📋 Overview ⚛ Physics Q1–45 🧪 Chemistry Q46–90 🧬 Biology Q91–180
📄 NEET 2025 · Official Paper

NEET 2025 Question Paper
Answer Key & Solutions

All 180 questions · Correct answers highlighted · Step-by-step solutions below every question

180Questions
45Physics
45Chemistry
90Biology
720Total Marks
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📘 NEET 2025 — Complete Paper with Step-by-Step Solutions

Ph

Physics

Questions 1–45

45 Qs · 180 Marks
PhysicsQ.
The current passing through the battery in the given circuit, is:
Options
A
1.5 A
B
2.0 A
C
0.5 A
✓
D
2.5 A
✦ Correct Answer
0.5 A
📐 Solution
1
The circuit forms a balanced Wheatstone bridge.
Equivalent resistance: R_eq = 8/3 + 1/3 + 1.5 + 5.5 = 10 Ω
Current i = V/R_eq = 5/10 = 0.5 A
PhysicsQ.
The electric field in a plane electromagnetic wave is given by Ez = 60 cos(5x + 1.5×10⁹t) V/m. The corresponding magnetic field is:
Options
A
By = 60 sin(5x + 1.5×10⁹t) T
B
By = 2×10−7 cos(5x + 1.5×10⁹t) T
✓
C
Bx = 2×10−7 cos(5x + 1.5×10⁹t) T
D
Bz = 60 cos(5x + 1.5×10⁹t) T
✦ Correct Answer
By = 2×10−7 cos(5x + 1.5×10⁹t) T
📐 Solution
1
E and B are in same phase; B₀ = E₀/c
By = 60/(3×10⁸) × cos(5x + 1.5×10⁹t)
= 2×10−7 cos(5x + 1.5×10⁹t) T
PhysicsQ.
A pipe open at both ends has fundamental frequency f in air. It is dipped vertically in water to half its length. The fundamental frequency of the air column is now:
Options
A
2f
B
f/2
C
f
✓
D
3f/2
✦ Correct Answer
f
📐 Solution
1
Open pipe: f = v/2L
Dipped half → acts as closed pipe of length L/2
f' = v/[4×(L/2)] = v/2L = f
PhysicsQ.
An electron (mass 9×10−31 kg) moves with speed c/100 in B = 9×10−4 T perpendicular to its motion. E applied so electron doesn't deflect:
Options
A
E ∥ B, 27×10⁴ V/m
B
E ⊥ B, 27×10⁴ V/m
C
E ⊥ B, 27×10² V/m
✓
D
E ∥ B, 27×10² V/m
✦ Correct Answer
E ⊥ B, 27×10² V/m
📐 Solution
1
For no deflection: eE = evB → E ⊥ B
E = vB = (3×10⁸/100) × 9×10−4 = 27×10² V/m
PhysicsQ.
Four similar thin convex lenses in contact. Power P and magnification M of combination vs single (p, m):
Options
A
p⁴ and m⁴
B
4p and 4m
C
p⁴ and 4m
D
4p and m⁴
✓
✦ Correct Answer
4p and m⁴
📐 Solution
1
Series: P_eff = 4p (additive)
m_eff = m₁×m₂×m₃×m₄ = m⁴
PhysicsQ.
2 A current in two circular coils; radii ratio 1:2. Ratio of magnetic moments:
Options
A
4:1
B
1:4
✓
C
1:2
D
2:1
✦ Correct Answer
1:4
📐 Solution
1
M = IA ∝ r² (same I)
M₁/M₂ = r₁²/r₂² = (1/2)² = 1/4
PhysicsQ.
Constant voltage 50 V between A and B. Current through branch CD:
Options
A
3.0 A
B
1.5 A
C
2.0 A
✓
D
2.5 A
✦ Correct Answer
2.0 A
📐 Solution
1
RAB = (1‖3) + (2‖4) = 3/4 + 8/6 = 25/12 Ω
Total I = 24 A; I1Ω=18A, I2Ω=16A
ICD = 18 − 16 = 2 A
PhysicsQ.
Two gases A and B at same pressure. Equal heat supplied, pistons displaced 16 cm and 9 cm, same ΔU. Ratio rA/rB:
Options
A
√3/2
B
4/3
C
3/4
✓
D
2/√3
✦ Correct Answer
3/4
📐 Solution
1
WA=WB → πrA²×16 = πrB²×9
rA/rB = √(9/16) = 3/4
PhysicsQ.
Container V₁=2L at 1 atm (5 mol), V₂=3L at 2 atm (4 mol). Partition removed. Equilibrium pressure:
Options
A
1.8 atm
B
1.3 atm
C
1.6 atm
✓
D
1.4 atm
✦ Correct Answer
1.6 atm
📐 Solution
1
P₁V₁ + P₂V₂ = P(V₁+V₂)
1(2) + 2(3) = P(5) → P = 8/5 = 1.6 atm
PhysicsQ.
Martian orbit ≈ 4× Mercury. Martian year = 687 days. Mercury year:
Options
A
124 days
B
88 days
✓
C
225 days
D
172 days
✦ Correct Answer
88 days
📐 Solution
1
Kepler's 3rd law: (T'/T)² = (4)³ = 64 → T'/T = 8
TMercury = 687/8 ≈ 88 days
PhysicsQ.
AC 220V, 50Hz; R=20Ω, XC=25Ω, XL=45Ω series. Current and phase angle:
Options
A
15.6 A and 45°
B
7.8 A and 30°
C
7.8 A and 45°
✓
D
15.6 A and 30°
✦ Correct Answer
7.8 A and 45°
📐 Solution
1
Z = √[(20)²+(20)²] = 20√2 Ω
I = 220/20√2 ≈ 7.8 A; tan φ = 1 → φ = 45°
PhysicsQ.
Wire resistance R cut into 8 equal pieces. Two sets (4 each) in parallel, then series:
Options
A
R/8
B
R/64
C
R/32
D
R/16
✓
✦ Correct Answer
R/16
📐 Solution
1
Each piece = R/8; Each parallel set = R/32
Two in series: R_eq = R/16
PhysicsQ.
Output Y of logic circuit (NOR + NAND combination):
Options
A
NOR
✓
B
AND
C
NAND
D
OR
✦ Correct Answer
NOR
📐 Solution
1
Y₁ = ‾(A+B), Y₂ = ‾(A·B)
Y = Y₁·Y₂ → simplifies to ‾(A+B) = NOR
PhysicsQ.
Two spheres A, B each charge q, force F. Third uncharged sphere touched A then B, removed. New force:
Options
A
3F/8
✓
B
3F/5
C
2F/3
D
F/2
✦ Correct Answer
3F/8
📐 Solution
1
After touching A: q/2 each. After touching B: 3q/4
F' = K(q/2)(3q/4)/r² = 3F/8
PhysicsQ.
Vernier calipers: 10VSD=9MSD, MSD=0.1cm, zero error=+0.1cm, M=5cm, coinciding div=8:
Options
A
5.00 cm
B
5.18 cm
C
5.08 cm
D
4.98 cm
✓
✦ Correct Answer
4.98 cm
📐 Solution
1
LC = 0.01 cm; Reading = 5 + 8×0.01 − 0.10 = 4.98 cm
PhysicsQ.
t = x² + x. Acceleration of particle:
Options
A
-2/(2x+1)
B
-2/(x+2)³
C
-2/(2x+1)³
✓
D
+2/(x+1)³
✦ Correct Answer
-2/(2x+1)³
📐 Solution
1
v = 1/(2x+1); a = v·dv/dx = [1/(2x+1)]×[-2/(2x+1)²] = -2/(2x+1)³
PhysicsQ.
Which option correctly shows variation of photoelectric current?
Options
A
B and D
B
A only
✓
C
A and C
D
A and D
✦ Correct Answer
A only
📐 Solution
1
Photoelectric current ∝ Intensity (linear).
Only Graph A (linear with intensity) is correct.
PhysicsQ.
Particle mass m, constant force F, Bohr model. Radius r and speed v depend on n:
Options
A
r ∝ n^(4/3); v ∝ n^(−1/3)
B
r ∝ n^(1/3); v ∝ n^(1/3)
C
r ∝ n^(1/3); v ∝ n^(2/3)
D
r ∝ n^(2/3); v ∝ n^(1/3)
✓
✦ Correct Answer
r ∝ n^(2/3); v ∝ n^(1/3)
📐 Solution
1
F=mv²/r=const → r∝v². mvr=nh/2π
Solving: v∝n^(1/3), r∝n^(2/3)
PhysicsQ.
Bob on string length l, horizontal velocity v₀. String slacks at angle θ from horizontal. v/v₀:
Options
A
[sinθ/(2+3sinθ)]^(1/2)
✓
B
(sinθ)^(1/2)
C
[1/(2+3sinθ)]^(1/2)
D
[cosθ/(2+3sinθ)]^(1/2)
✦ Correct Answer
[sinθ/(2+3sinθ)]^(1/2)
📐 Solution
1
mg sinθ = mv²/l; Energy conservation: v₀²/2 = v²/2 + gl(1+sinθ)
v/v₀ = [sinθ/(2+3sinθ)]^(1/2)
PhysicsQ.
Full wave rectifier, Vin=220sin(100πt). At t=15 ms:
Options
A
D₁ and D₂ both reverse biased
B
D₁ forward, D₂ reverse
C
D₁ reverse, D₂ forward
✓
D
D₁ and D₂ both forward
✦ Correct Answer
D₁ reverse, D₂ forward
📐 Solution
1
T=0.02s; t=15ms=3T/4 → negative half cycle
D₁ reverse biased, D₂ forward biased
PhysicsQ.
Balloon S, area A, density ρ, radius R→0 in time T. v(r)∝ra, T∝SαAβργRδ:
Options
A
a=1/2,α=1/2,β=−1/2,γ=1/2,δ=7/2
B
a=1/2,α=1/2,β=−1,γ=+1,δ=3/2
C
a=−1/2,α=−1/2,β=−1,γ=−1/2,δ=5/2
D
a=−1/2,α=−1/2,β=−1,γ=1/2,δ=7/2
✓
✦ Correct Answer
a=−1/2,α=−1/2,β=−1,γ=1/2,δ=7/2
📐 Solution
1
Dimensional analysis: α=−1/2, γ=1/2, β=−1, δ=7/2, a=−1/2
a=−1/2, α=−1/2, β=−1, γ=1/2, δ=7/2
PhysicsQ.
Microscope f₀=2cm, fe=4cm, L=40cm, D=25cm. Magnification:
Options
A
250
B
100
C
125
✓
D
150
✦ Correct Answer
125
📐 Solution
1
m = (L/f₀)×(D/fe) = (40/2)×(25/4) = 20×6.25 = 125
PhysicsQ.
Two masses P, Q on springs k₁, k₂. Same max speed. Ratio AQ/AP:
Options
A
√(k₁/k₂)
✓
B
k₂/k₁
C
k₁/k₂
D
√(k₂/k₁)
✦ Correct Answer
√(k₁/k₂)
📐 Solution
1
vmax=Aω=A√(k/m). AP√k₁=AQ√k₂
AQ/AP = √(k₁/k₂)
PhysicsQ.
Circular plate capacitor, charge density increasing at constant rate. Magnetic field due to displacement current:
Options
A
Zero between plates, non-zero outside
B
Zero everywhere
C
Constant between, zero outside
D
Non-zero everywhere, max at periphery
✓
✦ Correct Answer
Non-zero everywhere, max at periphery
📐 Solution
1
Acts like cylindrical wire → B increases linearly inside, 1/r outside.
Maximum at periphery (edge of plates)
PhysicsQ.
Dipole p=5×10−6 Cm in E=4×10⁵ N/C. Rotated 60°. ΔPE:
Options
A
1.5 J
B
0.8 J
C
1.0 J
✓
D
1.2 J
✦ Correct Answer
1.0 J
📐 Solution
1
ΔU = PE(cos0°−cos60°) = 5×10−6×4×10⁵×(1−½) = 1.0 J
PhysicsQ.
Two inclined surfaces 45°; one rough, one smooth. Body takes 2× longer on rough. μk:
Options
A
0.75
✓
B
0.25
C
0.40
D
0.5
✦ Correct Answer
0.75
📐 Solution
1
t∝1/√a; ratio=2 → a ratio=4
1/(1−μk)=4 → μk=0.75
PhysicsQ.
De-Broglie wavelength of electron in n=2 of H atom (a₀=0.052 nm):
Options
A
2.67 nm
B
0.067 nm
C
0.67 nm
✓
D
1.67 nm
✦ Correct Answer
0.67 nm
📐 Solution
1
r=0.052×4=0.208nm; nλ=2πr
λ=π×0.208 ≈ 0.67 nm
PhysicsQ.
Sun rotates in 27 days. Expands to 2R. New period:
Options
A
108 days
✓
B
100 days
C
105 days
D
115 days
✦ Correct Answer
108 days
📐 Solution
1
Conservation of L: Iω=I'ω' → T'=4T=108 days
PhysicsQ.
P=a³b²/c√d. Errors: a=1%, b=3%, c=2%, d=4%. % error in P:
Options
A
15%
B
10%
C
2%
D
13%
✓
✦ Correct Answer
13%
📐 Solution
1
ΔP/P=3(1)+2(3)+(2)+(½)(4)=3+6+2+2=13%
PhysicsQ.
Parallel plate capacitor d. Slabs K₁ (3d/8), K₂ (d/2). C=2C₀. K₁=1.25K₂. Value of K₁:
Options
A
1.33
B
2.66
✓
C
2.33
D
1.60
✦ Correct Answer
2.66
📐 Solution
1
Solving C_eq equation with air gap d/8: K₁ = 8/3 ≈ 2.66
PhysicsQ.
Ball 0.5 kg dropped 40m, rises 10m after collision. Impulse:
Options
A
84 NS
B
21 NS
✓
C
7 NS
D
0
✦ Correct Answer
21 NS
📐 Solution
1
v₁=28m/s (down), v₂=14m/s (up)
J=m(v₂+v₁)=0.5×42=21 NS
PhysicsQ.
Girl on scooty 60km/h. Bus every 30min (same dir), 10min (opposite). T and bus speed:
Options
A
15 min, 120 km/h
✓
B
9 min, 40 km/h
C
25 min, 100 km/h
D
10 min, 90 km/h
✦ Correct Answer
15 min, 120 km/h
📐 Solution
1
(v−60)×30=(v+60)×10 → v=120 km/h, T=15 min
PhysicsQ.
O₂ cylinder 30L, 18.20 mol. Gauge P drops to 11atm at 27°C. Mass withdrawn:
Options
A
0.156 kg
B
0.125 kg
C
0.144 kg
D
0.116 kg
✓
✦ Correct Answer
0.116 kg
📐 Solution
1
n_remaining=14.54 mol; removed=3.66 mol
Mass=3.66×32≈0.116 kg
PhysicsQ.
Circuit AB: 1H, 5V, 2Ω. i=2A, di/dt=+1A/s. VA−VB:
Options
A
10 volt
✓
B
5 volt
C
6 volt
D
9 volt
✦ Correct Answer
10 volt
📐 Solution
1
VA−L(di/dt)−5−iR=VB
VA−VB=1+5+4=10 V
PhysicsQ.
Sand leaks from spring-mass box. ω(t) and A(t) change as:
Options
A
ω decreases, A constant
B
ω increases, A constant
C
ω increases, A decreases
✓
D
ω decreases, A decreases
✦ Correct Answer
ω increases, A decreases
📐 Solution
1
m decreases → ω=√(k/m) increases; x₀=mg/k decreases → A decreases
PhysicsQ.
Electron in uniform B, flux=n(h/e). Magnetic moment at lowest state (n=1):
Options
A
heB/2πm
B
he/πm
C
he/2πm
✓
D
heB/πm
✦ Correct Answer
he/2πm
📐 Solution
1
μ=evr/2; using flux quantization: μ=neh/2πm
At n=1: μ=eh/2πm
PhysicsQ.
Body 48N on Earth surface. Force at h=R/3:
Options
A
36 N
B
16 N
C
27 N
✓
D
32 N
✦ Correct Answer
27 N
📐 Solution
1
W_h/W=[R/(R+R/3)]²=9/16
W_h=(9/16)×48=27 N
PhysicsQ.
Liquid surface tension S, density ρ. Equation for y(x):
Options
A
dy/dx=√(ρg/S)·x
B
d²y/dx²=(ρg/S)·x
C
d²y/dx²=(ρg/S)·y
✓
D
d²y/dx²=√(ρg/S)
✦ Correct Answer
d²y/dx²=(ρg/S)·y
📐 Solution
1
ρgy=S·d²y/dx² → d²y/dx²=(ρg/S)·y
PhysicsQ.
Polaroid at 22.5° between two crossed polaroids. Transmitted intensity:
Options
A
I₀/16
B
I₀/2
C
I₀/4
D
I₀/8
✓
✦ Correct Answer
I₀/8
📐 Solution
1
I=I₀cos²(22.5°)sin²(22.5°)=(I₀/4)×sin²(45°)=I₀/8
PhysicsQ.
Photon and electron (mass m) same energy E. Ratio λphoton/λelectron:
Options
A
(1/c)√(E/2m)
B
√(E/2m)
C
c√(2mE)
D
c√(2m/E)
✓
✦ Correct Answer
c√(2m/E)
📐 Solution
1
λphoton=hc/E; λe=h/√(2mE)
Ratio=c√(2m/E)
PhysicsQ.
Unpolarized light at Brewster's angle on medium μ=1.73:
Options
A
Transmitted completely polarized, refraction≈30°
B
Reflected completely polarized, reflection≈60°
✓
C
Reflected partially polarized, reflection≈30°
D
Both polarized at 60° and 30°
✦ Correct Answer
Reflected completely polarized, reflection≈60°
📐 Solution
1
μ=tan60°=1.73 → θP=60°
Reflected: completely polarized at 60°
PhysicsQ.
Uniform rod 20kg, 5m, on smooth wall at 60° to wall, rough floor. Friction force (g=10):
Options
A
200√3 N
B
100 N
C
100√3 N
✓
D
200 N
✦ Correct Answer
100√3 N
📐 Solution
1
Torque about foot: f=Mg·cosθ/(2sinθ)=(200/2)×cot60°×... = 100√3 N
PhysicsQ.
Three rods series: side rods 2K, middle K. Ends at 3T and T. T₁/T₂:
Options
A
5/4
B
3/2
C
4/3
D
5/3
✓
✦ Correct Answer
5/3
📐 Solution
1
Equal heat flow; solving: T₁=5T/2, T₂=3T/2
T₁/T₂=5/3
PhysicsQ.
Cars A(KE=100J, stops 1000m) and B(KE=225J, stops 1500m). FA/FB:
Options
A
1/2
B
3/2
C
2/3
✓
D
1/3
✦ Correct Answer
2/3
📐 Solution
1
F=KE/S → FA/FB=(100/1000)/(225/1500)=0.1/0.15=2/3
PhysicsQ.
Sphere R from sphere 2R. Ratio MOI small sphere to rest about Y-axis:
Options
A
7/64
B
7/8
C
7/40
D
7/57
✓
✦ Correct Answer
7/57
📐 Solution
1
Ismall=7MR²/40; Irest=8MR²/5−7MR²/40
Ratio=7/57
Ch

Chemistry

Questions 46–90

45 Qs · 180 Marks
ChemistryQ.
Λm=90 S cm² mol⁻¹ for 0.05 mol/L weak acid. Λ⁰₊=349.6, Λ⁰₋=50.4. Degree of dissociation α:
Options
A
0.215
B
0.115
C
0.125
D
0.225
✓
✦ Correct Answer
0.225
📐 Solution
1
α=Λm/Λ⁰m=90/(349.6+50.4)=90/400=0.225
ChemistryQ.
Statement I: Bond order zero → stable. Statement II: Higher bond order → longer bond length.
Options
A
I false, II true
B
Both true
C
Both false
✓
D
I true, II false
✦ Correct Answer
Both false
📐 Solution
1
Bond order zero → unstable. Higher bond order → shorter length. Both statements false.
ChemistryQ.
Ratio of wavelengths for n=2→3 and n=4→6 transitions in H atom:
Options
A
1/4
✓
B
1/36
C
1/16
D
1/9
✦ Correct Answer
1/4
📐 Solution
1
λ₂→₃=36hc/5RH; λ₄→₆=576hc/20RH
Ratio=(36/5)÷(576/20)=1/4
ChemistryQ.
Correct wavelength order: A=[Co(NH₃)₆]³⁺, B=[Co(CN)₆]³⁻, C=[Cu(H₂O)₄]²⁺, D=[Ti(H₂O)₆]³⁺:
Options
A
C
B
B
C
B✓
D
C
✦ Correct Answer
B
📐 Solution
1
Stronger field → smaller λ: CN⁻>NH₃>H₂O
Order: B<A<D<C
ChemistryQ.
k=0.03 s⁻¹ (1st order). Time for [A] from 7.2 to 0.9 mol/L:
Options
A
21.0 s
B
69.3 s
✓
C
23.1 s
D
210 s
✦ Correct Answer
69.3 s
📐 Solution
1
t=(2.303/0.03)×log(7.2/0.9)=(2.303/0.03)×3log2=69.3 s
ChemistryQ.
Match separation methods: A.CHCl₃+Aniline, B.Crude oil, C.Glycerol from spent-lye, D.Aniline-water. I.Distil. under reduced P, II.Steam distil., III.Fractional distil., IV.Simple distil.:
Options
A
A-III,B-IV,C-II,D-I
B
A-IV,B-III,C-I,D-II
✓
C
A-IV,B-III,C-II,D-I
D
A-III,B-IV,C-I,D-II
✦ Correct Answer
A-IV,B-III,C-I,D-II
📐 Solution
1
A→Simple distil.(IV), B→Fractional(III), C→Reduced pressure(I), D→Steam(II)
∴ A-IV, B-III, C-I, D-II
ChemistryQ.
Major product: PhCH₂CN + CH₃MgBr (excess) then H₃O⁺:
Options
A
Ph-CO-CH₃
B
Ph-C(CH₃)(OH)-CH₂CH₃
C
Ph-C(OH)(CH₃)-CH₂-CO-CH₃
✓
D
Ph-CH(OH)-CH₃
✦ Correct Answer
Ph-C(OH)(CH₃)-CH₂-CO-CH₃
📐 Solution
1
Excess Grignard adds twice to C≡N; H₃O⁺ hydrolysis gives β-hydroxy ketone
ChemistryQ.
Which can exist as cis-trans isomers?
Options
A
1,2-Dimethylcyclohexane
✓
B
Pent-1-ene
C
2-Methylhex-2-ene
D
1,1-Dimethylcyclopropane
✦ Correct Answer
1,2-Dimethylcyclohexane
📐 Solution
1
Each C bearing CH₃ has two different substituents → cis and trans isomers possible
ChemistryQ.
Equal atoms in: A.212g Na₂CO₃, B.248g Na₂O, C.240g NaOH, D.12g H₂, E.220g CO₂:
Options
A
B,D,E only
B
A,B,C only
C
A,B,D only
✓
D
B,C,D only
✦ Correct Answer
A,B,D only
📐 Solution
1
A:12Nₐ, B:12Nₐ, C:18Nₐ, D:12Nₐ, E:15Nₐ
A, B and D have equal atoms
ChemistryQ.
Correct bond dissociation energy order of C-H* in: I.benzene(sp²), II.phenylacetylene(sp), III.cyclohexane(sp³):
Options
A
II>III>I
B
II>I>III
✓
C
I>II>III
D
III>II>I
✦ Correct Answer
II>I>III
📐 Solution
1
% s-character: sp(50%)>sp²(33%)>sp³(25%) → stronger C-H bond
Order: II > I > III
ChemistryQ.
ΔHf(Ba²⁺, aq): ΔHf(SO₄²⁻)=−216, ΔHcryst(BaSO₄)=−4.5, ΔHf(BaSO₄)=−349 kcal/mol:
Options
A
+220.5
B
−128.5
✓
C
−133.0
D
+133.0
✦ Correct Answer
−128.5
📐 Solution
1
Hess's law: ΔHf(Ba²⁺)=−349−(−216)−(−4.5)=−128.5 kcal/mol
ChemistryQ.
Oxidation states of underlined: KO₂, H₂O₂, H₂SO₄:
Options
A
+4,−4,+6
B
+1,−1,+6
✓
C
+2,−2,+6
D
+1,−2,+4
✦ Correct Answer
+1,−1,+6
📐 Solution
1
KO₂: K=+1. H₂O₂: O=−1 (peroxide). H₂SO₄: S=+6
Answer: +1, −1, +6
ChemistryQ.
Minimum conductance in solution:
Options
A
[Co(NH₃)₅Cl]Cl
B
[Co(NH₃)₃Cl₃]
✓
C
[Co(NH₃)₄Cl₂]
D
[Co(NH₃)₆]Cl₃
✦ Correct Answer
[Co(NH₃)₃Cl₃]
📐 Solution
1
[Co(NH₃)₃Cl₃] is neutral complex → no ions in solution → minimum conductance
ChemistryQ.
Reaction NOT giving benzene as product:
Options
A
PhN₂Cl + H₂O (warm)
✓
B
Sodium benzoate + sodalime
C
n-hexane + Mo₂O₃ (773K)
D
Acetylene + red hot Fe (873K)
✦ Correct Answer
PhN₂Cl + H₂O (warm)
📐 Solution
1
PhN₂Cl + H₂O (warm) → Phenol + N₂ + HCl (NOT benzene)
ChemistryQ.
Paramagnetic species: A.[NiCl₄]²⁻, B.Ni(CO)₄, C.[Ni(CN)₄]²⁻, D.[Ni(H₂O)₆]²⁺, E.Ni(PPh₃)₄:
Options
A
A,D,E only
B
A,C only
C
B,E only
D
A,D only
✓
✦ Correct Answer
A,D only
📐 Solution
1
A: sp³, 2 unpaired ✓. B: sp³, 0 unpaired. C: dsp², 0 unpaired. D: sp³d², 2 unpaired ✓. E: sp³, 0 unpaired
∴ A and D only
ChemistryQ.
Does NOT decolourise bromine water:
Options
A
Para-aminophenol
B
Cyclohexane
✓
C
Para-hydroxybenzaldehyde
D
Styrene
✦ Correct Answer
Cyclohexane
📐 Solution
1
Cyclohexane (saturated, no activating groups) → no reaction with Br₂
ChemistryQ.
Match: A.Haber, B.Wacker, C.Wilkinson, D.Ziegler. I.Fe, II.PdCl₂, III.[(PPh₃)₃RhCl], IV.TiCl₄+Al(CH₃)₃:
Options
A
A-I,B-IV,C-III,D-II
B
A-I,B-II,C-IV,D-III
C
A-II,B-III,C-I,D-IV
D
A-I,B-II,C-III,D-IV
✓
✦ Correct Answer
A-I,B-II,C-III,D-IV
📐 Solution
1
Haber→Fe(I), Wacker→PdCl₂(II), Wilkinson→RhCl(III), Ziegler→TiCl₄(IV)
∴ A-I, B-II, C-III, D-IV
ChemistryQ.
Match vitamins-deficiency: A.B₁₂, B.Vit D, C.B₂, D.B₆. I.Cheilosis, II.Convulsions, III.Rickets, IV.Pernicious anaemia:
Options
A
A-IV,B-III,C-II,D-I
B
A-I,B-III,C-II,D-IV
C
A-IV,B-III,C-I,D-II
✓
D
A-II,B-III,C-I,D-IV
✦ Correct Answer
A-IV,B-III,C-I,D-II
📐 Solution
1
B₁₂→Pernicious anaemia(IV), D→Rickets(III), B₂→Cheilosis(I), B₆→Convulsions(II)
∴ A-IV, B-III, C-I, D-II
ChemistryQ.
Statement I: Ferromagnetism is extreme form of paramagnetism. Statement II: Unpaired e⁻ in Cr²⁺ = Nd³⁺.
Options
A
I false, II true
B
Both true
C
Both false
D
I true, II false
✓
✦ Correct Answer
I true, II false
📐 Solution
1
Ferromagnetism IS extreme paramagnetism ✓. Cr²⁺=4 unpaired; Nd³⁺=3 unpaired → NOT equal ✗
∴ I true, II false
ChemistryQ.
t½=1 min for 1st order. Time for 99.9% completion:
Options
A
10 minutes
✓
B
2 minutes
C
4 minutes
D
5 minutes
✦ Correct Answer
10 minutes
📐 Solution
1
t₉₉.₉% = 10 × t½ = 10 minutes
ChemistryQ.
Decreasing basic strength order:
Options
A
benzenamine>ethanamine>N-methylaniline>N-ethylethanamine
B
N-methylaniline>benzenamine>ethanamine>N-ethylethanamine
C
N-ethylethanamine>ethanamine>benzenamine>N-methylaniline
D
N-ethylethanamine>ethanamine>N-methylaniline>benzenamine
✓
✦ Correct Answer
N-ethylethanamine>ethanamine>N-methylaniline>benzenamine
📐 Solution
1
Aliphatic > Aromatic; Di > Mono (aliphatic)
Order: N-ethylethanamine > ethanamine > N-methylaniline > benzenamine
ChemistryQ.
Match cations to groups: A.Co²⁺, B.Mg²⁺, C.Pb²⁺, D.Al³⁺. I.Group-I, II.Group-III, III.Group-IV, IV.Group-VI:
Options
A
A-III,B-II,C-I,D-IV
B
A-III,B-IV,C-II,D-I
C
A-III,B-IV,C-I,D-II
✓
D
A-III,B-II,C-IV,D-I
✦ Correct Answer
A-III,B-IV,C-I,D-II
📐 Solution
1
Co²⁺→IV(III), Mg²⁺→VI(IV), Pb²⁺→I(I), Al³⁺→III(II)
∴ A-III, B-IV, C-I, D-II
ChemistryQ.
True about H₃PO₄ ionization: A.logK=logKa₁+Ka₂+Ka₃, B.H₃PO₄>H₂PO₄⁻>HPO₄²⁻, C.Ka₁>Ka₂>Ka₃, D.Ka₁=(Ka₂+Ka₃)/2:
Options
A
A,B,C only
✓
B
A,B only
C
A,C only
D
B,C,D only
✦ Correct Answer
A,B,C only
📐 Solution
1
A✓(K=Ka₁×Ka₂×Ka₃), B✓(neutral molecule strongest acid), C✓(successive ionization)
D✗. ∴ A, B and C only
ChemistryQ.
True statements: A.Ga high mp, Cs low mp. B.N and Cl same EN. C.Ar,K⁺,Cl⁻,Ca²⁺,S²⁻ isoelectronic. D.IE: Si>Al>Mg>Na. E.Cs>Li and Rb radius:
Options
A
A,C,E only
B
A,B,E only
C
C,E only
✓
D
C,D only
✦ Correct Answer
C,E only
📐 Solution
1
A✗(both low mp). B✗(N=3.0, Cl=3.0 same but statement context wrong). C✓(all 18e⁻). D✗(Mg anomaly). E✓
∴ C and E only
ChemistryQ.
Statement I: As forms AsH₃ like N forms NH₃. Statement II: Sb cannot form Sb₂O₅.
Options
A
I incorrect, II correct
B
Both correct
C
Both incorrect
D
I correct, II incorrect
✓
✦ Correct Answer
I correct, II incorrect
📐 Solution
1
As does form AsH₃ (arsine) ✓. Sb DOES form Sb₂O₅ → II incorrect ✗
∴ I correct, II incorrect
ChemistryQ.
Highest boiling point: A.0.015M C₆H₁₂O₆, B.0.01M Urea, C.0.01M KNO₃, D.0.01M Na₂SO₄:
Options
A
0.015M C₆H₁₂O₆
B
0.01M Urea
C
0.01M KNO₃
D
0.01M Na₂SO₄
✓
✦ Correct Answer
0.01M Na₂SO₄
📐 Solution
1
i×m: A=0.015, B=0.01, C=0.02, D=0.03
0.01M Na₂SO₄ (i=3) has highest value
ChemistryQ.
Benzenediazonium salt: I.Prepared at 273-278K, unstable dry. II.Iodobenzene via diazonium+KI (direct iodination difficult).
Options
A
I incorrect, II correct
B
Both correct
✓
C
Both incorrect
D
I correct, II incorrect
✦ Correct Answer
Both correct
📐 Solution
1
Both statements are correct. Diazonium salt explosive when dry; KI route needed for iodobenzene.
ChemistryQ.
Reagent to convert ester ArCO-OCH₃ → Ar-CHO:
Options
A
H₂/Pd-BaSO₄
B
(i)LiAlH₄ (ii)H⁺
C
(i)DIBAL-H (ii)H₂O
✓
D
(i)NaBH₄ (ii)H⁺
✦ Correct Answer
(i)DIBAL-H (ii)H₂O
📐 Solution
1
DIBAL-H at −78°C reduces ester to aldehyde (stops at aldehyde stage)
ChemistryQ.
Alkyl iodide SN2 faster than chloride. Reason: I is better leaving group (large size).
Options
A
A false, R true
B
Both true; R explains A
✓
C
Both true; R doesn't explain
D
A true, R false
✦ Correct Answer
Both true; R explains A
📐 Solution
1
RI > RCl in SN2 because I⁻ is better leaving group → Both correct, R explains A
ChemistryQ.
Correct decreasing acidity of aliphatic acids:
Options
A
HCOOH>(CH₃)₃CCOOH>(CH₃)₂CHCOOH>CH₃COOH
B
(CH₃)₃CCOOH>(CH₃)₂CHCOOH>CH₃COOH>HCOOH
C
CH₃COOH>(CH₃)₂CHCOOH>(CH₃)₃CCOOH>HCOOH
D
HCOOH>CH₃COOH>(CH₃)₂CHCOOH>(CH₃)₃CCOOH
✓
✦ Correct Answer
HCOOH>CH₃COOH>(CH₃)₂CHCOOH>(CH₃)₃CCOOH
📐 Solution
1
More alkyl groups → more +I effect → less acidic
HCOOH > CH₃COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH
ChemistryQ.
Reaction NOT part of Lassaigne's test:
Options
A
2CuO+C→2Cu+CO₂
✓
B
Na+C+N→NaCN
C
2Na+S→Na₂S
D
Na+X→NaX
✦ Correct Answer
2CuO+C→2Cu+CO₂
📐 Solution
1
2CuO+C→2Cu+CO₂ is not part of Lassaigne's test
ChemistryQ.
Monochlorination products of (CH₃)₂CH-CH₂-CH₃ (including stereoisomers):
Options
A
6
✓
B
2
C
3
D
5
✦ Correct Answer
6
📐 Solution
1
Counting all positions with stereoisomers: 1+2+1+2 = 6 isomers
ChemistryQ.
Sugar X: found in honey, keto sugar, laevorotatory. X is:
Options
A
Sucrose
B
D-Glucose
C
D-Fructose
✓
D
Maltose
✦ Correct Answer
D-Fructose
📐 Solution
1
D-Fructose: ketohexose, found in honey, laevorotatory (−93°), shows anomers
ChemistryQ.
Dalton's atomic theory could NOT explain:
Options
A
Law of gaseous volumes
✓
B
Law of conservation of mass
C
Law of constant proportion
D
Law of multiple proportion
✦ Correct Answer
Law of gaseous volumes
📐 Solution
1
Dalton's theory couldn't explain Gay-Lussac's law of gaseous volumes
ChemistryQ.
Higher yield of NO in N₂+O₂⇌2NO (ΔH=+180.7 kJ) at: A.Higher T, B.Lower T, C.Higher [N₂], D.Higher [O₂]:
Options
A
A,C,D only
✓
B
A,D only
C
B,C only
D
B,C,D only
✦ Correct Answer
A,C,D only
📐 Solution
1
Endothermic → higher T. Higher reactant concentrations shift forward
∴ A, C, D only
ChemistryQ.
Match Xe compounds with hybridization: A.XeO₃, B.XeF₂, C.XeOF₄, D.XeF₆. I.sp³d-linear, II.sp³-pyramidal, III.sp³d³-distorted oct, IV.sp³d²-sq pyramidal:
Options
A
A-IV,B-II,C-I,D-III
B
A-II,B-I,C-IV,D-III
✓
C
A-II,B-I,C-III,D-IV
D
A-IV,B-II,C-III,D-I
✦ Correct Answer
A-II,B-I,C-IV,D-III
📐 Solution
1
XeO₃→sp³ pyramidal(II), XeF₂→sp³d linear(I), XeOF₄→sp³d² sq pyramidal(IV), XeF₆→sp³d³ distorted(III)
∴ A-II, B-I, C-IV, D-III
ChemistryQ.
Match: A.Humidity, B.Alloys, C.Amalgams, D.Smoke. I.Solid in solid, II.Liquid in gas, III.Solid in gas, IV.Liquid in solid:
Options
A
A-III,B-II,C-I,D-IV
B
A-II,B-IV,C-I,D-III
C
A-II,B-I,C-IV,D-III
✓
D
A-III,B-I,C-IV,D-II
✦ Correct Answer
A-II,B-I,C-IV,D-III
📐 Solution
1
Humidity→Liq in gas(II), Alloys→Solid in solid(I), Amalgam→Liq in solid(IV), Smoke→Solid in gas(III)
∴ A-II, B-I, C-IV, D-III
ChemistryQ.
Energy and radius of first Bohr orbit of He⁺ and Li²⁺:
Options
A
E(Li²⁺)=−8.72×10⁻¹⁶J,r=17.6pm; E(He⁺)=−19.62×10⁻¹⁶J,r=17.6pm
B
E(Li²⁺)=−19.62×10⁻¹⁸J,r=17.6pm; E(He⁺)=−8.72×10⁻¹⁸J,r=26.4pm
✓
C
E(Li²⁺)=−8.72×10⁻¹⁸J,r=26.4pm; E(He⁺)=−19.62×10⁻¹⁸J,r=17.6pm
D
E(Li²⁺)=−19.62×10⁻¹⁶J,r=17.6pm; E(He⁺)=−8.72×10⁻¹⁶J,r=26.4pm
✦ Correct Answer
E(Li²⁺)=−19.62×10⁻¹⁸J,r=17.6pm; E(He⁺)=−8.72×10⁻¹⁸J,r=26.4pm
📐 Solution
1
He⁺(Z=2): E=−8.72×10⁻¹⁸J, r=26.4pm
Li²⁺(Z=3): E=−19.62×10⁻¹⁸J, r=17.6pm
ChemistryQ.
Main group elements: A.[Ne]3s¹, B.[Ar]3d³4s², C.[Kr]4d¹⁰5s²5p⁵, D.[Ar]3d¹⁰4s¹, E.[Rn]5f⁰6d²7s²:
Options
A
A,C,D only
B
B,E only
C
A,C only
✓
D
D,E only
✦ Correct Answer
A,C only
📐 Solution
1
A=Na(s-block)✓, B=V(d-block)✗, C=I(p-block)✓, D=Cu(d-block)✗, E=Th(f-block)✗
∴ A and C only
ChemistryQ.
C(s)+2H₂(g)→CH₄(g), ΔH=−74.8 kJ/mol. Correct energy diagram:
Options
A
R lower than P
B
R higher, P lower (exothermic)
✓
C
P and R same level
D
R lower, hump only
✦ Correct Answer
R higher, P lower (exothermic)
📐 Solution
1
ΔH<0 → exothermic → Products lower than Reactants
ChemistryQ.
Alkene+HBr(peroxide)→KCN→Na(Hg)/EtOH. Major product P:
Options
A
Nitrile on 2° C
B
Amine on anti-Markovnikov C
✓
C
Ketone on 2° C
D
Nitrile on 2° C
✦ Correct Answer
Amine on anti-Markovnikov C
📐 Solution
1
Anti-Markovnikov HBr→Br on terminal C; KCN→CN; reduction→primary amine (anti-Markovnikov)
ChemistryQ.
Correct orders: A.μ: H₂O>NH₃>CHCl₃. B.LP on Xe: XeF₄>XeO₃>XeF₂. C.Bond length: O-H>C-H>N-O. D.Bond enthalpy: N₂>O₂>H₂.
Options
A
B,C only
B
A,D only
✓
C
B,D only
D
A,C only
✦ Correct Answer
A,D only
📐 Solution
1
A✓(1.85>1.47>1.04D), B✗(XeF₂=3>XeF₄=2>XeO₃=1), C✗(N-O>C-H>O-H), D✓(triple>double>single)
∴ A and D only
ChemistryQ.
Total isomers (structural+stereo) of cyclic ethers C₄H₈O:
Options
A
11
B
6
C
8
D
10
✓
✦ Correct Answer
10
📐 Solution
1
Counting all cyclic ether structures and stereoisomers: 10
ChemistryQ.
A(g)⇌2B(g). kb=2500×kf. KP at 1000K:
Options
A
0.021
B
83.1
C
2.077×10⁵
D
0.033
✓
✦ Correct Answer
0.033
📐 Solution
1
KC=1/2500; Δn=1; KP=KC(RT)=(1/2500)×0.0831×1000=0.033
ChemistryQ.
5mol X + 10mol Y → VP=70 torr. P°X=63, P°Y=78 torr. Solution shows:
Options
A
Volume > sum
B
Positive deviation
C
Negative deviation
✓
D
Ideal solution
✦ Correct Answer
Negative deviation
📐 Solution
1
Pideal=73 torr; Pobs=70<73 → Pobs < Pideal
∴ Negative deviation from Raoult's law
Bio

Biology

Questions 91–180

90 Qs · 360 Marks
BiologyQ.
Unit of productivity of an Ecosystem:
Options
A
(KCal m⁻²)yr⁻¹
✓
B
gm⁻²
C
KCal m⁻²
D
KCal m⁻³
✦ Correct Answer
(KCal m⁻²)yr⁻¹
📐 Solution
1
Productivity = rate of biomass/energy production
Unit: (KCal m⁻²)yr⁻¹ or gm⁻²yr⁻¹
BiologyQ.
First menstruation is called:
Options
A
Ovulation
B
Menopause
C
Menarche
✓
D
Diapause
✦ Correct Answer
Menarche
📐 Solution
1
Menarche = first menstruation at puberty
BiologyQ.
A: All vertebrates are chordates but not all chordates are vertebrates. R: Vertebrata have notochord during embryonic period, replaced by vertebral column.
Options
A
A false, R true
B
Both A and R; R explains A
✓
C
Both A and R; R doesn't explain
D
A true, R false
✦ Correct Answer
Both A and R; R explains A
📐 Solution
1
Both A and R correct; R correctly explains A → vertebrates are subset of chordates. Both true, R explains A
BiologyQ.
Genes R and Y: independent assortment. RRYY×rryy. F₂ phenotypic ratio:
Options
A
9:7
B
1:2:1
C
3:1
D
9:3:3:1
✓
✦ Correct Answer
9:3:3:1
📐 Solution
1
Dihybrid cross → F₂ = 9:3:3:1
BiologyQ.
I: DNA fragments from gel can be used in rDNA. II: Smaller fragments near anode, larger near wells.
Options
A
I incorrect, II correct
B
Both correct
✓
C
Both incorrect
D
I correct, II incorrect
✦ Correct Answer
Both correct
📐 Solution
1
Both correct. Smaller fragments migrate farther (toward anode). Both I and II correct
BiologyQ.
Main function of spindle fibres during mitosis:
Options
A
Regulate cell growth
B
Separate chromosomes
✓
C
Synthesize DNA
D
Repair DNA
✦ Correct Answer
Separate chromosomes
📐 Solution
1
Spindle fibres attach to kinetochores → pull chromatids to poles → separate chromosomes
BiologyQ.
Meiotic + mitotic divisions from MMC to mature female gametophyte:
Options
A
No meiosis, 2 mitosis
B
2 meiosis, 3 mitosis
C
1 meiosis, 2 mitosis
D
1 meiosis, 3 mitosis
✓
✦ Correct Answer
1 meiosis, 3 mitosis
📐 Solution
1
MMC→meiosis(1)→functional megaspore→3 mitoses→8-nucleate embryo sac
∴ 1 meiosis + 3 mitosis
BiologyQ.
NOT correct about antibody structure:
Options
A
Constant regions at C-terminus
B
Two light and two heavy chains
C
H and L chains by disulfide bonds
D
Antigen binding at C-terminal region
✓
✦ Correct Answer
Antigen binding at C-terminal region
📐 Solution
1
Antigen binding site is at N-terminal (variable region), NOT C-terminal
BiologyQ.
True about gametogenesis: A.Female reductive div. earlier than male. B.Gap meiosis I-II shorter in males. C.1st polar body from primary oocyte. D.LH surge causes menstrual bleeding.
Options
A
B and C
B
A and B
✓
C
A and C
D
B and D
✦ Correct Answer
A and B
📐 Solution
1
A✓(female meiosis begins in fetal life), B✓(shorter gap in males), C✗(1st PB from 2° oocyte), D✗(LH→ovulation)
∴ A and B
BiologyQ.
A: Tapetum dense cytoplasm, >1 nucleus. R: Multiple nuclei nourish developing microspore mother cells.
Options
A
A false, R true
B
Both A and R; R explains
C
Both A and R; R doesn't explain
D
A true, R false
✓
✦ Correct Answer
A true, R false
📐 Solution
1
A✓(tapetal cells multinucleate). R✗(they nourish pollen grains, not MMC)
∴ A true, R false
BiologyQ.
Blue-white selection: I: Blue colonies have insert, are recombinant. II: White colonies have insert, are recombinant.
Options
A
I incorrect, II correct
✓
B
Both correct
C
Both incorrect
D
I correct, II incorrect
✦ Correct Answer
I incorrect, II correct
📐 Solution
1
Insert→insertional inactivation of β-gal→white = recombinant; blue = non-recombinant
∴ I incorrect, II correct
BiologyQ.
Gemmae in bryophytes help in:
Options
A
Gaseous exchange
B
Sexual reproduction
C
Asexual reproduction
✓
D
Nutrient absorption
✦ Correct Answer
Asexual reproduction
📐 Solution
1
Gemmae = asexual buds in gemma cups → asexual reproduction
BiologyQ.
Match: A.Adenosine, B.Adenylic acid, C.Adenine, D.Alanine. I.N-base, II.Nucleotide, III.Nucleoside, IV.Amino acid:
Options
A
A-II,B-III,C-I,D-IV
B
A-III,B-IV,C-II,D-I
C
A-III,B-II,C-IV,D-I
D
A-III,B-II,C-I,D-IV
✓
✦ Correct Answer
A-III,B-II,C-I,D-IV
📐 Solution
1
Adenosine→nucleoside(III), Adenylic acid→nucleotide(II), Adenine→N-base(I), Alanine→amino acid(IV)
∴ A-III, B-II, C-I, D-IV
BiologyQ.
Sex-linked recessive. Carrier female × affected male. Probability of carrier child:
Options
A
Zero
B
1/4
✓
C
1/2
D
1/8
✦ Correct Answer
1/4
📐 Solution
1
XᶜX × XᶜY → XᶜXᶜ, XᶜX(carrier), XᶜY, XY → carrier = 1/4
BiologyQ.
True about adrenal medullary hormones: A.Pupillary constriction, B.Hyperglycemic, C.Piloerection, D.Increases heart contraction:
Options
A
D only
B
C and D only
C
B,C,D only
✓
D
A,C,D only
✦ Correct Answer
B,C,D only
📐 Solution
1
A✗(causes dilation). B✓, C✓, D✓
∴ B, C and D only
BiologyQ.
Zygomorphic flower:
Options
A
Chilli
B
Petunia
C
Datura
D
Pea
✓
✦ Correct Answer
Pea
📐 Solution
1
Pea has bilateral (zygomorphic) symmetry; others are actinomorphic
BiologyQ.
Proposed triplet codon for amino acids:
Options
A
Franklin Stahl
B
George Gamow
✓
C
Francis Crick
D
Jacque Monod
✦ Correct Answer
George Gamow
📐 Solution
1
George Gamow proposed triplet codon system
BiologyQ.
I: Unidirectional energy flow sun→producers→consumers. II: Ecosystems exempt from 2nd law.
Options
A
I incorrect, II correct
B
Both correct
C
Both incorrect
D
I correct, II incorrect
✓
✦ Correct Answer
I correct, II incorrect
📐 Solution
1
I✓(energy flow unidirectional). II✗(ecosystems NOT exempt from 2nd law)
∴ I correct, II incorrect
BiologyQ.
Sweet potato (root) and potato (stem) represent:
Options
A
Analogy, divergent
B
Analogy, convergent
✓
C
Homology, divergent
D
Homology, convergent
✦ Correct Answer
Analogy, convergent
📐 Solution
1
Same function, different structure → analogous → convergent evolution
BiologyQ.
All living Cyclostomata are:
Options
A
Ectoparasites
✓
B
Free living
C
Endoparasites
D
Symbiotic
✦ Correct Answer
Ectoparasites
📐 Solution
1
All living cyclostomes (lampreys, hagfishes) are ectoparasites
BiologyQ.
Histones are enriched with:
Options
A
Phenylalanine & Arginine
B
Lysine & Arginine
✓
C
Leucine & Lysine
D
Phenylalanine & Leucine
✦ Correct Answer
Lysine & Arginine
📐 Solution
1
Histones are basic proteins rich in lysine and arginine
BiologyQ.
Verhulst-Pearl Logistic Growth equation:
Options
A
dN/dt=N(r−K/K)
B
dN/dt=r(K−N/K)
C
dN/dt=rN(K−N/K)
✓
D
dN/dt=rN(N−K/N)
✦ Correct Answer
dN/dt=rN(K−N/K)
📐 Solution
1
dN/dt = rN(K−N)/K
BiologyQ.
A: Golgi packages ER materials, delivers to targets. R: Vesicles from ER fuse at cis face, released from trans face.
Options
A
A false, R true
B
Both A and R; R explains
C
Both A and R; R doesn't explain
✓
D
A true, R false
✦ Correct Answer
Both A and R; R doesn't explain
📐 Solution
1
Both correct. But mechanism (R) doesn't explain the primary function (A) → Both true, R doesn't explain A
BiologyQ.
True statement about RuBisCO:
Options
A
Catalyzes carboxylation of RuBP
✓
B
Active only in dark
C
Higher affinity for O₂
D
Involved in photolysis
✦ Correct Answer
Catalyzes carboxylation of RuBP
📐 Solution
1
RuBisCO catalyzes carboxylation of RuBP in Calvin cycle
BiologyQ.
Match: A.Progesterone, B.Relaxin, C.MSH, D.Catecholamines. I.Pars intermedia, II.Ovary, III.Adrenal medulla, IV.Corpus luteum:
Options
A
A-III,B-II,C-IV,D-I
B
A-IV,B-II,C-I,D-III
✓
C
A-IV,B-II,C-III,D-I
D
A-II,B-IV,C-I,D-III
✦ Correct Answer
A-IV,B-II,C-I,D-III
📐 Solution
1
Progesterone→Corpus luteum(IV), Relaxin→Ovary(II), MSH→Pars intermedia(I), Catecholamines→Adrenal medulla(III)
∴ A-IV, B-II, C-I, D-III
BiologyQ.
Protein portion of an enzyme is called:
Options
A
Prosthetic group
B
Cofactor
C
Coenzyme
D
Apoenzyme
✓
✦ Correct Answer
Apoenzyme
📐 Solution
1
Apoenzyme + cofactor = holoenzyme. Apoenzyme = protein part
BiologyQ.
NOT essential for gene cloning: A.Restriction enzymes, B.DNA ligase, C.DNA mutase, D.DNA recombinase, E.DNA polymerase:
Options
A
B and C
B
C and D
✓
C
A and B
D
D and E
✦ Correct Answer
C and D
📐 Solution
1
Standard cloning needs A, B, E. C (DNA mutase) and D (DNA recombinase) are not essential
BiologyQ.
Immunity present at birth, non-specific:
Options
A
Humoral
B
Acquired
C
Innate
✓
D
Cell-mediated
✦ Correct Answer
Innate
📐 Solution
1
Innate immunity = present at birth, non-specific
BiologyQ.
Factor important for transcription termination:
Options
A
γ (gamma)
B
α (alpha)
C
σ (sigma)
D
ρ (rho)
✓
✦ Correct Answer
ρ (rho)
📐 Solution
1
σ factor → initiation; ρ (rho) factor → termination in prokaryotes
BiologyQ.
Pituitary hormone actually synthesized in hypothalamus:
Options
A
ACTH
B
LH
C
ADH
✓
D
FSH
✦ Correct Answer
ADH
📐 Solution
1
ADH (vasopressin) and oxytocin synthesized in hypothalamus, stored in posterior pituitary
BiologyQ.
Microbes NOT involved in household products: A.Aspergillus niger, B.Lactobacillus, C.Trichoderma polysporum, D.Saccharomyces cerevisiae, E.Propionibacterium:
Options
A
C and E
B
A and B
C
A and C
✓
D
C and D
✦ Correct Answer
A and C
📐 Solution
1
A→industrial citric acid; C→cyclosporin A (industrial)
∴ A and C not household
BiologyQ.
I: Fig non-vegetarian (contains wasps). II: Fig-wasp mutualism as wasp lives in fig, fig gets pollinated.
Options
A
I incorrect, II correct
B
Both correct
C
Both incorrect
✓
D
I correct, II incorrect
✦ Correct Answer
Both incorrect
📐 Solution
1
I✗(fig is vegetarian). II✗(flower, not fruit, gets pollinated)
∴ Both incorrect
BiologyQ.
Water vascular system in Echinoderms: A.Respiration & Locomotion, B.Excretion & Locomotion, C.Food capture & transport, D.Digestion & Respiration:
Options
A
B,D,E only
B
A and B
C
A and C only
✓
D
B and C
✦ Correct Answer
A and C only
📐 Solution
1
Water vascular system: locomotion, food capture/transport, respiration
∴ A and C only
BiologyQ.
Secondary lymphoid organs: A.Thymus, B.Bone marrow, C.Spleen, D.Lymph nodes, E.Peyer's patches:
Options
A
C,D,E only
✓
B
B,C,D only
C
A,B,C only
D
E,A,B only
✦ Correct Answer
C,D,E only
📐 Solution
1
Primary: Thymus(A) & Bone marrow(B). Secondary: Spleen(C), Lymph nodes(D), Peyer's patches(E)
BiologyQ.
Match: A.Evil Quartet, B.Ex situ, C.Lantana camara, D.Dodo. I.Cryopreservation, II.Alien species, III.Biodiversity loss causes, IV.Extinction:
Options
A
A-III,B-II,C-IV,D-I
B
A-III,B-II,C-I,D-IV
C
A-III,B-I,C-II,D-IV
✓
D
A-III,B-IV,C-II,D-I
✦ Correct Answer
A-III,B-I,C-II,D-IV
📐 Solution
1
Evil Quartet→III, Ex situ→I, Lantana→II(invasive), Dodo→IV
∴ A-III, B-I, C-II, D-IV
BiologyQ.
Correct about plant growth: A.Parthenocarpy by auxins, B.PGRs promote/inhibit, C.Dedifferentiation before redifferentiation, D.ABA=promoter, E.Apical dominance promotes lateral buds:
Options
A
B,D,E only
B
A,B,C only
✓
C
A,C,E only
D
A,D,E only
✦ Correct Answer
A,B,C only
📐 Solution
1
A✓, B✓, C✓, D✗(ABA=inhibitor), E✗(suppresses lateral)
∴ A, B, C only
BiologyQ.
Match: A.Pteridophyte, B.Bryophyte, C.Angiosperm, D.Gymnosperm. I.Salvia, II.Ginkgo, III.Polytrichum, IV.Salvinia:
Options
A
A-IV,B-III,C-II,D-I
B
A-III,B-IV,C-II,D-I
C
A-IV,B-III,C-I,D-II
✓
D
A-III,B-IV,C-I,D-II
✦ Correct Answer
A-IV,B-III,C-I,D-II
📐 Solution
1
Pteridophyte→Salvinia(IV), Bryophyte→Polytrichum(III), Angiosperm→Salvia(I), Gymnosperm→Ginkgo(II)
∴ A-IV, B-III, C-I, D-II
BiologyQ.
Why can't insulin be given orally?
Options
A
Increases bioavailability
B
Immune response
C
Digested in GI tract
✓
D
Structural variation
✦ Correct Answer
Digested in GI tract
📐 Solution
1
Insulin = protein → digested by proteases in GI tract
BiologyQ.
Characteristic feature of gymnosperms:
Options
A
Have flowers
B
Seeds enclosed in fruits
C
Seeds are naked
✓
D
Seeds absent
✦ Correct Answer
Seeds are naked
📐 Solution
1
Gymnos = naked → seeds are naked, not enclosed in fruits
BiologyQ.
Frogs respire in water by skin and buccal cavity; on land by all three:
Options
A
False for both
B
True for water, false for land
C
True for both
D
False for water, true for land
✓
✦ Correct Answer
False for water, true for land
📐 Solution
1
In water: only skin (cutaneous). On land: skin + buccal cavity + lungs
∴ False for water, true for land
BiologyQ.
Silencing of specific mRNA via RNAi is due to:
Options
A
Non-complementary ssRNA
B
Complementary dsRNA
✓
C
Inhibitory ssRNA
D
Complementary tRNA
✦ Correct Answer
Complementary dsRNA
📐 Solution
1
Complementary dsRNA → RISC complex → mRNA degradation
BiologyQ.
Twins are boy and girl. Which must be true?
Options
A
75% identical genetic content
B
Monozygotic twins
C
Fraternal twins
✓
D
Conceived via IVF
✦ Correct Answer
Fraternal twins
📐 Solution
1
Different sexes → two eggs fertilized → fraternal (dizygotic) twins
BiologyQ.
Match: A.Scutellum, B.Non-albuminous seed, C.Epiblast, D.Perisperm. I.Persistent nucellus, II.Monocot cotyledon, III.Groundnut, IV.Rudimentary cotyledon:
Options
A
A-II,B-IV,C-III,D-I
B
A-II,B-III,C-IV,D-I
✓
C
A-IV,B-III,C-II,D-I
D
A-IV,B-III,C-I,D-II
✦ Correct Answer
A-II,B-III,C-IV,D-I
📐 Solution
1
Scutellum→monocot cotyledon(II), Groundnut→non-albuminous(III), Epiblast→rudimentary cotyledon(IV), Perisperm→persistent nucellus(I)
∴ A-II, B-III, C-IV, D-I
BiologyQ.
Renal portal system in frog links:
Options
A
Kidney and lower body
✓
B
Liver and intestine
C
Liver and kidney
D
Kidney and intestine
✦ Correct Answer
Kidney and lower body
📐 Solution
1
Renal portal system: brings blood from hind limbs to kidney
BiologyQ.
Match: A.Heart, B.Kidney, C.GI tract, D.Adrenal cortex. I.Erythropoietin, II.Aldosterone, III.ANF, IV.Secretin:
Options
A
A-III,B-I,C-IV,D-II
✓
B
A-II,B-I,C-III,D-IV
C
A-IV,B-III,C-II,D-I
D
A-I,B-III,C-IV,D-II
✦ Correct Answer
A-III,B-I,C-IV,D-II
📐 Solution
1
Heart→ANF(III), Kidney→Erythropoietin(I), GI→Secretin(IV), Adrenal cortex→Aldosterone(II)
∴ A-III, B-I, C-IV, D-II
BiologyQ.
Cardiac activities regulated by: A.Nodal tissue, B.Medulla oblongata, C.Adrenal medullary hormones, D.Adrenal cortical hormones:
Options
A
A,B,D only
B
A,B,C only
✓
C
A,B,C,D
D
A,C,D only
✦ Correct Answer
A,B,C only
📐 Solution
1
A✓, B✓, C✓(adrenaline), D✗(cortical don't directly regulate heart)
∴ A, B and C only
BiologyQ.
Streptokinase used for:
Options
A
Removing blood clots
✓
B
Curd production
C
Ethanol production
D
Liver treatment
✦ Correct Answer
Removing blood clots
📐 Solution
1
Streptokinase = clot buster for myocardial infarction
BiologyQ.
Father of Ecology in India:
Options
A
Birbal Sahni
B
S.R. Kashyap
C
Ramdeo Misra
✓
D
Ram Udar
✦ Correct Answer
Ramdeo Misra
📐 Solution
1
Ramdeo Misra is the father of Ecology in India
BiologyQ.
A: Embryo sac at maturity is 8-nucleate, 7-celled. R: Egg apparatus has 2 polar nuclei.
Options
A
A false, R true
B
Both; R explains
C
Both; R doesn't explain
D
A true, R false
✓
✦ Correct Answer
A true, R false
📐 Solution
1
A✓(correct). R✗(polar nuclei in central cell, NOT egg apparatus)
∴ A true, R false
BiologyQ.
Neoplastic features: A.Mass of proliferating cells, B.Rapid growth, C.Invasion/damage, D.Confined to original location:
Options
A
B,C,D only
B
A,B only
C
A,B,C only
✓
D
A,B,D only
✦ Correct Answer
A,B,C only
📐 Solution
1
D = benign tumor feature. Neoplastic: A✓, B✓, C✓
∴ A, B, C only
BiologyQ.
Correct pteridophyte lifecycle sequence: A.Prothallus, B.Meiosis in spore mother cells, C.Fertilization, D.Archegonia/antheridia, E.Antherozoids to archegonia:
Options
A
E,D,C,B,A
B
B,A,D,E,C
✓
C
B,A,E,C,D
D
D,E,C,A,B
✦ Correct Answer
B,A,D,E,C
📐 Solution
1
B→A→D→E→C
∴ B, A, D, E, C
BiologyQ.
A: Wind/water pollinated flowers not colourful, no nectar. R: These flowers produce enormous pollen.
Options
A
A false, R true
B
Both; R explains
C
Both; R doesn't explain
✓
D
A true, R false
✦ Correct Answer
Both; R doesn't explain
📐 Solution
1
Both correct. But enormous pollen production is NOT why they lack colour — they lack colour because no pollinators needed
∴ Both correct, R doesn't explain A
BiologyQ.
Enzyme with 'Haem' as prosthetic group:
Options
A
Catalase
✓
B
RuBisCO
C
Carbonic anhydrase
D
Succinate dehydrogenase
✦ Correct Answer
Catalase
📐 Solution
1
Catalase contains haem prosthetic group
BiologyQ.
Match: A.Emphysema, B.Angina Pectoris, C.Glomerulonephritis, D.Tetany. I.Rapid muscle spasms, II.Damaged alveolar walls, III.Acute chest pain, IV.Kidney glomeruli inflammation:
Options
A
A-II,B-III,C-IV,D-I
✓
B
A-III,B-I,C-IV,D-II
C
A-III,B-I,C-II,D-IV
D
A-II,B-IV,C-III,D-I
✦ Correct Answer
A-II,B-III,C-IV,D-I
📐 Solution
1
Emphysema→II, Angina→III, Glomerulonephritis→IV, Tetany→I
∴ A-II, B-III, C-IV, D-I
BiologyQ.
NOT correct about monocot stem:
Options
A
Phloem parenchyma absent
B
Hypodermis parenchymatous
✓
C
Vascular bundles scattered
D
Vascular bundles conjoint and closed
✦ Correct Answer
Hypodermis parenchymatous
📐 Solution
1
Monocot hypodermis is sclerenchymatous, not parenchymatous
BiologyQ.
Male frog copulatory pad location:
Options
A
First digit of fore limb
✓
B
First and second digit of fore limb
C
First digit of hind limb
D
Second digit of fore limb
✦ Correct Answer
First digit of fore limb
📐 Solution
1
Copulatory (nuptial) pad on first digit of forelimb in male frogs
BiologyQ.
I: Primary energy source = solar. II: Rate of photosynthesis = Net Primary Productivity.
Options
A
I incorrect, II correct
B
Both correct
C
Both incorrect
D
I correct, II incorrect
✓
✦ Correct Answer
I correct, II incorrect
📐 Solution
1
I✓. II✗ — rate of photosynthesis = Gross Primary Productivity (GPP); NPP = GPP − R
∴ I correct, II incorrect
BiologyQ.
Foam breaker in bioreactor diagram:
Options
A
C
✓
B
A
C
B
D
D
✦ Correct Answer
C
📐 Solution
1
In standard bioreactor: C = foam breaker
BiologyQ.
PCR amplification after n cycles:
Options
A
2N²
B
N²
C
2ⁿ
✓
D
2n+1
✦ Correct Answer
2ⁿ
📐 Solution
1
After n cycles: 2ⁿ copies produced
BiologyQ.
Match sperm parts: A.Head, B.Middle piece, C.Acrosome, D.Tail. I.Enzymes, II.Motility, III.Energy, IV.Genetic material:
Options
A
A-III,B-II,C-I,D-IV
B
A-IV,B-III,C-I,D-II
✓
C
A-IV,B-III,C-II,D-I
D
A-III,B-IV,C-II,D-I
✦ Correct Answer
A-IV,B-III,C-I,D-II
📐 Solution
1
Head→IV(nucleus), Middle→III(mitochondria/energy), Acrosome→I(enzymes), Tail→II(motility)
∴ A-IV, B-III, C-I, D-II
BiologyQ.
I: ⊕=zygomorphic, G̲=inferior ovary. II: ⊕=actinomorphic, G=superior ovary.
Options
A
I incorrect, II correct
✓
B
Both correct
C
Both incorrect
D
I correct, II incorrect
✦ Correct Answer
I incorrect, II correct
📐 Solution
1
⊕=actinomorphic(not zygomorphic) → I incorrect. II correct
∴ I incorrect, II correct
BiologyQ.
True about ribosomes: A.Euk=80S, Prok=70S. B.Two subunits each. C.80S=60S+40S; 70S=50S+30S. D.80S=60S+20S. E.80S=60S+30S:
Options
A
B,D,E true
B
A,B,C true
✓
C
A,B,D true
D
A,B,E true
✦ Correct Answer
A,B,C true
📐 Solution
1
A✓, B✓, C✓, D✗, E✗
∴ A, B, C true
BiologyQ.
Increasing complexity (Whittaker): A.Fungi, B.Animalia, C.Monera, D.Plantae, E.Protista:
Options
A
C,E,A,B,D
B
A,C,E,B,D
C
C,E,A,D,B
✓
D
A,C,E,D,B
✦ Correct Answer
C,E,A,D,B
📐 Solution
1
Monera→Protista→Fungi→Plantae→Animalia = C, E, A, D, B
BiologyQ.
Correct bryophyte lifecycle sequence: A.Fusion(fertilization), B.Gametophyte attaches, C.Meiosis→spores, D.Sporophyte forms, E.Antherozoids released:
Options
A
D,E,A,B,C
B
D,E,A,C,B
C
B,E,A,C,D
D
B,E,A,D,C
✓
✦ Correct Answer
B,E,A,D,C
📐 Solution
1
B→E→A→D→C = B, E, A, D, C
BiologyQ.
Correct about cancer: A.CT/MRI detect internal. B.Chemo kills non-cancerous. C.α-interferons activate immune. D.Chemo drugs=biological response modifiers. E.Leukaemia=decreased blood cells:
Options
A
A and C only
✓
B
B and D only
C
D and E only
D
C and D only
✦ Correct Answer
A and C only
📐 Solution
1
A✓, C✓, B✗(kills cancerous), D✗, E✗(leukaemia=increased)
∴ A and C only
BiologyQ.
Enzyme class for: S−G + S# → S + S#−G:
Options
A
Ligase
B
Hydrolase
C
Lyase
D
Transferase
✓
✦ Correct Answer
Transferase
📐 Solution
1
Group transfer from one substrate to another = Transferase
BiologyQ.
Correct about human pregnancy: A.Major systems by 12 weeks. B.Formed by 8 weeks. C.Heart after 1 month. D.Limbs/digits by 2nd month. E.Hair in 5th month:
Options
A
A,C,D,E only
✓
B
A and E only
C
B and C only
D
B,C,D,E only
✦ Correct Answer
A,C,D,E only
📐 Solution
1
A✓, B✗, C✓, D✓, E✓
∴ A, C, D and E only
BiologyQ.
Non-distilled alcoholic beverage by yeast:
Options
A
Rum
B
Whisky
C
Brandy
D
Beer
✓
✦ Correct Answer
Beer
📐 Solution
1
Beer and wine = fermentation only (no distillation)
BiologyQ.
I: RNA was first genetic material, acts as catalyst, unstable. II: DNA evolved from RNA, more stable, dsDNA, has repair.
Options
A
I incorrect, II correct
B
Both correct
✓
C
Both incorrect
D
I correct, II incorrect
✦ Correct Answer
Both correct
📐 Solution
1
Both statements are correct (RNA world hypothesis)
BiologyQ.
I: tRNA and rRNA don't interact with mRNA. II: RNAi in all eukaryotes as cellular defense.
Options
A
I incorrect, II correct
✓
B
Both correct
C
Both incorrect
D
I correct, II incorrect
✦ Correct Answer
I incorrect, II correct
📐 Solution
1
I✗(both interact during translation), II✓
∴ I incorrect, II correct
BiologyQ.
Correct diagram for PCT and DCT secretion/reabsorption:
Options
A
Option 1
B
Option 2
C
Option 3
✓
D
Option 4
✦ Correct Answer
Option 3
📐 Solution
1
PCT: reabsorbs NaCl, HCO₃⁻, H₂O; secretes H⁺, NH₃, K⁺
DCT: reabsorbs HCO₃⁻; secretes H⁺, K⁺, NH₃ → Option 3
BiologyQ.
Pattern of inheritance for polygenic trait:
Options
A
X-linked recessive
B
Mendelian
C
Non-Mendelian
✓
D
Autosomal dominant
✦ Correct Answer
Non-Mendelian
📐 Solution
1
Multiple genes → quantitative variation → Non-Mendelian inheritance
BiologyQ.
Protein-rich layer in cereal seeds separating endosperm from embryo:
Options
A
Aleurone layer
✓
B
Coleoptile
C
Coleorhiza
D
Integument
✦ Correct Answer
Aleurone layer
📐 Solution
1
Aleurone layer = peripheral protein-rich layer of endosperm
BiologyQ.
Match leaf pigments: A.Chl a, B.Chl b, C.Xanthophylls, D.Carotenoids. I.Yellow-green, II.Yellow, III.Blue-green, IV.Yellow to yellow-orange:
Options
A
A-I,B-IV,C-III,D-II
B
A-III,B-IV,C-II,D-I
C
A-III,B-I,C-II,D-IV
✓
D
A-I,B-II,C-IV,D-III
✦ Correct Answer
A-III,B-I,C-II,D-IV
📐 Solution
1
Chl a→blue-green(III), Chl b→yellow-green(I), Xanthophylls→yellow(II), Carotenoids→yellow-orange(IV)
∴ A-III, B-I, C-II, D-IV
BiologyQ.
Organism used by Eli Lilly to produce human insulin:
Options
A
Phage
B
Bacterium (E. coli)
✓
C
Yeast
D
Virus
✦ Correct Answer
Bacterium (E. coli)
📐 Solution
1
Eli Lilly introduced insulin genes into E. coli plasmid → first recombinant insulin
BiologyQ.
Post-transcriptional events in eukaryotes: A.Transport before splicing, B.Intron removal/exon joining, C.Methyl group at 5' end, D.Adenine residues at 3' end, E.Base pairing of complementary RNAs:
Options
A
C,D,E only
B
A,B,C only
C
B,C,D only
✓
D
B,C,E only
✦ Correct Answer
B,C,D only
📐 Solution
1
B(splicing)✓, C(5' capping)✓, D(3' polyadenylation)✓, A✗, E✗
∴ B, C, D only
BiologyQ.
Match: A.Centromere, B.Cilium, C.Cristae, D.Cell membrane. I.Mitochondrion, II.Cell division, III.Cell movement, IV.Phospholipid bilayer:
Options
A
A-II,B-III,C-I,D-IV
✓
B
A-I,B-II,C-III,D-IV
C
A-II,B-I,C-IV,D-III
D
A-IV,B-II,C-III,D-I
✦ Correct Answer
A-II,B-III,C-I,D-IV
📐 Solution
1
Centromere→II, Cilium→III, Cristae→I, Cell membrane→IV
∴ A-II, B-III, C-I, D-IV
BiologyQ.
Match: A.Hershey & Chase, B.Euchromatin, C.Griffith, D.Heterochromatin. I.Streptococcus, II.Dark-stained, III.Light-stained, IV.DNA as genetic material:
Options
A
A-III,B-II,C-IV,D-I
B
A-II,B-IV,C-I,D-III
C
A-IV,B-II,C-I,D-III
D
A-IV,B-III,C-I,D-II
✓
✦ Correct Answer
A-IV,B-III,C-I,D-II
📐 Solution
1
Hershey&Chase→DNA(IV), Euchromatin→light-stained(III), Griffith→Streptococcus(I), Heterochromatin→dark(II)
∴ A-IV, B-III, C-I, D-II
BiologyQ.
Human chromosome with most genes:
Options
A
Chromosome 10
B
Chromosome X
C
Chromosome Y
D
Chromosome 1
✓
✦ Correct Answer
Chromosome 1
📐 Solution
1
Chromosome 1 = largest, ~2968 genes
BiologyQ.
Drawbacks of IVF: A.High fatality to mother, B.Expensive, C.Husband/wife must be donors, D.Less adoption, E.Not in India, F.Embryo may not survive:
Options
A
A,B,C,E,F only
B
B,D,F only
✓
C
A,C,D,F only
D
A,B,C,D only
✦ Correct Answer
B,D,F only
📐 Solution
1
B✓, D✓, F✓; A✗, C✗, E✗
∴ B, D, F only
BiologyQ.
Example of ex-situ conservation:
Options
A
Protected areas
B
National Park
C
Wildlife Sanctuary
D
Zoos and botanical gardens
✓
✦ Correct Answer
Zoos and botanical gardens
📐 Solution
1
Ex-situ = outside natural habitat: Zoos, botanical gardens, seed banks
BiologyQ.
Membranous structure in prokaryote for cell wall, DNA replication, respiration:
Options
A
ER
B
Mesosome
✓
C
Chromatophores
D
Cristae
✦ Correct Answer
Mesosome
📐 Solution
1
Mesosome = infolding of plasma membrane in bacteria
BiologyQ.
Alien DNA inserted at EcoRI site (within β-gal gene). Select recombinants:
Options
A
Blue on ampicillin
B
Ampicillin + tetracycline
C
Blue colonies
D
White colonies
✓
✦ Correct Answer
White colonies
📐 Solution
1
Insertion→insertional inactivation→no β-gal→white colonies = recombinant
BiologyQ.
Blood vessel carrying deoxygenated blood from body to heart in frog:
Options
A
Vena cava
✓
B
Aorta
C
Pulmonary artery
D
Pulmonary vein
✦ Correct Answer
Vena cava
📐 Solution
1
Vena cava → right auricle (deoxygenated blood from body)
BiologyQ.
Cannot fix nitrogen: A.Azotobacter, B.Oscillatoria, C.Anabaena, D.Volvox, E.Nostoc:
Options
A
E only
B
A only
C
D only
✓
D
B only
✦ Correct Answer
D only
📐 Solution
1
Azotobacter, Oscillatoria, Anabaena, Nostoc fix N₂. Volvox cannot
∴ D only
BiologyQ.
Body cavity with mesoderm toward body wall but NOT toward gut:
Options
A
Spongocoelomate
B
Acoelomate
C
Pseudocoelomate
✓
D
Schizocoelomate
✦ Correct Answer
Pseudocoelomate
📐 Solution
1
Pseudocoelom: mesoderm only on body wall side (e.g., Ascaris/Nematoda)
BiologyQ.
Reductionist biology refers to:
Options
A
Behavioural approach
B
Physico-chemical approach
✓
C
Physiological approach
D
Chemical approach
✦ Correct Answer
Physico-chemical approach
📐 Solution
1
Reductionist biology = studying life using physics and chemistry principles
BiologyQ.
Epiphytes on mango branch is example of:
Options
A
Amensalism
B
Commensalism
✓
C
Mutualism
D
Predation
✦ Correct Answer
Commensalism
📐 Solution
1
Epiphyte benefits, mango unaffected → Commensalism
BiologyQ.
Phytohormone promoting nutrient mobilization, delaying leaf senescence:
Options
A
Cytokinin
✓
B
Ethylene
C
Abscisic acid
D
Gibberellin
✦ Correct Answer
Cytokinin
📐 Solution
1
Cytokinin promotes nutrient mobilization and delays senescence
BiologyQ.
Complex II of mitochondrial ETC also known as:
Options
A
NADH dehydrogenase
B
Cytochrome bc₁
C
Succinate dehydrogenase
✓
D
Cytochrome c oxidase
✦ Correct Answer
Succinate dehydrogenase
📐 Solution
1
ETC: I=NADH dehydrogenase, II=Succinate dehydrogenase, III=Cyt bc₁, IV=Cyt c oxidase
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