NEET-UG 2021 Question Paper Solutions

Displacement of particle executing

SHM y = Asin(ωt + ϕ) ...(i)

Potential energy U = 1/2 ky²

or, U = 1/2 kA² sin²(ωt + ϕ) ...(ii)

From eq. (i) and (ii),

Time period of y = Asin(ωt + ϕ)

T₁ = 2π/ω

∴ Frequency n₁ = ω/2π and time period of y² = A²sin²(ωt + ϕ) T₂ = π/ω

∴ Frequency n₂ = ω/π

∴ n₂ = 2n₁ = 2n

Potential energy varies periodically with double the frequency of SHM. It is maximum and equal to total energy at extreme position and minimum at mean position.

In polar molecules, like HCl → Hᵟ⁺ Clᵟ⁻ the centre of positive charges does not coincide with the centre of negative charges. Hence, permanent electric dipole exists.

Drift velocity, v Subscript (d)= I/neA = eEτ/m

Electrical resistivity, ρ = 1/σ = E/J (∵ σ = J/E)

Relaxation period, τ = m/(ne²ρ)

Current density, J = I/A = nev₍d₎

(A) - (iii), (B) - (iv), (C) - (i), (D) - (ii)

Relation between drift velocity, v₍d₎, electrical resistivity (ρ), relaxation period (τ) and current density (J)

v_d = i / neA = J/ne = σE/ne = E/ρne = V/ρlne ;

R = ρl/A = m/(ne²τ) · l/A

Charge density σ = charge / area When two conductors are connected by a conducting wire, then they have same potential i.e., V₁ = V₂ ∴ 1/(4πε₀) · Q₁/R₁ = 1/(4πε₀) · Q₂/R₂ ⇒ 1/(4πε₀) · Q₁/R₁ × R₁/R₁ = 1/(4πε₀) · Q₂/R₂ × R₂/R₂ ⇒ Q₁R₁ / (4πR₁²ε₀) = Q₂R₂ / (4πR₂²ε₀) ∴ [Q₁ / (4πR₁²)] / [Q₂ / (4πR₂²)] = R₂ / R₁ or, σ₁/σ₂ = R₂ / R₁

Energy density i.e. energy / volume with electric field [Plate Area = A] u = dU / dV = 1/2 ε₀E² ⇒ dU = 1/2 ε₀E² dV ∴ Energy stored in the capacitor U = ∫ dU = ∫ 1/2 ε₀E² dV = 1/2 ε₀E² ∫ dV = 1/2 ε₀E² V ∴ U = 1/2 ε₀E² Ad [∵ V = Ad]

Magnetic field due to a current carrying wire at a distance r from the wire. or, B = (10⁻⁷ × 2 × 5) / (20 × 10⁻²) = 1/2 × 10⁻⁵ B = (μ₀ / 4π) · (2I / r) = (4π × 10⁻⁷ × 2 × 5) / (4π × 20 × 10⁻²) or, B = (10⁻⁷ × 2 × 5) / (20 × 10⁻²) = 1/2 × 10⁻⁵ B upward to the plane of paper Therefore, force acting on electron Fvector = q(vvector × Bvector) |Fvector| = 1.6 × 10⁻¹⁹ × 10⁵ × 1/2 × 10⁻⁵ = 8 × 10⁻²⁰ N

Dimensional formula of energy E = 1/2 mv² [E] = [M¹ L² T⁻²] ...(i) Dimensional formula of gravitational constant G = Fr² / (m₁m₂) [G] = [M⁻¹ L³ T⁻²] ...(ii) From eq. (i) & (ii) ∴ E/G = [M¹ L² T⁻²] / [M⁻¹ L³ T⁻²] = [M² L⁻¹ T⁰]

Numerical constant like, ½, 1, 2 or 2π has no dimension.

A large focal length and large aperture of an objective of an astronomical telescope, ensures better light gathering power, and resolution so contributes to the quality and visibility of the images of the far-off objects. Resolving power of a telescope RP = a / (1.22λ) , a = aperture of the objective. And resolving limit of telescope dθ = 1.22λ / a

Root mean square RMS speed of gas molecule vsubscript(RMS) = √(3 RT / M) Pressure exerted by ideal gas P = 1/3 nmv̄² Average kinetic energy of a molecule = 3/2 ksubscript(B) T Total internal energy of a gas is (U) = 1/2 nfRT Here, for one mole n = 1 and f = 5 for diatomic ∴ U = 5/2 RT

• A zener diode is connected in reverse biased when used as voltage regulator as in reverse biased after breakdown, voltage becomes constant.

• Potential barrier of silicon diode is nearly 0.7V Statement A is correct and statement B is incorrect.

Potential energy of electric dipole in external electric field U = −Pvector · Evector = −PE cos θ = −PE cos 180° Since angle between Evector and Pvector is 180° ∴ U = +PE On moving towards right electric field strength decrease ∴ U decreases. Net force on electric dipole is towards right and net torque acting on it is zero so move towards right.

From the ray diagram light refracted from convex lens virtually meet at focus of concave lens ∴ d = f subscript A - f subscript B = 20 - 5 = 15 cm

Escape velocity from the earth's surface v subscript e = √[2GM / R] = √[ (2Gρ · 4/3 πR³) / R ] [ ∵ ρ = M/V ] or, v_e = √[ (8Gρπ / 3) R² ] or, v_e ∝ R (∵ same density) ∴ v / v₁ = R / 4R or, v₁ = 4v

Impedance Z = V_RMS / I_RMS V subscript (RMS) = √[ V subscript R² + (V subscript L − V subscript C)² ] = √[ (40)² + (40 − 10)² ] = 50 V I_RMS = I₀ / √2 = 10√2 / √2 = 10 A ∴ Z = V subscript RMS / I subscript RMS = 50 / 10 = 5 Ω

Let the height from the surface of the earth be x. When body from rest falls through height (S − x) Then from v² = u² + 2gs v² = 0² + 2g(S − x) ⇒ v = √[ 2g(S − x) ] ...(i) At this point, U = mgx From question, kinetic energy, K = 3U ∴ 1/2 m(v)² = 3 × mg(x) ⇒ 1/2 × m × 2g(S − x) = 3 × mgx (using (i)) ⇒ S − x = 3x ∴ x = S/4 ...(ii) ∴ v = √[ 2 × g(S − S/4) ] = √[ 3gS / 2 ]

Acceleration along inclined plane, a = g sin θ S subscript n = distance travelled by object during nᵗʰ second. Initial speed u = 0 S subscript n = u + a/2 (2n − 1) ⇒ S subscript n = 0 + (g sin θ)/2 (2n − 1) = (g sin θ)/2 (2n − 1) ...(i) Ssubscript (n+1) = 0 + (g sin θ)/2 [ 2(n + 1) − 1 ] = (g sin θ)/2 (2n + 1) ...(ii) Dividing equations (i) and (ii) S subscript n / S subscript (n+1) = [ (g sin θ)/2 (2n − 1) ] / [ (g sin θ)/2 (2n + 1) ] = (2n − 1) / (2n + 1)

activity of a radioactive substance A = A₀ (1/2)ᵗᐟᵀ₁ᐟ₂⇒ A/A₀ = (1/2)ᵗᐟᵀ₁ᐟ₂ Given, T₁/₂ = 100 hours & t = 150 hours ⇒ A/A₀ = (1/2)¹⁵⁰ᐟ¹⁰⁰ ⇒ A/A₀ = (1/2)³ᐟ² ∴ A/A₀ = 1 / 2√2 Activity of a radioactive substance is the rate of disintegration or the number of atoms of any material decaying per second. Activity ∝ 1 / half life

The circuit is as shown is figure, Points 1, 2, 3 are connected by conducting wire so these points are at same potential. So the capacitor is short circuited. So it does not store any charge. Therefore equivalent capacitance C subscript (AB) = C + C = 2C

Resistance of each wire in this case will be same as l & A are same and made of same material. ∴ R₁ = R₂ = R₃ = R₄ = R In parallel, 1/R subscript p = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄ ∴ R subscript p = R/4 ⇒ 0.25 = R/4 ∴ R = 1 Ω When arranged in series then equivalent resistance. R subscript s = R + R + R + R = 4R ∴ R subscript s = 4 × 1 = 4Ω

Gain in BE = (BE) of products – (BE) of reactants. = [120 + 120] × 8.5 – [240] × 7.6 = (240) × 8.5 – 240 × 7.6 = (2040 – 1824) MeV = 216 MeV

Power, P = E/t = n/t (hc/λ) ∴ No. of photons / second, n/t = P / (hc/λ) ⇒ n/t = (3.3 × 10⁻³ × 6 × 10⁻⁷) / (6.6 × 10⁻³⁴ × 3 × 10⁸) = 10¹⁶ The number of photons emitted per second from a source of monochromatic radiation of wavelength λ and power P and if E = energy of each photon then, n/t = P/E = P/hν = P / (hc/λ)

From Ampere's circuital law, magnetic field B = (μ₀I / 2πR²).r If r < R ⇒ B Subscript (inside) ∝ r B = μ₀I / 2πr If r ≥ R ⇒ B Subscript (outside) ∝ 1/r Hence the correct magnetic field B versus distance r graph is shown in the option.

Displacement current I Subscript (d) = C dV/dt = C d/dt(V₀ sin ωt) (∵ V = V₀ sin ωt given) = C(V₀ω)cos ωt ∴ I Subscript (d) = V₀ωCcos ωt

Let Fᵥ = viscous force and F Subscript (B) = Bouyant force acting on the ball. When body moves with constant velocity 𝐌𝐠 = 𝐅_𝐁 + 𝐅ᵥ [𝐚 = 𝟎] 𝐅ᵥ = 𝐌𝐠 − 𝐅 Subscript (B) = 𝐝𝐕𝐠 − 𝐝/𝟐 . 𝐕𝐠 (𝐌 = 𝐝𝐕𝐠) 𝐕 = volume of ball. or 𝐅ᵥ = 𝐝/𝟐 𝐕𝐠 = 𝐌/𝟐 𝐠 [∵ 𝐌 = 𝐝𝐕] Due to motion of a body through the fluid, the fluid in contact with the body is dragged with it causes relative motion in fluid layers near the body and hence viscous force starts.

From Snell's law at point P. sin i / sin r = μ Subscript (air) / μ Subscript (Prism) ⇒ sin 30° / sin e = 1 / √3 (∵ ∠r = ∠e emergent angle) ⇒ sin e = √3 · 1/2 ∴ ∠e = 60°

If a light ray is incident normally on first surface i.e., ∠i = 0° so refraction of first surface ∠r₁ = 0°. When angle of prism A > C (critical angle) then no ray emerges from second surface of the prism.

Energy, E ∝ Fᵃ Aᵇ Tᶜ [E] = [Fᵃ][Aᵇ][Tᶜ] ⇒ [ML²T⁻²] = [MLT⁻²]ᵃ [LT⁻²]ᵇ [T]ᶜ [ML²T⁻²] = [𝐌ᵃ𝐋ᵃ ⁺ ᵇ 𝐓⁻²ᵃ ⁻ ²ᵇ ⁺ ᶜ] Comparing dimensions on both sides. a = 1; a + b = 2 and −2 = −2a − 2b + c ∴ b = 1 and −2 = −2 − 2 + c or, c = 2 ∴ Dimensions of energy = [FAT²]

If ℓ₁ and ℓ₂ are balancing lengths then E₁ / E₂ = ℓ₁ / ℓ₂ ∴ ℓ₂ = ℓ₁ (E₂ / E₁) = 36 (2.5 / 1.5) = 60 cm

Spring constant k = F / x = 10 / (5 × 10⁻²) = 200 N / m And for spring-mass system undergoing SHM T = 2π √(m / k) = 2π √(2 / 200) = 2π / 10 = 0.628 s (because m = 2 kg given)

Direction of propagation along x-axis. (-ĵ + k̂) × (-ĵ - k̂) = 2î [NOTES] E vector × B vector is always in the direction of propagation of wave. The EM wave propagating in the positive x-direction E = E Subscript (y) = E₀ sin(kx - ωt) and B = B Subscript (z) = B₀ sin (kx - ωt)

From Einstein's photoelectric equation hc / λ = ϕ₀ + k Where ϕ₀ = work function k = maximum kinetic energy of photoelectrons ∴ hc / λ = k = P² / 2m ⇒ P = √(2mhc / λ) [∵ ϕ = negligible] de-Broglie wavelength, λ Subscript d = h / P = h / √(2mhc / λ) ⇒ √λ = λ Subscript d √(2mc / h) ∴ λ = (2mc / h) λ Subscript d²

Power on turbine P = d(mgh) / dt = gh (dm / dt) = 10 × 60 × 15 = 9000 W Due to frictional force, lossses are 10% ∴ Power generated = (1 - 10 / 100) × 9000 = 8.1 kW

Let charge and radius of smaller drop is q and r respectively and R = radius of bigger drop For smaller drop, electric potential, V = kq / r = 220 V As volume remains the same (4/3 πr³) × 27 = 4/3 πR³ ⇒ R = ∛(27r) = 3r And from conservation of charge, Q = 27q ∴ Potential of bigger drop, V Subscript (big drop) = kQ / R = k(27q) / 3r = 9(kq / r) = 9 × 220 = 1980 V

For first refraction from convex lens v₁ = ?, u = -60 cm, f = 30 cm From lens formula, 1/v₁ - 1/u = 1/f ⇒ 1/v₁ + 1/60 = 1/30 ⇒ v₁ = 60 cm v₁ = first image by lens Image produced by plane mirror at distance 20 cm left of it. Therefore for second refraction from convex lens, u = -20 cm, v = ?, f = 30 cm 1/v - 1/u = 1/f ⇒ 1/v + 1/20 = 1/30 ⇒ 1/v = 1/30 - 1/20 ⇒ v = -60 cm Hence the final image which is virtual formed at a distance, 60 - 40 = 20 cm from plane mirror.

Lorentz force, F vector = q (v vector × B vector) = q v vector × (B î + B ĵ + B₀ k̂) ∴ q = 1, v vector = 2î + 4ĵ + 6k̂ and F vector = 4î - 20ĵ + 12k̂ (given) ∴ (4î - 20ĵ + 12k̂) = 1 × [(2î + 4ĵ + 6k̂) × (B î + B ĵ + B₀ k̂)] or (4î - 20ĵ + 12k̂) = î (4B₀ - 6B) - ĵ (2B₀ - 6B) + k̂ (2B - 4B) Comparing L.H.S and R.H.S, we get 4B₀ - 6B = 4 ⇒ 2B₀ - 3B = 2 ...(i) -(2B₀ - 6B) = -20 ⇒ B₀ - 3B = 10 ...(ii) 2B - 4B = 12 ⇒ B = -6 ...(iii) From eq. (ii) and (iii) B = -6 and B₀ = -8 ∴ Complete expression for B vector = -6î - 6ĵ - 8k̂ [NOTES] In a magnetic field a moving charged particle experiences magnetic force F Subscript (m) = q (v vector × B vector) F Subscript (m) = q (v vector × B vector) is called Lorentz force.

Let i be the current in outer loop Magnetic field at centre B = μ₀i / 2R₁ Magnetic flux through inner coil ϕ Subscript (i) = B × πR₂² ϕ Subscript (i) = μ₀i / 2R₁ × πR₂² ϕ Subscript (i) = μ₀i / 2 × πR₂² / R₁ ∴ ϕ = Mi ∴ M = (μ₀π / 2) R₂² / R₁ or, M ∝ R₂² / R₁

Boolean expression for combination of logic gates Output, y = A . B + Bar (B . C)

Output y is high (1) so output at the terminal y, v₀ = 5V

As per question, 90° arc is removed from circular ring, ∴ Mass removed = M / 4 ∴ Mass remaining M - M/4 = 3/4 M

Moment of inertia of remaining I_remaining = ∫ dmr² = R² ∫ dm (∵ r = R)

∴ I Subscript (remaining) = 3MR² / 4 = KMR² ∴ K = 3 / 4 [NOTES] Moment of inertia of a continuous distribution of mass, treating the element of mass dm at position r as particle then dI = dmr² so I = ∫ r²dm

Resistance r₂ and r₃ are in parallel so potential difference across r₂ = r₃, = V ∴ i₂r₂ = i₃r₃ ⇒ i₂ = i₃r₃ / r₂ ...(i) Current i₁ = i₂ + i₃ ⇒ i₁ = (r₃ / r₂ + 1) i₃ ∴ i₃ / i₁ = r₂ / (r₂ + r₃)

Velocity of car at t = 4 s = velocity of the ball dropped out of a window v = u + at ⇒ v = 0 + 5 × 4 = 20 ms⁻¹ ∴ Velocity of ball along the horizontal and is 20ms⁻¹. After 2 seconds of motion: Vertical velocity of ball (v Subscript (y)) = u Subscript (y) + a Subscript (y) t v Subscript (y)= 0 + 10 × 2 = 20 ms⁻¹ (∵ a Subscript (y) = g = 10 m/s²) ∴ Magnitude of velocity of ball (v) = √(v Subscript (x²) + v Subscript (y²)) = 20√2 m/s Since the ball is under free fall. Acceleration of ball (a) at t = 6s a = g = 10 m/s² [NOTES] All bodies, in the absence of air resistance, irrespective of the size, weight or composition fall with same acceleration near the surface of the earth is called free fall.

Power P = V × I For an ideal transformer Input power = Output power VₚIₚ = VₛIₛ or, 220 × Iₚ = 44 ∴ Iₚ = 0.2 A

Impulse, 𝐈⃗ = change in linear momentum or, I = Δ𝐏⃗ = 𝐏⃗ Subscript (𝐟) − 𝐏⃗ Subscript (i) or, I = m₂v₂ − m₁v₁ Velocity of ball at ground (v) = √(2gh) [∵ v² − u² = 2gh] = √(2 × 10 × 10) = 10√2 m/s ∴ 𝐈⃗ = 0.15 × 10√2 (−𝒋̂) − 0.15 × 10√2 (𝒋̂) ⇒ 𝐈⃗ = 2 × 0.15 × 10√2 (−𝒋̂) = 4.2 (𝒋̂) = 4.2 kg m/s [NOTES] Impulse I = Δ𝐏⃗ = 𝐏⃗ Subscript (𝐟) − 𝐏⃗ Subscript (i) = change in momentum is called impulse momentum theorem. Impulse of a force is a measure of total effect of force.

For LCR series circuit resonance frequency ω₀ = 1 / √(LC) = 1 / √(5 × 80 × 10⁻⁶) = 50 rad / s ∴ Half power frequency, ω = ω₀ ± R / 2L i.e. ω Subscript L = 50 - 40 / (2 × 5) = 46 rad / s and ω Subscript H = 50 + 40 / (2 × 5) = 54 rad / s

Rod will be in equilibrium, if 𝛕⃗ Subscrip𝐭 (net) = 0 about point O. Torque at point O due to 2 kg mass 𝛕⃗ = 𝐫⃗ × 𝐅⃗ = rF sin θ (𝐧̂) 𝛕 ₁ = 20 × 20 × 10⁻² × sin 90° (𝐤̂) = 4 Nm (𝐤̂) Torque due to mass of rod 𝛕 ₂ = 5 × 60 × 10⁻² × sin 90° (−𝐤̂) = 3 Nm (−𝐤̂) Torque due to mass m 𝛕 ₃ = mg × 120 × 10⁻² × sin 90° (−𝐤̂) = 12m Nm (−𝐤̂) So 𝛕⃗ ₁ + 𝛕⃗₂ + 𝛕⃗ ₃ = 0 (∵ 𝛕⃗ Subscrip𝐭 (net) = 0) ⇒ 4 − 3 − 12m = 0 ⇒ 12m = 1 ∴ m = 1 / 12 kg

Current in the loop I = V / R equal for both loops. Magnetic moment of equilateral triangle of side a M Subscript (T) = (12a / 3a) · I · (√3 / 4) a² = √3 I a² [∵ M = NIA] Similarly magnetic moment of square of side a = (12a / 4a) · I · a² [∵ M = NIA] M Subscript (S) = 3 I a² [NOTES] A current carrying coil behaves as a bar magnet and its magnetic moment M = NiA, N = no. of turns, i = current through the coil, A = area of the coil. For a given perimeter circular shape has maximum area hence maximum magnetic moment.

Number of octahedral and tetrahedral voids formed by N closed packed atoms are N and 2N respectively. Hexagonal unit cell contains 6 atoms therefore, number of tetrahedral void =N=6x2=12 Number of octahedral void = N = 6

At constant volume, 𝐪 ᵥ = 𝐂 ᵥ Δ𝐓 = Δ𝐄 ...(i) At constant pressure,

𝐪 ₚ = 𝐂 Subscript (p) Δ𝐓 = Δ𝐇 ...(ii) For a mole of an ideal gas, Δ𝐇 = Δ𝐄 + Δ(𝐏𝐕) = Δ𝐄 + 𝐑Δ𝐓 ...(iii) On putting the values of Δ𝐇 and Δ𝐄 in eqn. (iii), we get 𝐂ₚ Δ𝐓 = 𝐂 ᵥ Δ𝐓 + 𝐑Δ𝐓 𝐂ₚ = 𝐂ᵥ + 𝐑 𝐂ₚ − 𝐂ᵥ = 𝐑

In vapour phase Beryllium chloride tends to form a chloro-bridged dimer.

[NOTES] In the solid, BeCl₂ molecules polymerise to make long chains. They do this by forming coordinate bonds between one pairs on Cl and adjacent Be atoms.

According to Saytzeff's rule, in dehydrohalogenation reactions, highly substituted alkene is the major product.

In BF₃, hybridisation is sp² and number of electrons are 6.

2-methylbutan-2-ol

According to Kohlrausch law of independent migration of ions. Λ° ₘ (CH₃COOH) = Λ°ₘ (CH₃COONa) + Λ° ₘ (HCl) − Λ° ₘ(NaCl) = 91.0 S cm² mol⁻¹ + 426.16 S cm² mol⁻¹ − 126.45 S cm² mol⁻¹ = 390.71 S cm² mol⁻¹

Most of the trivalent lanthanoids ions are coloured both in the solid state and in aqueous solutions. [NOTES] Actinoid contraction is greater from element to element than lanthanoid contraction due to the poor shielding by 5f electrons.

In the modern periodic table, moving down the group as the size of halogen atom increases, the H − X bond length also increases as a result the bond enthalpy decreases. Hence, The acidic strength also increases.

2,6-Dimethyl-dec-4-ene:

The size of halogen atom increases from F to I hence bond length from C - F to C - I increases. ∴ Bond enthalpy from CH₃ - F to CH₃ - I decreases.

Given ν = 1,368 kHz = 1368 × 10³ Hz Wavelength = c / ν = 3 × 10⁸ / (1368 × 10³) = 219.3 m

Dimethylammonium acetate is a salt of weak acid and weak base whose pH can be calculated as pH = 7 + 1/2 (pK ₐ− pK Subscript (b)) = 7 + 1/2 (4.77 − 3.27) = 7.75

Order of molar mass of solute decreases as Sucrose > Glucose > Urea So, correct order of osmotic pressure of solution is P₃ < P₁ < P₂

[NOTE] Osmotic pressure (π) = iCRT where C is molar concentration of the solution. With increase in molar concentration of solution osmotic pressure increases. Since, weight of all solutes and its solution volume are equal, so higher will be the molar mass of solute, smaller will be molar concentration and smaller will be the osmotic pressure.

ΔHᵣₓₙ = (Eₐ)f − (Eₐ)b −4.2 = 9.6 − (Eₐ)b (Eₐ)b = 9.6 + 4.2 = 13.8 kJ mol⁻¹ Since reaction is exothermic, so possible graph is (c) only.

Compounds with formula C₄H₁₀O can be ethers which may exhibit metamerism. For example CH₃–CH₂–O–CH₂–CH₃ CH₃–O–CH(CH₃)–CH₃ CH₃–O–CH₂–CH₂–CH₃ are metamers.

The correct option is (CH₃)₂CH–CH₂–CH₂Br

Reactions(a)and (b) are decomposition reactions. Reaction (d) s displacement reaction, while reaction (c) is an example of metal displacement reaction.

EDTA ion is a hexadentate ligand with four oxygen and two nitrogen donor atoms.

The reaction of Hinsberg's reagent with primary amine (CH₃CH₂NH₂) yields N-ethylbenzene sulphonamide which is soluble in alkali. Benzenesulphonyl chloride (C₆H₅SO₂Cl) is also known as Hinsberg's reagent. Primary and Secondary amines reacts to give sulfonamide derivatives. Whereas tertiary amines reacts to produce quaternary salt which is a intermediate but this rapidly breakdowns to original tertiary amine.

Eclipsed conformer is least stable while staggered conformer is most stable. In eclipsed conformer the dihedral angle is 0°.

NOTES:

The C–C bond is not completely free to rotate. There is a small 3 kcal/mol barrier to rotation that must be overcome for the bond to rotate. This rotational barrier is not high enough to prevent constant rotation except at extremely cold temperature.

Possible empirical formula is CH₃.

Noble gases have weak dispersion forces hence they have low melting and boiling points.

Teflon is an addition polymer. Monomer of teflon is tetrafluoroethene.

Notes; Dacron, Nylon-66 and Novolac are prepared by condensation polymerisation.

A-(iv), B-(iii), C-(i), D-(ii)

For a spontaneous process, ΔSₜₒₜₐₗ > 0 and since irreversible process is always spontaneous, therefore ΔSₜₒₜₐₗ > 0. Since ΔU = nCᵥΔT and ΔT = 0 for isothermal process, therefore ΔU = 0.

Magnetic moment, μ = √[n(n + 2)] BM

NaBH₄ is a reducing agent. It reduces carbonyl group into alcohols but does not reduce esters. NaBH₄ is a selective reducing agent for carbonyl group while LiAlH₄ can reduce both carbonyl group and ester group.

On moving down the group, bond dissociation enthalpy of hydrides of Group 16 elements decreases; hence acidity increases and pKₐ value decreases. Correct order of pKₐ value will be H₂O > H₂S > H₂Se > H₂Te

CH₃CH₂COO⁻Na⁺ ⟶(NaOH + CaO, Heat) CH₃CH₃ + Na₂CO₃

Note: Alkane is produced by heating sodium salt of carboxylic acid with sodalime. The ratio of NaOH : CaO = 3 : 1.

SbCl₅ : Net vector summation of bond moments will be zero so SbCl₅ is a non-polar molecule.

• NO₂ ⇒ polar molecule.

• POCl₃ ⇒ polar molecule.

• CH₂O ⇒ polar molecule.

A - (iv), B - (iii), C - (i), D - (ii) A chain reaction occurs from interaction of NO with sunlight in which NO is converted to NO₂ which absorb energy from sunlight and breaks into NO and O, which causes photochemical smog. NO₂(g) ──hv──> NO(g) + O(g)

ⁿO₂ = 4/32 = 1/8 ⁿH₂ = 2/2 = 1 nₜₒₜₐₗ = 1/8 + 1 = 9/8 PₜₒₜₐₗV = nₜₒₜₐₗRT

P₁ = (9/8 × 0.082 × 273)/1 = 25.18 atm

Given : n subscript (benzene) : n subscript (octane) = 3 : 2 So, X Subscript (benzene) = 3/5 X Subscript (octane) = 2/5 Pₛ = P° Subscript (benzene) X Subscript (benzene) + P° Subscript (octane) X Subscript (octane) = 280 × (3/5) + 420 × (2/5) = 168 + 168 = 336 mm of Hg

Species Number of electrons

Fe²⁺ 26 − 2 = 24

Mn²⁺ 25 − 2 = 23

O²⁻ 8 + 2 = 10

F⁻ 9 + 1 = 10

Na⁺ 11 − 1 = 10

Mg²⁺ 12 − 2 = 10

Fe³⁺ 26 − 3 = 23

Etard’s reaction

Reagent R is C₂H₅OH with diazonium salt. Ar–N₂⁺X⁻ + RCH₂–OH → Ar–O–CH₂R + N₂ + HX (A) Ar–N₂⁺X⁻ + RCH₂–OH ⇌ Ar–N=N–O–CH₂R Ar–N=N–O–CH₂R → ArH + RCHO + N₂ + HX (B) Where process A proceeds through an S Subscript (N) 2 displacement of the diazonium group, while process B involves homolytic fission of the bonds to produce radicals which ultimately lead to aromatic hydrocarbon and aldehyde.

Arrhenius equation k = Ae⁻ᴱᵃ/ᴿᵀ ln k = ln A + ln e⁻ᴱᵃ/ᴿᵀ ln k = ln A − (Eₐ/R)(1/T)  ...(1) Slope of ln k vs 1/T curve, m = −Eₐ/R −5 × 10³ = −Eₐ/R Eₐ = 5 × 10³ × 8.314 J mol⁻¹ = 41.57 × 10³ J mol⁻¹ = 41.5 kJ mol⁻¹

A - (ii), B - (iii), C - (iv), D - (i) Gattermann Koch reaction is inapplicable to phenol substrate.

Pyramid of biomass in sea is inverted because the amount of biomass is least at the base of the pyramid and the amount of biomass is maximum at the apex of the pyramid. For example, biomass of zooplanktons is higher than that of phytoplanktons because zooplanktons have a low weight and are smaller in size. Phytoplankton has a shorter life span when compared to zooplankton. An ecological pyramid shows the relationship of biomass, productivity or energy at different trophic level. There are three types of Ecological Pyramid. • Pyramid of number – It can be upright or inverted. • Pyramid of biomass – It can be upright or inverted. • Pyramid of energy – It is always upright.

The term monoecious refers to plants that bear male and female sex organ on same plant body. This feature can be seen in Chara. A dioecious plant is one where the male and female sex organs occur on separate plant body. Cycas circinalis, Carica papaya and Marchantia polymorpha are dioecious.

Most of the plants that are adapted to dry tropical regions have the C₄ pathway. These plants are called C₄ plants. Sugarcane, sorghum, maize etc. are examples of these plants. The first stable product of CO₂ fixation in sorghum is oxaloacetic acid. The first stable product of C₃ cycle is 3-phosphoglyceric acid.Pyruvic acid is the end product of glycolysis. Succinic acid is an intermediate product in Krebs cycle.

In diadelphous condition, stamens are arranged in two bundles. For example, pea. Monoadelphous stamens are grouped in single bundle whereas polyadelphous stamens occur in more than two bundles. China rose has monoadelphous stamens while citrus has polyadelphous stamens. The filaments are united forming two bundles e.g. in Lathyrus (sweet pea), Pisum (edible pea) there are 10 stamens; filaments of 9 stamens are fused forming one bundle and the other bundle is represented by the filament of the tenth free stamen (Ai + (9)).

After the bands are stained, they.are viewed in UV light. The bands appear bright orange in colour. Ethidium brofnide is the intercalating agent that stacks in between the nitrogenous bases.

The ability of an organism to change its structure in response to different environments is termed as plasticity. It is a particularly important characteristic to enable sessile plants to adapt to rapid changes in their surroundings. A special case of plasticity is represented by heterophylly, the ability of semi-aquatic plants to produce different types of leaves below and above water.

The overall productivity of a system can be found in an equation where the Net Primary Production (NPP) is equal to the Gross Primary Production (GPP) minus the Respiration loss (R). Formula: NPP = GPP − R The NPP is the overall efficiency of the plants in the ecosystem.Through having a constant efficiency in NPP, the ecosystem is more sustainable. The GPP refers to the rate of energy stored by photosynthesis in plants. The R refers to the maintenance and reproduction of plants from the energy expended.

The production of gametes (n) by the parents (2n), the formation of the zygote (2n), the F₁ and F₂ plants can be understood from a diagram called Punnett square. It is used to predict the genotypes of a particular cross or breeding experiment. It is named after Reginald C.

Mature sieve tube elements are deep , tubular and do not contain nucleus in the cells with no secondary wall and terminal sieve plate. They lack nucleus at the maturity so that they are associated with companion cells for support. Their features are ' the same as prokaryotic cells.Rest all statements are correct.

The amount of all the inorganic substances or nutrients such as carbon, nitrogen, phosphorus and calcium present in soil at any given time is referred as standing state. ‘Each trophic level has a certain mass of living material at a particular time called the standing crop.It is measured as the mass of living organism (biomass) or the number in a unit area. ‘Climax community is the last community in biotic succession which is relatively stable and is in near equilibrium with the environment of that area.

Xenogamy refers to the transfer of pollen grains from anthers of one plant to stigma of a different plant which, during pollination, brings genetically different types of pollen grains to stigma. Cleistogamy is a type of automatic self-pollination of certain plants that can propagate by using non opening, self-pollinating flowers. Geitonogamy refers to the transfer of pollen grain from anther to stigma of another flower of the same plant. Chasmogamy is a condition in which flowers remain open.

• Division of centromere occurs in anaphase II.

• Telophase II is the last stage of meiosis II.

• During this phase, the chromatids reach the poles and start uncoiling.

• Chromosomes form two parallel plates in metaphase I and one plate in metaphase II.

Polymerase chain reaction or PCR consists of the following three steps: Denaturation- The two DNA strands of template DNA separate from each other when heated to 94°C. Annealing- The primers anneal to the 3’ end of single strands of DNA. Extension- The primers are extended by DNA polymerase by the addition of nucleotides to form complete strands of DNA. Hence the sequence of steps is denaturation, annealing, extension.

Gene therapy is a technique that modifies a person’s genes to treat or cure disease. Biopiracy is the term used to refer to the use of bioresources by multinational companies and other organisation without proper authorisation from the countries and people concerned without compensatory payment. Bioresources by multinational companies and other organisations without proper authorisation from the countries and people concerned without compensatory payment. Molecular diagnosis refers to the act or process of determining the nature and cause of disease.

The inner membrane of mitochondria forms infoldings called cristae. Thylakoids are flattened membranous sacs in stroma of plastids. Cisternae are disc shaped sacs in Golgi apparatus. Primary constriction in chromosome that holds two chromatids together is called centromere.

Genetic drift is a random process that can lead to large changes in populations over a short period of time. Random drift is caused by recurring small population sizes, severe reductions in population size called “bottlenecks” and founder events where a new population starts from a small number of individuals. Crossing over which occurs during gamete formation results in genetic recombination. Mutations are random and directionless.

Chromosomes are classified into four types based on the position of the centromere. These are metacentric, sub-metacentric, telocentric and acrocentric. A chromosome having centromere at the center is called as metacentric chromosome. In this, centromere divides chromosome into two equal arms. A chromosome having centromere slightly away from the center is called as sub-metacentric chromosome. A chromosome having centromere close to one of the end is called as acrocentric chromosome. A chromosome having a terminal centromere is called as telocentric chromosome.

The gemmae are green, small discs of haploid tissue, and they directly give rise to new gametophytes. They are produced by some liverworts like marchantia. Mosses reproduce vegetatively by fragmentation and budding of protonema. Pteridophytes and Gymnosperms normally do not reproduce asexually.

Various enzymes like protease, RNase, etc. are added to break down substances like proteins, RNA, etc. Once all these substances are broken down, DNA is left which is precipitated out by adding chilled ethanol. Histones are basic proteins that help condense DNA in a cell.

Cohesion is mutual attraction among water molecules.

Adhesion is attraction towards polar surfaces.

Surface tention explains water molecules are more attracted in

liquid phase than gaseous phase. Guttation is loss of water is liquid form the leaf margins

Photoperiodism refers to the responses related to the lengths of dark and light periods. The site of perception of photoperiod is leaves. The hormone florigen, which is responsible for flowering, induces flowering as it migrates from leaves to shoot apices on induction of required photoperiods. The site of perception of low temperature stimulus during vernalisation is shoot apex and embryo. Axillary bud are not sites of perception of photoperiod.

The compound 2,4-dichlorophenoxyacetic acid, or 2,4-D, is a weedicide used to remove broad leaved weeds or dicotyledonous weeds in cereal crops or monocotyledonous plants. IAA and IBA are natural auxins. NAA is a synthetic auxin.

Algae are eukaryotic organisms with different chemical components in their cell wall. The cell wall of red algae is composed of agar, carrageen and funori along with cellulose.

Cell wall of brown algae contains algin while in green algae it is composed of cellulose and pectin.

In blue green algae cell wall is composed of mucopeptides.

Carrageen are a family of natural linear sulphated polysaccharide that are extracted from red edible sea weeds.

DNA replication is the process by which DNA makes a copy of itself during cell division. The process by which DNA is copied to RNA is called transcription, and that by which RNA is used to produce proteins is called translation. Transduction is transfer of genetic material from one bacterium to another with the help of virus or a bacteriophages.

Interspecific competition is the competition for resources by individuals of different species. Inspite of interspecific competition the competing species may co-exist by doing resource partitioning. • Mutualism is defined as an interaction between individuals of different species that results in positive (beneficial) effects on per capita reproduction and/or survival of the interacting populations. • Predation is a biological interaction where one organism, the predator, kills and eats another organism, its prey. • In competitive release there occurs dramatic increase in population of a less distributed species when its superior competitor is removed.

PCR is Polymerase Chain Reaction. It is used for making multiple copies of the gene. Hence PCR is used for • Gene amplification. • PCR-based assays have been developed that detect the presence of gene sequences of the infectious agents. • It is also used in detecting mutations. • Protein is not the target of PCR. Hence, plays no role in its purification.

Meristematic tissues are those tissues which have cells with active cell division capacity. Simple tissues are those tissues which have all the cells similar in structure and function. Vascular tissues are complex permanent tissues hence they have different types of cells. Sclereids are sclerenchymatous cells which are dead with highly thickened walls and narrow lumen.

The embryo sac of angiosperms consists of seven cells. A typical angiosperm embryo at maturity is called 7 celled 8 nucleates. The cells are arranged in three sets. In the centre, there are two polar nuclei which are a part of the large central cell. The polar nuclei are present at the centre of the cell. The polar nuclei are not separated by a membrane. There are two ends of the yolk sac. One of the ends is called the chalazal end. This end has 3 antipodal cells.

Ectocarpus belongs to class-Phaeophyceae, in which reserve food is found in form of laminarin, mannitol and oil. Ulothrix and Volvox belong to chlorophyceae (green algae). Members of this class have starch as reserve food material. Gracilaria is a member of red algae (Rhodophyceae). This class is characterised by having Floridean starch as storeg. food material.

Genera like Selaginella and Salvinia which produce two kinds of spores i.e. macro (large) and micro (small) spores, are known as heterosporous. Most of the pteridophytes produce single type of spores and are called homosporous. Sorus are brownish or yellowish cluster of spores producing structures located on the lower surface of fern leaves. Selaginella and Salvinia are considered the first plants to exhibit a significant step towards the evolution of seed habit because embryo develops in female gametophyte which is retained on parent sporophyte.

Amensalism refers to association wherein one partner is inhibited while other remains nearly unaffected and is denoted by − 0. Commensalism refers to relationship wherein one organism is benefitted while other remains unaffected and is denoted by + 0. Mutualism is association of organisms wherein both are benefitted and is denoted by + +. Association of organisms which benefits one of the partners at the expense of other is called as parasitism and is denoted by + −. A prime example of amensalism is penicillin killing bacteria. The bread mould Penicillium secretes penicillin that ultimately kills bacteria. In this contact between two organisms, one is destroyed or inhibited, and other remains unaffected.

In eukaryote cells, RNA polymerase III (also called Pol III) transcribes DNA to synthesize ribosomal 5S rRNA, tRNA and other small RNAs. The genes transcribed by RNA Pol III fall in the category of "housekeeping" genes whose expression is required in all cell types and most environmental conditions. Therefore, the regulation of Pol III transcription is primarily tied to the regulation of cell growth and the cell cycle, thus requiring fewer regulatory proteins than RNA polymerase II.

The correct pair is when the cells of medullary rays differentiated they give rise to the new cambium called interfascicular cambium. Other pairs are incorrect and can be corrected as. Loose parenchyma cells rupturing the epidermis and forming a lens shaped opening in bark are called complementary cells. Large colorless empty cells in the epidermis of grass leaves are called Bulliform cells. In dicot leaves, vascular bundles are surrounded by large thick walled cells called bundle sheath cells.

pBR322 is a commonly used cloning vector. When the gene for β-galactoside is inserted in the ampicillin resistance gene by using Pst I, the recombinant E.coli will lose ampicillin resistance due to insertional inactivation of the antibiotic resistance gene. The host (recombinant) cell will produce β-galactoside which is not a novel protein nor does it have dual ability. The transformed cells cannot resist ampicillin as they have lost ampicillin resistance. A recombinant E. coli is produced and the host cell will not undergo lysis due to insertion of β-galactoside gene. pBR322 is a plasmid and was one of the first widely used E. coli cloning vectors. Created in 1977 in the laboratory of Herbert Boyer at the University of California, San Francisco, it was named after Francisco Bolivar Zapata and Raymond L. Rodriguez. The p stands for "plasmid" and BR for "Bolivar" and "Rodriguez". pBR322 is 4361 base pairs in length and has two antibiotic resistance genes - the gene bla encoding the ampicillin resistance (Ampᴿ) protein, and the gene tetA encoding the tetracycline resistance (Tetᴿ) protein.

DNA fingerprinting involves identifying differences in some specific regions in DNA sequence called as repetitive DNA, because in these sequences, a small stretch of DNA is repeated many times. These sequences normally do not code for any proteins, but they form a large portion of human genome. The basis of DNA fingerprinting is VNTR (Variable number tandem repeats) a satellite DNA as a probe that show very high degree of polymorphism. Polymorphism is the variation at genetic level. Allelic sequence variation has traditionally been described as a DNA Polymorphism.

The integral equation for exponential growth is Nₜ = N₀eʳᵗ. Where Nₜ = Population density after time t, N₀ = Population density at time zero, r = intrinsic rate of natural increase, e = the base of natural logarithms (2.71828)

Statement 1 is incorrect because cyclic photophosphorylation involves only PSI. Both PSI and PSII are involved in non cyclic photophosphorylation where both ATP and NADPH + H⁺ are synthesised. Both PSI and PSII are found on grana lamellae whereas stroma lamellae have PSI only and lack NADP reductase.

In some blue-green algae specialised cells called heterocyst fixes atmospheric nitrogen into ammonia.

• Fusion of two nuclei is called Karyogamy.

• Organisms that depend on living plants are parasites, saprophytes grow on dead material.

• Fusion of protoplasts of two cells is called plasmogamy.

During respiration, process of ATP synthesis is explained by chemiosmotic model. It says that a proton gradient is required for ATP synthesis that is established by oxidation-reduction reactions.

In ETC, one NADH + H⁺ produces 3 ATP while one FADH₂ produces 2 ATP molecules.

ATP is synthesised via complex V.

In ETS, oxygen acts as terminal electron acceptor.

In members of some plant families like Solanaceae, Rosaceae and Leguminosae the pollen grains retain their viability for several months.

In cereals (Poaceae) pollen grains retain viability for around 30 minutes.

Split gene arrangement is characteristic of eukaryotes.

In capping 5-methyl guanosine triphosphate is added at 5′ end of hn RNA.

At 3′ end poly-A tail is added.

The non-coding or template strand is copied to an mRNA.

RNA polymerase associate with rho factor and it alters the specificity of the RNA polymerase to terminate the processes.

The floral formula of

In a polypeptide or a protein, amino acids are linked by a peptide bond which is formed when the carboxyl (–COOH) group of one amino acid reacts with amino (–NH₂) group of the next amino acid with the elimination of a water moiety.

Unsaturated fatty acids are with one or more C = C double bonds.

In nucleic acids, a phosphate moiety links the 3′-carbon of one sugar of one nucleotide to the 5′-carbon of the sugar of the succeeding nucleotide. The bond between the phosphate and hydroxyl group is an ester bond. As there is one such ester bond on either side, it is called phosphodiester bond.

In a polysaccharide, the individual monosaccharides are linked by a glycosidic bond.

• In S phase DNA replication takes place.

• In G₂ phase there is synthesis of proteins, RNA etc.

• Quiescent stage is inactive stage of cell cycle but cells remain metabolically active in this stage.

• G₁ phase is the interval between mitosis and initiation of DNA replication.

Autoradiography allows the detection/localisation of radioactive isotope within a biological sample. Probe is a radiolabelled ss DNA or ss RNA depending on the technique.

To identify the mutated gene probe is allowed to hybridise to its complementary DNA in a clone of cells followed by detection using autoradiography. The mutated gene will not appear on the photographic film, because the probe does not have complementarity with the mutated gene.

The juxtaglomerular cells that are present in the kidney produce a specific peptide hormone, erythropoetin. This hormone plays a important role in the stimulation of RBC formation. Glucagon is produced by alpha cells of pancreas, formed elements are synthesised from the cells of bone marrow and the cells of rostral adenohypophysis are involved in the formation of hormones of the anterior lobe of the pituitary gland.

A ring of gastric caecae is located at the junction of foregut and midgut so option 2 is incorrect. At the junction of midgut and hindgut a ring of 100-150 thin filaments are present and called as Malpighian tubule.

The secretion of brush border cells of the mucosa and the goblet cells constitute the intestinal juice that is also called as succus entericus. Different type of enzymes such as maltase, lipase, dipeptidases etc are present in this juice.

When food is mixed with the acidic juice of the stomach then called as chyme. Pancreatic juice and gastric juice are the secretion of pancreas and the stomach respectively.

During the process of meiosis I an X shaped structure is formed that is called as chiasmata. The formation of the chiasmata takes place in the Diplotene stage of the meiosis I while its termination occurs in the stage of Diakinesis. In zygotene stage the formation of bivalents occurs while in pachytene stage crossing over takes place. Chromosomal condensation occurs during the Leptotene stage of meiotic prophase.

The partial pressure of O₂ and CO₂ in alveoli are 104 mmHg and 40 mmHg respectively. So the correct option is 2) pO₂ = 104 and pCO₂ = 40.

159 mmHg and 0.3 mmHg are the partial pressure of O₂ and CO₂ respectively at atmospheric level. In the case of deoxygenated blood the partial pressures of O₂ and CO₂ are 40 mmHg and 45 mmHg.

Oxygenated blood has 95 mmHg of pO₂ and 40 mmHg of pCO₂.

Physalia is a marine hydrozoans and is generally found in the atlantic ocean and the Indian ocean. It is also known as Portuguese man of war or blue bottle jelly fish. Limulus is known as king crab and is considered as living fossil. Pinctata is named as pearl oyster and the Ancylostoma is called hookworm.

Denaturation of the DNA is a process in which the separation of double stranded DNA into single strands takes place. For this process high temperature is needed. So when high temperature is not maintained during PCR then denaturation process is affected.

DNA Ligase is responsible for the ligation of the DNA fragments.

Annealing and the extension are the second and third steps of PCR.

Aspergillus niger is a fungus that is used for the production of citric acid. Acetic acid is produced with the help of Acetobacter aceti. Clostridium butylicum and lactobacillus are involved in the production of butyric acid and lactic acid respectively.

Intercalated disc is a type of junction that is present only in the cardiac muscle tissue so this is not a characteristic of smooth muscle. Thus option (d) has incorrect statement.

The smooth muscle fibres have no striations and are involuntary in nature. They are generally present in the wall of internal organs like stomach, intestine and the blood vessels.

Myasthenia gravis is a type of chronic auto immune disorder that affects the neuromuscular junction and causes weakness and paralysis of the skeletal muscle.

Muscular dystrophy is a type of genetic disorder that leads to the progressive degeneration of skeletal muscle.

When uric acid crystal is deposited in the joints and causes inflammation is called as Gout disease.

Blood group AB have antigen A and antigen B and do not contain antibodies A and B in the plasma. So the person who has blood group AB can receive the blood from persons with any type of blood group. Thus person with AB blood group is called as Universal recipients.

S phase of the cell cycle is a synthesis phase in which amount of DNA increases but the number of chromosomes remain constant. So the number of chromosomes will be 8 in S phase because the number of chromosome is 8 at the time of G₁ phase.

In mammals during the process of fertilisation sperm interacts with thick extracellular coat of an egg that is called as the zona pellucida receptors are present. The space that is present between the vitelline membrane and the zona pellucida is called as Perivitelline space. Corona radiata is present in the brain. It is a layer that is made up of radially arranged cells of membrane

Muscidae are a family of flies such as house flies. Firefly belongs to the family Lampyridae, grasshopper is present in the family of Acrididae and cockroach is kept in Blattidae family.

Ricin is a type of toxin that are obtained from Ricinus plant.

Opioids are drugs, codeine is a type of alkaloids. Concanavalin A is lectin and the abrin that is obtained from plant Abrus is a toxin.

Such type of infections or disease that are spreaded through sexual intercourse are known as venereal disease or sexually transmitted disease. These diseases are transmitted by the sharing of infected needles, transfusion of blood from an infected person or from an infected mother to foetus.

Endoplasmic reticulum, lysosomes, golgi complex and vacuoles are considered as an endomembrane system. These organelles show co-ordination in their functions. Ribosomes does not contain any membrane and play important role in the process of protein synthesis.

Enzyme Linked Immuno sorbent assay (ELISA) is a technique that is used to early diagnosis of the disease. This technique identify the presence of pathogenic antigen or the antibodies that are formed against the pathogen.
Hybridisation technique is used to detect the mutated gene, Southern blotting is used to identify the specific DN sequence that is present in the given samples and a specific protein is detected by the technique of Western blotting.

Insulin is an important hormone that is made up of two short polypeptide chains in which one is named as A and the other is B. These two chains are attached by disulphide bridges.
C peptide is an extra stretch that is not present in the mature insulin.

Endonuclease is a type of restriction enzyme that is responsible to make a cut in a specific position within a DNA. It recognises a specific sequence of the DNA that is called as palindromic nucleotide sequence.

The primer sequence of the PCR is called as degenerate primer sequence due to presence of several possible bases on of its some positions.

The DNA-dependent RNA polymerase is an important enzyme. In prokaryotic organisms, this enzyme is responsible for the process of initiation, elongation and termination during the process of transcription.

DNase is an enzyme that is involved in the cleavage of phosphodiester linkages of the DNA backbone.

DNA dependent DNA polymerase is responsible for the process of DNA replication. DNA ligase joins the fragments of the DNA.

DNA ligase joins the fragments of the DNA.

Pneumatic bones are hollow and contain many air cells. These bones are present in the animals that are kept in the class of Aves such as Neophron. These bones are helpful in the flying of the birds because they defy gravity during flying process.

Hemidactylus belongs to the class of Reptilia, Ornithorhynchus and Macropus are grouped in the class of Mammalia.

The process of reproduction in which there is alternation of an asexual with a sexual generation is present is known as metagenesis such as in a Cnidarian.

Echinoderms have water vascular system that is involved in the locomotion, capturing and transportation of the food. It also helps in the process of respiration.

Comb plates are responsible for locomotion and are generally found in Ctenophores.

• Diaphragms, cervical caps and vaults are barrier methods of contraception for female which works by blocking the entry of sperms through the cervix.

• IUDs increase phagocytosis of sperms within the uterus.

• Vasectomy is a surgical method of contraception in males in which a small part of the vas deferens is removed or tied up through a small incision on the scrotum.

• Tubectomy is a surgical method of contraception in females where a small part of the fallopian tube is removed or tied up through a small incision in the abdomen or through vagina.

The sphincter of oddi is a type of muscular valve that regulates the common hepato-pancreatic duct.

The junction of ileum and caecum is guarded by Ileo-caecal valve that inhibit the back flow of faecal matter into the ileum of the humans.

The opening of Oesophagus into the stomach is regulated by the gastro-oesophageal sphincter.

According to the Chargaff’s rule, the amount of adenine

(A) should be equal to thymine (T) and the cytosine (C) should be equal to guanine (G) in a double stranded DNA.

In the question Adenine is given 30% so thymine will be 30%. Thymine and adenine will be 60% and rest 40% will be guanine and cytosine. Since cytosine and guanine are equal in amount so each will be 20%.

Thus, A = 30%, T = 30%, G = 20% and C = 20%.

Metamerism is commonly seen in the members of phylum Annelida where the body is externally and internally divided intosegments with a serial repetition of atleast some organs.

Water canal system is present in the members of phylum Porifera.

The body of Ctenophores bears 8 external rows of ciliated comb plates which help in locomotion.

Cnidoblasts or cnidocytes are characteristic feature of cnidarians (coelentrata).

During coagulation of blood, an enzyme complex thrombokinase helps in the conversion of prothrombin (present in plasma) into thrombin.

Thrombin further helps in the conversion of inactive fibrinogens into fibrins which form network of threads.

Renin is secreted by JG cells in response to fall in glomerular blood flow, which converts angiotensinogen in blood to angiotensin-I.

Epinephrine or adrenaline is secreted by adrenal medulla in response to stress of any kind and during emergency.

Biofortification is a process that is used to increase the bioavailability and the amount of nutrients in the crops with the help of plant breeding and recombinant DNA technology.

Generally, it is used to increase the content and quality of the protein, oil, vitamin, micronutrient and the mineral.

It is not used to create resistance in plants against diseases.

Dobson unit is used to measure the thickness of ozone layer. Troposphere is the lowermost layer of the atmosphere. Chlorofluorocarbons (CFCs) are chemicals that are involved inthe depletion of ozone layer.

During respiration, process of ATP synthesis is explained

by chemiosmotic model. It says that a proton gradient is required

for ATP synthesis that is established by oxidation-reduction

In ETC, one NADH + H⁺ produces 3 ATP while one FADH,

produces 2 ATP molecules.

LNG 20 is hormone releasing IUD that releases levonorgestrel. It is responsible to increase the sperm phagocytosis.

Multiload 375, Cu T and Cu 7 are copper releasing IUDs.

Centriole is a cylindrical organelle that is involved in the cell division and in the formation of flagella and cilia. In animal cells, the duplication of centriole takes place during the S-phase of the cell cycle.

In the G₂ phase of the cell cycle, many important organelles such as mitochondria, golgi bodies, chloroplast etc are duplicated.

Chromatin starts to condense in prophase and the alignment of the chromosomes occurs during metaphase.

siRNA is silencing RNA that is not involved in the process of protein synthesis.

mRNA, tRNA and rRNA are the types of RNA that are respon- sible for the protein synthesis.

As given in the question

HbᴬHbˢ × HbᴬHbˢ – Parents

↓


HbˢHbˢ, HbᴬHbˢ, HbᴬHbˢ, HbᴬHbᴬ – Progenies

Since total number of affected progenies = 1

So Percentage of diseased/affected progenies

= 1/4 × 10 = 25%

There are certain factors such as high pO₂, Low pCO₂, Less H⁺ concentration and lower temperature that favour the formation of oxyhaemoglobin of the alveoli.

Polar seal contains short ears and limbs that help to minimise heat loss. This follows the Allen's rule. Physiological adaptation is shown by Kangaroo rat. Behavioural adaptation is exhibited by desert lizard. Generally Marine fishes are found in great depths in the ocean so they follow biochemical adaptation to survive there.

The process of evolution in which different species in a given geographical area starting from a point and literally radiating to other areas of geography is known as adaptive radiation such as Darwin’s finches.

The organs whose anatomical structures are not similar but show similar functions are known as analogous organs such as wings of butterfly and Bird. This is the result of convergent evolution.

Homologous organs shows different functions but shows similarities in their anatomical structure such as bones of forelimbs in man and whale. It is the result of divergent evolution.

The evolution that occurs due to human interference is called as evolution by anthropogenic action such as antibiotic resistant microbes, herbicides resistant varieties and pesticide resistant varieties.

Prolactin is a lactation hormone that is involved in the development of mammary glands and in milk production. This hormone is not involved in the process of parturition in humans.
Oxytocin is known as birth hormone because it plays importantrole during parturition and causes strong uterine contractions.
Prostaglandins also help in the uterine contraction so they are released during this process.

Histones contain basic amino acids such as lysine and arginine and their pH is basic in nature. The histone proteins are classified into five types such as H₁, H₂A, H₂B, H₃ and H₄. Four of them are paired and produce a unit of 8 molecules that is called as histone octamer.

Muscle contraction is initiated by the signal that is sent by the central nervous system. In this process certain events occurs such as
• the head of the myosin hydrolyses ATP molecule and binds with the active sites of actin and forms the cross bridge.

• the Z-line that is connected with actins are pulled inwards and thus causes the shortening of the sarcomeres.

• During the contraction of muscle, 'I' bands reduces and A bands remain the same.

• ADP + Pi are released by myosin and goes back to the relaxed condition.

Scapula is a large triangular flat bone situated in the dorsal part of the thorax between the second and the seventh ribs.
Fibrous joint is shown by the flat skull bones which fuse end-to- end with the help of dense fibrous connective tissues in the form of sutures, to form cranium.
Sternum is a flat bone on the ventral midline of thorax.
Cartilaginous joints between the adjacent vertebrae in the vertebral column permits limited movement.

Altitude sickness can be experienced at high altitude where body does not get enough oxygen due to low atmospheric pressure and causes nausea, fatigue and heart palpitations.
Hence correct option is [R] is correct explanation of [A].

Adenosine deaminase (ADA) enzyme is crucial for the immune system to function. Hence, its deficiency results in the dysfunction of immune system.

• Hyposecretion of hormones of the adrenal cortex causes Addison's disease.

• Parkinson's disease is a long-term degenerative disorder of the central nervous system.

• Disorders which affect GIT & associated glands are called digestive disorders.

Multiple Ovulation Embryo Transfer Technology is used for herd improvement in short time.

• Cows are administered hormones, with FSH-like activity for superovulation.

• 8-32 celled embryos are transferred to surrogate mothers.
• 6-8 eggs are produced per cycle.
• Cows can be fertilised by artificial insemination.

Wuchereria bancrofti is a type of worm that causes a disease named as filariasis.
Amoebiasis is protozoal disease that is caused by Entamoeba histolytica.
Pneumonia is a type of bacterial disease that occurs due to the infection of Streptococcus pneumoniae and Haemophilus influenzae bacteria.
Ringworm is a fungal disease that is caused by Trichophyton.

The lipids in which only single bonds are present are known as saturated fatty acids. Unsaturated fatty acids contain atleast one double bond between two carbon atoms in the molecule.

Lecithin is an important phospholipid that is present in the cell membrane. Palmitic acid is a type of saturated fatty acid that contains 16 carbon atoms in the molecule.

A small fleshy lobe like structure is present on the mouth of earthworm is known as prostomium. It covers the mouth and help to create crackles in the soil. It also act as a sensory organ of the earthworm.

Relaxin is a type of hormone that helps in the process of childbirth by relaxing the ligaments in the pelvis. It also helps in the softening and widening of the cervix. This hormone is released by corpus luteum.
During the pregnancy of the women, graafian follicle is not formed so it is not involved in relaxin secretion.
Relaxin is not produced by uterus and foetus.

Statement I is incorrect because phenylalanine is not coded by AUG, only methionine is coded b AUG.
Statement II is correct because lysine is coded by both 'AAA' and 'AAG' codons.

Tight junction and gap junctions are the type of cell junction and are generally present in the epithelium and some other tissues.

Tight junction is responsible to stop the leakage of the substances across the tissues. Gap junctions play important role in the process of communication among the cells or the tissues.