CBSE EXAMINATION PAPER-2013
NEET-ZOOLOGY
(Solved)
General Instructions :
Read the following instructions carefully and follow them :
- This question paper contains 180 questions. All questions are compulsory.
- This question paper is divided into 1 sections.
- Section A – questions number 1 to 180 are multiple choice questions
- There is no overall choice given in the question paper. However, an internal choice has been provided in few questions.
- Use of calculator is NOT allowed.
Section A
In an experiment four quantities a, b, c and d are measured with percentage error 1%, 2%, 3% and 4% respectively.
Quantity P is calculated as follows P = a³b²/cd % error in P is:
Given, P=a³b²/cd
Therefore, ∆P/P×100%= 3∆a/a x100%
+ 2 ∆b/b×100%+∆c/c ×100%+ ∆d/d ×100% =3x1%+2×2%+3%+4%=14%
A stone falls freely under gravity. It covers distances h₁, h₂ and h₃ in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between h₁, h₂ and h₃ is
∴ h=1/2gt²
∴ h₁=1/2g(5)²=125
h₁+h₂=1/2g(10)²=500
⇒ h₂=375
h₁+h₂+h₃ =1/2g(15)²=1125
⇒ h₃ =625
h₂= 3h₁,h₃ =5h₁
or h₁=h₂/3=h₃/5
The velocity of a projectile at the initial point A is (2î + 3ĵ) m/s. It's velocity (in m/s) at point B is
At point B the direction of velocity component of the projectile along y-component reverses, while the x-component remains unchanged.
Hence, 𝐯⃗_B = 2î - 3ĵ
The upper half of an inclined plane of inclination θ is perfectly smooth while lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by
For upper half of inclined plane
v² = u² + 2a S/2 = 2 (g sin θ) S/2 = gS sin θ
For lower half of inclined plane
0 = u² + 2 g (sin θ – μ cos θ) S/2
⇒ – gS sin θ = gS ( sinθ – μ cos θ)
⇒ 2 sin θ = μ cos θ
⇒ μ = 2 sin θ / cos θ = 2 tan θ
A uniform force of (3î + ĵ) newton acts on a particle of mass 2 kg. The particle is displaced from position (2î+ k̂) meter to position (4î + 3 ĵ - k̂) meter. The work done by the force on the particle is
Given : 𝐅⃗ = 3î + ĵ
𝐫⃗ = (2î + k̂), 𝐫⃗_2 = (4î + 3ĵ -𝐤⃗ )
𝐫⃗ = 𝐫⃗₂ - 𝐫⃗₁ = (4î + 3ĵ -𝐤⃗ ) - (2î + k̂)
or 𝐫⃗ = 2î + 3ĵ - 2k̂
So work done by the given force, w = 𝐟⃗ ·𝐫⃗
= (3î + ĵ) · (2î + 3ĵ - 2k̂) = 6 + 3 = 9J
An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass 1 kg moves with a speed of 12 ms⁻¹ and the second part of mass 2 kg moves with speed 8 ms⁻¹. If the third part flies off with speed 4 ms⁻¹ then its mass is
P_resultant = √12² + 16²
= √144 + 256 = 20
m₃v₃ = 20 (momentum of third part)
or, m₃ = 20/4 = 5 kg
A rod PQ of mass M and length Lis hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in figure. When string is cut, the initial angular acceleration of the rod is
Weight of the rod will produce the torque
τ = mg L / 2 = I α = m L^2 / 3) α [ ∵ Irod = M L^2 / 3 ]
Hence, angular acceleration, α = 3g / 2L
A small object of uniform density rolls up a curved surface with an initial velocity 'v'. It reaches upto a maximum height of 3v² / 4g with respect to the initial position. The object is a
From law of conservation of mechanical energy,
=> 1/2 Iω² + 0 + 1/2 mv² = mg × 3v²/4g
=> 1/2 Iω² = 3/4 mv² - 1/2 mv²
=> 1/2 Iω² = mv²/2 (3/2 - 1)
or, 1/2 I (V²/R²) = mv²/4 or, I = 1/2 mR²
Hence, object is a disc.
A body of mass 'm' is taken from the earth's surface to the height equal to twice the radius (R) of the earth. The change in potential energy of body will be
Initial P.E., U_i = -GMm / R ,
Final P.E., U_f = -GMm / 3R [∴ R' = R + 2R = 3R]
∴ Change in potential energy,
ΔU = -GMm / 3R + GMm / R
= GMm / R (1 - 1/3) = 2/3 GMm / R = 2/3 mgR
(∴ GMm / R = mgR)
Infinite number of bodies, each of mass 2 kg are situated on x-axis at distances 1m, 2m, 4m, 8m, ..... respectively, from the origin. The resulting gravitational potential due to this system at the origin will be
Gravitational potential V = -Gm / r
V₀ = - Gx2 / 1 - Gx2 / 2 - Gx2 / 4 - Gx2 / 8 ----∞
V₀ = -2G [ 1 + 1/2 + 1/4 + 1/8 + ...∞]
V₀= -2G x 1 / (1 - 1/2) = -2G x 1 / (1/2) = -4G
The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?
F = YA / L × ΔL
So, extension, ΔL ∝ L / A ∝ L / D²
[∵ F and Y are constant]
ΔL₁ ∝ 100 / 1² ∝ 100 and ΔL₂ ∝ 200 / 2² ∝ 50
ΔL₃ ∝ 300 / 3² ∝ 100 / 3 and ΔL₄ ∝ 50 / (1/4) ∝ 200
The ratio of L / D² is maximum for case (d).
The wetability of a surface by a liquid depends primarily on
Wetability of a surface by a liquid primarily depends on angle of contact between the surface and liquid.
If angle of contact is acute liquids wet the solid and vice-versa
A piece of iron is heated in a flame. It first becomes dull red, then becomes reddish yellow and finally turns to white hot. The correct explanation for the above observation is possible by using
Wein's displacement law
According to this law
λmax ∝ 1/T
or, λmax × T = constant
So, as the temperature increases λ decreases.
A gas is taken through the cycle A →B → C → A, as shown in figure. What is the net work done by the gas?
W_net = Area of triangle ABC
= 1/2 AC × BC
= 1/2 × 5 × 10^-3 × 4 × 10^5 = 1000 J
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its temperature. The ratio of Cp / Cv for the gas is
According to question P ∝ T^3
But as we know for an adiabatic process the pressure P
∝ T^γ/γ-1.
So, γ/γ-1= 3 ⇒ γ = 3/2 or, C_p/C_v = 3/2
In the given (V - T) diagram, what is the relation between pressure P₁ and P₂?
As V = constant ⇒ P ∝ T
Hence from V–T graph P1 > P2
The amount of heat energy required to raise the temperature of 1g of Helium at NTP, from T₁ K to T₂ K is
From first law of thermodynamics,
ΔQ = ΔU + ΔW = 3/2 . 1/4 R(T2 - T1) + 0
= 3/8 NₐK_B(T2 - T1) [∵ K = R/N]
The molar specific heats of an ideal gas at constant pressure and volume are denoted by Cₚ and Cᵥ, respectively. If γ = Cp / Cv and R is the universal gas constant, then Cᵥ is equal to
Cₚ - Cᵥ = R ⇒ Cₚ = Cᵥ + R
∴ γ = Cₚ / Cᵥ = (Cᵥ + R) / Cᵥ = Cᵥ / Cᵥ + R / Cᵥ
⇒ γ = 1 + R / Cᵥ ⇒ R / Cᵥ = γ - 1
⇒ Cᵥ = R / (γ - 1)
A wave travelling in the +ve x-direction having displacement along y-direction as 1m, wavelength 2π m and frequency 1/π Hz is represented by
The standard equation of a wave travelling along x-axis (+ve direction) is given by, Y = A sin (kx - ωt) where,
angular frequency, ω= 2πf
⇒ 2π/π = 2 [∴f = 1/π]
angular wave number, k = 2π/λ ⇒ 2π/2π= 1 [∴λ= 2π]
∴ Y = 1 sin (x - 2t) [∴ Amplitude, A = 1 m]
If we study the vibration of a pipe open at both ends, then which of the following statements is not true?
Pressure change will be minimum at both ends. In fact, pressure variation is maximum at l/2 because the displacement node is pressure antinode.
A source of unknown frequency gives 4 beats/s, when sounded with a source of known frequency 250 Hz. The second harmonic of the source of unknown frequency gives five beats per second, when sounded with a source of frequency 513 Hz. The unknown frequency is
When sounded with a source of known frequency fundamental frequency = 250 ±4 Hz = 254 Hz or 246 Hz
2ⁿᵈ harmonic if unknown frequency (suppose) 254 Hz = 2 x 254 = 508 Hz
As it gives 5 beats
∴508 + 5 = 513 Hz
Hence, unknown frequency is 254 Hz
Two pith balls carrying equal charges are suspended from a common point by strings of equal length. The equilibrium separation between them is r. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls now become
T cos θ = mg
T sin θ = F
∴ tan θ = F / mg
From figure, tan θ = Fe / mg
=> (r / 2) / y = (kq^2 / r^2) / mg
[∵ F = kq^2 / r^2 from coulomb's law]
=> r^3 ∝ y
When string is clamped at half the height
r'^3 ∝ y / 2
Dividing the above two equations, we have
r' / r = 1 / 2^(1/3) => r' = r / ³√2
A, B and C are three points in a uniform electric field. The electric potential is
Potential at B, V_B is maximum
V_B > V_C > V_A
As in the direction of electric field potential decreases.
A wire of resistance 4 Ω is stretched to twice its original length. The resistance of stretched wire would be
Resistance R = ρℓ / A = 4 Ω
∴ ℓ' = 2 ℓ
On stretching, volume of wire remains constant
∴ ℓA' = ℓ'A'
∴ A' = A / 2
∴ R' = ρ(2ℓ) / (A / 2) = 4R = 4 × 4 Ω = 16 Ω ⇒ ℓA = 2ℓA'
The resistance of the four arms P, Q, R and S in a Wheatstone's bridge are 10 ohm, 30 ohm, 10 ohm and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If the galvanometer resistance is 50 ohm, the current drawn from the cell will be
Given : V = 7V
r = 5Ω
R_eq = (40 ×120) / (40 + 120) Ω
I = V / R = 7 / (5 + (40 ×120) / (40 + 120))
= 7 / (5 + 30) = 1 / 5 = 0.2 A.
When a proton is released from rest in a room, it starts with an initial acceleration a₀ towards west. When it is projected towards north with a speed v₀ it moves with an initial acceleration 3a₀towards west. The electric and magnetic fields in the room are respectively
When moves with an acceleration a₀ towards west, electric field
E = F/q = ma₀/e (West)
When moves with an acceleration 3a₀ towards east, magnetic field
B = 2ma₀/ev₀ (downward)
A current loop in a magnetic field
A current loop in a magnetic field is in equilibrium in two orientations one is stable and another unstable.
A bar magnet of length 'l' and magnetic dipole moment 'M' is bent in the form of an arc as shown in figure. The new magnetic dipole moment will be
Magnetic dipole moment
M = m × ℓ M' = m × r
From figure
ℓ = πr / 3 or r = 3ℓ / π
so, M' = m × r = m × 3ℓ / π = 3 / π M
A wire loop is rotated in a magnetic field. The frequency of change of direction of the induced e.m.f. is
This is the case of periodic EMI
From graph, it is clear that direction is changing once in 1/2 cycle.
A coil of self-inductance L is connected in series with a bulb B and an AC source. Brightness of the bulb decreases when
By inserting iron rod in the coil,
L ↑ Z ↑ I ↓ so brightness ↓
The condition under which a microwave oven heats up a food item containing water molecules most efficiently is
Required condition : Frequency of microwaves =Resonant frequency of water molecules.
A plano convex lens fits exactly into a plano concave lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices μ₁ and μ₂ and R is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is
From the question,
1/f = 1/f1 + 1/f2
= (μ₁ - 1) (1/∞ - 1/-R) + (μ₂ - 1) (1/∞ - 1/R)
= (μ₁ - 1)/R - ( μ₂- 1)/R => 1/f = (μ₁ - μ₂)/R
=> f = R / (μ₁ - μ₂ )
Hence, focal length of the combination is=R/ μ₁ - μ₂
For a normal eye, the cornea of eye provides a converging power of 40D and the least converging power of the eye lens behind the cornea is 20D. Using this information, the distance between the retina and the eye lens of the eye can be estimated to be
P_cornea = +40 D
P_e = +20 D
Total power of combination = 40 + 20 = 60 D
Focal length of combination = 1/60 × 100 cm
= 5/3 cm
For minimum converging state of eye lens,
u = -∞ v = ? f = 5/3
From lens formula,
1/f = 1/v - 1/u ⇒ v = 5/3 cm
Distance between retina and cornea-eye lens
= 5/3 = 1.67 m
In Young's double slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths λ_1 = 12000Å and λ_2 = 10000Å. At what minimum distance from the common central bright fringe on the screen 2 m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other ?
∵ Y = nλD / d
∴ n₁λ₁ = n₂λ₂
⇒ n₁ × 12000 × 10⁻¹⁰ = n₂ × 10000 × 10⁻¹⁰
or, n (12000 × 10⁻¹⁰) = (n + 1) (10000 × 10⁻¹⁰)
⇒ n = 5
(∵ λ₁ = 12000 × 10⁻¹⁰ m; λ₂ = 10000 × 10⁻¹⁰ m)
Hence, Y_common = nλ₁D / d
5(12000 × 10⁻¹⁰) × 2
= ---------------------
2 × 10⁻³
(∵ d = 2 mm and D = 2m)
= 5 × 12 × 10⁻⁴ m
= 60 × 10⁻⁴ m
= 6 × 10⁻³ m = 6 mm
A parallel beam of fast moving electrons is incident normally on a narrow slit. A fluorescent screen is placed at a large distance from the slit. If the speed of the electrons is increased, which of the following statements is correct ?
Angular width, θ = Y/D = nλD/dD [ ∵ Y = Dλ/d ]
so, θ = λ/d , v ↑ λ ↓ θ ↓
[For central maxima n = 1]
Hence, with increase in speed of electrons angular width of central maximum decreases.
The wavelength λ_e of an electron and λ_p of a photon are of same energy E are related by
As P = E/c
λp = hc/E ...(i)
λe = h / √(2mE) ⇒ λe² = h² / 2mE ...(ii)
From equations (i) and (ii)
λp ∝ λe²
For photoelectric emission from certain metal the cut-off frequency is ν. If radiation of frequency 2ν impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass)
From photo-electric equation,
hv' = hv + K_max ...(i)
h.2v = hv + 1/2 mV^2_max [ .'. v' = 2 v]
⇒ hv = 1/2 mV^2_max ⇒ V_max = √(2hv/m)
In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is
For Lyman series (2 → 1)
1/λ_L = R [ 1 - 1/2^2 ] = 3R/4
For Balmer series (3 → 2)
1/λ_B = R [ 1/4 - 1/9 ] = 5R/36
⇒ λ_L / λ_B = (4 / 3R) / (36 / 5R) = 4/36 (5/3) = 5/27
A certain mass of Hydrogen is changed to Helium by the process of fusion. The mass defect in fusion reaction is 0.02866 a.m.u. The energy liberated per a.m.u. is (Given : 1 a.m.u = 931 MeV)
Mass defect Δm = 0.02866 a.m.u.
Energy = 0.02866 × 931 = 26.7 MeV
As_1H² + _1H² → _2He⁴
Energy liberated per a.m.u = 13.35/2 MeV
= 6.675 MeV
The half life of a radioactive isotope 'X' is 20 years. It decays to another element 'Y' which is stable. The two elements 'X' and 'Y' were found to be in the ratio of 1 : 7 in a sample of a the given rock. The age of the rock is estimated to be
The value of x is 1/8
= x₀/8 = x₀/2³ ⇒ t = 3T = 3 × 20 = 60 years
Hence the estimated age of the rock is 60 years
X → Y₀
at t = 0 N₀ 0
at t = t N N₀-N
In a n-type semiconductor, which of the following statement is true?
In a n-type semiconductor holes are minority carriers and pentavalent atoms are dopants.
In a common emitter (CE) amplifier having a voltage gain G, the transistor used has transconductance 0.03 mho and current gain 25. If the above transistor is replaced with another one with transconductance 0.02 mho and current gain 20, the voltage gain will be
Voltage gain Av = β (R_out / R_in)
⇒ G = 25 R_out / R_in ...(i)
Transconductance g_m = β / R_in
⇒R_in = β / g_m = 25 / 0.03
Putting this value of R_in in eqn. (i)
G = 25 R_out / 25 × 0.03 ...(ii)
∴ G' = 20 R_out / 20× 0.02 ...(iii)
From eqs. (ii) and (iii)
Voltage gain of new transistor G' = 2/3 G
The output(X) of the logic circuit shown in figure will be
Three blocks with masses m, 2 m and 3 m are connected by strings as shown in the figure. After an upward force F is applied on block m, the masses move upward at constant speed v. What is the net force on the block of mass 2m? (g is the acceleration due to gravity)
From figure
F = 6 mg,
As speed is constant, acceleration a = 0
∴ 6 mg = 6ma = 0, F = 6 mg
∴ T = 5 mg , T' = 3 mg
T" = 0
F_net on block of mass 2 m
= T - T' - 2 mg = 0
The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω is
Given : emf ε = 2.1 V
I = 0.2 A, R = 10Ω
Internal resistance r = ?
From formula.
ε – Ir = V = IR
2.1 – 0.2r = 0.2 × 10
2.1 – 0.2r = 2 or 0.2r = 0.1
⇒ r = 0.1 / 0.2 = 0.5Ω
6.02 × 10²⁰ molecules of urea are present in 100 mL of its
solution. The concentration of solution is :
M = (6.02 × 10²⁰ × 1000) / (100 × 6.02 × 10²³) = (6.02 × 10²³) / (6.02 × 10²¹)
= 0.01 M
The value of Planck's constant is 6.63 × 10⁻³⁴ Js. The speed of light is 3 × 10⁻¹⁷ nm s⁻¹. Which value is closest to the
wavelength in nanometer of a quantum of light with frequency of 6 × 10¹⁵ s⁻¹?
c = υλ
λ = c / υ = (3 × 10¹⁷ ) / (6 × 10¹⁵) = 50 nm
Based on equation E = -2.178 × 10⁻¹⁸ (Z² / n²) J, certain conclusions are written. Which of them is not correct ?
The electrons has more negative energy in lower that in higher orbit and also the electrons are more: bound in the smallest allowed orbit.
What is the maximum numbers of electrons that can be associated with the following set of quantum numbers?
n = 3, l = 1 and m = -1
n = 3 → 3rd shell
l = 1 → p sub shell.
m = -1 is possible for two electrons present in an
orbital.
Which of the following is electron - deficient?
(BH₃)₂ or (B₂H₆)
It contains two, 3 centre - 2 electron bonds that are present above and below the plane of molecules. Compounds which do not have sufficient number of electrons to form normal covalent bonds are called electron deficient molecules.
Which of the following is a polar molecule?
SF₄ has 4 bond pairs and 1 lone pair of electrons,
sp³d hybridisation leads to irregular shape and
resultant μ ≠ 0.
XeF₂ is isostructural with
Which of the following is paramagnetic?
Molecular orbital configuration of O₂⁻ is
O₂⁻ (17) = σ1s², σ*1s², σ2s², σ*2s²,
σ2p_z², π2p_x² = π2p_y², π*2p_x² = π*2p_y¹
Which one of the following molecules contains no pi bond?
Molecular orbital configuration of O₂⁻ is
O₂⁻ (17) = σ1s², σ*1s², σ2s², σ*2s²,
σ2p_z², π2p_x² = π2p_y², π*2p_x² = π*2p_y¹
Dipole-induced dipole interactions are present in which of the following pairs:
HCl is polar (μ ≠ 0) and He is non-polar (μ = 0), thus gives dipole-induced dipole interaction.
Maximum deviation from ideal gas is expected from:
Higher the critical temperature more easily will be the gas liquify. Now since most easily liquefiable gas show larger deviation, NH₃ will show maximum deviation from ideal behaviour.
Which of these is least likely to act as Lewis base?
BF3 acts as Lewis acid
Identify the correct order of solubility in aqueous medium:
Solubility of alkali metal is maximum among the following. Among ZnS (1.7 × 10⁻⁵) & CuS (8 × 10⁻³⁷) ZnS has higher value of Ksp.
Which of the following structure is similar to graphite?
Boron nitride (BN) is known as inorganic graphite. The most stable form is hexagonal one. It has layered structure similar to graphite.
Which of these is not a monomer for a high molecular mass silicone polymer?
Since Me₃SiCl contains only one Cl, therefore it can't form high molecular mass silicon polymer. It can form only dimer.
The basic structural unit of silicates is:
SiO4^4- is basic structural unit of silicates
Structure of the compound whose IUPAC name is 3-ethyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid is:
-
(A)
-
(B)
-
(C)
-
(D)
IUPAC name of the structure is 3-ethyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid
The structure of isobutyl group in an organic compound is:
-
(A)
- (B) CH₃-CH-CH₃-CH₃
- (C) CH₃-CH₂-CH₂-CH₂-
-
(D)
The radical, is aromatic because it has:
Presence of 6 p orbitals, each containing one unpaired electron, in a six membered cyclic structure is in accordance with Huckel rule of aromaticity.
Some meta-directing substituents in aromatic substitution are given. Which one is most deactivating?
Decreasing order of deactivating effect of the given m-directing group is
NO₂ > -CN > -SO₃H > -COOH
—NO₂ group is most deactivating group due to strong — E, —I and —M effects.
Which of the following compounds will not undergo Friedel-Craft's reaction easily:
—NO₂ is a powerful electron withdrawing group. Its presence on ring makes the ring less active.
Roasting of sulphides give the gas X as a by product. This is colorless gas with choking smell of burnt sulphur and caused great damage to respiratory organs as a result of acid rain. Its aqueous solution is acidic, acts as a reducing agent and its acid has never been isolated. The gas X is:
Based on the features given, gas must be SO₂.
Which of the following statements about the interstitial compounds is incorrect?
Interstitial compounds are chemically inert.
The number of carbon atoms per unit cell of diamond unit cell is:
Diamond is like ZnS. In diamond cubic unit cell, there are eight corner atoms, six face centered atoms and four more atoms inside the structure (tetrahedral voids).
Total no. of atoms present per unit cell
= 8 x 1/8 + 6 x 1/2 + 4 = 8
(corners) (face centered) (inside body)
A metal has an fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm⁻³. The molar mass of the metal is: (Na, Avogadro's constant = 6.02 * 10²³ mol⁻¹)
Density is given by d=Z x M/ N_Aᵃ³
where Z = number of formula units present in
unit cell, which is 4 for fcc
a = edge length of unit cell. M = Molecular mass
2.72 = 4 x M / 6.02 x 10²³ x (404 x 10⁻¹⁰)³
(∵ 1pm = 10⁻¹⁰ cm)
2.72 x 6.02 x (404)³/ 4 x 10⁷= 26.99 = 27 gmol ⁻¹
How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0M HNO₃ ? The concentrated
acid is 70% HNO₃
Molarity (M) = (wt x 1000) / (mol. wt. x vol (mL))
2 = (wt. / 63) x (1000 / 250)
wt. = 63 / 2 g
wt. of 70% acid = 100/70 × 31.5 = 45 g
At 25°C molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is 9.54 ohm⁻¹ cm²mol⁻¹ and at
infinite dilution its molar conductance is 238 ohm⁻¹ cm² mol⁻¹ . The degree or ionisation of ammonium hydroxide at
the same concentration and temperature is
α=Λm/Λₘ∞ = 9.54 / 238 = 0.04008 = 4.008%
A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be ?
H₂ → 2H+ + 2e⁻
1 atm 10⁻¹⁰
E_(H2/H+) = 0 - (0.059 / 2) log((10⁻¹⁰)² / 1)
E_(H₂/H+) = +0.59 V
A button cell used in watches functions as following:
Zn(s) + Ag₂O(s) + H₂O(l) ⇄ 2Ag(s) + Zn²⁺(aq) + 2OH⁻(aq)
If half cell potentials are:
Zn²⁺(aq) + 2e⁻→Zn(s); E° = -0.76 V
Ag2O(s) + H₂O(l) + 2e⁻→ 2Ag(s) + 2OH⁻(aq); E° = 0.34 V
The cell potential will be:
E°_Cell = E°_OP + E°_RP = 0.76 + 0.34 = 1.10 V
What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20°C to 35°C?
(R = 8.314 J mol⁻¹ K⁻¹)
log k₂/k₁ = Ea/2.303R (1/T₁ - 1/T₂)
log 2 = Eₐ / (2.303 × 8.314) [1/293 - 1/308]
0.3 = Eₐ / (2.303 × 8.314) × 15 / (293 × 308)
Eₐ = (0.3 × 2.303 × 8.314 × 293 × 308) / 15
= 34673 J mol⁻¹ = 34.7 kJ mol⁻¹
A reaction having equal energies of activation for forward and reverse reaction has:
ΔH = E_af - E_ab = 0
Which of the following does not give oxygen on heating?
(NH₄)₂Cr₂O₇⎯⎯Δ→ N₂ + Cr₂O₃ + 4H₂O
Zn(ClO₃)₂ ⎯⎯Δ→ ZnCl₂ + 3O₂
2 KClO₃ ⎯⎯Δ→ 2KCl + 3O₂
4K₂Cr₂O₇ ⎯⎯Δ→ 4K₂CrO₄+ 2Cr₂O₃ + 3O₂
Which is the strongest acid in the following:
HClO₄ is the strongest acid amongst all because the oxidation state of Cl is maximum (+7).
Which of the following lanthanoid ions is diamagnetic?
(At. nos. Ce = 58, Sm = 62, Eu = 63, Yb = 70)
Sm²⁺(Z=62) [Xe]4f⁶6s²- 6 unpaired e⁻
Eu²⁺(Z=63) [Xe]4f⁷6s² - 7 unpaired e-
Yb²⁺(Z=70) [Xe]4f¹⁴ 6s²- 0 unpaired e-
Ce²⁺(Z=58) [Xe]4f¹ 5d¹ 6s² - 2 unpaired e-
Only Yb²⁺ is diamagnetic.
KMnO₄ can be prepared from K₂MnO₄ as per the reaction:
3MnO₄²- + 2H₂O <=> 2MnO₄⁻ + MnO₂ + 4OH⁻
The reaction can go to completion by removing OH^- ions by adding:
HCl and SO₂ are reducing agents and can reduce MnO₄⁻. CO₂ which is neither oxidising and nor reducing will provide only acidic medium. It can shift reaction in forward direction and reaction can go to completion
An excess of AgNO₃ is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium (III) chloride. The number of moles of AgCl precipitated would be:
[Cr(H₂O)₄Cl₂]Cl + AgNO₃ →AgCl + [Cr(H₂O)₄Cl₂]NO₃
Molarity = wt / mol. mass x 1000 / vol.
wt / mol. mass = molarity x vol. / 1000 = 0.01 x 100 / 1000
= 0.001
A magnetic moment of 1.73 BM will be shown by one among the following:
[Cu(NH₃)4]²⁺ hybridisation dsp²
Cu⁺² - 3d⁹has one unpaired e-
So, magnetic moment
μ = √n(n+2) = √1(1+2)
= √3 = 1.73
Among the following ethers, which one will produce methyl alcohol on treatment with hot concentrated HI?
-
(A)
-
(B)
-
(C)
-
(D)
The reaction will proceed via S_N1 mechanism because alkyl group attached is 3°.
Reaction by which Benzaldehyde cannot be prepared:
-
(A)
-
(B)
-
(C)
-
(D)
Zn/Hg and conc. HCl reduce carboxyl group to methylene group (Clemmensen reduction).
The order of stability of the following tautomeric compounds is:
Enolic form predominates in compounds containing two carbonyl groups separated by a -CH₂ group. This is due to following two factors.
(i) Presence of conjugation which increases stability.
(ii) Formation of intramolecular hydrogen bond between enolic hydroxyl group and second carbonyl group which leads to stabilisation of the molecule. Hence, the correct answer is III > II > I.
In the reaction
A is:
Enolic form predominates in compounds containing two carbonyl groups separated by a -CH₂ group. This is due to following two factors.
(i) Presence of conjugation which increases stability.
(ii) Formation of intramolecular hydrogen bond between enolic hydroxyl group and second carbonyl group which leads to stabilisation of the molecule. Hence, the correct answer is III > II > I.
Nitrobenzene on reaction with conc. HNO₃ / H₂SO₄ at 80-100°C forms which one of the following products?
Enolic form predominates in compounds containing two carbonyl groups separated by a -CH₂ group. This is due to following two factors.
(i) Presence of conjugation which increases stability.
(ii) Formation of intramolecular hydrogen bond between enolic hydroxyl group and second carbonyl group which leads to stabilisation of the molecule. Hence, the correct answer is III > II > I.
Nylon is an example of:
Nylon is a synthetic polyamide polymer.
Which is the monomer of Neoprene in the following?
- (A) CH₂ = CH - CH = CH₂
- (B) CH₂ = CH - C ≡ CH
-
(C)
-
(D)
(chloroprene) is the monomer of neoprene
Antiseptics and disinfectants either kill or prevent growth of microorganisms. Identify which of the following statements is not true
Dilute solutions of boric acid and hydrogen peroxide are weak antiseptics.
Which one of the following is not a correct statement?
Museums- are generally set up in educational institutes such as schools and colleges. Museums have collections of preserved plant and animal specimens for study and reference.
Which of the following are likely to be present in deep sea water?
Archaebacteria live in some of the most harsh habitats such as extreme salty area (halophiles), hot springs (thermoacidophiles) and marshy areas (methanogens) and in deep sea water.
Isogamous condition with non-flagellated gametes is found in
In Spirogyra, sexual reproduction occurs through conjugation. Gametes are non-flagellated, morphologically similar. But physiologically different (isogamy with physiological anisogamy). Volvox and Fucus are examples of oogamous and Chlamydomonas contains isogamous flagellated gametes.
Which of the following is not correctly matched for the organism and its cell wall degrading enzyme?
Algae is a plant and so its cell wall is made up of cellulose. Cellulase enzyme is needed for degradation of its cell wall.
Besides paddy fields cyanobacteria are also found inside vegetative part of
In Cycas specialised root called coralloid roots are associated with N₂- fixing cyanobacteria either Nostoc or Anabaena. Coralloid roots lie near the soil surface. They are irregular and often dichotomously branched. Root hair and root cap are absent in these roots.
Read the following statements (A-E) and answer the question which follows them.
A. In liverworts, mosses and ferns gametophytes are free-living
B. Gymnosperms and some ferns are heterosporous.
C. Sexual reproduction in Fucus, Volvox and Albugo is oogamous
D. The sporophyte in liverworts is more elaborate than that in mosses
E. Both, Pinus and Marchantia are dioecious
How many of the above statements are correct?
Statement (A), (B) and (C) are correct. In liverworts and ferns gametophytes are free living while in fern, sporophytes are free living. Gymnosperms and genera like Selaginella and Salvinia are heterosporous. The sporophyte in mosses are more elaborate than that of liverworts, Pinus is monoecious and heterosporous. Marchantia is dioecious.
Which group of animals belong to the same phylum?
Prawn , Scorpion and Locusta belong to phylum Arthropoda. This is the largest phylum of Animalia, over two-thirds of all named species on earth are Arthropoda. Phylum is named so due to presence of jointed appendages (arthros - jointed, poda - appendages) in body of members of this phylum.
One of the representatives of phylum Arthropoda is
Representative of Phylum Arthropoda is silverfish. Arthropoda is the largest phylum of Animalia, which covers two-thirds of all named species
Which of the following are correctly matched with respect to their taxonomic classification
House fly, butterfly, tse tse fly, silverfish all belongs to insecta
Match the name of the animal (column I), with one characteristics (column II), and the phylum/class (column III) to which it belongs:
Ichthyophis - Amphibian
Limulus - Arthropoda
Adamsia - Cnidaria
Petromyzon - Jawless vertebrate & ectoparasite, cyclostomate
Among bitter gourd, mustard, brinjal, pumpkin, china rose, lupin, cucumber, sunhemp, gram, guava, bean, chilli, plum, petunia, tomato, rose, withania, potato, onion, aloe and tulip how many plants have hypogynous flower?
Mustard (cruciferae) , brinjal (solanaceae) china rose (malvaceae), lupin (leguminosae), gram (leguminosae), bean (leguminosae), chilli, Petunia, potato, tomato, Withania (solanaceae), onion, Aloe, tulip (liliaceae) have hypogynous flower while bittergourd, cucumber (cucurbitaceae), guava (myrtaceae) have epigynous flower and rose has perigynous flower.
In china rose the flowers are:
In china rose (Hibiscus rosa-sinensis) flowers are actinomorphic , hypogynous with twisted aestivation in corolla.
Age of a tree can be estimated by:
Age of a tree can be estimated by number of annual rings. Annual ring constitute alternate concentric rings of spring wood and autumn wood.
Lenticels are involved in:
Lenticels are lens shaped openings occuring in stems of most of the woody trees. It permits the exchange of gases between the outer atmosphere and the internal tissue of the stem.
Interfascicular cambium develops from the cells of:
At the time of secondary growth interfascicular cambium is formed by parenchymatous medullary rays. Interfascicular cambium along with intrafascicular cambium (formed from cambium cells present between xylem and phloem) constitute continuous cambium ring. If new cells are cut off in both directions it causes secondary growth in most dicotyledonous plants.
What external changes are visible after the last moult of a cockroach nymph?
The development of cockroach is paurometabolous meaning that their development is through nymphal stage. The nymphs look very much like adults. The nymph grows by moulting about 13 times to reach adult form. The last nymphal stage has wingpads while adult cockroaches have wings.
Which one of the following organelle in the figure correctly matches with its function?
Rough endoplasmic reticulum is a network or reticulum of tiny tubular structures scattered in the cytoplasm and bear ribosomes on their outer surface. These are involved in protein synthesis and secretion. They are extensive and continuous with the outer membrane of the nucleus
The Golgi complex plays a major role :
Golgi apparatus plays a major role in post translational modification of proteins forming glycoprotein and glycosidation of lipid forming glycolipids. A number of proteins and lipids synthesised on endoplasmic reticulum (rough and smooth respectively) are modified in the cisternae of the Golgi apparatus before they are released from the trans face.
A major site for synthesis of lipids is :
The smooth endoplasmic reticulum is the major site for synthesis of lipid. In animal cells lipid like steroidal hormones are synthesised in SER.
A phosphoglyceride is always made up of :
A saturated or unsaturated fatty acid esterified to a glycerol molecule to which a phosphate group is also attached forms phosphoglycerides.
The essential chemical components of many coenzymes are :
The essential chemical components of many coenzymes are vitamins, e.g., coenzyme nicotinamide adenine dinucleotide (NAD) and NADP contain vitamin niacin
Transition state structure of the substrate formed during an enzymatic reaction is :
Transition state structure formed during an enzymatic reaction is transient and unstable.
Macro molecule chitin is :
Chitin or fungal cellulose is a nitrogen containing polysaccharide and hetero-polymer of NAG.
A stage in cell division is shown in the figure. Select the answer which gives correct identification of the stage with its characteristics.
Telophase is the stage of reconstitution of nuclei. The chromosomes that have reached their respective poles decondense and lose their individuality and collect in a mass in the two poles. Nuclear envelope assemble around chromatin mass. Nucleolus, Golgi complex and ER reform.
The complex formed by a pair of synapsed homologous chromosomes is called :
During zygotene, second stage of meiosis I, homologous chromosomes start pairing together forming a complex structure called synaptonemal complex. The complex formed by a pair of synapsed homologous chromosome is called a bivalent or a tetrad. Crossing over occurs between non-sister chromatids of bivalent in the next stage.
Which of the following criteria does not pertain to facilitated transport?
Substances that have a hydrophilic moiety, find it difficult to pass through the membrane, their movement has to be facilitated by the membrane protein without expenditure of ATP energy. Facilitated Diffusion cannot cause net transport of molecules from a low to a high concentration or uphill transport.
The first stable product of fixation of atmospheric nitrogen in leguminous plants is :
Ammonia is the first stable compound produced in fixation of atmospheric nitrogen in leguminous plant. A molecule of nitrogen is changed into two molecules of ammonia. It requires 16 ATP, a source of electron donor ferredoxin, hydrogen donor NADPH or FMNH2 and dinitrogenase enzyme
Pigment-containing membranous extensions in some cyanobacteria are :
In some photosynthetic bacteria, bacteriochlorophyll and bacteriophaeophytin or bacteriorhodopsin pigments containing membranous extensions called chromatophores are present. These are sites of bacterial photosynthesis.
Which of the metabolites is common to respiration-mediated breakdown of fats, carbohydrates and proteins
Acetyl CoA is common to respiration mediated breakdown of fats, carbohydrates and proteins. Fats are broken down to fatty acid and glycerol and finally we degraded to acetyl co - A. Protein first degraded by proteases to individual amino acids which are deaminated to pyruvic acid and further decarboxylated to acetyl co-A.
The three boxes in this diagram represent the three major biosynthetic pathways in aerobic respiration. Arrows represent reactants or products.
Arrows numbered 4, 8 and 12 can all be :
In the given diagram of aerobic respiration - pathway A is glycolysis, pathway B is Kreb's cycle and pathway C is ETS , thus 4, 8 & 12 are ATP. ATP act as energy currency. The energy is trapped in form of ATP and it is broken down whenever and wherever it needs to be utilised.
During seed germination its stored food is mobilized by :
Gibberellins stimulate the synthesis of α-amylase and protease enzyme in germinating grains of cereals. They are involved in the conversion of starch into sugar. The proteases convert an inactive β-amylase to the active form. The active β-amylase and α-amylase together digest starch to glucose which is mobilised to meet the metabolic demands of embryo.
Select the correct match of the digested products in humans given in column - I with their absorption site and mechanism in column-II
Small intestine is major area of absorbtion of nutrients. Approximately 90% of absorbtion take place here. Glucose, fructose, fatty acids, amino acids (Glycine etc) are absorbed through mucosa into blood and lymph by active absorbtion.
The figure shows a diagrammatic view of human respiratory system with labels A, B, C and D. Select the option which gives correct identification and main function and/or characteristics.
Alveoli are very thin, irregular walled bag like structures for gaseous exchange. Tracheae bronchi and bronchioles are supported by incomplete cartilaginous rings. Double layered pleural membrane surrounds the lungs with pleural fluid between them. It reduces friction on the lung surface.
Figure shows schematic plan of blood circulation in humans with labels A to D. Identify the label and give its function's.
A is pulmonary vein which takes pure blood from lungs to heart. B is dorsal aorta which takes pure blood from heart to various body parts. C is vena cava which takes impure blood from various body parts to right auricle of the heart. D is pulmonary artery which takes impure blood from heart to lungs.
The diagram given here is the standard ECG of a normal person the P-wave represents the
In the given ECG of a normal person, P-wave represents the electrical excitation (or depolarization) of the atria, which leads to the contraction of both the atria. The QRS complex represents the depolarization of the ventricles, which initiates the ventricular contraction. The contraction starts shortly after Q and marks the beginning of the systole.
Figure shown human urinary system with structures labelled A to D. Select option which correctly identifies them and gives their characteristics and/or functions.
A - Adrenal gland - located at the anterior part of kidney, secretes Catecholamines which stimulate glycogen breakdown.
The H-zone in the skeletal muscle fibre is due to:
Central part of thick filament, not overlapped by thin filaments is called the 'H' zone. 'H' zone is also called Hensen's Line.
The characteristics and an example of a synovial joint in humans is :
Characteristics | Examples
Synovial joints are characterized by the presence of a fluid filled synovial cavity between the articulating surfaces of the two bones. Such an arrangement allows considerable movement. These joints help in locomotion and many other movements. Ball and socket joint (between humerus and pectoral girdle), Hinge joint (knee joint), Pivot joint (between atlas and axis), Gliding joint (between the carpals) and Saddle joint (between carpal and metacarpal of thumb) are some examples.
Select the correct statement with respect to locomotion in humans :
Vertebral column has 12 thoracic vertebrae. The joints between adjacent vertebrae is cartilaginous joint which permits limited movements. Progesterone is secreted by corpus luteum which supports in pregnancy in females. Deposition of uric acid in form of urate crystal in the joint are responsible for the inflammation and painful symptoms of gout
The most abundant intracellular cation is :
The most abundant intracellular cation is K+.
A diagram showing axon terminal and synapse is given. Identify correctly at least two of A-D.
A-Receptor, B-Synaptic vesicles B' is synaptic cleft. A synapse is formed by the membranes of a synaptic neuron and post synaptic neuron, which may or may not separated by a gap called synaptic cleft. It is filled by fluid called neurotransmitter which are involved in transmission of impulse at these synapses.
Parts A, B, C and D of the human eye are shown in the diagram. Select the option which gives correct identification along with its functions/characteristics :
Retina is the innermost layer of eyeball and it contains three layers of cells - ganglion cells, bipolar cells and photoreceptor cells. Photoreceptor cells are not present in blind spot. The space between the cornea and the lens is called the aqueous chamber and contains a thin watery fluid called aqueous humor.
Choroid is anterior part of external sclera layer of eyeball while ciliary body is anterior part of middle choroid layer.
Which of the following statements is correct in relation to the endocrine system?
Hormones are non-nutrient chemicals which act as intercellular messengers and are produced in trace amounts. Gastrointestinal tracts secretes four major peptide hormones - gastrin, secretin, cholecystokinin (CCK) and gastric inhibitory peptide (GIP) while juxtaglomerular cells of kidney secretes erythropoietin, a peptide hormone. Releasing and inhibitory hormones are produced by hypothalamus. Neurohypophysis as posterior pituitary is under direct neural regulation of the hypothalamus.
A pregnant female delivers a baby who suffers from stunted growth, mental retardation, low intelligence quotient and abnormal skin. This is the result of :
Iodine is essential for the normal rate of hormone synthesis in the thyroid. Deficiency of iodine in our diet results in hypothyroidism and enlargement of the thyroid gland, commonly called goitre.
Select the answer which correctly matches the endocrine gland with the hormone it secretes and its function/deficiency symptom:
• Growth hormone secreted by Anterior pituitary
• Corpus luteum secreted Progesterone
• Oxytocin is secreted by Posterior pituitary
Select the wrong statement:
In oomycetes female gamete is large and non-motile while male gamete is small and motile
Monoecious plant of Chara shows occurrence of :
Male sex organ is called antheridium or globule while female sex organ is called oogonium. They develop on the same branchlet in the same plant in Chara.
Meiosis takes place in :
Meiosis takes place in meiocyte while conidia and gemmule are asexual structures and megaspore is female gametophyte i.e., embryosac.
Megasporangium is equivalent to
Ovule is also called integumented megasporangium. It develops into seed after fertilisation in spermatophytes. It occurs singly or in a cluster inside ovary with parenchymatous cushions called placenta.
Advantage of cleistogamy is :
Cleistogamy favours no dependence on pollinator because flowers never open. In such flowers, the anthers and stigma lie close to each other. When anthers dehisce in flower buds pollen grains come in contact with the stigma to effect pollination.
Product of sexual reproduction generally generates :
Sexual reproduction leads to formation of new progeny with appearance of variations by Genetic recombination, of two different organisms interaction etc. During sexual reproduction provides vigour and vitality to the offsprings. They better adapt themselves in changing environmental conditions and also plays an important role in evolution.
Which one of the following statements is correct?
Sporogenous tissue is always diploid, endothecium is second layer of anther wall and perform the function of protection and help in dehiscence of anther to release the pollen. Inner most layer of pollen is called exine but tapetum always nourishes the developing pollen.
Cells of the tapetum possess dense cytoplasm and generally have more than one nucleus (polyploid).
Perisperm differs from endosperm in :
Perisperm is remnants of nucellus which is diploid (2n) but endosperm is triploid (3n). Perisperm occurs in the seeds of Black pepper, coffee, castor, cardamum, Nymphaea. Endosperm is the food laden tissue which is meant for nourishing the embryo in seed plants. In angiosperms the endosperm is formed as a result of vegetative fertilization, triple fusion or fusion of a male gamete with diploid secondary nucleus of the central cell.
Seed coat is not thin, membranous in:
The seed coat develops from integuments originally surrounding the ovule. It is thick and hard in coconut which protect the embryo from mechanical injury and from drying out.
Menstrual flow occurs due to lack of :
During menstrual or bleeding phase which comprises of 3 - 5 days out of 28 days menstrual cycle, the production of LH from the anterior lobe of the pituitary gland is considerably reduced. The withdrawal of this hormone causes degeneration of corpus luteum and therefore, progesterone production is reduced. Production of estrogens is also reduced in this phase.
Which one of the following is not the function of placenta? It :
Function of placenta - The placenta facilitate the supply of oxygen and nutrients to the embryo and also removal of carbon dioxide and excretory/waste materials produced by the embryo. The placenta is connected to the embryo through an umbilical cord which helps in the transport of substances to and from the embryo. Placenta also acts as an endocrine tissue and produces several hormones like human chorionic gonadotropin (hCG), human placental lactogen (hPL), estrogens, progesterone etc.
Which of the following cannot be detected in a developing foetus by amniocentesis?
Jaundice cannot be detected in a developing foetus by amniocentesis. It is a foetal sex determination test based on chromosomal pattern in the amniotic fluid surrounding the developing embryo. It is now legally ban in India.
One of the legal methods of birth control is:
Medical termination of pregnancy (MTP) or induced abortion become legal in India from 1971 with some strict conditions to avoid its misuse. Such restrictions are important to check indiscriminate and illegal female foeticides which are reported to be high in India. All the other options are of natural methods of contraception.
Artificial insemination means:
In Artificial Insemination (AI) technique, the semen collected either from the husband or a healthy donor is artificially introduced either into the vagina or into the uterus (IUI - intrauterine insemination) of the female.
If two persons with 'AB' blood group marry and have sufficiently large number of children these children could be classified as 'A' blood group: 'AB' blood group: 'B' blood group in 1 : 2 : 1 ratio. Modern technique of protein electrophoresis reveals presence of both 'A' and 'B' type proteins in 'AB' blood group individuals. This is an example of:
ABO blood group system in human beings is an example of codominant, dominant recessive and multiple alleles. Blood groups are controlled by the gene I located on 9th chromosome that has 3 multiple alleles, out of which any two are found in a person. In codominance, both genes express themselves completely.
Which Mendelian idea is depicted by a cross in which the F₁ generation resembles both the parents?
In co-dominance, F₁ generation resembles both the parents. Ex : Blood group inheritance.
Which of the following statements is not true of two genes that show 50% recombination frequency?
Tightly linked genes show more linkage than crossing over.
If both parents are carriers for thalassemia, which is an autosomal recessive disorder, what are the chances of pregnancy resulting in an affected child?
Genotype of carrier parents is - Aa (male parent) x Aa (female parent)
AA → normal child (25%)
Aa → carrier child (50%)
aa → affected child (25%)
The incorrect statement with regard to Haemophilia is:
Haemophilia is sex linked recessive disease in which a simple protein that is a part of protein cascade involved in clotting of blood is affected. Due to this, in an affected individual, a simple cut will result in non-stop bleeding.
The diagram shows an important concept in the genetic implication of DNA. Fill in the blanks A to C.
In this question A is transcription, B - translation and C is Francis Crick (central dogma). It is unidirectional flow of information from DNA to mRNA (transcription) and then decoding the information present in mRNA in the formation of polypeptide chain or protein (translation).
Which enzymes will be produced in a cell in which there is a nonsense mutation in the lac Y gene?
A nonsense mutation is the one which stops polypeptide synthesis due to formation of a terminating or nonsense codon e.g. ATT (UAA), ATC (UAG), ACT (UGA). The lactose or lac operon of Escherichia coli contains structural genes (z, y, a). If y codes for termination of polypeptide chain then only the product of 'z' gene transcribe to form beta galactosidase.
The eye of octopus and eye of cat show different patterns of structure, yet they perform similar function. This is an example of
The eye of octopus and eye of cat show different patterns of structure, yet they perform similar functions. This is an example of analogous organs.
Analogous organs have evolved due to convergent evolution. Analogous organs have developed in the evolutionary process through adaptation of quite different organisms to similar mode of life.
The process by which organisms with different evolutionary history evolve similar phenotypic adaptations in response to a common environmental challenge, is called:
The wings is a classic example of convergent evolution. Flying insects, birds and bats all evolved the capacity of flight independently. All four serve the same function and are similar in structure but anatomically different.
Variation in gene frequencies within populations can occur by chance rather than by natural selection. This is referred to as:
Genetic drift is variation in gene frequencies within populations can occur by chance rather than by natural selection, sometimes the change in allele frequency is so
According to Darwin, the organic evolution is due to:
According to Darwin the organic evolution is due to interspecific competition. Individuals of same species have variation in characteristics. Those characteristics which enable some to survive better in natural conditions (climate, food, physical factors etc.) would outbreed others that are less-endowed to survive under such natural conditions. These, therefore, will survive more and hence are selected by nature. He called it natural selection and implied it as a mechanism of evolution.
Infection of Ascaris usually occurs by:
Ascaris, an intestinal parasite causes ascariasis. Symptoms of these disease include internal bleeding, muscular pain, fever, anemia and blockage of the intestinal passage. The eggs of the parasite are excreted along with the feces of infected persons which contaminate soil, water, plants, etc. A healthy person acquires this infection through contaminated water vegetables, fruits, etc.
The cell-mediated immunity inside the human body is carried out by:
T - lymphocytes mediate cell mediated immunity (CMI). It is one of the two types of acquired immunity responsible for graft or transplant rejection.
In plant breeding programmes, the entire collection (of plants/seeds) having all the diverse alleles for all genes in a given crop is called:
Germplasm collection is the first step of plant breeding programmes. As genetic variability is the root of any breeding programmes. In many crops pre-existing genetic variability is available from wild relatives of the crop.
During sewage treatment, biogases are produced which include :
The major component of biogas is methane (about 50-68%) which is highly inflammable. The other gases are carbon dioxide (25-35%), hydrogen (1-5%), nitrogen (2-7%), Oxygen (0-0.1%) and rarely hydrogen sulfide. Biogas is a "mixture of gases" produced from degradable organic matter by the activity of various anaerobic bacteria that offers a low cost alternative for energy requirements.
A good producer of citric acid is:
A good source of citric acid is Aspergillus niger (a fungus). Apart from citric acid, oxalic acid, gallic acid, gluconic acid are extracted from fungus.
The colonies of recombinant bacteria appear white in contrast to blue colonies of non-recombinant bacteria because of:
Alternative selectable markers have been developed
which differentiate recombinant from non-recombinants on the basis of their ability to produce colour in the presence of chromogenic substrate. It this, a recombinant DNA is inserted within the coding sequence of an enzyme β-galactosidase. This results into inactivation of the enzyme, which is referred to as insertional inactivation. The presence of chromogenic substrate gives blue coloured colonies of the plasmid in the bacteria that does not have an insert. Presence of insert results into insertional inactivation of the galactosidase and the colonies do not produce any colour; these are identified as recombinant colonies.
DNA fragments generated by the restriction endonucleases in a chemical reaction can be separated by:
In a chemical reaction DNA fragments generated by restriction endonucleases, can be separated by gel electrophoresis. Since DNA fragments are negatively charged molecules, they can be separated by forcing them to move towards the anode under an electric field through a medium/matrix. The DNA fragments separate according to their size through sieving effect provided by matrix.
Which of the following Bt crops is being grown in India by the farmers?
Bt cotton is being grown in India by the farmers. Bt cotton is pest resistant plant which could decrease the pesticides used. Bt toxin is produced by a bacterium Bacillus thuringiensis (BT for short). Bt toxin gene has been cloned from the bacteria and been expressed in plants to provide resistance to insects without the need for insecticides.
A sedentary sea anemone gets attached to the shell lining of hermit crab. The association is:
Commensalism is an interaction where one species is benefitted and other is neither benefitted nor harmed.
A biologist studied the population of rats in a barn. He found that the average natality was 250, average mortality 240, immigration 20 and emigration 30. The net increase in population is:
Net increase in population:
(Natality + Immigration) – (Mortality + Emigration)
= (250 + 20) – (240 + 30) = 270 – 270 = 0
Which one of the following processes during decomposition is correctly described?
Humification, catabolism, leaching and fragmentation are the steps of decomposition which operate simultaneously on the detritus. Fragmentation is breaking down detritus into smaller particles by detritivores like earthworm. By the process of leaching, water soluble inorganic nutrients go down into soil horizon and get precipitated as unavailable salts. Humification occurs at very slow rate.
Secondary productivity is rate of formation of new organic matter by:
The rate of resynthesis of organic matter by the consumers is known as secondary productivity. It depends upon the loss while transferring energy containing organic matter from the previous trophic level plus the consumption due to respiration and predation. Respiration loss is about 20% for autotrophs, 30% for herbivores and up to 60% in case of carnivores. Therefore, net productivity decreases with each trophic level.
Natural reservoir of phosphorus is:
The natural reservoir of phosphate is rock, which contains phosphorus in the form of phosphates. When rocks are weathered, minute amounts of these phosphates dissolve in the soil solution and are absorbed by the roots of plants. Herbivores and other animals obtain this element from plants.
Which of the following represent maximum number of species among global biodiversity?
The number of species of fungi is maximum with respect to other options. The number of fungal species in the world is more than the combined total of species of fishes, amphibians, reptiles and mammals.
Which one of the following is not used for ex-situ plant conservation?
Shifting or Jhum cultivation contributes to deforestation.
Ex-situ conservation is a biodiversity conservation method in which threatened animals and plants are taken out of their natural habitat and placed in protected environments such as zoological parks, botanical gardens, wildlife safari parks, seed banks, and gene banks.
The Air (Prevention and Control of Pollution) Act came into force in:
In lndia, the Air (Prevention and control of ‘pollution)
Act came into force in 1981 but was amended in 1987 to
include noise as an air pollutant.
Kyoto Protocol was endorsed at:
In December 1997, at the Conference of Parties III (COP-3) in Kyoto, Japan, the Kyoto Conference on Climate Change took place. During this conference, developed countries agreed to specific targets for cutting their emissions of greenhouse gases. A general framework was defined, with specific details to be finalised in the following years. This agreement became known as the Kyoto Protocol.
Global warming can be controlled by:
In the first three options, CO₂ concentration increases, whereas it decreases in option (d). Global warming is the increase in the Earth's average temperature. It is caused by the greenhouse effect, a naturally occurring phenomenon responsible for heating the Earth's surface and atmosphere. The major greenhouse gas is carbon dioxide (CO₂).
The first stable product of fixation of atmospheric nitrogen in leguminous plants is:
The process of conversion of nitrogen (N₂) to ammonia is termed nitrogen fixation.
N₂ → N₂H₂ → N₂H₄ → 2NH₃
(Nitrogen) → (Diimide) → (Hydrazine) → (Ammonia)
2NH₃ + 3O₂ → 2NO₂⁻ + 2H⁺ + 2H₂O
2NO₂⁻ + O₂ → 2NO₃⁻
In reductive amination, ammonia reacts with alpha-ketoglutaric acid and forms glutamic acid.
α-Ketoglutaric acid + NH₄⁺ + NADPH → Glutamate + H₂O + NADP⁺
Which of the following metabolites is common to respiration-mediated breakdown of fats, carbohydrates and proteins?
Acetyl-CoA is common to respiration-mediated breakdown of fats, carbohydrates, and proteins.
Glucose and fructose are phosphorylated to give glucose-6-phosphate by the activity of the enzyme hexokinase. Glucose-6-phosphate is then converted into fructose-6-phosphate, and further into fructose-1,6-bisphosphate.
Pyruvic acid is the end product of glycolysis.
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