NEET 2025 Question PaperPhysics, Chemistry and Biology, with checked answers and solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
154 questions from the NEET 2025 paper, conducted by NTA. Tap an option to check your answer and read why it is right. Each subject has a free PDF with the answer key and explanations.
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- NEET 2025
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- 31 questions
- Chemistry
- 41 questions
- Biology
- 82 questions
NEET 2025 Physics questions
31 questions. Tap an option to check your answer.
Q1NEET 2025
A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to displacement current is:
Not quite — the answer is A.
A steady rate of change of σ gives a uniform displacement current density between the plates. By the Ampere–Maxwell law, B = μ₀ε₀(dE/dt)r/2 inside (r ≤ R), growing linearly with r, and B ∝ 1/r outside, so the field is largest at r = R, the cylindrical surface joining the plate edges. It is neither zero between the plates nor constant there.
Q2NEET 2025
An electric dipole with dipole moment 5 × 10⁻⁶ C m is aligned with the direction of a uniform electric field of magnitude 4 × 10⁵ N/C. The dipole is then rotated through an angle of 60° with respect to the electric field. The change in the potential energy of the dipole is:
Not quite — the answer is D.
U = −pE cos θ, so ΔU = pE(cos 0° − cos 60°) = 5 × 10⁻⁶ × 4 × 10⁵ × (1 − 0.5) = 2 × 0.5 = 1.0 J. Using sin 60° or forgetting the (1 − cos θ) factor gives the other values.
Q3NEET 2025
A ball of mass 0.5 kg is dropped from a height of 40 m. The ball hits the ground and rises to a height of 10 m. The impulse imparted to the ball during its collision with the ground is (Take g = 9.8 m/s²)
Not quite — the answer is C.
Speed on impact = √(2 × 9.8 × 40) = 28 m/s downward; speed on rebound = √(2 × 9.8 × 10) = 14 m/s upward. Impulse = change in momentum = 0.5 × (14 + 28) = 21 N s upward. Subtracting the speeds (ignoring the reversal of direction) gives 7 N s.
Q4NEET 2025
The intensity of transmitted light when a polaroid sheet, placed between two crossed polaroids at 22.5° from the polarization axis of one of the polaroids, is (I₀ is the intensity of polarised light after passing through the first polaroid):
Not quite — the answer is A.
By Malus's law, the middle sheet passes I₀cos²22.5° and the last polaroid, at 90° − 22.5° = 67.5° from the middle sheet, passes a further cos²67.5° = sin²22.5°. I = I₀(sin 22.5° cos 22.5°)² = I₀(sin 45°/2)² = I₀/8.
Q5NEET 2025
The kinetic energies of two similar cars A and B are 100 J and 225 J respectively. On applying brakes, car A stops after 1000 m and car B stops after 1500 m. If F_A and F_B are the forces applied by the brakes on cars A and B respectively, then the ratio of F_A/F_B is
Not quite — the answer is D.
By the work–energy theorem F × d = KE. F_A = 100/1000 = 0.1 N and F_B = 225/1500 = 0.15 N, so F_A/F_B = 0.1/0.15 = 2/3. Inverting the ratio gives 3/2.
Q6NEET 2025
A particle of mass m is moving around the origin with a constant force F pulling it towards the origin. If Bohr model is used to describe its motion, the radius of the nth orbit and the particle's speed v in the orbit depend on n as
Not quite — the answer is A.
Centripetal condition: mv²/r = F, so v ∝ r^(1/2). Bohr quantisation mvr = nh/2π gives r^(3/2) ∝ n, so r ∝ n^(2/3) and v ∝ r^(1/2) ∝ n^(1/3).
Q7NEET 2025
In some appropriate units, time (t) and position (x) relation of a moving particle is given by t = x² + x. The acceleration of the particle is
Not quite — the answer is D.
dt/dx = 2x + 1, so v = 1/(2x + 1). Then a = v dv/dx = [1/(2x + 1)] × [−2/(2x + 1)²] = −2/(2x + 1)³. The particle decelerates, so the positive options are wrong.
Q8NEET 2025
A model for quantized motion of an electron in a uniform magnetic field B states that the flux passing through the orbit of the electron is n(h/e) where n is an integer, h is Planck's constant and e is the magnitude of electron's charge. According to the model, the magnetic moment of an electron in its lowest energy state will be (m is the mass of the electron)
Not quite — the answer is D.
For circular motion r = mv/eB, so v = eBr/m. Magnetic moment μ = (e/2)vr = e²Br²/2m. The quantisation Bπr² = nh/e gives Br² = nh/(πe), so μ = e²/(2m) × nh/(πe) = nhe/(2πm); for n = 1, μ = he/2πm, independent of B.
Q9NEET 2025
A microscope has an objective of focal length 2 cm, eyepiece of focal length 4 cm and the tube length of 40 cm. If the distance of distinct vision of eye is 25 cm, the magnification in the microscope is
Not quite — the answer is D.
For final image at infinity, m = (L/f₀)(D/fₑ) = (40/2) × (25/4) = 20 × 6.25 = 125. Using (1 + D/fₑ) for the near-point setting would give 145, which is not offered.
Q10NEET 2025
There are two inclined surfaces of equal length (L) and same angle of inclination 45° with the horizontal. One of them is rough and the other is perfectly smooth. A given body takes 2 times as much time to slide down on rough surface than on the smooth surface. The coefficient of kinetic friction (μk) between the object and the rough surface is close to
Not quite — the answer is B.
From L = ½at², t ∝ 1/√a, so doubling the time means a(rough) = a(smooth)/4. g(sin 45° − μ cos 45°) = g sin 45°/4 gives 1 − μ = 1/4, so μ = 0.75. Halving instead of quartering the acceleration gives 0.5.
Q11NEET 2025
A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of 60° with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (Take g = 10 m/s²)
Not quite — the answer is D.
Friction balances the horizontal wall reaction N. Torques about the foot: N × L cos 60° = mg × (L/2) sin 60°, so N = (mg/2) tan 60° = (200/2) × √3 = 100√3 N. Using the angle with the floor instead of the wall gives 100/√3 type values.
Q12NEET 2025
Two identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is F. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B and finally removed from both. New force of repulsion between spheres A and B (Radii of A and B are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as:
Not quite — the answer is B.
Touching A shares q equally: A and the third sphere each have q/2. Touching B then shares q + q/2, leaving B with 3q/4. New force ∝ (q/2)(3q/4) = 3q²/8, so F' = 3F/8.
Q13NEET 2025
Two cities X and Y are connected by a regular bus service with a bus leaving in either direction every T min. A girl is driving scooty with a speed of 60 km/h in the direction X to Y notices that a bus goes past her every 30 minutes in the direction of her motion, and every 10 minutes in the opposite direction. Choose the correct option for the period T of the bus service and the speed (assumed constant) of the buses.
Not quite — the answer is B.
Buses in each direction are spaced vT apart. Same direction: vT/(v − 60) = 30; opposite: vT/(v + 60) = 10. Dividing, (v + 60)/(v − 60) = 3, so v = 120 km/h, and then 120T/60 = 30 gives T = 15 min.
Q14NEET 2025
De-Broglie wavelength of an electron orbiting in the n = 2 state of hydrogen atom is close to (Given Bohr radius = 0.052 nm)
Not quite — the answer is D.
Bohr's condition 2πr = nλ gives λ = 2πr/n with r = n²a₀ = 4 × 0.052 = 0.208 nm. λ = 2π × 0.208/2 = π × 0.208 ≈ 0.65 nm, closest to 0.67 nm (0.665 nm with a₀ = 0.0529 nm).
Q15NEET 2025
To an ac power supply of 220 V at 50 Hz, a resistor of 20 Ω, a capacitor of reactance 25 Ω and an inductor of reactance 45 Ω are connected in series. The corresponding current in the circuit and the phase angle between the current and the voltage is, respectively
Not quite — the answer is D.
Net reactance = 45 − 25 = 20 Ω, so Z = √(20² + 20²) = 20√2 ≈ 28.3 Ω and I = 220/28.3 ≈ 7.8 A. tan φ = 20/20 = 1, so φ = 45°. 15.6 A comes from using R alone in place of Z after halving.
Q16NEET 2025
A pipe open at both ends has a fundamental frequency f in air. The pipe is now dipped vertically in a water drum to half of its length. The fundamental frequency of the air column is now equal to:
Not quite — the answer is D.
Open pipe: f = v/2L. Dipping half of it in water leaves a closed pipe of length L/2, whose fundamental is v/(4 × L/2) = v/2L = f. Halving the length without noting the change to a closed pipe gives 2f.
Q17NEET 2025
Two identical point masses P and Q, suspended from two separate massless springs of spring constants k₁ and k₂, respectively, oscillate vertically. If their maximum speeds are the same, the ratio (A_Q/A_P) of the amplitude A_Q of mass Q to the amplitude A_P of mass P is
Not quite — the answer is B.
v_max = Aω with ω = √(k/m). Equal v_max gives A_Q√k₂ = A_P√k₁, so A_Q/A_P = √(k₁/k₂). The inverse ratio comes from matching amplitudes to ω the wrong way.
Q18NEET 2025
Two gases A and B are filled at the same pressure in separate cylinders with movable pistons of radius r_A and r_B, respectively. On supplying an equal amount of heat to both the systems reversibly under constant pressure, the pistons of gas A and B are displaced by 16 cm and 9 cm, respectively. If the change in their internal energy is the same, then the ratio r_A/r_B is equal to
Not quite — the answer is D.
Equal Q and equal ΔU mean equal work W = pΔV, and the pressures are equal, so ΔV is equal: πr_A² × 16 = πr_B² × 9. Hence r_A/r_B = √(9/16) = 3/4. Inverting gives 4/3.
Q19NEET 2025
A balloon is made of a material of surface tension S and its inflation outlet (from where gas is filled in it) has small area A. It is filled with a gas of density ρ and takes a spherical shape of radius R. When the gas is allowed to flow freely out of it, its radius r changes from R to 0 (zero) in time T. If the speed v(r) of gas coming out of the balloon depends on r as r^a and T ∝ S^α A^β ρ^γ R^δ then
Not quite — the answer is A.
Excess pressure ∝ S/r drives the flow, so by Bernoulli ½ρv² ∝ S/r and v ∝ (S/ρ)^(1/2) r^(−1/2), i.e. a = −1/2. Volume outflow: 4πr² dr/dt = −Av ∝ −A(S/ρ)^(1/2) r^(−1/2), so r^(5/2) dr ∝ dt and T ∝ R^(7/2) ρ^(1/2) S^(−1/2) A^(−1). Hence α = −1/2, β = −1, γ = 1/2, δ = 7/2.
Q20NEET 2025
A physical quantity P is related to four observations a, b, c and d as follows: P = a³b²/(c√d) The percentage errors of measurement in a, b, c and d are 1%, 3%, 2%, and 4% respectively. The percentage error in the quantity P is
Not quite — the answer is A.
Maximum relative error adds with powers as weights: 3(1%) + 2(3%) + 1(2%) + ½(4%) = 3 + 6 + 2 + 2 = 13%. Subtracting the errors of the quantities in the denominator is the usual mistake.
Q21NEET 2025
The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sphere of uniform density.
Not quite — the answer is B.
With no external torque, angular momentum Iω is conserved and I = (2/5)MR². Doubling R makes I four times larger, so ω falls to a quarter and the period becomes 4 × 27 = 108 days.
Q22NEET 2025
The radius of Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 Earth days. Then which of the following is the length of 1 year on Mercury?
Not quite — the answer is C.
Kepler's third law: T ∝ r^(3/2), so T_Mars/T_Mercury = 4^(3/2) = 8. T_Mercury = 687/8 ≈ 86 days, closest to 88 earth days. Using T ∝ r gives 172 days.
Q23NEET 2025
A wire of resistance R is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net effective resistance of the combination is:
Not quite — the answer is A.
Each piece is R/8. Four in parallel give (R/8)/4 = R/32, and two such groups in series give 2 × R/32 = R/16. R/32 is the value of only one group.
Q24NEET 2025
A photon and an electron (mass m) have the same energy E. The ratio (λ_photon/λ_electron) of their de Broglie wavelengths is: (c is the speed of light)
Not quite — the answer is A.
Photon: λ = hc/E. Electron: p = √(2mE), so λ = h/√(2mE). Ratio = (hc/E) × √(2mE)/h = c√(2m/E). The inverted expression is the trap.
Q25NEET 2025
An electron (mass 9 × 10⁻³¹ kg and charge 1.6 × 10⁻¹⁹ C) moving with speed c/100 (c = speed of light) is injected into a magnetic field B⃗ of magnitude 9 × 10⁻⁴ T perpendicular to its direction of motion. We wish to apply a uniform electric field E⃗ together with the magnetic field so that the electron does not deflect from its path. Then (speed of light c = 3 × 10⁸ m s⁻¹)
Not quite — the answer is D.
For no deflection the electric force must cancel the magnetic force, so E⃗ must be perpendicular to both v⃗ and B⃗ with eE = evB. E = vB = (3 × 10⁸/100) × 9 × 10⁻⁴ = 3 × 10⁶ × 9 × 10⁻⁴ = 2700 = 27 × 10² V m⁻¹. A field parallel to B⃗ could not cancel the magnetic force.
Q26NEET 2025
The electric field in a plane electromagnetic wave is given by E_z = 60 cos(5x + 1.5 × 10⁹ t) V/m. Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field) :
Not quite — the answer is C.
The wave travels along the x-axis (ω/k = 1.5 × 10⁹/5 = 3 × 10⁸ m/s) with E along z, so B must be along y, in phase with E. B₀ = E₀/c = 60/(3 × 10⁸) = 2 × 10⁻⁷ T. B along x would be along the direction of propagation, which is impossible for an EM wave.
Q27NEET 2025
A body weighs 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is:
Not quite — the answer is D.
g' = g R²/(R + R/3)² = g/(4/3)² = 9g/16. Weight = 48 × 9/16 = 27 N. Using the approximation g(1 − 2h/R) gives 16 N, which is invalid for such a large height.
Q28NEET 2025
An unpolarized light beam travelling in air is incident on a medium of refractive index 1.73 at Brewster's angle. Then
Not quite — the answer is C.
Brewster's law tan i_B = μ = 1.73 = √3 gives i_B = 60°, which is also the angle of reflection. At this angle the reflected light is completely plane polarised, while the refracted light (at 30°) is only partially polarised.
Q29NEET 2025
The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K₁ and K₂ with thickness (3/8)d and d/2, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If K₁ = 1.25K₂, the value of K₁ is:
Not quite — the answer is C.
C = ε₀A/[d − t₁ − t₂ + t₁/K₁ + t₂/K₂] = 2ε₀A/d, so the bracket equals d/2. With 1 − 3/8 − 1/2 = 1/8: 1/8 + 3/(8K₁) + 1/(2K₂) = 1/2. Using K₂ = 0.8K₁, 1/(2K₂) = 0.625/K₁, so (0.375 + 0.625)/K₁ = 3/8 and K₁ = 8/3 ≈ 2.66.
Q30NEET 2025
Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its 10 Vernier Scale Divisions (V.S.D.) are equal to its 9 Main Scale Divisions (M.S.D.). The least division in the M.S. is 0.1 cm and the zero of V.S. is at x = 0.1 cm when the jaws of Vernier callipers are closed. If the main scale reading for the diameter is M = 5 cm and the number of coinciding vernier division is 8, the measured diameter after zero error correction, is
Not quite — the answer is A.
Least count = 1 MSD − 1 VSD = 0.1 − 0.09 = 0.01 cm. Observed reading = 5 + 8 × 0.01 = 5.08 cm. The vernier zero lies 0.1 cm to the right of the main-scale zero, a positive zero error of 0.1 cm, so the corrected diameter is 5.08 − 0.10 = 4.98 cm. 5.08 cm is the uncorrected reading.
Q31NEET 2025
A 2 amp current is flowing through two different small circular copper coils having radii ratio 1 : 2. The ratio of their respective magnetic moments will be
Not quite — the answer is C.
Magnetic moment of a loop m = I × πr². With the same current, m₁ : m₂ = r₁² : r₂² = 1 : 4. Taking m ∝ r gives 1 : 2.
Chapter-wise split of NEET 2025 Physics
- Moving Charges and Magnetism3 Q
- Electromagnetic Waves2 Q
- Electrostatic Potential and Capacitance2 Q
- Laws of Motion2 Q
- Wave Optics2 Q
- Atoms2 Q
- Motion in a Straight Line2 Q
- System of Particles and Rotational Motion2 Q
- Units and Measurements2 Q
- Gravitation2 Q
- Work, Energy and Power1 Q
- Ray Optics and Optical Instruments1 Q
- Electric Charges and Fields1 Q
- Alternating Current1 Q
- Waves1 Q
- Oscillations1 Q
- Thermodynamics1 Q
- Mechanical Properties of Fluids1 Q
- Current Electricity1 Q
- Dual Nature of Radiation and Matter1 Q
NEET 2025 Chemistry questions
41 questions. Tap an option to check your answer.
Q1NEET 2025
Identify the suitable reagent for the following conversion. C₆H₅COOCH₃ (methyl benzoate) → C₆H₅CHO (benzaldehyde)
Not quite — the answer is D.
DIBAL-H (AlH(iBu)₂) at low temperature reduces an ester only as far as the aldehyde stage, and hydrolysis then releases benzaldehyde. LiAlH₄ reduces the ester all the way to benzyl alcohol, NaBH₄ hardly reduces esters at all, and H₂/Pd–BaSO₄ (Rosenmund) is used for acid chlorides, not esters.
Q2NEET 2025
The correct order of decreasing acidity of the following aliphatic acids is
Not quite — the answer is A.
Alkyl groups are electron-donating (+I) and destabilise the carboxylate anion, so acidity falls as more methyl groups are attached to the α-carbon. Formic acid, with no alkyl group, is the strongest, followed by acetic, isobutyric and finally pivalic acid; the reversed orders ignore this +I effect.
Q3NEET 2025
Which one of the following reactions does NOT belong to "Lassaigne's test"?
Not quite — the answer is B.
In Lassaigne's test the organic compound is fused with sodium so that N, S and halogens become NaCN, Na₂S and NaX, which are then tested in the sodium fusion extract. Heating with copper(II) oxide to oxidise carbon to CO₂ is the test for carbon (and the basis of Duma's/Liebig methods), not part of the sodium fusion test.
Q4NEET 2025
If the rate constant of a reaction is 0.03 s⁻¹, how much time does it take for 7.2 mol L⁻¹ concentration of the reactant to get reduced to 0.9 mol L⁻¹? (Given: log 2 = 0.301)
Not quite — the answer is C.
The unit s⁻¹ shows a first-order reaction. t = (2.303/k) log(7.2/0.9) = (2.303/0.03) × log 8 = 76.77 × 0.903 ≈ 69.3 s; equivalently 7.2 → 0.9 is three half-lives of 0.693/0.03 = 23.1 s each. 23.1 s is only one half-life.
Q5NEET 2025
Given below are two statements : Statement I: A hypothetical diatomic molecule with bond order zero is quite stable. Statement II: As bond order increases, the bond length increases. In the light of the above statements, choose the most appropriate answer from the options given below:
Not quite — the answer is D.
A bond order of zero means equal numbers of bonding and antibonding electrons, so no net bond forms and the molecule (like He₂) does not exist stably. Higher bond order means stronger, shorter bonds, so bond length decreases with bond order. Both statements are false.
Q6NEET 2025
Which of the following aqueous solution will exhibit highest boiling point?
Not quite — the answer is A.
Elevation of boiling point is proportional to i × concentration: Na₂SO₄ gives 3 × 0.01 = 0.03, KNO₃ 2 × 0.01 = 0.02, glucose 0.015 and urea 0.01. Sodium sulphate has the largest number of solute particles and so the highest boiling point; glucose looks larger only if the van't Hoff factor is ignored.
Q7NEET 2025
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : CH₃CH₂CH₂CH₂CH₂–I undergoes S_N2 reaction faster than CH₃CH₂CH₂CH₂CH₂–Cl Reason (R) : Iodine is a better leaving group because of its large size. In the light of the above statements, choose the correct answer from the options given below:
Not quite — the answer is C.
For the same primary alkyl group, the rate of S_N2 depends on how easily the C–X bond breaks. Iodide is large and highly polarisable, the C–I bond is the weakest, and I⁻ is a weak base, so it leaves most easily; this is exactly why the iodide reacts faster than the chloride. Both statements are true and the reason explains the assertion.
Q8NEET 2025
Consider the following compounds : K̲O₂, H₂O̲₂ and H₂S̲O₄ The oxidation state of the underlined elements in them are, respectively,
Not quite — the answer is C.
Potassium, an alkali metal, is always +1 (KO₂ is potassium superoxide, O₂⁻). In the peroxide H₂O₂, 2(+1) + 2x = 0 gives oxygen −1. In H₂SO₄, 2(+1) + x + 4(−2) = 0 gives sulphur +6. Taking oxygen as −2 in a peroxide is the usual trap.
Q9NEET 2025
Match List - I with List - II. List - I: A. Haber process; B. Wacker oxidation; C. Wilkinson catalyst; D. Ziegler catalyst List - II: I. Fe catalyst; II. PdCl₂; III. [(PPh₃)₃RhCl]; IV. TiCl₄ with Al(CH₃)₃ Choose the correct answer from the options given below :
Not quite — the answer is A.
The Haber synthesis of ammonia uses an iron catalyst; Wacker oxidation of ethene to ethanal uses PdCl₂; Wilkinson's catalyst for hydrogenation is [(PPh₃)₃RhCl]; and the Ziegler(–Natta) catalyst for polymerising alkenes is TiCl₄ with an organoaluminium compound. Each catalyst matches its own process in order.
Q10NEET 2025
Given below are two statements : Statement I: Like nitrogen that can form ammonia, arsenic can form arsine. Statement II: Antimony cannot form antimony pentoxide. In the light of the above statements, choose the most appropriate answer from the options given below :
Not quite — the answer is A.
All group 15 elements form hydrides EH₃, so arsenic forms arsine (AsH₃) just as nitrogen forms NH₃. Group 15 elements also form pentoxides E₂O₅, and Sb₂O₅ is a known oxide, so the second statement is wrong.
Q11NEET 2025
Given below are two statements : Statement I: Ferromagnetism is considered as an extreme form of paramagnetism. Statement II : The number of unpaired electrons in a Cr²⁺ ion (Z = 24) is the same as that of a Nd³⁺ ion (Z = 60). In the light of the above statements, choose the correct answer from the options given below :
Not quite — the answer is A.
Ferromagnetism arises when unpaired spins align co-operatively in domains, so it is treated as an extreme form of paramagnetism. Cr²⁺ is 3d⁴ with 4 unpaired electrons, whereas Nd³⁺ is 4f³ with only 3, so the second statement is false.
Q12NEET 2025
Which one of the following reactions does NOT give benzene as the product?
Not quite — the answer is B.
Warming benzenediazonium chloride with water replaces the –N₂⁺ group by –OH, giving phenol, not benzene. Ethyne trimerises to benzene in a red-hot iron tube, decarboxylation of sodium benzoate with soda lime gives benzene, and n-hexane is aromatised to benzene over a Mo₂O₃-type catalyst at 773 K and 10–20 atm.
Q13NEET 2025
Match List-I with List-II List - I: A. XeO₃; B. XeF₂; C. XeOF₄; D. XeF₆ List - II: I. sp³d; linear; II. sp³; pyramidal; III. sp³d³; distorted octahedral; IV. sp³d²; square pyramidal Choose the correct answer from the options given below :
Not quite — the answer is C.
XeO₃ has 3 bond pairs and 1 lone pair (sp³, pyramidal); XeF₂ has 2 bond pairs and 3 lone pairs (sp³d, linear); XeOF₄ has 5 bond pairs and 1 lone pair (sp³d², square pyramidal); XeF₆ has 6 bond pairs and 1 lone pair (sp³d³, distorted octahedral).
Q14NEET 2025
How many products (including stereoisomers) are expected from monochlorination of the following compound? (CH₃)₂CH–CH₂–CH₃
Not quite — the answer is B.
2-Methylbutane has four kinds of hydrogen. Substitution at a CH₃ of the isopropyl end gives 1-chloro-2-methylbutane (chiral, 2 enantiomers), at the CH gives 2-chloro-2-methylbutane (achiral, 1), at the CH₂ gives 2-chloro-3-methylbutane (chiral, 2) and at the terminal CH₃ gives 1-chloro-3-methylbutane (achiral, 1): 2 + 1 + 2 + 1 = 6. Counting only the 4 constitutional isomers misses the enantiomers.
Q15NEET 2025
Which of the following statements are true? A. Unlike Ga that has a very high melting point, Cs has a very low melting point. B. On Pauling scale, the electronegativity values of N and Cl are not the same. C. Ar, K⁺, Cl⁻, Ca²⁺, and S²⁻ are all isoelectronic species. D. The correct order of the first ionization enthalpies of Na, Mg, Al, and Si is Si > Al > Mg > Na. E. The atomic radius of Cs is greater than that of Li and Rb. Choose the correct answer from the options given below :
Not quite — the answer is D.
Gallium itself melts at only about 303 K, so A is false; N and Cl both have electronegativity 3.0 on the Pauling scale, so B is false. All five species in C have 18 electrons (true). The ionisation enthalpy order is Si > Mg > Al > Na because Mg's filled 3s² is extra stable, so D is false. Cs is larger than both Li and Rb, so E is true.
Q16NEET 2025
The standard heat of formation, in kcal/mol of Ba²⁺ is : [Given : standard heat of formation of SO₄²⁻ ion (aq) = −216 kcal/mol, standard heat of crystallisation of BaSO₄(s) = −4.5 kcal/mol, standard heat of formation of BaSO₄(s) = −349 kcal/mol]
Not quite — the answer is C.
For Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s), ΔH(crystallisation) = ΔfH(BaSO₄) − ΔfH(Ba²⁺) − ΔfH(SO₄²⁻). So −4.5 = −349 − x − (−216), giving x = −349 + 216 + 4.5 = −128.5 kcal/mol. Ignoring the crystallisation enthalpy gives the −133.0 trap.
Q17NEET 2025
Match List-I with List-II List - I (Example): A. Humidity; B. Alloys; C. Amalgams; D. Smoke List - II (Type of Solution): I. Solid in solid; II. Liquid in gas; III. Solid in gas; IV. Liquid in solid Choose the correct answer from the options given below:
Not quite — the answer is D.
Humidity is water (liquid) dispersed in air (gas); alloys such as brass are solid in solid; amalgams are mercury (liquid) dissolved in a solid metal; smoke is solid particles in a gas. Swapping alloys and amalgams is the usual mistake.
Q18NEET 2025
Sugar 'X' A. is found in honey B. is a keto sugar C. exists in α and β-anomeric forms. D. Is laevorotatory. 'X' is :
Not quite — the answer is D.
Fructose is a ketohexose, occurs free in honey and fruits, forms α and β anomeric furanose rings and is strongly laevorotatory. Glucose is an aldose and dextrorotatory, while sucrose and maltose are disaccharides and not keto sugars.
Q19NEET 2025
Total number of possible isomers (both structural as well as stereoisomers) of cyclic ethers of molecular formula C₄H₈O is :
Not quite — the answer is A.
Five-membered ring: tetrahydrofuran, 1 isomer. Four-membered ring: 2-methyloxetane (chiral, 2 forms) and 3-methyloxetane (1 form). Three-membered ring: 2-ethyloxirane (chiral, 2 forms), 2,2-dimethyloxirane (1 form) and 2,3-dimethyloxirane (cis meso form plus a pair of trans enantiomers, 3 forms). Total = 1 + 3 + 6 = 10; counting only constitutional isomers gives 6.
Q20NEET 2025
For the reaction A(g) ⇌ 2B(g), the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K. [Given : R = 0.0831 L atm mol⁻¹ K⁻¹] Kp for the reaction at 1000 K is
Not quite — the answer is A.
Kc = kf/kb = 1/2500 = 4 × 10⁻⁴. With Δn(g) = 2 − 1 = 1, Kp = Kc(RT) = 4 × 10⁻⁴ × 0.0831 × 1000 = 4 × 10⁻⁴ × 83.1 ≈ 0.033. Multiplying by 2500 instead of dividing gives the 2.077 × 10⁵ trap.
Q21NEET 2025
The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes n = 2 → n = 3 and n = 4 → n = 6 transitions, respectively, is
Not quite — the answer is B.
1/λ ∝ (1/n₁² − 1/n₂²). For 2 → 3: 1/4 − 1/9 = 5/36; for 4 → 6: 1/16 − 1/36 = 5/144. So λ₁/λ₂ = (5/144)/(5/36) = 1/4. Taking the ratio of the wave numbers instead gives 4, not among the options.
Q22NEET 2025
If the molar conductivity (Λm) of a 0.050 mol L⁻¹ solution of a monobasic weak acid is 90 S cm² mol⁻¹, its extent (degree) of dissociation will be [Assume Λ°₊ = 349.6 S cm² mol⁻¹ and Λ°₋ = 50.4 S cm² mol⁻¹.]
Not quite — the answer is A.
By Kohlrausch's law Λ°m = 349.6 + 50.4 = 400 S cm² mol⁻¹. Degree of dissociation α = Λm/Λ°m = 90/400 = 0.225. The concentration is not needed for α itself.
Q23NEET 2025
5 moles of liquid X and 10 moles of liquid Y make a solution having a vapour pressure of 70 torr. The vapour pressures of pure X and Y are 63 torr and 78 torr respectively. Which of the following is true regarding the described solution?
Not quite — the answer is D.
x(X) = 5/15 = 1/3 and x(Y) = 2/3, so Raoult's law predicts p = 63/3 + 78 × 2/3 = 21 + 52 = 73 torr. The observed 70 torr is lower, which is negative deviation (stronger X–Y attractions, volume contraction), so a larger volume or positive deviation is wrong.
Q24NEET 2025
Among the following, choose the ones with equal number of atoms. A. 212 g of Na₂CO₃(s) [molar mass = 106 g] B. 248 g of Na₂O(s) [molar mass = 62 g] C. 240 g of NaOH(s) [molar mass = 40 g] D. 12 g of H₂(g) [molar mass = 2 g] E. 220 g of CO₂(g) [molar mass = 44 g] Choose the correct answer from the options given below:
Not quite — the answer is D.
Moles of atoms: Na₂CO₃ 2 mol × 6 = 12; Na₂O 4 mol × 3 = 12; NaOH 6 mol × 3 = 18; H₂ 6 mol × 2 = 12; CO₂ 5 mol × 3 = 15. Only sodium carbonate, sodium oxide and hydrogen contain the same number (12 N_A) of atoms; comparing moles of molecules instead of atoms is the trap.
Q25NEET 2025
Which of the following are paramagnetic? A. [NiCl₄]²⁻ B. Ni(CO)₄ C. [Ni(CN)₄]²⁻ D. [Ni(H₂O)₆]²⁺ E. Ni(PPh₃)₄ Choose the correct answer from the options given below :
Not quite — the answer is A.
Ni²⁺ (3d⁸) with weak-field Cl⁻ is tetrahedral with 2 unpaired electrons, and octahedral [Ni(H₂O)₆]²⁺ also has 2 unpaired electrons, so both are paramagnetic. Strong-field CN⁻ pairs the electrons in square-planar [Ni(CN)₄]²⁻, and Ni(0) in Ni(CO)₄ and Ni(PPh₃)₄ is d¹⁰, so these three are diamagnetic.
Q26NEET 2025
If the half-life (t₁/₂) for a first order reaction is 1 minute, then the time required for 99.9% completion of the reaction is closest to :
Not quite — the answer is B.
t = (2.303/k) log(100/0.1) = (2.303 × 3/0.693) × t₁/₂ = 9.97 × 1 min ≈ 10 minutes. In other words about ten half-lives leave 1/1024 ≈ 0.1% of the reactant.
Q27NEET 2025
Energy and radius of first Bohr orbit of He⁺ and Li²⁺ are [Given R_H = 2.18 × 10⁻¹⁸ J, a₀ = 52.9 pm]
Not quite — the answer is C.
For n = 1, E = −R_H Z² and r = a₀/Z. He⁺ (Z = 2): E = −2.18 × 10⁻¹⁸ × 4 = −8.72 × 10⁻¹⁸ J, r = 52.9/2 ≈ 26.4 pm. Li²⁺ (Z = 3): E = −2.18 × 10⁻¹⁸ × 9 = −19.62 × 10⁻¹⁸ J, r = 52.9/3 ≈ 17.6 pm. The 10⁻¹⁶ options have the wrong power of ten.
Q28NEET 2025
Dalton's Atomic theory could not explain which of the following?
Not quite — the answer is B.
Dalton's theory of indivisible atoms combining in simple whole-number ratios accounts for conservation of mass, constant composition and multiple proportions. Gay-Lussac's law of combining gas volumes needed Avogadro's idea of molecules (e.g. H₂, O₂ splitting into atoms), which Dalton's theory could not provide.
Q29NEET 2025
Identify the correct orders against the property mentioned A. H₂O > NH₃ > CHCl₃ - dipole moment B. XeF₄ > XeO₃ > XeF₂ - number of lone pairs on central atom C. O–H > C–H > N–O - bond length D. N₂ > O₂ > H₂ - bond enthalpy Choose the correct answer from the options given below:
Not quite — the answer is C.
Dipole moments are about 1.85 D (H₂O), 1.47 D (NH₃) and 1.04 D (CHCl₃), so A is right. Lone pairs on Xe are 2 (XeF₄), 1 (XeO₃) and 3 (XeF₂), so B is wrong. O–H (~96 pm) is the shortest of the three bonds, not the longest, so C is wrong. Bond enthalpies are N₂ 946 > O₂ 498 > H₂ 436 kJ mol⁻¹, so D is right.
Q30NEET 2025
Match List-I with List-II. List - I (Name of Vitamin): A. Vitamin B₁₂; B. Vitamin D; C. Vitamin B₂; D. Vitamin B₆ List - II (Deficiency disease): I. Cheilosis; II. Convulsions; III. Rickets; IV. Pernicious anaemia Choose the correct answer from the options given below:
Not quite — the answer is D.
Vitamin B₁₂ deficiency causes pernicious anaemia, vitamin D deficiency causes rickets, riboflavin (B₂) deficiency causes cheilosis (cracked lips) and pyridoxine (B₆) deficiency causes convulsions. Swapping B₂ and B₆ is the common slip.
Q31NEET 2025
The correct order of decreasing basic strength of the given amines is:
Not quite — the answer is A.
Aliphatic amines are much stronger bases than aryl amines because the aryl ring delocalises the nitrogen lone pair. Diethylamine (secondary, two +I groups) is more basic than ethylamine, and N-methylaniline is slightly more basic than aniline because of the extra methyl +I effect (pKb about 9.3 against 9.4).
Q32NEET 2025
The correct order of the wavelength of light absorbed by the following complexes is, A. [Co(NH₃)₆]³⁺ B. [Co(CN)₆]³⁻ C. [Cu(H₂O)₄]²⁺ D. [Ti(H₂O)₆]³⁺ Choose the correct answer from the options given below:
Not quite — the answer is D.
Absorbed wavelength is inversely related to the crystal-field splitting. CN⁻ is the strongest-field ligand, so [Co(CN)₆]³⁻ absorbs at the shortest wavelength, followed by [Co(NH₃)₆]³⁺ (Co³⁺ with NH₃). [Ti(H₂O)₆]³⁺ absorbs around 500 nm, and [Cu(H₂O)₄]²⁺, with a weak field and a +2 ion, absorbs in the red (~600–800 nm), the longest wavelength.
Q33NEET 2025
Which one of the following compounds does not decolourize bromine water?
Not quite — the answer is C.
Styrene adds bromine across its C=C bond, while aniline and phenol are so strongly activated that they rapidly give 2,4,6-tribromo products; all three decolourise bromine water. Cyclohexane is a saturated alkane with no activated ring or double bond, so it does not react with bromine water.
Q34NEET 2025
Match List-I with List-II List - I (Mixture): A. CHCl₃ + C₆H₅NH₂; B. Crude oil in petroleum industry; C. Glycerol from spent-lye; D. Aniline - water List - II (Method of separation): I. Distillation under reduced pressure; II. Steam distillation; III. Fractional distillation; IV. Simple distillation Choose the correct answer from the options given below:
Not quite — the answer is C.
Chloroform (b.p. 334 K) and aniline (b.p. 457 K) differ widely in boiling point, so simple distillation works. Crude oil fractions with close boiling points need fractional distillation. Glycerol decomposes at its boiling point and is recovered from spent lye by distillation under reduced pressure. Aniline is steam-volatile and immiscible with water, so steam distillation is used.
Q35NEET 2025
Which among the following electronic configurations belong to main group elements? A. [Ne]3s¹ B. [Ar]3d³4s² C. [Kr]4d¹⁰5s²5p⁵ D. [Ar]3d¹⁰4s¹ E. [Rn]5f⁰6d²7s² Choose the correct answer from the option given below :
Not quite — the answer is D.
Main-group (s- and p-block) elements have the last electron in an s or p orbital: [Ne]3s¹ is sodium and [Kr]4d¹⁰5s²5p⁵ is iodine. [Ar]3d³4s² (V) and [Ar]3d¹⁰4s¹ (Cu) are d-block, and [Rn]6d²7s² (Th) is an f-block actinoid; copper is the usual trap because its outer electron is 4s.
Q36NEET 2025
Which one of the following compounds can exist as cis-trans isomers?
Not quite — the answer is B.
In 1,2-dimethylcyclohexane the two methyl groups on different ring carbons can be on the same or opposite faces of the ring, giving cis and trans forms. 1,1-Dimethylcyclopropane has both methyls on one carbon, and pent-1-ene and 2-methylhex-2-ene each have one alkene carbon carrying two identical groups, so none of these shows geometrical isomerism.
Q37NEET 2025
Phosphoric acid ionizes in three steps with their ionization constant values Ka₁, Ka₂ and Ka₃, respectively, while K is the overall ionization constant. Which of the following statements are true? A. log K = log Ka₁ + log Ka₂ + log Ka₃ B. H₃PO₄ is a stronger acid than H₂PO₄⁻ and HPO₄²⁻ C. Ka₁ > Ka₂ > Ka₃ D. Ka₁ = (Ka₃ + Ka₂)/2 Choose the correct answer from the options given below :
Not quite — the answer is B.
Adding the three steps multiplies their constants, K = Ka₁·Ka₂·Ka₃, so log K is the sum of the three logs. Removing H⁺ from an increasingly negative species gets harder, so Ka₁ > Ka₂ > Ka₃ and H₃PO₄ is the strongest of the three acids. There is no averaging relation between the constants, so the last statement is false.
Q38NEET 2025
Match List-I with List-II List - I (Ion): A. Co²⁺; B. Mg²⁺; C. Pb²⁺; D. Al³⁺ List - II (Group Number in Cation Analysis): I. Group-I; II. Group-III; III. Group-IV; IV. Group-VI Choose the correct answer from the options given below :
Not quite — the answer is D.
In the classical scheme Pb²⁺ is precipitated in group I as PbCl₂, Al³⁺ in group III as Al(OH)₃, Co²⁺ in group IV as CoS (H₂S in ammoniacal medium) and Mg²⁺ is left for group VI. So Co²⁺ goes with Group-IV, Mg²⁺ with Group-VI, Pb²⁺ with Group-I and Al³⁺ with Group-III.
Q39NEET 2025
Higher yield of NO in N₂(g) + O₂(g) ⇌ 2NO(g) can be obtained at [ΔH of the reaction = +180.7 kJ mol⁻¹] A. Higher temperature B. Lower temperature C. Higher concentration of N₂ D. Higher concentration of O₂ Choose the correct answer from the options given below :
Not quite — the answer is B.
The forward reaction is endothermic, so by Le Chatelier's principle a higher temperature shifts the equilibrium towards NO. Adding more of either reactant, N₂ or O₂, also shifts it forward. Lower temperature would favour the reactants.
Q40NEET 2025
Given below are two statements : Statement-I : Benzenediazonium salt is prepared by the reaction of aniline with nitrous acid at 273 − 278 K. It decomposes easily in the dry state. Statement-II : Insertion of iodine into the benzene ring is difficult and hence iodobenzene is prepared through the reaction of benzenediazonium salt with KI. In the light of the above statements, choose the most appropriate answer from the options given below:
Not quite — the answer is C.
Diazotisation of aniline with NaNO₂/HCl is carried out at 273–278 K because the diazonium salt is unstable; it is used in solution since the dry salt decomposes (even explosively). Direct iodination of benzene is reversible and inefficient, so iodobenzene is conveniently made by warming the diazonium salt with KI. Both statements are correct.
Q41NEET 2025
The major product of the following reaction is C₆H₅–CO–CH₂–CH₂–CN →[(i) CH₃MgBr (excess), (ii) H₃O⁺]
Not quite — the answer is D.
Excess CH₃MgBr adds to the ketone to give a tertiary alkoxide and also adds once to the nitrile to give an imine magnesium salt, which cannot react further. Acid work-up protonates the alkoxide to a tertiary alcohol and hydrolyses the imine to a methyl ketone. The diol would need a second Grignard addition to the ketone formed from the nitrile, which only appears after hydrolysis.
Chapter-wise split of NEET 2025 Chemistry
- Hydrocarbons4 Q
- Aldehydes, Ketones and Carboxylic Acids3 Q
- Organic Chemistry: Some Basic Principles and Techniques3 Q
- Solutions3 Q
- Coordination Compounds3 Q
- Equilibrium3 Q
- Chemical Kinetics2 Q
- Chemical Bonding and Molecular Structure2 Q
- Classification of Elements and Periodicity in Properties2 Q
- Biomolecules2 Q
- Structure of Atom2 Q
- Some Basic Concepts of Chemistry2 Q
- Amines2 Q
- Haloalkanes and Haloarenes1 Q
- Redox Reactions1 Q
- The d- and f-Block Elements1 Q
- Thermodynamics1 Q
- Electrochemistry1 Q
NEET 2025 Biology questions
82 questions. Tap an option to check your answer.
Q1NEET 2025
The protein portion of an enzyme is called:
Not quite — the answer is A.
The protein part of a conjugated enzyme is the apoenzyme; together with its non-protein cofactor it forms the active holoenzyme. Prosthetic groups and coenzymes are kinds of non-protein cofactors, not the protein portion.
Q2NEET 2025
Given below are two statements: Statement I: The primary source of energy in an ecosystem is solar energy. Statement II : The rate of production of organic matter during photosynthesis in an ecosystem is called net primary productivity (NPP). In the light of the above statements, choose the most appropriate answer from the options given below:
Not quite — the answer is A.
Sunlight is the primary energy source of nearly all ecosystems, so Statement I is correct. The total rate of production of organic matter during photosynthesis is gross primary productivity; net primary productivity is what remains after respiratory losses, so Statement II is incorrect.
Q3NEET 2025
Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : A typical unfertilised, angiosperm embryo sac at maturity is 8 nucleate and 7-celled. Reason (R) : The egg apparatus has 2 polar nuclei. In the light of the above statements, choose the correct answer from the options given below :
Not quite — the answer is A.
A mature embryo sac has 8 nuclei in 7 cells: three antipodals, two synergids, one egg and one binucleate central cell, so the Assertion is true. The egg apparatus consists of the egg cell and two synergids; the two polar nuclei lie in the central cell, so the Reason is false.
Q4NEET 2025
Neoplastic characteristics of cells refer to: A. A mass of proliferating cell B. Rapid growth of cells C. Invasion and damage to the surrounding tissue D. Those confined to original location Choose the correct answer from the options given below:
Not quite — the answer is D.
Neoplastic (malignant) cells form a mass of proliferating cells, grow rapidly and invade and damage surrounding tissue. Cells that remain confined to their original location describe a benign tumour, not neoplastic malignancy.
Q5NEET 2025
Which one of the following is the characteristic feature of gymnosperms?
Not quite — the answer is D.
Gymnosperms bear ovules that are not enclosed by an ovary wall, so their seeds remain naked. Flowers and fruits enclosing seeds are features of angiosperms.
Q6NEET 2025
Match List-I with List-II List - I List - II A Progesterone I Pars intermedia B Relaxin II Ovary C Melanocyte stimulating hormone III Adrenal Medulla D Catecholamines IV Corpus luteum Choose the correct answer from the options given below :
Not quite — the answer is C.
Progesterone is secreted mainly by the corpus luteum, relaxin by the ovary, melanocyte stimulating hormone by the pars intermedia, and catecholamines by the adrenal medulla. The trap is pairing progesterone with the ovary in general instead of the corpus luteum.
Q7NEET 2025
Which chromosome in the human genome has the highest number of genes?
Not quite — the answer is A.
The Human Genome Project found chromosome 1 to carry the most genes (about 2968). Chromosome Y has the fewest genes (about 231).
Q8NEET 2025
Which of the following statements about RuBisCO is true?
Not quite — the answer is B.
RuBisCO catalyses the carboxylation of RuBP, the first step of the Calvin cycle. It does not split water, works in light-driven conditions as part of the dark reactions, and has a higher affinity for CO₂ than for O₂ (it binds O₂ only when CO₂ is low, causing photorespiration).
Q9NEET 2025
The first menstruation is called :
Not quite — the answer is D.
The first menstruation at puberty is called menarche. Menopause is the cessation of menstrual cycles around 50 years of age.
Q10NEET 2025
Which of the following genetically engineered organisms was used by Eli Lilly to prepare human insulin?
Not quite — the answer is C.
Eli Lilly prepared DNA sequences for the A and B chains of insulin and introduced them into plasmids of the bacterium E. coli, then joined the separately produced chains. No virus, phage or yeast was used as the producing organism.
Q11NEET 2025
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : All vertebrates are chordates but all chordates are not vertebrate. Reason (R) : The members of subphylum vertebrata possess notochord during the embryonic period, the notochord is replaced by cartilaginous or bony vertebral column in adults. In the light of the above statements, choose the correct answer from the options given below:
Not quite — the answer is C.
Vertebrates have a notochord only in the embryo, which is replaced by a vertebral column in adults; chordates without this replacement are not vertebrates. This is exactly why every vertebrate is a chordate but not every chordate is a vertebrate, so the Reason correctly explains the Assertion.
Q12NEET 2025
What is the main function of the spindle fibers during mitosis?
Not quite — the answer is C.
Spindle fibres attach to kinetochores and pull sister chromatids to opposite poles at anaphase, separating the chromosomes. DNA synthesis and repair happen at other times and do not involve the spindle.
Q13NEET 2025
Match List-I with List-II List - I List - II A Alfred Hershey and Martha Chase I Streptococcus pneumoniae B Euchromatin II Densely packed and dark-stained C Frederick Griffith III Loosely packed and light-stained D Heterochromatin IV DNA as genetic material confirmation Choose the correct answer from the options given below:
Not quite — the answer is A.
Hershey and Chase confirmed DNA as the genetic material; euchromatin is loosely packed and light-stained; Griffith worked with Streptococcus pneumoniae; heterochromatin is densely packed and dark-stained. Swapping the two chromatin types is the common trap.
Q14NEET 2025
Match List-I with List-II List - I List - II A Adenosine I Nitrogen base B Adenylic acid II Nucleotide C Adenine III Nucleoside D Alanine IV Amino acid Choose the option with all correct matches.
Not quite — the answer is A.
Adenosine is a nucleoside (adenine + sugar), adenylic acid is a nucleotide (nucleoside + phosphate), adenine is a nitrogen base, and alanine is an amino acid. Confusing nucleoside and nucleotide is the main trap.
Q15NEET 2025
In frog, the Renal portal system is a special venous connection that acts to link:
Not quite — the answer is B.
In frog, the renal portal system carries venous blood from the lower parts of the body to the kidneys. The connection between intestine and liver is the hepatic portal system.
Q16NEET 2025
Which of the following are the post-transcriptional events in an eukaryotic cell? A. Transport of pre-mRNA to cytoplasm prior to splicing. B. Removal of introns and joining of exons. C. Addition of methyl group at 5' end of hnRNA. D. Addition of adenine residues at 3' end of hnRNA. E. Base pairing of two complementary RNAs. Choose the correct answer from the options given below :
Not quite — the answer is D.
Processing of hnRNA in eukaryotes includes splicing (removal of introns and joining of exons), capping at the 5′ end with methyl guanosine triphosphate, and tailing with adenylate residues at the 3′ end. Splicing occurs before mRNA leaves the nucleus, so transport of unspliced pre-mRNA to the cytoplasm is wrong, and pairing of two complementary RNAs is RNA interference, not processing.
Q17NEET 2025
Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Both wind and water pollinated flowers are not very colourful and do not produce nectar. Reason (R) : The flowers produce enormous amount of pollen grains in wind and water pollinated flowers. In the light of the above statements, choose the correct answer from the options given below:
Not quite — the answer is D.
Wind- and water-pollinated flowers do not need to attract animals, so they are not colourful and produce no nectar; they also produce enormous numbers of pollen grains. Both statements are true, but the lack of colour and nectar is explained by the absence of animal pollinators, not by the large pollen output.
Q18NEET 2025
Epiphytes that are growing on a mango branch is an example of which of the following?
Not quite — the answer is C.
An epiphyte gets support and access to light by growing on the mango branch while the mango is neither harmed nor benefited, which is commensalism. It is not mutualism because the host gains nothing.
Q19NEET 2025
Find the correct statement : (A) In human pregnancy, the major organ systems are formed at the end of 12 weeks. (B) In human pregnancy the major organ systems are formed at the end of 8 weeks. (C) In human pregnancy heart is formed after one month of gestation. (D) In human pregnancy, limbs and digits develop by the end of second month. (E) In human pregnancy the appearance of hair is usually observed in the fifth month. Choose the correct answer from the options given below :
Not quite — the answer is B.
By the end of 12 weeks (first trimester) most major organ systems are formed; the heart forms after one month; limbs and digits develop by the end of the second month; hair appears and first movements are felt in the fifth month. The claim that major organ systems are formed by the end of 8 weeks is the incorrect one.
Q20NEET 2025
Consider the following statements regarding function of adrenal medullary hormones: (A) It causes pupilary constriction. (B) It is a hyperglycemic hormone. (C) It causes piloerection. (D) It increases strength of heart contraction. Choose the correct answer from the options given below :
Not quite — the answer is D.
Adrenaline and noradrenaline raise blood glucose by glycogenolysis, cause piloerection, and increase the strength of heart contraction. They dilate rather than constrict the pupil, so pupillary constriction is wrong.
Q21NEET 2025
Read the following statements on plant growth and development. (A) Parthenocarpy can be induced by auxins. (B) Plant growth regulators can be involved in promotion as well as inhibition of growth. (C) Dedifferentiation is a pre-requisite for re-differentiation. (D) Abscisic acid is a plant growth promoter. (E) Apical dominance promotes the growth of lateral buds. Choose the option with all correct statements.
Not quite — the answer is C.
Auxins can induce parthenocarpy, growth regulators can both promote and inhibit growth, and redifferentiation follows dedifferentiation. Abscisic acid is a growth inhibitor, and apical dominance suppresses rather than promotes lateral bud growth.
Q22NEET 2025
Which of the following hormones released from the pituitary is actually synthesized in the hypothalamus?
Not quite — the answer is D.
ADH (vasopressin) is synthesised by hypothalamic neurosecretory cells and transported along axons to the posterior pituitary for release. FSH, LH and ACTH are synthesised in the anterior pituitary itself.
Q23NEET 2025
Which of the following is an example of non-distilled alcoholic beverage produced by yeast?
Not quite — the answer is A.
Wine and beer are produced by yeast fermentation without distillation. Whisky, brandy and rum are made by distilling the fermented broth.
Q24NEET 2025
What is the pattern of inheritance for polygenic trait?
Not quite — the answer is D.
A polygenic trait is controlled by three or more genes with additive effects, giving continuous variation such as human skin colour. It does not show simple Mendelian dominant/recessive ratios, so it is treated as a non-Mendelian pattern; it is neither autosomal dominant nor X-linked.
Q25NEET 2025
Match List-I with List-II. List - I List - II A Head I Enzymes B Middle piece II Sperm motility C Acrosome III Energy D Tail IV Genetic material Choose the correct answer from the options given below :
Not quite — the answer is C.
The sperm head carries the nucleus (genetic material), the middle piece has mitochondria that provide energy, the acrosome holds enzymes for penetrating the ovum, and the tail gives motility. Matching the head with energy is the trap.
Q26NEET 2025
Which of the following is an example of a zygomorphic flower?
Not quite — the answer is A.
Pea has a papilionaceous corolla that can be cut into two equal halves along only one plane, so it is zygomorphic. Chilli, petunia and datura flowers are actinomorphic.
Q27NEET 2025
Which of following organisms cannot fix nitrogen? A. Azotobacter B. Oscillatoria C. Anabaena D. Volvox E. Nostoc Choose the correct answer from the options given below:
Not quite — the answer is D.
Volvox is a colonial green alga with no nitrogenase and cannot fix nitrogen. Azotobacter is a free-living nitrogen-fixing bacterium, and Oscillatoria, Anabaena and Nostoc are nitrogen-fixing cyanobacteria.
Q28NEET 2025
Which one of the following is an example of ex-situ conservation?
Not quite — the answer is A.
Ex situ conservation protects species outside their natural habitat, as in zoos, botanical gardens and seed banks. Protected areas, national parks and wildlife sanctuaries conserve species in their natural habitats, which is in situ conservation.
Q29NEET 2025
Who is known as the father of Ecology in India?
Not quite — the answer is D.
Ramdeo Misra laid the foundations of ecology teaching and research in India and is called the father of Indian ecology. Birbal Sahni is known for palaeobotany and S.R. Kashyap for bryology.
Q30NEET 2025
Given below are two statements: Statement I: In the RNA world, RNA is considered the first genetic material evolved to carry out essential life processes. RNA acts as a genetic material and also as a catalyst for some important biochemical reactions in living systems. Being reactive, RNA is unstable. Statement II: DNA evolved from RNA and is a more stable genetic material. Its double helical strands being complementary, resist changes by evolving repairing mechanism. In the light of the above statements, choose the most appropriate answer from the options given below :
Not quite — the answer is C.
In the RNA world, RNA acted both as genetic material and as a catalyst, and being reactive it was unstable, so Statement I is correct. DNA arose from RNA with chemical changes that make it more stable, and its complementary double strands resist change through repair mechanisms, so Statement II is also correct.
Q31NEET 2025
Given below are two statements : Statement I : Transfer RNAs and ribosomal RNA do not interact with mRNA. Statement II : RNA interference (RNAi) takes place in all eukaryotic organisms as a method of cellular defence. In the light of the above statements, choose the most appropriate answer from the options given below :
Not quite — the answer is B.
tRNA and rRNA interact with mRNA during translation: tRNA anticodons pair with mRNA codons on the ribosome, so Statement I is incorrect. RNA interference occurs in all eukaryotes as a method of cellular defence, so Statement II is correct.
Q32NEET 2025
Match List-I with List-II. List - I List - II A Heart I Erythropoietin B Kidney II Aldosterone C Gastro-intestinal tract III Atrial natriuretic factor D Adrenal Cortex IV Secretin Choose the correct answer from the options given below :
Not quite — the answer is B.
The heart secretes atrial natriuretic factor, the kidney erythropoietin, the gastro-intestinal tract secretin, and the adrenal cortex aldosterone. Pairing the heart with erythropoietin is the trap.
Q33NEET 2025
All living members of the class Cyclostomata are :
Not quite — the answer is B.
All living cyclostomes (lampreys and hagfishes) are ectoparasites on some fishes. They migrate for spawning but are not free-living or endoparasitic.
Q34NEET 2025
Streptokinase produced by bacterium Streptococcus is used for
Not quite — the answer is B.
Streptokinase from Streptococcus, modified by genetic engineering, is used as a clot buster to remove clots from blood vessels of patients with myocardial infarction. Curd and ethanol are made by other microbes.
Q35NEET 2025
Role of the water vascular system in Echinoderms is : A. Respiration and Locomotion B. Excretion and Locomotion C. Capture and transport of food D. Digestion and Respiration E. Digestion and Excretion Choose the correct answer from the options given below :
Not quite — the answer is D.
The water vascular system of echinoderms helps in locomotion, capture and transport of food, and respiration. It is not an organ of excretion or digestion.
Q36NEET 2025
Match List-I with List-II. List - I List - II A Pteridophyte I Salvia B Bryophyte II Ginkgo C Angiosperm III Polytrichum D Gymnosperm IV Salvinia Choose the option with all correct matches.
Not quite — the answer is D.
Salvinia is a pteridophyte, Polytrichum a bryophyte, Salvia an angiosperm and Ginkgo a gymnosperm. The trap is confusing Salvinia (a water fern) with Salvia (a flowering plant).
Q37NEET 2025
Which are correct: A. Computed tomography and magnetic resonance imaging detect cancers of internal organs. B. Chemotherapeutics drugs are used to kill non-cancerous cells. C. α-interferon activate the cancer patients' immune system and helps in destroying the tumour. D. Chemotherapeutic drugs are biological response modifiers. E. In the case of leukaemia blood cell counts are decreased. Choose the correct answer from the options given below:
Not quite — the answer is B.
CT and MRI detect cancers of internal organs, and α-interferon activates the immune system to help destroy the tumour. Chemotherapeutic drugs kill cancerous cells and are not biological response modifiers (α-interferon is), and leukaemia raises rather than lowers blood cell counts.
Q38NEET 2025
What are the potential drawbacks in adoption of the IVF method? A. High fatality risk to mother B. Expensive instruments and reagents C. Husband/wife necessary for being donors D. Less adoption of orphans E. Not available in India F. Possibility that the early embryo does not survive Choose the correct answer from the options given below:
Not quite — the answer is C.
IVF needs expensive instruments and reagents, the early embryo may not survive after transfer, and couples choosing IVF reduce the adoption of orphans. It carries no high fatality risk to the mother, donor gametes need not come from the husband or wife, and it is available in India.
Q39NEET 2025
Consider the following : A. The reductive division for the human female gametogenesis starts earlier than that of the male gametogenesis. B. The gap between the first meiotic division and the second meiotic division is much shorter for males compared to females. C. The first polar body is associated with the formation of the primary oocyte. D. Luteinizing Hormone (LH) surge leads to disintegration of the endometrium and onset of menstrual bleeding. Choose the correct answer from the options given below:
Not quite — the answer is C.
Meiosis in females begins in the foetal ovary, long before males begin it at puberty, and the gap between the two meiotic divisions is short in males but very long in females (meiosis II finishes only at fertilisation). The first polar body forms when the primary oocyte becomes the secondary oocyte, and the LH surge triggers ovulation, not menstruation.
Q40NEET 2025
In bryophytes, the gemmae help in which one of the following?
Not quite — the answer is D.
Gemmae are green, multicellular asexual buds formed in gemma cups of liverworts; they detach and grow into new plants. They are means of vegetative (asexual) reproduction.
Q41NEET 2025
Which one of the following statements refers to Reductionist Biology?
Not quite — the answer is C.
Reductionist biology studies and understands living organisms by a physico-chemical approach, breaking life processes down to molecules and reactions. A purely chemical, behavioural or physiological approach is not what the term refers to.
Q42NEET 2025
After maturation, in primary lymphoid organs, the lymphocytes migrate for interaction with antigens to secondary lymphoid organ(s) / tissue(s) like A. thymus B. bone marrow C. spleen D. lymph nodes E. Peyer's patches Choose the correct answer from the options given below
Not quite — the answer is B.
Secondary lymphoid organs include the spleen, lymph nodes, tonsils, Peyer's patches and MALT, where mature lymphocytes meet antigens. The thymus and bone marrow are primary lymphoid organs where lymphocytes mature.
Q43NEET 2025
Match List-I with List-II List - I List - II A The Evil Quartet I Cryopreservation B Ex situ conservation II Alien species invasion C Lantana camara III Causes of biodiversity losses D Dodo IV Extinction Choose the option with all correct matches.
Not quite — the answer is D.
The evil quartet lists the causes of biodiversity loss, cryopreservation is an ex situ method, Lantana camara is an invasive alien species, and the dodo is an example of extinction. Pairing ex situ conservation with alien species is the trap.
Q44NEET 2025
How many meiotic and mitotic divisions need to occur for the development of a mature female gametophyte from the megaspore mother cell in an angiosperm plant?
Not quite — the answer is A.
The megaspore mother cell undergoes one meiosis to form four megaspores; the functional megaspore then goes through three successive mitotic divisions to give the 8-nucleate embryo sac. Two mitoses would give only four nuclei.
Q45NEET 2025
Which of the following type of immunity is present at the time of birth and is a non-specific type of defence in the human body?
Not quite — the answer is D.
Innate immunity is present from birth and gives non-specific defence through barriers such as skin, mucus, phagocytes and fever. Acquired immunity, including humoral and cell-mediated responses, is specific and develops after exposure to antigens.
Q46NEET 2025
Given below are two statements : One is labelled as Assertion (A) and other is labelled as Reason (R). Assertion (A) : Cells of the tapetum possess dense cytoplasm and generally have more than one nucleus. Reason (R) : Presence of more than one nucleus in the tapetum increases the efficiency of nourishing the developing microspore mother cells. In light of the above statements, choose the most appropriate answer from the options given below:
Not quite — the answer is C.
Tapetal cells have dense cytoplasm and generally more than one nucleus, and the tapetum nourishes the developing microspores. Being multinucleate raises their metabolic activity, which makes this nourishing role more efficient, so the Reason explains the Assertion.
Q47NEET 2025
From the statements given below choose the correct option : A. The eukaryotic ribosomes are 80S and prokaryotic ribosomes are 70S. B. Each ribosome has two sub-units. C. The two sub-units of 80 S ribosome are 60 S and 40 S while that of 70 S are 50 S and 30 S . D. The two sub-units of 80 S ribosome are 60 S and 20 S and that of 70 S are 50 S and 20 S . E. The two sub-units of 80 S are 60 S and 30 S and that of 70 S are 50 S and 30 S .
Not quite — the answer is C.
Eukaryotic ribosomes are 80S and prokaryotic ones 70S; each has two subunits, 60S + 40S and 50S + 30S. Subunits of 20S or 30S for the 80S ribosome are wrong.
Q48NEET 2025
Which one of the following enzymes contains 'Haem' as the prosthetic group?
Not quite — the answer is B.
Haem is the prosthetic group of peroxidase and catalase, which break down hydrogen peroxide. Succinate dehydrogenase uses FAD, carbonic anhydrase uses zinc and RuBisCO has no haem.
Q49NEET 2025
What is the name of the blood vessel that carries deoxygenated blood from the body to the heart in a frog?
Not quite — the answer is B.
In frog, the vena cavae bring deoxygenated blood from the body to the sinus venosus, which opens into the right atrium. Pulmonary veins carry oxygenated blood from the lungs and the aorta carries blood away from the heart.
Q50NEET 2025
Given below are the stages in the life cycle of pteridophytes. Arrange the following stages in the correct sequence. A. Prothallus stage B. Meiosis in spore mother cells C. Fertilisation D. Formation of archegonia and antheridia in gametophyte. E. Transfer of antherozoids to the archegonia in presence of water. Choose the correct answer from the options given below:
Not quite — the answer is C.
In pteridophytes, spore mother cells undergo meiosis to form spores, which germinate into the prothallus; the prothallus bears antheridia and archegonia, antherozoids reach the archegonia through water, and fertilisation follows. Putting fertilisation before the transfer of antherozoids is the trap.
Q51NEET 2025
The blue and white selectable markers have been developed which differentiate recombinant colonies from nonrecombinant colonies on the basis of their ability to produce colour in the presence of a chromogenic substrate. Given below are two statements about this method: Statement I : The blue coloured colonies have DNA insert in the plasmid and they are identified as recombinant colonies. Statement II : The colonies without blue colour have DNA insert in the plasmid and are identified as recombinant colonies. In the light of the above statements, choose the most appropriate answer from the options given below:
Not quite — the answer is B.
An insert in the β-galactosidase gene inactivates it, so recombinant colonies cannot act on the chromogenic substrate and remain colourless (white), while non-recombinants turn blue. So the blue colonies lack the insert, Statement I is incorrect and Statement II is correct.
Q52NEET 2025
Which of the following microbes is NOT involved in the preparation of household products? A. Aspergillus niger B. Lactobacillus C. Trichoderma polysporum D. Saccharomyces cerevisiae E. Propionibacterium sharmanii Choose the correct answer from the options given below:
Not quite — the answer is D.
Lactobacillus (curd), Saccharomyces cerevisiae (bread dough) and Propionibacterium sharmanii (Swiss cheese) are used for household or food products. Aspergillus niger is used industrially for citric acid and Trichoderma polysporum for the drug cyclosporin A.
Q53NEET 2025
Silencing of specific mRNA is possible via RNAi because of
Not quite — the answer is C.
In RNA interference, a complementary double-stranded RNA binds the specific mRNA and prevents its translation, silencing it. Single-stranded or non-complementary RNA cannot target a particular mRNA.
Q54NEET 2025
The complex II of mitochondrial electron transport chain is also known as
Not quite — the answer is D.
Complex II of the mitochondrial electron transport chain is succinate dehydrogenase, which passes electrons from FADH₂ to ubiquinone. Complex I is NADH dehydrogenase, complex III is cytochrome bc₁, and complex IV is cytochrome c oxidase.
Q55NEET 2025
While trying to find out the characteristic of a newly found animal, a researcher did the histology of adult animal and observed a cavity with presence of mesodermal tissue towards the body wall but no mesodermal tissue was observed towards the alimentary canal. What could be the possible coelom of that animal?
Not quite — the answer is D.
A body cavity lined by mesoderm only on the body-wall side, with mesoderm scattered as pouches rather than covering the gut, is a pseudocoelom, as in aschelminthes. In a true coelom (schizocoelom) mesoderm lines both the body wall and the gut.
Q56NEET 2025
Given below are two statements: Statement I: In a floral formula ⊕ stands for zygomorphic nature of the flower, and G̲ stands for inferior ovary. Statement II: In a floral formula ⊕ stands for actinomorphic nature of the flower and G̲ stands for superior ovary. In the light of the above statements, choose the correct answer from the options given below:
Not quite — the answer is B.
In a floral formula, ⊕ means actinomorphic and G̲ (G with a line below) means a superior ovary. So Statement I is wrong on both counts and Statement II is correct.
Q57NEET 2025
Given below are two statements: Statement I: In ecosystem, there is unidirectional flow of energy of sun from producers to consumers. Statement II: Ecosystems are exempted from 2nd law of thermodynamics. In the light of the above statements, choose the most appropriate answer from the options given below:
Not quite — the answer is A.
Energy enters as sunlight, passes from producers to consumers and is lost as heat, so its flow is one-way. Ecosystems obey the second law of thermodynamics and need a constant energy input to stay organised, so Statement II is wrong.
Q58NEET 2025
Which of the following is the unit of productivity of an Ecosystem?
Not quite — the answer is B.
Productivity is the rate of biomass or energy production per unit area over a time period, expressed as g m⁻² yr⁻¹ or (kcal m⁻²) yr⁻¹. kcal m⁻² and g m⁻² have no time component and describe standing energy or biomass, not productivity.
Q59NEET 2025
In the seeds of cereals, the outer covering of endosperm separates the embryo by a protein-rich layer called:
Not quite — the answer is B.
In cereal grains, the outer layer of the endosperm is the protein-rich aleurone layer, which separates the endosperm from the embryo. Coleoptile and coleorhiza are sheaths of the plumule and radicle.
Q60NEET 2025
Match List-I with List-II List - I List - II A Chlorophyll a I Yellow-green B Chlorophyll b II Yellow C Xanthophylls III Blue-green D Carotenoids IV Yellow to Yellow-orange Choose the option with all correct matches.
Not quite — the answer is D.
On a chromatogram, chlorophyll a is bright or blue-green, chlorophyll b yellow-green, xanthophylls yellow and carotenoids yellow to yellow-orange. Swapping the two chlorophyll colours is the trap.
Q61NEET 2025
Who proposed that the genetic code for amino acids should be made up of three nucleotides?
Not quite — the answer is C.
George Gamow argued that 4 bases must code for 20 amino acids, so the code needs combinations of three nucleotides (4³ = 64). Crick later helped confirm the triplet nature experimentally, but the proposal came from Gamow.
Q62NEET 2025
Histones are enriched with -
Not quite — the answer is C.
Histones are rich in the basic amino acids lysine and arginine, whose positive charges bind the negatively charged DNA. Phenylalanine and leucine are not basic.
Q63NEET 2025
Which of the following enzyme(s) are NOT essential for gene cloning? A. Restriction enzymes B. DNA ligase C. DNA mutase D. DNA recombinase E. DNA polymerase Choose the correct answer from the options given below:
Not quite — the answer is C.
Gene cloning needs restriction enzymes to cut DNA, DNA ligase to join fragments and DNA polymerase for amplification. 'DNA mutase' and 'DNA recombinase' are not tools of recombinant DNA technology.
Q64NEET 2025
A specialised membranous structure in a prokaryotic cell which helps in cell wall formation, DNA replication and respiration is
Not quite — the answer is C.
Mesosomes are infoldings of the prokaryotic plasma membrane that help in cell wall formation, DNA replication and its distribution to daughter cells, and respiration. Chromatophores are pigment-bearing membranes for photosynthesis, and cristae and ER are not found in prokaryotes.
Q65NEET 2025
Which factor is important for termination of transcription?
Not quite — the answer is A.
In bacteria, the rho (ρ) factor binds RNA polymerase and causes transcription to stop. The sigma (σ) factor is needed for initiation.
Q66NEET 2025
Which of the following statement is correct about location of the male frog copulatory pad?
Not quite — the answer is B.
Male frogs have a copulatory (nuptial) pad on the first digit of each forelimb, used to grip the female in amplexus. It is absent from the hind limb and other digits.
Q67NEET 2025
Identify the statement that is NOT correct.
Not quite — the answer is A.
The antigen-binding sites lie in the variable regions at the N-terminal ends of the heavy and light chains, so placing them at the C-terminal region is incorrect. Constant regions are at the C-terminal side, each antibody has two heavy and two light chains (H₂L₂), and disulphide bonds hold the chains together.
Q68NEET 2025
Match List I with List II : List - I List - II A Scutellum I Persistent nucellus B Non-albuminous seed II Cotyledon of Monocot seed C Epiblast III Groundnut D Perisperm IV Rudimentary cotyledon Choose the option with all correct matches.
Not quite — the answer is C.
The scutellum is the single cotyledon of a monocot (grass) seed, groundnut has a non-albuminous seed, the epiblast is a rudimentary cotyledon, and the perisperm is persistent nucellus. Pairing the scutellum with the epiblast's role is the trap.
Q69NEET 2025
Find the statement that is NOT correct with regard to the structure of monocot stem.
Not quite — the answer is C.
The monocot stem has a sclerenchymatous hypodermis, so calling it parenchymatous is wrong. Its vascular bundles are conjoint, closed and scattered, and phloem parenchyma is absent.
Q70NEET 2025
Twins are born to a family that lives next door to you. The twins are a boy and a girl. Which of the following must be true?
Not quite — the answer is D.
A boy and a girl cannot come from a single zygote, so the twins must have formed from two separate eggs fertilised by two sperms (fraternal or dizygotic). Fraternal twins share on average about 50% of their genes, not 75%, and IVF is not required.
Q71NEET 2025
Sweet potato and potato represent a certain type of evolution. Select the correct combination of terms to explain the evolution.
Not quite — the answer is C.
The sweet potato is a modified root and the potato a modified stem; both store food, so they are analogous organs. Different structures evolving a similar function reflects convergent evolution.
Q72NEET 2025
Which one of the following phytohormones promotes nutrient mobilization which helps in the delay of leaf senescence in plants?
Not quite — the answer is B.
Cytokinins promote nutrient mobilisation, which delays leaf senescence. Ethylene and abscisic acid speed up senescence.
Q73NEET 2025
Why can't insulin be given orally to diabetic patients?
Not quite — the answer is D.
Insulin is a polypeptide hormone, so if swallowed it would be broken down by proteolytic enzymes in the gut before it could act; it must be injected. The issue is digestion, not immunity or structure.
Q74NEET 2025
Name the class of enzyme that usually catalyze the following reaction : S − G + S# → S + S# − G Where, G → a group other than hydrogen S → a substrate S# → another substrate
Not quite — the answer is A.
The reaction moves a group G (other than hydrogen) from one substrate to another, which is what transferases catalyse. Ligases join two compounds, hydrolases add water to break bonds, and lyases remove groups to form double bonds.
Q75NEET 2025
Given below are two statements: Statement I : The DNA fragments extracted from gel electrophoresis can be used in construction of recombinant DNA. Statement II : Smaller size DNA fragments are observed near anode while larger fragments are found near the wells in an agarose gel. In the light of the above statements, choose the most appropriate answer from the options given below :
Not quite — the answer is C.
Separated DNA bands are cut from the gel and eluted, then used to construct recombinant DNA, so Statement I is correct. DNA is negatively charged and moves toward the anode, smaller fragments moving farther, so larger fragments stay near the wells and Statement II is also correct.
Q76NEET 2025
The correct sequence of events in the life cycle of bryophytes is A. Fusion of antherozoid with egg. B. Attachment of gametophyte to substratum. C. Reduction division to produce haploid spores. D. Formation of sporophyte. E. Release of antherozoids into water. Choose the correct answer from the options given below :
Not quite — the answer is A.
The gametophyte attaches to the substratum by rhizoids, antherozoids are released into water, an antherozoid fuses with the egg, the zygote forms the sporophyte, and meiosis in the capsule forms haploid spores. Sequences starting with the sporophyte ignore that the gametophyte is the dominant independent phase.
Q77NEET 2025
Genes R and Y follow independent assortment. If RRYY produce round yellow seeds and rryy produce wrinkled green seeds, what will be the phenotypic ratio of the F2 generation?
Not quite — the answer is A.
RRYY × rryy gives RrYy in F₁; with independent assortment, F₂ shows round yellow, round green, wrinkled yellow and wrinkled green in a 9 : 3 : 3 : 1 ratio. The 9 : 7 ratio needs complementary gene interaction, and 3 : 1 and 1 : 2 : 1 are monohybrid ratios.
Q78NEET 2025
Each of the following characteristics represent a Kingdom proposed by Whittaker. Arrange the following in increasing order of complexity of body organization. A. Multicellular heterotrophs with cell wall made of chitin. B. Heterotrophs with tissue/organ/organ system level of body organization. C. Prokaryotes with cell wall made of polysaccharides and amino acids. D. Eukaryotic autotrophs with tissue/organ level of body organization. E. Eukaryotes with cellular body organization. Choose the correct answer from the options given below :
Not quite — the answer is D.
Monera (prokaryotes with a cell wall of polysaccharides and amino acids) < Protista (eukaryotes with cellular organisation) < Fungi (multicellular heterotrophs with chitin walls) < Plantae (eukaryotic autotrophs with tissue/organ level) < Animalia (heterotrophs with tissue/organ/organ-system level). Putting animals before plants is the trap.
Q79NEET 2025
Match List-I with List- II. List - I List - II A Centromere I Mitochondrion B Cilium II Cell division C Cristae III Cell movement D Cell membrane IV Phospholipid Bilayer Choose the correct answer from the options given below :
Not quite — the answer is B.
The centromere holds sister chromatids and is needed in cell division, cilia cause cell movement, cristae are infoldings of the inner mitochondrial membrane, and the cell membrane is a phospholipid bilayer.
Q80NEET 2025
Which one of the following equations represents the Verhulst-Pearl Logistic Growth of population?
Not quite — the answer is D.
Verhulst-Pearl logistic growth is dN/dt = rN[(K − N)/K], where growth slows as N approaches the carrying capacity K. The other forms either reverse the sign of the limiting term or drop N or r.
Q81NEET 2025
Match List-I with List-II List - I List - II A Emphysema I Rapid spasms in muscle due to low Ca⁺⁺ in body fluid B Angina Pectoris II Damaged alveolar walls and decreased respiratory surface C Glomerulonephritis III Acute chest pain when not enough oxygen is reaching to heart muscle D Tetany IV Inflammation of glomeruli of kidney Choose the correct answer from the options given below :
Not quite — the answer is B.
Emphysema damages alveolar walls and reduces the respiratory surface; angina pectoris is acute chest pain when heart muscle gets too little oxygen; glomerulonephritis is inflammation of kidney glomeruli; tetany is rapid muscle spasms due to low Ca⁺⁺ in body fluid.
Q82NEET 2025
Cardiac activities of the heart are regulated by: A. Nodal tissue B. A special neural centre in the medulla oblongata C. Adrenal medullary hormones D. Adrenal cortical hormones Choose the correct answer from the options given below :
Not quite — the answer is C.
Heart beat starts in the auto-excitable nodal tissue, a neural centre in the medulla oblongata moderates it through autonomic nerves, and adrenal medullary hormones raise cardiac output. Adrenal cortical hormones are not listed as regulators of cardiac activity.
Chapter-wise split of NEET 2025 Biology
- Molecular Basis of Inheritance7 Q
- Sexual Reproduction in Flowering Plants6 Q
- Human Health and Disease5 Q
- Plant Kingdom5 Q
- Human Reproduction5 Q
- Biomolecules4 Q
- Chemical Coordination and Integration4 Q
- Biotechnology and Its Applications4 Q
- Animal Kingdom4 Q
- Ecosystem3 Q
- Structural Organisation in Animals3 Q
- Organisms and Populations3 Q
- Microbes in Human Welfare3 Q
- Cell: The Unit of Life3 Q
- Biotechnology: Principles and Processes3 Q
- Photosynthesis in Higher Plants2 Q
- Plant Growth and Development2 Q
- Principles of Inheritance and Variation2 Q
- Morphology of Flowering Plants2 Q
- Biological Classification2 Q
- Biodiversity and Conservation2 Q
- Cell Cycle and Cell Division1 Q
- Reproductive Health1 Q
- The Living World1 Q
- Respiration in Plants1 Q
- Anatomy of Flowering Plants1 Q
- Evolution1 Q
- Breathing and Exchange of Gases1 Q
- Body Fluids and Circulation1 Q
About this paper
How many questions from NEET 2025 are on this page?
154 questions (Physics 31, Chemistry 41, Biology 82). Questions that need a diagram we could not reproduce, or whose wording in the source was unclear, are left out.
Are these the original questions?
Yes. The questions and options are as set in the NEET 2025 paper, conducted by NTA. The answer to every question was checked by the RankUp team, and the explanations are RankUp's own.
Is NEET 2025 useful for NEET 2027?
Yes. The NEET syllabus is built on the same NCERT books.
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