calyx and corolla
are called as accessory whorls. In some Column-I Column-II
plants they are not differentiated from one (a) Making multiple
copies of any template (iii) Cloning
another, in such case they are termed as DNA
perianth. Hence, the perianth is the term (b) Technique to alter
(iv) Genetic
used when calyx and corolla are similar. the chemistry of genetic
engineering
material
93. (c) Technique of using
live organisms or
(i) Biotechnology
Answer: 4 enzymes to produce
products
Sol: (d) Separate bands of
DNA are cut from (ii) Elution
Ecology is the study of living organisms agarose gel
(plants, animals, microbes) and their
interaction with the environment.
Physiological ecology ;Physiological ecology
refers to the study of physiology of different
organisms. This in turn helps to explore their
mechanism of adaptation to the respective
environment in terms of reproduction and
survival.
97. 99.
Answer: 2 Answer: 3
Sol: Sol:
ALVEOLAR PRESSURE- Nerves arising from brain are called cranial
nerves. There are 12 - pairs of cranial nerves
Inspiration is an active process whereas expiration is found in reptiles, birds and mammals but
a passive process. amphibians and fishes have only 10 - pairs of
It is the pressure of the air which is moving inside your cranial nerves (Accessory spinal and
lungs or outside. This pressure is negative at the time of hypoglossal nerves are absent).
Inspiration as the air moves out from the mouth to the
lungs where the alveoli are situated at a lower level. So Fifth cranial nerve of frog is
calledTrigeminal nerve.
And at the time of Expiration, this value of pressure is
positive as air flows out from the region of alveoli in the The vagus nerve, is the tenth
lungs at lower pressure to the higher pressure cranial nerve or CN X, and interfaces with the
outside. During expiration, muscles of the diaphragm parasympathetic control of the heart, lungs,
relax. and digestive tract. The vagus nerves are
normally referred to in the singular.
Inspiration occurs via the active contraction of muscles –
such as the diaphragm. The ophthalmic nerve (CNV1) is a terminal
branch of the trigeminal nerve (along with
Hence, the correct option is "2" - Inspiration is an the maxillary and mandibular nerves). It
active process whereas expiration is a passive process. provides sensory innervation to the skin,
98. mucous membranes and sinuses of the upper
face and scalp.
Answer: 1
The optic nerve connects the eye to the
Sol: brain. The optic nerve carries the impulses
formed by the retina, the nerve layer that
(A) Family - (5) Solanaceae
lines the back of the eye and senses light
(B) Kingdom - (4) Plantae and creates impulses. These impulses are
dispatched through the optic nerve to the
(C) Order - (2) Polymoniales brain, which interprets them as images.
(D) Species - (1) nigrum 100.
Answer: 2
Sol:
In solanaceae family, ovary is superior,
oblique ovary. The gynoecium is located in an
oblique position relative to the flowers
median plane. Carpels are obliquely placed
and ovules on swollen axile placenta.
In Cruciferae family, parietal placentation are
occurs.
In Malvaceae family, axial placentation are
occurs.
In Poaceae family, basal placentation are
occurs.
Hence, the correct answer is Solanaceae.
101. 104.
Answer: 4 Answer: 4
Sol: Sol:
same ecological niche, Explanation: An Biotechnology is the branch of biology that
ecological niche is the role and position comprises of living systems and organisms
a species has in its environment and how it for developing new and useful products.
meets its needs of food and shelter,
how it surviveand reproduce. Two different The definition given by the European
species cannot share same Federation of Biotechnology is a
ecological niche as theirrequirement is comprehensive one and comprise of
different from each other. traditional and modern view.

102. The definition is as follows- Integration of

natural science and organism cells, parts
Answer: 2 thereof and molecular analogies for products
and services.
Sol:
Hence, the correct answer is option "4"
Photosynthetically active radiation - The integration of natural science and
(PAR) is light of wavelengths 400-700 nm organisms cells, parts thereof and
and is the portion of the light spectrum molecular analogies for products and
utilised by plants for photosynthesis. Hence services.
statement I is false.
105.
Mineral ions required for photolysis of water
are manganese, calcium and chlorine. Answer: 2
Oxygen is released as the byproduct of
photosynthesis in the photolysis process. Sol:
Hence statement II is false. The amine group of haemoglobin is reacted
In Cyclic Photophosphorylation plant by about 20–25% of CO2 resulting in the
cells only accomplish the ADP to ATP for formation of carboxyhaemoglobin. The
immediate energy for the cells. This process deoxygenated blood, when it reaches the
usually takes place in the thylakoid alveoli of the lungs, then there occurs the
membrane and uses Photosystem I and the dissociation of the carbaminohaemoglobin
chlorophyll P700.Hence statement III is true. and sodium bicarbon­ate due to low PCO2 and
high level of PO2 in the alveoli.
103.
The stimulation of this dissociation is done by
Answer: 3 oxyhaemoglobin. The CO2, thus freed from
the blood, goes into the atmosphere.
Sol: Haldane effect is the effect of
12th- Ncert , Pag.no.37 oxyhaemoglobin on the dissociation of these
compounds. The oxyhaemoglobin acts like a
strong acid in this reaction (i.e., it frees H+
ion in the medium).
106.
Answer: 1
Sol:
In binomial nomenclature, the components
"X" and "Y" are the generic name and the
specific epithet. 'Z' indicates Carolus
Linnaeus that introduced the binomial
system of nomenclature in 1751.
107. 110.
Answer: 3 Answer: 4
Sol: Sol:

108.
Answer: 3
Sol:
Sensory nerves found in the epidermis
mediate touch reception, pain, and thermal
sensation. The skin is innervated by two
types of nerve fibers, sensory and
autonomic. Nerve fibers innervating the skin
originate from dorsal root ganglia nerve cell
bodies of sensory nerves .The sensory
nerves elongate from the dorsal root
ganglion and migrate toward the In cyclic photophosphorylation only
skin.Sensory nerves penetrate the basement photosystem I is active. movement of the
membrane and innervate the epidermis by electrons in a cyclic manner for synthesizing
moving vertically, terminating at the granular ATP molecules.occur in both stroma and
layer of the epidermis. grana thylakoid.
109. 111.
Answer: 2 Answer: 1
Sol: Sol:
E.P. Odum (1969) defined ecology as Parturition (giving birth after the full
“the study of structure and function of development of the fetus) is induced by the
nature” . signals released through the developing fetus
American biologist at the University of when the oxytocin hormone is released from
Georgia known for his pioneering work on the pituitary gland. This is known as fetal
ecosystem ecology ejection reflex.
This causes contractions in the uterine walls
that trigger the release of oxytocin hormone
which further increases the contraction rate.
These contractions with the help of oxytocin
hormone helps in the expulsion of the fetus
out of the uterine cavity of the mother.
Hence, the correct answer is option "1"
- II→I →IV→ V→ III
112. 115.
Answer: 4 Answer: 3
Sol: Sol:
Correct statements Here the hormone can be ADH . Because,
here the target tissue is kidney cell and the
Origin of replication' is a sequence from ADH mainly acts on collecting ducts of the
where replication starts by binding of kidney and increases the water reabsorption
polymerase enzyme. in this part.
DNA ligase acts on cut DNA molecules and 116.
joins their ends.
Answer: 2
Plasmid is autonomously replicating circular
extra-chromosomal DNA. Sol:
More than 900 restriction endonucleases A is desert.
have been discovered from over 230 strains
of bacteria. B is grassland.

113. C stands for tropical forest.

Answer: 3 D is a temperate forest.

Sol: E stands for coniferous forest.

The amount of blood pumped by heart per F stands for Arctic and Alpine Tundra.
minute is called cardiac output or heart 117.
output.
Answer: 2
114.
Sol:
Answer: 2
The above given figure is norplant.
Sol:
Norplant is placed subcutenous and having
Mucor and Rhizopus are fungi that belong to levonorgestral.
the class Phycomycetes.
These having 6 match stick like cyclinder
While other option are not :- which is effective up to 5 years.
Ascomycetes → Includes sac Norplant having 2 cyclinder are effective till 2
fungi like Penicillium and Saccharomyces; years.
they reproduce sexually with ascospores in
an ascus. 118.

Basidiomycetes → Includes club Answer: 4

fungi like Agaricus (mushrooms)
Sol:
and Puccinia (rust fungi); reproduce
by basidiospores. Dicot stem
Deuteromycetes → Also called Fungi 11th NCERT PAGE NO.- 75
Imperfecti, these fungi (e.g., Alternaria)
reproduce only asexually by conidia.
119. 121.
Answer: 2 Answer: 2
Sol: Sol:
In anaerobic respiration, sugar (typically The total cardiac cycle is for 0.8 seconds
glucose) undergoes incomplete breakdown
due to the absence of oxygen. The products Atrial systole is for 0.1 sec.
differ depending on the organism: Atrial diastole is 0.7 sec.
In yeast and some bacteria: The The closure of semilunar valve is at end of
process produces alcohol and CO₂ ventricular systole (0.3 sec).
(alcoholic fermentation).
Ventricular diastole is 0.5 sec.
Equation:
C6 H12 O6 → 2 C2 H5 OH + 2 CO2 ​ AV valves close at the end of atrial systole
In muscle cells during intense and beginning of ventricular systole (0.1
exercise: Lactic acid is produced (lactic second) now the time gap between these two
acid fermentation). events is 0.5 seconds.
120. 122.
Answer: 4 Answer: 2
Sol: Sol:
B–galactosidase acts on it, Explanation: X- Euglenoids are single celled organism i.e.
gal is an analog of lactose, and therefore mostly autotrophic (is the organism which
may be hydrolyzed by the β-galactosidase behaved like plants in the presence of light)
enzyme which cleaves the β-glycosidic bond but they can be heterotrophic (in the
in D-lactose. absence of organic food, it behaves like a
X-gal, when cleaved by β-galactosidase, heterotrophs). They share the characteristics
yields galactose and 5-bromo-4-chloro-3- of both plants and animals.
hydroxyindole. The latter then spontaneously
dimerizes and is oxidized into 5,5'-dibromo- Archaebacteria live in some of the harshest
4,4'-dichloro-indigo, an intensely blue habitats such as extreme salty areas, hot
product which is insoluble. X-gal itself is springs etc. Nostoc is true bacteria that help
colorless, so the presence of blue-colored in atmospheric nitrogen. Paramecium has
product may therefore be used as a test for cilia on the surface of the body included in
the presence of active β-galactosidase. This protozoa.
easy identification of an active enzyme Hence the correct answer is Euglena.
allows the gene for β-galactosidase (the lacZ
gene) to be used as a reporter gene in
various applications.
123. 126.
Answer: 2 Answer: 4
Sol: Sol:
C – Generates ATP through substrate Tropical rainforests have a very dense plant
level phosphorylation cover and experience a large amount of
precipitation, thus the forest floor is always
During the conversion of succinyl-CoA to damp.
succinic acid a molecule of GTP is
synthesised. This is a substrate level Thus, the conditions there have led animals
phosphorylation. In a coupled reaction GTP is to get adapted to arboreal habitats. Most
converted to GDP with the simultaneous animals found there are tree dwellers as
synthesis of ATP from ADP. Also there are almost every space on the forest floor is
three points in the cycle where NAD+ is occupied by vegetation.
reduced to NADH + H+ and one point where
FAD+ is reduced to FADH2. Hence, the correct answer is option "4".
127.
Answer: 1
Sol:
12th NCERT Page No. 119
128.
Answer: 1
Sol:
Golden rice is a genetically modified crop
which is highly nutritive and have high
124. content of vitamin-A
Answer: 1
Sol:
NEW 11th NCERT, PAGE NO- 245
125.
Answer: 2
Sol:
Vascular bundles of monocot stem are
scattered in the ground tissue and V.B. are
generally oval shape. Vascular bundles lie
towards the center are large in size and less
in number. Vascular bundles situated towards
the periphery are small in size but more in
number. Each vascular bundle is conjoint
collateral and closed (cambium is absent
between the xylem and phloem) and xylem
is endarch (xylem is centrifugal).
129. 131.
Answer: 4 Answer: 3
Sol: Sol:
11th NCERT, Page No.- 212 LH and FSH are collectively called
gonadotropins. Luteinizing hormone
Correct statements (LH) plays a key role in gonadal function.
4. Angiotensin II also activates the adrenal LH, in synergy with follicle-stimulating
cortex to release Aldosterone. Aldosterone hormone (FSH), stimulates follicular growth
causes reabsorption of Na+ and water from and ovulation.
the distal parts of the tubule. This also leads Growth hormone (GH), also called
to an increase in blood pressure and GFR. somatotropin, It stimulates the growth of
Incorrect statements essentially all tissues of the body, including
bones.
1. ADH facilitates water reabsorption from
latter parts of the tubule, thereby preventing Oxytocin is a hormone that acts on organs
diuresis. in the body (including the breast and uterus)
and as a chemical messenger in the brain.
2. The JGA plays a complex regulatory role.
A fall in glomerular blood flow/glomerular Prolactin, also called luteotropic hormone
blood pressure/GFR can activate the JG cells (LTH) or luteotropin, is a protein hormone
to release renin which converts produced by the pituitary gland of mammals
angiotensinogen in blood to angiotensin I that acts with other hormones to initiate the
and further to angiotensin II. secretion of milk by the mammary glands.

3. An increase in blood flow to the atria of 132.

the heart can cause the release of Atrial Answer: 4
Natriuretic Factor (ANF) and It acts as a
check on the renin angiotensin mechanism. Sol:
130. Bulliform cells(motor cells) are found in the
epidermis.These cells work to reduce the rate
Answer: 2 of transpiration.Their main function is to
Sol: store the water. It results in rolling of leaf to
prevent the water loss due to transcription
Heterocysts, which are specialized cells found under stress. When water is abundant, water
in the cyanobacterium Nostoc, are and bulge are absorbed and shrink when less
specifically adapted for nitrogen fixation. water is present, curling the leaf which helps
Nitrogen fixation is the process by which to reduce water loss due to evaporation. This
atmospheric nitrogen is converted into a is important because folding the leaf changes
form that can be used by living organisms, its exposure to light and the amount of water
such as ammonia or nitrate. it retains.
Heterocysts are specialized cells within 133.
Nostoc that are involved in nitrogen fixation.
They have thick cell walls and lack Answer: 2
photosynthetic pigments like chlorophyll, Sol:
making them different from the other cells in
the filamentous cyanobacteria. The Tropical Savanna biome is located in
South America and Australia. It is also
called tropical grassland. The rainfall is
seasonal and very high. In this biome, grass
is found along with a few scattered trees.
134. 137.
Answer: 3 Answer: 4
Sol: Sol:
Oxygen is required in cellular respiration in Distal convoluted tubule (DCT) and collecting
the electron transport chain. The presence of duct allow extensive reabsorption of water
oxygen is vital since it drives the whole and certain electrolytes, which help in
process by removing hydrogen from the
osmoregulation.
system.
It serves as the final electron acceptor of the Approximately 7 –10 % of filtered calcium
electron transport chain, facilitating the is reabsorbed in the DCT.
movement of electrons down the chain and
The proximal convoluted tubule helps in
hence producing ATP, or adenosine
maintaining the pH and ionic balance by
triphosphate.
reabsorbing hydrogen ions, ammonia,
Carbon dioxide is a waste byproduct of and potassium ions into the filtrate.
cellular respiration. This is removed from the
body by respiration. All plants and animals
respire by reducing carbohydrates in their
cells to produce energy and give out carbon
dioxide into the atmosphere.
135.
Answer: 1
Sol:
12th- Ncert, Pag.no.113
136.
Answer: 4
Sol:
Insulin consists of two short polypeptide 138.
chains: chain A and chain B, that are linked
Answer: 2
together by disulphide bridges.
Sol:
Insulin is in pro-hormone form( it
contains extra stretch called c-peptide) and Most of the members of green algae have
needs to be processed before it becomes one or more storage bodies called Pyrenoids
fully functional hormone(C-peptide is not located in the chloroplasts. It contains
present). protein besides starch.
The main challenge for production of insulin
using rDNA techniques was getting insulin
assembled into a mature form.
Hence according to the given situation it will
not be a functional insulin and will not
treat pateint sucessfully and in addition may
cause allergic reactions.
Hence, the correct answer is option "4"
- Only (iii).
Hence, the correct answer is algae.
139. 141.
Answer: 2 Answer: 3
Sol: Sol:
Geometric growth is the growth where the Geographical location, Explanation: The
successive change in population differs by marine biome is the biggest biome in the
constant ration. It is characterised by a slow world. It covers about 70% of the earth. It
growth in the initial stages and a rapid includes five main oceans: the Pacific,
growth during the later stages. The daughter Atlantic, Indian, Arctic, and Southern, as well
cells derived from mitosis retain the ability to as many smaller Gulfs and Bays. Marine
divide, but slow down because of a limited biomes remain almost same for all
nutrient supply. geographical location. It is affected by
temperature, salinity and types of sea floor
The exponential growth can be expressed as: or stratification.
W1 = W0 ert 142.
W1 = final size (weight, height, number etc.) Answer: 4
W0 = initial size at the beginning of the Sol:
period
All the given options are correct for Bt
r = growth rate transgenic plant except option (d). It can be
t = time of growth corrected as follows Bttransgenic plants are
genetically engineeredplants that contain
e = base of natural logarithms genes from Bacillus thuringiensis. These are
140. resistant to various diseases, pests, insects
and possess other important characteristics.
Answer: 4
143.
Sol:
Answer: 1
YR Yr yR yr
YR YYRR YYRr YyRR YyRr Sol:
Yr YYRr YYrr YyRr Yyrr
yR YyRR YyRr yyRR yyRr
i) Process of tubular secretion helps to
yr YyRr Yyrr yyRr yyrr secrete the urea from the blood to the
Phenotypic ratio : collecting duct which is finally excreted in
form of urine. The purified blood comes from
Yellow Yellow Green Green the kidney through the renal vein which has
the blood with the least amount of urea.
Round Wrinkled Round Wrinkled
iii) The kidney conserves water by first
9 : 3 : 3 : 1 diluting urine as it moves through the loop of
Genoypic ratio : Henle and then concentrating urine in the
distal tubules and collecting ducts.
YYRR YYRr YyRR YyRr YYrr Yyrr
yyRR yyRr yyrr iv) Glomerular filtrate contains all the
components of blood plasma except the
1 : 2 : 2 : 4 : 1 proteins.
: 2 : 1 : 2 : 1
144.
So as in this above-given dihybrid cross, we
can see that YYRR, YYrr, yyRR and yyrr is Answer: 3
coming only one time so overall 4 genotype Sol:
is there which represented only once in a
dihybrid cross. Haploid
145. 149.
Answer: 3 Answer: 1
Sol: Sol:
For one of the progeny to have O blood Cytotaxonomy that is based on cytological
group parents will have have heterozygous information like chromosome number,
blood group for A and B. structure, behaviour and chemotaxonomy
that uses the chemical constituents of the
In ABO blood grouping A and B are dominant plant to resolve confusions, are also used by
and O is recessive hence need to be in pair if taxonomists these days.
expressed.
150.
Hence, the correct answer is option "3".
Answer: 3
blood group of parents A O
B AB BO Sol:
O AO OO
xaxa × xAy
146. xA y
Answer: 1 xa xAxa xay
Sol: 50% 50%
We must remember that the physico- 151.
chemical (abiotic) components alone do
not characterise the habitat of an Answer: 2
organism completely; the habitat Sol:
includes biotic components also –
pathogens, parasites, predators and A larger protected area meant for
competitors – of the organism with which conservation of biodiversity and culture of
they interact constantly. that area is called Biosphere Reserve.

147. A national park is a park in use for

conservation purposes, created and
Answer: 1 protected by national governments.
Sol: A wildlife sanctuary is an area where animals
Cytokinin is a plant hormone which promote and birds can live protected and safe in their
nutrient mobilisation which helps in the delay natural habitats, away from poaching or
of leaf senescence. trafficking.

148. To take care of natural heritage of India,

government has set-up 90 national
Answer: 1 parks and 448 wild life sanctuaries.
The Indian government has established 14
Sol: biosphere reserves to protect larger areas of
A human protein that is obtained from natural habitat.
transgenic animals and is widely used to 152.
treat emphysema is α −1-antitrypsin.
Transgenic animals produce a lot of useful Answer: 3
biological products which are used to create
several protein molecules. This is done by Sol:
the introduction of a portion of DNA which Ethylene promotes the elongation of
codes for particular products such as human internodes in deep-water rice plants so that
proteins. the leaves and upper parts of the plant
remain above water.
153. 156.
Answer: 3 Answer: 1
Sol: Sol:
The squamous epithelium is made of a single Class 12th NCERT Page No. 259
thin layer of flattened cells with irregular
boundaries. They are found in the walls of 157.
blood vessels and air sacs of lungs and are Answer: 3
involved in functions like forming a diffusion
boundary. Sol:

Cuboidal/Columnar involved in functions like Two male gametes are produced from a
secretion and absorption. single pollen grain in flowering plants. Pollen
grains released from pollen-sacs at the two-
154. celled stage, wherein the generative cell
Answer: 2 divides further to form two male gametes.
They are then released into the embryo sac.
Sol: They fuse with the female gametes for the
formation of the embryo (egg) and
Dragonflies are the natural predators of endosperm (central cells). Once pollination is
mosquitoes. Hence, they can be used as achieved, the pollen tube grows up through
biocontrol agents to protect the crops from the style and stigma and towards the ovules
mosquitoes. in the ovaries. In the pollen grains, the germ
155. cells divide. This releases two sperm cells
that move down the pollen tube.
Answer: 1
One meiosis gives rise to four pollen grains.
Sol: Each of these pollen grain gives rise to 2
male gametes. Consequently, the overall
Crossing over is the process of exchange of
outcome of meiosis is 8 male gametes.
genetic material between non-sister
In given question, there are 21MMC in an
chromatids of homologous chromosomes
anther then the number of male gametes
which produce new genetics combinations.
produced from them, is -
Further, the independent assortment is
21 x 8 = 168.
segregation of factors for a trait independent
of other factors during gamete formation 158.
followed by their random rearrangement in
progeny thereby producing both parental and Answer: 4
new combinations. The linkage is the Sol:
tendency of closely placed genes on a
chromosome to stay together during NCERT 11th Page No.116
inheritance (no crossover and independent
assortment); it produces more parental The body of the frog is divisible into head
combination and less/no new combinations and trunk. Neck and tail are absent in a frog.
of the gene. Frog's forelimbs and hind limbs help in
Completely linked genes do not show leaping, swimming, burrowing, and walking.
crossover and independent assortment, The hind limbs of frogs have five digits while
thereby produce only parental the forelimbs end in four digits. Hind limbs
combinations; are more muscular and larger when
compared to forelimbs.
A number of recombinant types in a
population depends on upon crossover
frequency between the genes; the higher the
crossover frequency, the higher is the
number of recombinant types.
159. 163.
Answer: 2 Answer: 2
Sol: Sol:
Lysosomes release the enzymes on activation An informosome is a particle found in animal
by intracellular environment. In plants cells, cells that is made up of a special protein and
particularly at the time of seed macromolecular ribonucleic acid (RNA). The
germination, lysosomal enzymes degrade protein in informosomes may help move
macromolecules like starch and reserve mRNA from the nucleus to the cytoplasm,
proteins into glucose and amino acids protect mRNA from destruction, and control
respectively. the rate of protein synthesis.
160. 164.
Answer: 4 Answer: 2
Sol: Sol:
12th NCERT PAGE NO.- 106 Ribosomes are the cell organelle which
is non- membranous and found in both
161. Prokaryotes and Eukaryotes. Prokaryotes and
Answer: 3 eukaryotes are the two different types of
cells.
Sol:
Eukaryotic cells contain membrane-
In seed plants, fertilization is called bound organelles, such as the nucleus,
Siphonogamy because the male gametes are endoplasmic reticulum, mitochondria
brought to the egg present in female while prokaryotic cells do not.
gametophyte by a pollen tube.
165.
Internal fertilization: Syngamy occurs inside
the body of organisms. It is present in the Answer: 3
majority of plants like Bryophytes Sol:
to Angiosperms. In all these organisms egg
is formed inside the female body where Hybrid varieties of several of our food and
syngamy occurs. The male gametes either vegetable crops are being extensively
through water or pollen tube, are transferred cultivated. Cultivation of hybrids has
to female gametes. In order to enhance the tremendously increased productivity. One of
chances of syngamy large number of sperms the problems of hybrids is that hybrid seeds
are produced in these organisms and to have to be produced every year.
compensate for this, there is a significant
reduction in the number of eggs produced. If the seeds collected from hybrids are sown,
the plants in the progeny will segregate
162. and do not maintain hybrid characters.
Production of hybrid seeds is costly and
Answer: 3 hence the cost of hybrid seeds become too
Sol: expensive for the farmers.Apomictics can
also help preventing seggregation.
Class 11th NCERT Page No. 118
166.
Answer: 1
Sol:
12th NCERT Page No. 130-131
167. 172.
Answer: 4 Answer: 4
Sol: Sol:
Synthesis of DNA from RNA occurs by 12th NCERT Page No. - 156, 157, 158
reverse transcriptase enzyme in reverse
Column I Column II
transcription, discovered by Temin &
Don’t die of
Baltimore (a) AIDS (iii)
ignorance
(b) Cancer (iv) Metastasis
168.
Biological Response
(c) (i) α -Interferon
modifier
Answer: 3
(d) HIV factory (ii) Macrophages
Sol:
173.
The main difference is -
Answer: 3
The endoplasmic reticulun bearing ribosomes
Sol:
on their surface is called rough endoplasmic
reticulum (RER). In the absence of Each strand has a backbone made of
ribosomes they appear smooth and are alternating groups of sugar (deoxyribose)
called smooth endoplasmic reticulum (SER). and phosphate groups. Attached to each
169. sugar is one of four bases: adenine (A),
cytosine (C), guanine (G), and thymine (T).
Answer: 3 The two strands are held together by bonds
between the bases, adenine forming a base
Sol: pair with thymine, and cytosine forming a
12th NCERT Page No. 138 base pair with guanine. If a DNA nucleotide
chain has AGCTTCGA sequence, then
170. nucleotide sequence of other chain would be
TCGAAGCT.
Answer: 4
The correct answer is option C
Sol:
174.
Class 12th NCER Page No. 90
Answer: 2
171.
Sol:
Answer: 3
Class 11th NCERT Page No. 169
Sol:
In meiosis, the centromere divides during
Class 11th NCERT Page No. 168
anaphase II. During anaphase II, the
During the pachytene stage of prophase I in centromere of each chromosome splits,
meiosis, homologous chromosomes pair up allowing the sister chromatids to move to
to form bivalents (also called tetrads). opposite poles of the cell.

Each bivalent consists of: 175.

Four chromatids: Two chromatids from Answer: 2

each homologous chromosome.
Sol:
Two centromeres: One centromere from
11th NCERT, Page No.- 54
each homologous chromosome.
176. 179.
Answer: 4 Answer: 1
Sol: Sol:
Anaphase, Anaphase-II 11th NCERT Page No – 46
11th NCERT PAGE NO.- 127 The members of subphylum Vertebrata
possess notochord during the embryonic
177. period. The notochord is replaced by a
Answer: 4 cartilaginous or bony vertebral column in the
adult.
Sol:
All vertebrates are chordates, but not all
Anopheles is the vector of malaria. chordates are vertebrates. This is because
vertebrates are a subphylum within the
Lac insect (Laccifer) is not a vector.
phylum Chordata, meaning all vertebrates
Culex. Culex, a large group of mosquitoes are classified as chordates, but there are
also known as common house mosquitoes, other chordates (like tunicates and lancelets)
are the principal vectors that spread the that are not vertebrates.
viruses that cause West Nile fever, St. Louis
A is true but R is false
encephalitis, and Japanese encephalitis,
180.
Aedes aegypti is a known vector of several
viruses including yellow fever virus, dengue Answer: 2
virus chikungunya virus and Zika virus.
Sol:
178.
Ostrich and Corvus = Scales on hind limbs
Answer: 4
NCERT : Page No.- 48 to 51
Sol:
Chromosome synapsis is accompanied by the
formation of complex structure called
synaptonemal complex.
The complex formed by a pair of synapsed
homologous chromosomes( one from
paternal and one from maternal) is called
a bivalent or a tetrad.