NEET 2024 Question paper with answer key pdf T4 is available for download. NEET 2024 T4 question paper has been conducted by the NTA on May 5, 2024, in pen-paper mode. NEET 2024 question paper code T4 consists of 200 MCQs- 180 to be attempted in 200 minutes. Each of the 4 subjects (Zoology, Botany, Chemistry, Physics) in NEET T4 question paper 2024 have 50 MCQs (45 to be attempted). You can download NEET 2024 question paper with answer key with solutions PDF for T4 using the links given below. Check NEET Cutoff
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NEET 2024 Question Paper with Answer Key PDF T4 in English
| NEET 2024 Question Paper with Answer Key | Check Solutions |
The moment of inertia of a thin rod about an axis passing through its midpoint and perpendicular to the rod is 2400 g cm². The length of the 400 g rod is nearly:
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A bob is whirled in a horizontal plane by means of a string with an initial speed of \( \omega \) rpm. The tension in the string is \( T \). If speed becomes \( 2\omega \) while keeping the same radius, the tension in the string becomes:
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A thermodynamic system is taken through the cycle \( abcda \). The work done by the gas along the path \( bc \) is:
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In the nuclear reaction: \[ {}^{290}_{82}X \xrightarrow{\alpha} {}Y \xrightarrow{\beta^-} {}Z \xrightarrow{\beta^-} {}P \xrightarrow{e} {}Q \]
The mass number and atomic number of the product \( Q \) respectively, are:
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An unpolarised light beam strikes a glass surface at Brewster's angle. Then:
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If \( c \) is the velocity of light in free space, the correct statements about photon among the following are:
A. The energy of a photon is E = hv.
B. The velocity of a photon is c.
C. The momentum of a photon, v = hpc .
D. In a photon-electron collision, both total energy and total momentum are conserved.
E. Photon possesses positive charge.
Choose the correct answer from the options given below:
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Two bodies A and B of same mass undergo completely inelastic one dimensional collision. The body A moves with velocity \( v_1 \) while body B is at rest before collision. The velocity of the system after collision is \( v_2 \). The ratio \( v_1 : v_2 \) is
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A light ray enters through a right angled prism at point P with the angle of incidence 30° as shown in figure. It travels through the prism parallel to its base BC and emerges along the face AC. The refractive index of the prism is:
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If \( 5 \sin x = \pi + 3x \) represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are:
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At any instant of time \( t \), the displacement of any particle is given by \( 2t - 1 \) (SI unit) under the influence of force of 5 N. The value of instantaneous power is (in SI unit):
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A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as \( 4\pi \times 10^{-7} \) SI units):
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A particle moving with uniform speed in a circular path maintains:
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A logic circuit provides the output \( Y \) as per the following truth table:
| A | B | Y |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
The expression for the output \( Y \) is:
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Consider the following statements A and B and identify the correct answer:
A. For a solar-cell, the I-V characteristics lies in the IV quadrant of the given graph.
B. In a reverse biased pn junction diode, the current measured in \( \mu A \), is due to majority charge carriers.
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In an ideal transformer, the turns ratio is \( \frac{N_P}{N_S} = \frac{P}{S} \). The ratio \( V_S : V_P \) is equal to (the symbols carry their usual meaning):
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A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is \( \nu \) in the direction shown, which one of the following options is correct (P and Q are any highest and lowest points on the wheel, respectively)?
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If the monochromatic source in Young’s double slit experiment is replaced by white light, then
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In the given diagram, a strong bar magnet is moving towards solenoid-2 from solenoid-1. The direction of induced current in solenoid-1 and that in solenoid-2, respectively, are through the directions:
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In a vernier calipers, \( (N + 1) \) divisions of the vernier scale coincide with \( N \) divisions of the main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is:
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The output (Y) of the given logic gate is similar to the output of an/a
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Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: The potential (V) at any axial point, at 2 m distance (r) from the centre of the dipole of dipole moment vector \( \mathbf{P} \) of magnitude, \( 4 \times 10^{-6} \) C·m, is \( \pm 9 \times 10^3 \) V.
Reason R: The potential \( V = \pm \frac{2P}{4\pi \epsilon_0 r^2} \), where \( r \) is the distance of any axial point, situated at 2 m from the centre of the dipole.
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In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds. The moment of inertia of the needle is \( 9.8 \times 10^{-6} \) kg m². If the magnitude of the magnetic moment of the needle is \( x \times 10^{-5} \) Am², then the value of \( x \) is:
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Match List-I with List-II.
| List-I (Material) | List-II (Susceptibility (\( \chi \))) |
|---|---|
| A. Diamagnetic | I. \( \chi = 0 \) |
| B. Ferromagnetic | II. \( 0 > \chi \geq -1 \) |
| C. Paramagnetic | III. \( \chi \gg 1 \) |
| D. Non-magnetic | IV. \( 0 < \chi < \epsilon \) (a small positive number) |
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A horizontal force of 10 N is applied to a block A as shown in figure. The mass of blocks A and B are 2 kg and 3 kg, respectively. The blocks slide over a frictionless surface. The force exerted by block A on block B is:
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Given below are two statements:
Statement I: Atoms are electrically neutral as they contain equal numbers of positive and negative charges.
Statement II: Atoms of each element are stable and emit their characteristic spectrum.
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The terminal voltage of the battery, whose emf is 10 V and internal resistance 1 , when connected through an external resistance of 4 as shown in the figure, is:
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A wire of length ‘l’ and resistance 100 is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:
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The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young’s modulus, respectively, are \( 8 \times 10^8 \, N/m^2 \) and \( 2 \times 10^{11} \, N/m^2 \), is:
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A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If surface tension of water is 0.07 N m, then the excess force required to take it away from the surface is:
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Match List-I with List-II.
| List-I (Spectral Lines of Hydrogen for transitions from) | List-II (Wavelengths (nm)) |
|---|---|
| A. \( n_2 = 3 \) to \( n_1 = 2 \) | I. 410.2 |
| B. \( n_2 = 4 \) to \( n_1 = 2 \) | II. 434.1 |
| C. \( n_2 = 5 \) to \( n_1 = 2 \) | III. 656.3 |
| D. \( n_2 = 6 \) to \( n_1 = 2 \) | IV. 486.1 |
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In the following circuit, the equivalent capacitance between terminal A and terminal B is:
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The mass of a planet is \( \frac{1}{10} \)th that of the Earth and its diameter is half that of the Earth. The acceleration due to gravity on that planet is:
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The graph which shows the variation of \( \lambda \) and its kinetic energy, \( E \), is (where \( \lambda \) is the de Broglie wavelength of a free particle):
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The quantities which have the same dimensions as those of solid angle are:
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A thin spherical shell is charged by some source. The potential difference between the two points C and P (in V) shown in the figure is: (Take \( 9 \times 10^9 \) SI units)
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The following graph represents the T-V curves of an ideal gas (where T is the temperature and V the volume) at three pressures \( P_1 \), \( P_2 \), and \( P_3 \) compared with those of Charles’s law represented as dotted lines. Then the correct relation is:
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The property which is not of an electromagnetic wave traveling in free space is that:
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A small telescope has an objective of focal length 140 cm and an eye piece of focal length 5.0 cm. The magnifying power of telescope for viewing a distant object is:
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A parallel plate capacitor is charged by connecting it to a battery through a resistor. If \( I \) is the current in the circuit, then in the gap between the plates:
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A metallic bar of Young’s modulus, \( 0.5 \times 10^{11} \, N/m^2 \) and coefficient of linear thermal expansion \( 10^{-5} \, °C^{-1} \), length 1 m and area of cross-section \( 10^{-3} \, m^2 \) is heated from 0°C to 100°C without expansion or bending. The compressive force developed in it is:
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Two heaters A and B have power ratings of 1 kW and 2 kW, respectively. Those two are first connected in series and then in parallel to a fixed power source. The ratio of power outputs for these two cases is:
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An iron bar of length \( L \) has magnetic moment \( M \). It is bent at the middle of its length such that the two arms make an angle 60° with each other. The magnetic moment of this new magnet is:
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The velocity (v) – time (t) plot of the motion of a body is shown below:
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A 10 µF capacitor is connected to a 210 V, 50 Hz source as shown in figure. The peak current in the circuit is nearly (\(\pi\) = 3.14):
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A force defined by \( F = \alpha t^2 + \beta t \) acts on a particle at a given time t. The factor which is dimensionless, if \( \alpha \) and \( \beta \) are constants, is:
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Choose the correct circuit which can achieve the bridge balance.
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If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is \( 2^x \) times its original time period. Then the value of \( x \) is:
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If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then:
A. The charge stored in it increases.
B. The energy stored in it decreases.
C. Its capacitance increases.
D. The ratio of charge to its potential remains the same.
E. The product of charge and voltage increases.
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The minimum energy required to launch a satellite of mass \( m \) from the surface of Earth of mass \( M \) and radius \( R \) in a circular orbit at an altitude of \( 2R \) from the surface of the Earth is:
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A sheet is placed on a horizontal surface in front of a strong magnetic pole. A force is needed to:
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Match List-I with List-II.
| List-I (Process) | List-II (Conditions) |
|---|---|
| A. Isothermal process | II. Carried out at constant temperature |
| B. Isochoric process | III. Carried out at constant volume |
| C. Isobaric process | IV. Carried out at constant pressure |
| D. Adiabatic process | I. No heat exchange |
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Match List I with List II.
| List I (Complex) | List II (Type of Isomerism) |
|---|---|
| A. \([Co(NH_3)_5(NO_2)]Cl_2\) | II. Linkage isomerism |
| B. \([Co(NH_3)_5(SO_4)]Br\) | III. Ionization isomerism |
| C. \([Co(NH_3)_6][Cr(CN)_6]\) | IV. Coordination isomerism |
| D. \([Co(H_2O)_6]Cl_3\) | I. Solvate isomerism |
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The most stable carbocation among the following is:
On heating, some solid substances change from solid to vapour state without passing through the liquid state. The technique used for the purification of such solid substances based on the above principle is known as:
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Match List I with List II.
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Intramolecular hydrogen bonding is present in:
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The highest number of helium atoms is in:
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For the reaction \( 2A \rightleftharpoons B + C \), \( K_c = 4 \times 10^{-3} \). At a given time, the composition of the reaction mixture is: \( [A] = [B] = [C] = 2 \times 10^{-3} \) M. Then, which of the following is correct?
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The \( E^\circ \) value for the Mn\(^{3+}\)/Mn\(^{2+}\) couple is more positive than that of Cr\(^{3+}\)/Cr\(^{2+}\) or Fe\(^{3+}\)/Fe\(^{2+}\) due to change of:
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Fehling's solution 'A' is:
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Match List I with List II.
| List I (Compound) | List II (Shape/geometry) |
|---|---|
| A. NH3 | I. Trigonal Pyramidal |
| B. BrF5 | IV. Square Pyramidal |
| C. XeF4 | II. Square Planar |
| D. SF6 | III. Octahedral |
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Given below are two statements:
Statement I: Both \([Co(NH_3)_5]^{3+}\) and \([CoF_6]^{3-}\) complexes are octahedral but differ in their magnetic behaviour.
Statement II: \([Co(NH_3)_5]^{3+}\) is diamagnetic whereas \([CoF_6]^{3-}\) is paramagnetic.
In the light of the above statements, choose the correct answer from the options given below:
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Among Group 16 elements, which one does NOT show -2 oxidation state?
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Which plot of \( \ln k \) vs \( \frac{1}{T} \) is consistent with the Arrhenius equation?
Arrange the following elements in increasing order of electronegativity: N, O, F, C, Si
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Given below are two statements:
Statement I: The boiling point of three isomeric pentanes follows the order n-pentane \(>\) isopentane \(>\) neopentane.
Statement II: When branching increases, the molecule attains a shape of sphere. This results in smaller surface area for contact, due to which the intermolecular forces between the spherical molecules are weak, thereby lowering the boiling point.
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Which reaction is NOT a redox reaction?
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Arrange the following elements in increasing order of first ionization enthalpy: Li, Be, B, C, N
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Which one of the following alcohols reacts instantaneously with Lucas reagent?
Match List I with List II.
| List I (Molecule) | List II (Number and types of bonds between two carbon atoms) |
|---|---|
| A. ethane | III. one \( \sigma \)-bond |
| B. ethene | IV. one \( \sigma \)-bond and one \( \pi \)-bond |
| C. carbon molecule, C2 | II. two \( \pi \)-bonds |
| D. ethyne | I. one \( \sigma \)-bond and two \( \pi \)-bonds |
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Given below are two statements:
Statement I: The boiling point of hydrides of Group 16 elements follow the order
H2O > H2Te > H2Se > H2S
.Statement II: On the basis of molecular mass, \( \text{H_2O \) is expected to have lower boiling point than the other members of the group but due to the presence of extensive H-bonding in \( H_2O \), it has higher boiling point.
In the light of the above statements, choose the correct answer from the options given below:
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Given below are two statements:
Statement I: Aniline does not undergo Friedel-Crafts alkylation reaction.
Statement II: Aniline cannot be prepared through Gabriel synthesis.
In the light of the above statements, choose the correct answer from the options given below:
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Match List I with List II.
| List I (Conversion) | List II (Number of Faraday required) |
|---|---|
| A. 1 mol of H2O to O2 | I. 4F |
| B. 1 mol of MnO4 to Mn | III. 5F |
| C. 1.5 mol of Ca from molten CaCl2 | II. 3F |
| D. 1 mol of FeO to Fe2O3 | IV. 1F |
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In which of the following equilibria, Kp and Kc are NOT equal?
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The Henry’s law constant (Kh) values of three gases (A, B, C) in water are 145, \( 2 \times 10^{-5} \), and 35 kbar, respectively. The solubility of these gases in water follow the order:
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Identify the correct reagents that would bring about the following transformation.
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The compound that will undergo Sn1 reaction with the fastest rate is:
The energy of an electron in the ground state (\(n = 1\)) for \(\mathrm{He^+}\) ion is \(-x\) J, then that for an electron in \(n = 2\) state for \(\mathrm{Be^{3+}}\) ion in J is:
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A compound with a molecular formula of C\textsubscript{6}H\textsubscript{14} has two tertiary carbons. Its IUPAC name is:
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The reagents with which glucose does not react to give the corresponding tests/products are:
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'Spin only' magnetic moment is same for which of the following ions?
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Match List I with List II.
| List I (Quantum Number) | List II (Information provided) |
|---|---|
| A. ml | I. Shape of orbital |
| B. ms | II. Size of orbital |
| C. l | III. Orientation of orbital |
| D. n | IV. Orientation of spin of electron |
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1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution, the mass of sodium hydroxide left unreacted is equal to:
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In which of the following processes does entropy increase?
A. A liquid evaporates to vapor.
B. Temperature of a crystalline solid is lowered from 130 K to 0 K.
C. \( 2NaHCO_3 (s) \rightarrow Na_2CO_3 (s) + CO_2 (g) + H_2O (g) \)
D. \( Cl_2 (g) \rightarrow 2Cl(g) \)
Choose the correct answer from the options given below:
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Activation energy of any chemical reaction can be calculated if one knows the value of:
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Major products A and B formed in the following reaction sequence, are:
The work done during reversible isothermal expansion of one mole of hydrogen gas at 25°C from pressure of 20 atmosphere to 10 atmosphere is (Given R = 2.0 cal K mol)
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Consider the following reaction in a sealed vessel at equilibrium with concentrations of \[ N_2 = 3.0 \times 10^{-3} \, M, \quad O_2 = 4.2 \times 10^{-3} \, M, \quad NO = 2.8 \times 10^{-3} \, M. \]
If 0.1 mol/L of NO(g) is taken in a closed vessel, what will be the degree of dissociation (\( \alpha \)) of NO(g) at equilibrium?
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For the given reaction:
Identify the major product ('P').
The pair of lanthanide ions which are diamagnetic is:
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Identify the major product C formed in the following reaction sequence:
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The products A and B obtained in the following reactions, respectively, are: \[ 3ROH + PCl_5 \rightarrow 3RCl + A \] \[ ROH + PCl_5 \rightarrow RCl + HCl + B \]
Choose the correct answer from the options given below:
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Given below are certain cations. Using inorganic qualitative analysis,
arrange them in increasing group number from 0 to VI: \[ A. Al^{3+} \quad B. Cu^{2+} \quad C. Ba^{2+} \quad D. Co^{2+} \quad E. Mg^{2+} \]
Choose the correct answer from the options given below:
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A compound X contains 32% of A, 20% of B and the remaining percentage of C. Then, the empirical formula of X is: (Given atomic masses of A = 64; B = 40; C = 32 u)
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The rate of a reaction quadruples when temperature changes from 27°C to 57°C. Calculate the energy of activation. Given \(R = 8.314 \, J K^{-1} \, mol^{-1}\), \(log_4\) = 0.6021
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The plot of osmotic pressure (\(\Pi\)) vs concentration (mol L\(^{-1}\)) for a solution gives a straight line with slope 25.73 L bar mol\(^{-1}\). The temperature at which the osmotic pressure measurement is done is (Use \(R = 0.083 \, L bar mol^{-1} K^{-1}\)):
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During the preparation of Mohr's salt solution (Ferrous ammonium sulphate), which of the following acid is added to prevent hydrolysis of Fe\(^{2+}\) ion?
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Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 100 seconds is (Given: Molar mass of Cu: 63 g mol\(^{-1}\), 1 F = 96487 C):
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Identify the correct answer.
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Given below are two statements:
Statement I: \([Co(NH_3)_6]^{3+}\) is a homoleptic complex, whereas \([Co(NH_3)_4Cl_2]^+\) is a heteroleptic complex.
Statement II: Complex \([Co(NH_3)_6]^{3+}\) has only one kind of ligands but \([Co(NH_3)_4Cl_2]^+\) has more than one kind of ligands.
Choose the correct answer from the options given below:
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In the given figure, which component has thin outer walls and highly thickened inner walls?
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A transcription unit in DNA is defined primarily by the three regions in DNA and these are with respect to upstream and downstream ends;
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The equation of Verhulst-Pearl logistic growth is \[ \frac{dN}{dt} = rN \left( 1 - \frac{N}{K} \right) \]
From this equation, \( K \) indicates:
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Identify the part of the seed from the given figure which is destined to form the root when the seed germinates.
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Inhibition of Succinic dehydrogenase enzyme by malonate is a classical example of:
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A pink flowered Snapdragon plant was crossed with a red flowered Snapdragon plant. What type of phenotype/s is/are expected in the progeny?
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The type of conservation in which the threatened species are taken out from their natural habitat and placed in special settings where they can be protected and given special care is called
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These are regarded as major causes of biodiversity loss:
A. Over exploitation
B. Co-extinction
C. Mutation
D. Habitat loss and fragmentation
E. Migration
Choose the correct option:
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Which of the following are required for the dark reaction of photosynthesis?
A. Light
B. Chlorophyll
C. CO₂
D. ATP
E. NADPH
Choose the correct answer from the options given below:
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Bulliform cells are responsible for
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Identify the type of flowers based on the position of calyx, corolla, and androecium with respect to the ovary from the given figures (a) and (b)
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Which one of the following is not a criterion for classification of fungi?
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Hind II always cuts DNA molecules at a particular point called recognition sequence and it consists of:
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Auxin is used by gardeners to prepare weed-free lawns. But no damage is caused to grass as auxin
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Match List I with List II
| List I | List II | ||
|---|---|---|---|
| A. | Two or more alternative forms of a gene | I. | Back cross |
| B. | Cross of F1 progeny with homozygous recessive parent | II. | Ploidy |
| C. | Cross of F1 progeny with any of the parents | III. | Allele |
| D. | Number of chromosome sets in plant | IV. | Test cross |
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Spindle fibers attach to kinetochores of chromosomes during
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Match List I with List II
| List I | List II | ||
|---|---|---|---|
| A. | Clostridium butylicum | I. | Ethanol |
| B. | Saccharomyces cerevisiae | II. | Streptokinase |
| C. | Trichoderma polysporum | III. | Butyric acid |
| D. | Streptococcus sp. | IV. | Cyclosporin-A |
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Which one of the following can be explained on the basis of Mendel's Law of Dominance?
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What is the fate of a piece of DNA carrying only a gene of interest which is transferred into an alien organism?
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How many molecules of ATP and NADPH are required for every molecule of CO\(_2\) fixed in the Calvin cycle?
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In a plant, black seed color (BB/Bb) is dominant over white seed color (bb). In order to find out the genotype of the black seed plant, with which of the following genotypes will you cross it?
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Lecithin, a small molecular weight organic compound found in living tissues, is an example of:
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Match List I with List II
| List I | List II | ||
|---|---|---|---|
| A. | Rhizopus | III. | Bread mould |
| B. | Ustilago | II. | Smut fungus |
| C. | Puccinia | IV. | Rust fungus |
| D. | Agaricus | I. | Mushroom |
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Step 1: \textit{Rhizopus is a bread mould fungus commonly found on stale bread (Matching A \(\to\) III).
Step 2: \textit{Ustilago is a smut fungus, which affects cereal crops like maize (Matching B \(\to\) II).
Step 3: \textit{Puccinia is a rust fungus, known for causing rust diseases in wheat and other crops (Matching C \(\to\) IV).
Step 4: \textit{Agaricus includes mushrooms, which are edible fungi belonging to Basidiomycetes (Matching D \(\to\) I).
Thus, the correct answer is option (3), as the correct matches are A-III, B-II, C-IV, D-I. Quick Tip: Fungi are classified based on their reproductive structures and pathogenic effects—Rhizopus (bread mould), Ustilago (smut), Puccinia (rust), and Agaricus (mushroom).
Tropical regions show the greatest level of species richness because
A. Tropical latitudes have remained relatively undisturbed for millions of years, hence more time was available for species diversification.
B. Tropical environments are more seasonal.
C. More solar energy is available in tropics.
D. Constant environments promote niche specialization.
E. Tropical environments are constant and predictable.
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Step 1: Tropical regions show high species richness due to stable environmental conditions and longer evolutionary time.
Step 2: Statement A is correct as tropical regions have remained undisturbed for millions of years, allowing species diversification.
Step 3: Statement C is correct because higher solar energy in the tropics increases primary productivity, supporting diverse life forms.
Step 4: Statement D is valid as stable tropical climates promote niche specialization, allowing more species to coexist.
Step 5: Statement E is also correct because predictable tropical environments support species stability and survival.
Step 6: Statement B is incorrect because tropical environments are not more seasonal but rather stable compared to temperate regions.
Thus, the correct answer is option (3), as A, C, D, and E explain tropical species richness. Quick Tip: Tropical regions have high species richness due to evolutionary stability, abundant solar energy, and niche specialization.
Match List I with List II
| List I | List II | ||
|---|---|---|---|
| A. | Nucleolus | I. | Site of formation of glycolipid |
| B. | Centriole | II. | Organization like the cartwheel |
| C. | Leucoplasts | III. | Site for active ribosomal RNA synthesis |
| D. | Golgi apparatus | IV. | For storing nutrients |
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Step 1: The nucleolus is responsible for active ribosomal RNA synthesis, making A \(\to\) III correct.
Step 2: The centriole has a cartwheel-like organization, making B \(\to\) II correct.
Step 3: Leucoplasts store nutrients like starch, lipids, and proteins, making C \(\to\) IV correct.
Step 4: The Golgi apparatus is involved in the formation of glycolipids and glycoproteins, making D \(\to\) I correct.
Thus, the correct answer is option (3), as A-III, B-II, C-IV, D-I are the correct matches. Quick Tip: The nucleolus synthesizes rRNA, centrioles have a cartwheel structure, leucoplasts store nutrients, and the Golgi apparatus forms glycolipids.
The lactose present in the growth medium of bacteria is transported to the cell by the action of
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Step 1: Lactose transport in bacteria is facilitated by lactose permease, an enzyme encoded by the \textit{lacY gene in the lac operon.
Step 2: Lactose permease acts as a membrane transporter, enabling lactose to enter the bacterial cell.
Step 3: Beta-galactosidase (\textit{lacZ) is responsible for lactose hydrolysis, not transport. Acetylase (\textit{lacA) and polymerase do not play a role in lactose uptake.
Thus, the correct answer is option (1), as permease transports lactose into the bacterial cell. Quick Tip: Lactose permease enables lactose entry into bacterial cells, while beta-galactosidase hydrolyzes it into glucose and galactose.
Given below are two statements:
Statement I: Chromosomes become gradually visible under light microscope during leptotene stage.
Statement II: The beginning of diplotene stage is recognized by dissolution of synaptonemal complex.
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Step 1: Leptotene stage is the first stage of prophase I in meiosis, where chromosomes gradually become visible under the light microscope. (Statement I is true)
Step 2: Diplotene stage is marked by the dissolution of the synaptonemal complex, leading to the formation of chiasmata. (Statement II is true)
Thus, the correct answer is option (3), as both statements are true. Quick Tip: During leptotene, chromosomes condense and become visible, while during diplotene, the synaptonemal complex dissolves, forming chiasmata.
Formation of interfascicular cambium from fully developed parenchyma cells is an example of
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Step 1: Dedifferentiation refers to the process where mature, differentiated cells regain the ability to divide and form meristematic tissues.
Step 2: Interfascicular cambium is formed when parenchyma cells dedifferentiate and regain meristematic activity.
Step 3: Differentiation refers to cells becoming specialized, while redifferentiation applies to cells that regain function after dedifferentiation. Maturation does not apply here.
Thus, the correct answer is option (1), as formation of interfascicular cambium is an example of dedifferentiation. Quick Tip: Dedifferentiation allows mature cells to revert to a meristematic state, helping in secondary growth in plants.
Given below are two statements:
Statement I: Parenchyma is living but collenchyma is dead tissue.
Statement II: Gymnosperms lack xylem vessels but presence of xylem vessels is the characteristic of angiosperms.
View Solution
Step 1: Parenchyma and collenchyma are both living tissues in plants. Statement I is false because collenchyma is living, not dead.
Step 2: Gymnosperms lack xylem vessels and have only tracheids for water conduction, whereas angiosperms possess xylem vessels for efficient water transport. Statement II is true.
Thus, the correct answer is option (2), as Statement I is false but Statement II is true. Quick Tip: Parenchyma and collenchyma are both living tissues. Gymnosperms lack xylem vessels, while angiosperms have them for efficient conduction.
Identify the set of correct statements:
View Solution
Step 1: Vallisneria is a water-pollinated plant, but its flowers are not colourful and do not produce nectar. This makes Statement A incorrect.
Step 2: Water lily is an insect-pollinated plant (entomophilous), not pollinated by water. Statement B is correct.
Step 3: In water-pollinated species (hydrophily), pollen grains are covered with a mucilaginous coat to prevent wetting. Statement C is correct.
Step 4: Pollen grains of certain hydrophytes, such as \textit{Zostera, are long and ribbon-like, helping them move in water. Statement D is correct.
Step 5: In some hydrophytes, pollen grains are passively carried inside water to reach the stigma. Statement E is correct.
Thus, the correct answer is option (2), as B, C, D, and E are correct. Quick Tip: Hydrophilous plants like \textit{Vallisneria and Zostera rely on water for pollination, with adaptations like mucilaginous pollen and ribbon-like grains.
Which of the following is an example of an actinomorphic flower?
View Solution
Step 1: Actinomorphic flowers exhibit radial symmetry, meaning they can be divided into equal halves along multiple planes.
Step 2: Zygomorphic flowers exhibit bilateral symmetry, meaning they can only be divided into two equal halves along one plane.
Step 3: Datura is an actinomorphic flower, making option (3) correct.
Step 4: \textit{Pisum (\textit{pea), \textit{Sesbania, and \textit{Cassia are zygomorphic flowers, making them incorrect answers.
Thus, the correct answer is option (3), as \textit{Datura is actinomorphic. Quick Tip: Actinomorphic flowers like \textit{Datura have radial symmetry, while zygomorphic flowers like Pisum and Cassia have bilateral symmetry.
The capacity to generate a whole plant from any cell of the plant is called:
View Solution
Step 1: Totipotency is the ability of a single plant cell to regenerate into an entire plant under appropriate conditions.
Step 2: This property is widely used in plant tissue culture to generate plants from explants.
Step 3: Differentiation refers to the process by which cells become specialized. Somatic hybridization involves fusing somatic cells, while micropropagation is a technique to propagate plants using tissue culture.
Thus, the correct answer is option (3), as totipotency enables the regeneration of a whole plant from a single cell. Quick Tip: Totipotency is the basis of plant tissue culture, allowing entire plants to be regenerated from a single cell.
Given below are two statements:
Statement I: Bt toxins are insect group specific and coded by a gene \textit{cry IAc.
Statement II: Bt toxin exists as inactive protoxin in \textit{B. thuringiensis. However, after ingestion by the insect, the inactive protoxin gets converted into active form due to acidic pH of the insect gut.
View Solution
Step 1: Bt toxin is produced by the bacterium \textit{Bacillus thuringiensis and is encoded by the cry gene. Statement I is correct.
Step 2: The Bt toxin exists as an inactive protoxin in \textit{B. thuringiensis, but it gets activated in the insect gut due to alkaline pH, not acidic pH. Statement II is incorrect.
Thus, the correct answer is option (1), as Statement I is true but Statement II is false. Quick Tip: Bt toxin is a crystalline protein that remains inactive until activated in the alkaline pH of the insect gut, where it disrupts the midgut epithelium.
List of endangered species was released by
View Solution
Step 1: The International Union for Conservation of Nature (IUCN) maintains the Red List, which classifies species based on their extinction risk.
Step 2: The IUCN Red List categories include Extinct (EX), Critically Endangered (CR), Endangered (EN), Vulnerable (VU), Near Threatened (NT), and Least Concern (LC).
Step 3: WWF (World Wildlife Fund) is an environmental organization but does not maintain the Red List. GEAC (Genetic Engineering Approval Committee) deals with genetically modified organisms, and FOAM is unrelated.
Thus, the correct answer is option (2), as IUCN is responsible for listing endangered species. Quick Tip: The IUCN Red List is the global authority on species conservation status, assessing their risk of extinction.
The cofactor of the enzyme carboxypeptidase is:
View Solution
Step 1: Carboxypeptidase is a metalloenzyme involved in protein digestion, requiring zinc (Zn\(^{2+}\)) as a cofactor for its catalytic activity.
Step 2: The zinc ion in carboxypeptidase helps in stabilizing the enzyme-substrate complex and aids in hydrolysis of peptide bonds.
Step 3: Flavin is associated with flavoproteins in oxidation-reduction reactions, haem is found in hemoglobin and cytochromes, and niacin is a precursor for NAD\(^+\)/NADP\(^+\) coenzymes.
Thus, the correct answer is option (3), as zinc is the cofactor for carboxypeptidase. Quick Tip: Carboxypeptidase is a zinc-dependent metalloenzyme involved in the hydrolysis of peptide bonds during protein digestion.
Match List I with List II
| List I | List II | ||
|---|---|---|---|
| A. | Citric acid cycle | I. | Mitochondrial matrix |
| B. | Glycolysis | II. | Cytoplasm |
| C. | Electron transport system | III. | Inner mitochondrial membrane |
| D. | Proton gradient | IV. | Intermembrane space of mitochondria |
View Solution
Step 1: The citric acid cycle (Krebs cycle) occurs in the mitochondrial matrix. (Matching A \(\to\) II).
Step 2: Glycolysis is the breakdown of glucose and takes place in the cytoplasm. (Matching B \(\to\) I).
Step 3: The electron transport system (ETS) is located in the inner mitochondrial membrane. (Matching C \(\to\) IV).
Step 4: The proton gradient required for ATP synthesis is established in the intermembrane space of mitochondria. (Matching D \(\to\) III).
Thus, the correct answer is option (4), as the correct matches are A-II, B-I, C-IV, D-III. Quick Tip: The citric acid cycle occurs in the mitochondrial matrix, glycolysis in the cytoplasm, ETS in the inner mitochondrial membrane, and the proton gradient in the intermembrane space.
Which of the following statements is correct regarding the process of replication in E. coli?
View Solution
Step 1: DNA replication in E. coli follows the 5' \(\to\) 3' polymerization rule.
Step 2: DNA-dependent DNA polymerase adds nucleotides in the 5' \(\to\) 3' direction using a DNA template.
Step 3: The enzyme also has 3' \(\to\) 5' exonuclease activity, but polymerization does not occur in this direction.
Step 4: Option (1) is incorrect because DNA polymerase cannot catalyze polymerization in both directions.
Step 5: Option (3) is incorrect because polymerization does not occur in the 3' \(\to\) 5' direction. Option (4) is incorrect because RNA polymerase synthesizes RNA, not DNA.
Thus, the correct answer is option (2), as DNA polymerase catalyzes DNA synthesis in the 5' \(\to\) 3' direction. Quick Tip: DNA polymerase in \textit{E. coli synthesizes new DNA strands in the 5' \(\to\) 3' direction while also having 3' \(\to\) 5' exonuclease proofreading activity.
Match List I with List II
| List I | List II | ||
|---|---|---|---|
| A. | Robert May | I. | Species-Area relationship |
| B. | Alexander von Humboldt | III. | Global species diversity at about 7 million |
| C. | Paul Ehrlich | IV. | Rivet popper hypothesis |
| D. | David Tilman | II. | Long-term ecosystem experiment using outdoor plots |
View Solution
Step 1: Robert May estimated global species diversity to be around 7 million, making (A \(\to\) III) correct.
Step 2: Alexander von Humboldt proposed the Species-Area relationship, which describes how species richness increases with increasing habitat area, making (B \(\to\) I) correct.
Step 3: Paul Ehrlich gave the Rivet Popper Hypothesis, which compares species in an ecosystem to rivets in an airplane, emphasizing the importance of species conservation, making (C \(\to\) IV) correct.
Step 4: David Tilman conducted long-term ecosystem experiments using outdoor plots to study biodiversity and ecosystem functioning, making (D \(\to\) II) correct.
Thus, the correct answer is option (4), as the correct matches are A-III, B-I, C-IV, D-II. Quick Tip: Humboldt described the species-area relationship, May estimated global species diversity, Ehrlich proposed the rivet popper hypothesis, and Tilman conducted long-term ecosystem experiments.
Identify the correct description about the given figure:
View Solution
Wind-pollinated plants typically have exposed stamens and flowers that are adapted to release pollen into the air. This type of inflorescence is designed to facilitate wind pollination.
Thus, the correct answer is
(3) Wind-pollinated plant inflorescence showing flowers with well-exposed stamens. Quick Tip: Wind-pollinated flowers often have exposed stamens and reduced petals, facilitating the dispersal of pollen by wind.
Identify the step in tricarboxylic acid cycle, which does not involve oxidation of substrate.
View Solution
Step 1: The tricarboxylic acid (TCA) cycle, also known as the Krebs cycle, involves multiple oxidation reactions.
Step 2: The conversion of Succinyl-CoA to Succinic acid is catalyzed by Succinyl-CoA synthetase, where substrate-level phosphorylation occurs instead of oxidation. (Correct step)
Step 3: The reactions Isocitrate to \(\alpha\)-ketoglutarate and Malate to Oxaloacetate involve NAD\(^+\)-dependent oxidation, making them incorrect options.
Step 4: The step Succinic acid to Malic acid involves oxidation via FAD-dependent dehydrogenation, making it incorrect.
Thus, the correct answer is option (1), as Succinyl-CoA to Succinic acid is a phosphorylation step, not an oxidation step. Quick Tip: The conversion of Succinyl-CoA to Succinic acid in the Krebs cycle is a substrate-level phosphorylation step, unlike other oxidation reactions.
Given below are two statements:
Statement I: In C\(_3\) plants, some O\(_2\) binds to RuBisCO, hence CO\(_2\) fixation is decreased.
Statement II: In C\(_4\) plants, mesophyll cells show very little photorespiration while bundle sheath cells do not show photorespiration.
View Solution
Step 1: C\(_3\) plants exhibit photorespiration, where RuBisCO binds to O\(_2\) instead of CO\(_2\), reducing CO\(_2\) fixation efficiency. Statement I is correct.
Step 2: C\(_4\) plants have an adaptation to minimize photorespiration using the Hatch-Slack pathway, but both mesophyll and bundle sheath cells contribute to the process. Statement II is incorrect because mesophyll cells do not show high levels of photorespiration.
Thus, the correct answer is option (1), as Statement I is true but Statement II is false. Quick Tip: C\(_3\) plants undergo photorespiration due to RuBisCO's oxygenase activity, while C\(_4\) plants minimize it using a spatial separation of CO\(_2\) fixation.
In an ecosystem if the Net Primary Productivity (NPP) of first trophic level is 100\(x\) (kcal m\(^{-2}\) yr\(^{-1}\)), what would be the GPP (Gross Primary Productivity) of the third trophic level of the same ecosystem?
View Solution
Step 1: In an ecosystem, energy transfer follows the 10 law, meaning only 10 of energy is transferred to the next trophic level, while 90 is lost as heat.
Step 2: Given that NPP of the first trophic level is 100\(x\) kcal m\(^{-2}\) yr\(^{-1}\), the second trophic level would receive 10\(x\) kcal m\(^{-2}\) yr\(^{-1}\).
Step 3: The third trophic level would receive 10 of 10\(x\) kcal m\(^{-2}\) yr\(^{-1}\), which equals 1\(x\) kcal m\(^{-2}\) yr\(^{-1}\).
Thus, the correct answer is option (1), as the GPP of the third trophic level is 10\(x\) kcal m\(^{-2}\) yr\(^{-1}\). Quick Tip: According to Lindeman's 10 law, only 10 of energy is transferred to the next trophic level, while 90 is lost as heat.
Match List-I with List-II
| List I | List II | ||
|---|---|---|---|
| A. | GLUT-4 | I. | Enables glucose transport into cells |
| B. | Insulin | II. | Hormone |
| C. | Trypsin | III. | Enzyme |
| D. | Collagen | IV. | Intercellular ground substance |
View Solution
Step 1: GLUT-4 (Glucose Transporter 4) is responsible for glucose transport into cells in response to insulin. (Matching A \(\to\) IV).
Step 2: Insulin is a hormone that regulates blood glucose levels. (Matching B \(\to\) I).
Step 3: Trypsin is an enzyme involved in protein digestion in the small intestine. (Matching C \(\to\) II).
Step 4: Collagen is a structural protein found in the extracellular matrix and provides support. (Matching D \(\to\) III).
Thus, the correct answer is option (3), as the correct matches are A-IV, B-I, C-II, D-III. Quick Tip: GLUT-4 enables glucose uptake, insulin is a hormone, trypsin is a digestive enzyme, and collagen provides structural support in tissues.
Match List I with List II
| List I | List II | ||
|---|---|---|---|
| A. | Frederick Griffith | I. | Transformation |
| B. | Francois Jacob & Jacque Monod | II. | Lac operon |
| C. | Har Gobind Khorana | III. | Genetic code |
| D. | Meselson & Stahl | IV. | Semi-conservative mode of DNA replication |
View Solution
Step 1: Frederick Griffith discovered the process of transformation in bacteria using the pneumococcus experiment, making (A \(\to\) III) correct.
Step 2: Francois Jacob and Jacque Monod proposed the Lac operon model, which explains gene regulation in prokaryotes, making (B \(\to\) IV) correct.
Step 3: Har Gobind Khorana played a significant role in deciphering the genetic code, making (C \(\to\) I) correct.
Step 4: Meselson and Stahl experimentally proved the semi-conservative mode of DNA replication in \textit{E. coli, making (D \(\to\) II) correct.
Thus, the correct answer is option (4), as the correct matches are A-III, B-IV, C-I, D-II. Quick Tip: Griffith discovered transformation, Jacob and Monod explained the \textit{Lac operon, Khorana decoded the genetic code, and Meselson \& Stahl demonstrated semi-conservative replication.
The DNA present in chloroplast is:
View Solution
Step 1: Chloroplast DNA (cpDNA) resembles prokaryotic DNA in structure and organization.
Step 2: It is circular and double-stranded, similar to mitochondrial DNA (mtDNA) and bacterial DNA.
Step 3: Unlike nuclear DNA, which is linear, chloroplast DNA does not have histones and exists in multiple copies within the organelle.
Thus, the correct answer is option (4), as chloroplast DNA is circular and double-stranded. Quick Tip: Chloroplast DNA (cpDNA) is circular, double-stranded, and resembles prokaryotic DNA, supporting the endosymbiotic theory.
Spraying sugarcane crop with which of the following plant growth regulators, increases the length of stem, thus, increasing the yield?
View Solution
Step 1: Gibberellins (GA) are plant hormones that promote stem elongation, seed germination, and flowering.
Step 2: In sugarcane, gibberellins stimulate internode elongation, leading to an increase in yield.
Step 3: Cytokinins promote cell division, auxins regulate root growth, and abscisic acid induces dormancy rather than elongation.
Thus, the correct answer is option (4), as gibberellin increases stem length in sugarcane. Quick Tip: Gibberellins promote internode elongation in sugarcane, increasing yield by enhancing stem growth.
Read the following statements and choose the set of correct statements:
In the members of Phaeophyceae,
A. Asexual reproduction occurs usually by biflagellate zoospores.
B. Sexual reproduction is by oogamous method only.
C. Stored food is in the form of carbohydrates which is either mannitol or laminarin.
D. The major pigments found are chlorophyll \textit{a, \textit{c, carotenoids, and xanthophyll.
E. Vegetative cells have a cellulosic wall, usually covered on the outside by gelatinous coating of algin.
Choose the correct answer from the options given below:
View Solution
Step 1: Phaeophyceae (Brown Algae) reproduce asexually via biflagellate zoospores. (Statement A is correct)
Step 2: Sexual reproduction in Phaeophyceae can be isogamous, anisogamous, or oogamous, not just oogamous. (Statement B is incorrect)
Step 3: The stored food in Phaeophyceae is mannitol or laminarin. (Statement C is correct)
Step 4: Their pigments include chlorophyll a, c, carotenoids, and xanthophylls. (Statement D is correct)
Step 5: Their vegetative cells have a cellulosic wall covered by algin, a gelatinous coating. (Statement E is correct)
Thus, the correct answer is option (1), as A, C, D, and E are correct. Quick Tip: Phaeophyceae store food as laminarin or mannitol and reproduce via biflagellate zoospores, with a cell wall covered in algin.
Which of the following are fused in somatic hybridization involving two varieties of plants?
View Solution
Step 1: Somatic hybridization is a technique where protoplasts from different plant varieties are fused to produce a hybrid cell.
Step 2: Protoplast fusion is achieved using PEG (Polyethylene Glycol) or electrofusion, leading to the formation of somatic hybrids.
Step 3: Pollens are involved in sexual reproduction, not somatic hybridization. Callus is a mass of undifferentiated cells, and somatic embryos arise from tissue culture but are not fused in somatic hybridization.
Thus, the correct answer is option (1), as protoplast fusion is the key step in somatic hybridization. Quick Tip: Somatic hybridization involves protoplast fusion, enabling genetic recombination between two plant varieties.
Match List I with List II
| List I | List II | ||
|---|---|---|---|
| A. | Rose | II. | Perigynous flower |
| B. | Pea | IV. | Marginal placentation |
| C. | Cotton | I. | Twisted aestivation |
| D. | Mango | III. | Drupe |
View Solution
Step 1: Rose exhibits perigynous flowers, where the ovary is surrounded by floral parts at the same level. (Matching A \(\to\) II)
Step 2: Pea has marginal placentation, where ovules are attached along one side of the ovary. (Matching B \(\to\) IV)
Step 3: Cotton has twisted aestivation, where one petal overlaps another. (Matching C \(\to\) I)
Step 4: Mango is a drupe fruit, with a hard endocarp protecting the seed. (Matching D \(\to\) III)
Thus, the correct answer is option (3), as the correct matches are A-II, B-IV, C-I, D-III. Quick Tip: Rose has perigynous flowers, pea has marginal placentation, cotton shows twisted aestivation, and mango is a drupe fruit.
Match List I with List II
| List I (Types of Stamens) | List II (Example) | ||
|---|---|---|---|
| A. | Monoadelphous | I. | Citrus |
| B. | Diadelphous | II. | Pea |
| C. | Polyadelphous | IV. | China-rose |
| D. | Epiphyllous | III. | Lily |
View Solution
Step 1: Monoadelphous stamens have filaments fused into a single bundle, seen in China-rose (Hibiscus). (Matching A \(\to\) IV)
Step 2: Diadelphous stamens have filaments arranged in two bundles, as in Pea (Fabaceae family). (Matching B \(\to\) II)
Step 3: Polyadelphous stamens have multiple groups of fused filaments, as in Citrus. (Matching C \(\to\) I)
Step 4: Epiphyllous stamens are attached to petals, seen in Lily (Liliaceae family). (Matching D \(\to\) III)
Thus, the correct answer is option (3), as the correct matches are A-IV, B-II, C-I, D-III. Quick Tip: Monoadelphous stamens (one bundle) are found in China-rose, diadelphous (two bundles) in Pea, polyadelphous (multiple bundles) in Citrus, and epiphyllous (attached to petals) in Lily.
Match List I with List II
| List I | List II | ||
|---|---|---|---|
| A. | Common cold | III. | Rhinoviruses |
| B. | Haemozoin | I. | Plasmodium |
| C. | Widal test | II. | Typhoid |
| D. | Allergy | IV. | Dust mites |
View Solution
Step 1: Common cold is caused by Rhinoviruses. (Matching A \(\to\) III)
Step 2: Haemozoin is a malarial pigment produced by Plasmodium during its life cycle. (Matching B \(\to\) I)
Step 3: Widal test is a diagnostic test for Typhoid, caused by \textit{Salmonella typhi. (Matching C \(\to\) II)
Step 4: Allergies are triggered by allergens such as dust mites. (Matching D \(\to\) IV)
Thus, the correct answer is option (1), as the correct matches are A-III, B-I, C-II, D-IV. Quick Tip: Rhinoviruses cause the common cold, \textit{Plasmodium produces haemozoin, Widal test diagnoses typhoid, and dust mites trigger allergies.
The flippers of Penguins and Dolphins are an example of:
View Solution
Step 1: Convergent evolution occurs when unrelated organisms evolve similar traits due to similar environmental pressures.
Step 2: Penguins (birds) and Dolphins (mammals) belong to different groups but have flippers adapted for swimming, an example of convergent evolution.
Step 3: Divergent evolution leads to different traits in related species, adaptive radiation involves diversification from a common ancestor, and natural selection drives evolution but does not always lead to convergence.
Thus, the correct answer is option (1), as flippers in Penguins and Dolphins evolved due to convergent evolution. Quick Tip: Convergent evolution leads to similar structures in unrelated species due to similar environmental pressures, as seen in Penguins and Dolphins.
Given below are some stages of human evolution. Arrange them in correct sequence (Past to Recent).
A. \textit{Homo habilis
B. \textit{Homo sapiens
C. \textit{Homo neanderthalensis
D. \textit{Homo erectus
Choose the correct sequence of human evolution from the options given below:
View Solution
Step 1: The correct sequence of human evolution is:
- Homo habilis (earliest tool users) → \textit{Homo erectus (upright walkers) →
- \textit{Homo neanderthalensis (Neanderthals) → \textit{Homo sapiens (modern humans).
Step 2: \textit{Homo habilis was the earliest species, followed by \textit{Homo erectus, then \textit{Homo neanderthalensis, and finally \textit{Homo sapiens.
Thus, the correct answer is option (2), as A-D-C-B represents the correct evolutionary order. Quick Tip: The sequence of human evolution follows: \textit{Homo habilis → Homo erectus → Homo neanderthalensis → Homo sapiens.
Which one of the following factors will not affect the Hardy-Weinberg equilibrium?
View Solution
Step 1: The Hardy-Weinberg equilibrium states that allele frequencies remain constant in a population under stable conditions.
Step 2: Gene migration, genetic recombination, and genetic drift cause changes in allele frequencies, disturbing equilibrium.
Step 3: A constant gene pool means no changes in allele frequencies, keeping the population in equilibrium.
Thus, the correct answer is option (2), as a constant gene pool maintains Hardy-Weinberg equilibrium. Quick Tip: Hardy-Weinberg equilibrium remains stable if the gene pool is constant, meaning no evolutionary forces act on the population.
Which of the following factors are favourable for the formation of oxyhaemoglobin in alveoli?
View Solution
Step 1: Oxyhaemoglobin (HbO\(_2\)) formation occurs in the alveoli, where oxygen binds to hemoglobin.
Step 2: High pO\(_2\) (partial pressure of oxygen) favors oxygen binding to hemoglobin.
Step 3: Low pCO\(_2\) and lesser H\(^+\) concentration create an alkaline environment, promoting HbO\(_2\) formation.
Step 4: High pCO\(_2\) and high H\(^+\) concentration (acidic environment) promote oxygen release (Bohr effect), making options (1) and (3) incorrect.
Thus, the correct answer is option (4), as high pO\(_2\) and low H\(^+\) concentration favor oxyhaemoglobin formation. Quick Tip: In alveoli, high pO\(_2\) and lower H\(^+\) concentration favor oxyhaemoglobin formation, while high pCO\(_2\) promotes oxygen release.
Which of the following is not a natural/traditional contraceptive method?
View Solution
Step 1: Natural contraceptive methods involve preventing pregnancy without external devices or hormones.
Step 2: Lactational amenorrhea, coitus interruptus, and periodic abstinence are natural contraceptive methods.
Step 3: Vaults are barrier contraceptives (diaphragms), which physically prevent sperm entry, making them a non-natural method.
Thus, the correct answer is option (2), as vaults are not a natural contraceptive method. Quick Tip: Natural contraceptive methods rely on behavioral practices, while vaults (diaphragms) are physical barriers used for contraception.
Match List I with List II
| List I | List II | ||
|---|---|---|---|
| A. | Pons | III. | Connects different regions of the brain |
| B. | Hypothalamus | IV. | Neurosecretory cells |
| C. | Medulla | II. | Controls respiration and gastric secretions |
| D. | Cerebellum | I. | Provides space for neurons, regulates posture and balance |
View Solution
Step 1: Pons acts as a bridge connecting different brain regions. (Matching A \(\to\) III)
Step 2: Hypothalamus contains neurosecretory cells, playing a key role in hormonal regulation. (Matching B \(\to\) IV)
Step 3: Medulla oblongata controls respiration and gastric secretions. (Matching C \(\to\) II)
Step 4: Cerebellum regulates posture, balance, and coordination. (Matching D \(\to\) I)
Thus, the correct answer is option (4), as the correct matches are A-III, B-IV, C-II, D-I. Quick Tip: The cerebellum controls balance, the medulla regulates involuntary functions, the hypothalamus has neurosecretory cells, and the pons connects brain regions.
Given below are two statements:
Statement I: The presence or absence of hymen is not a reliable indicator of virginity.
Statement II: The hymen is torn during the first coitus only.
View Solution
Step 1: The hymen can be torn due to various non-sexual activities like sports, cycling, or physical exertion. Statement I is true.
Step 2: The belief that the hymen tears only during the first coitus is incorrect, as it may already be absent due to other factors. Statement II is false.
Thus, the correct answer is option (1), as Statement I is true but Statement II is false. Quick Tip: The hymen is not a definitive indicator of virginity, as it can be torn due to various physical activities unrelated to sexual intercourse.
Match List I with List II
| List I | List II | ||
|---|---|---|---|
| A. | Axoneme | I. | Cilia and flagella |
| B. | Cartwheel pattern | II. | Centriole |
| C. | Crista | III. | Mitochondria |
| D. | Satellite | IV. | Chromosome |
View Solution
Step 1: Axoneme is the structural core of cilia and flagella, consisting of microtubules. (Matching A \(\to\) II)
Step 2: Cartwheel pattern is a feature of centrioles, seen in their early formation stage. (Matching B \(\to\) I)
Step 3: Crista are the folds of the inner mitochondrial membrane, increasing the surface area for ATP production. (Matching C \(\to\) IV)
Step 4: Satellite DNA refers to repetitive DNA sequences found in chromosomes. (Matching D \(\to\) III)
Thus, the correct answer is option (2), as the correct matches are A-II, B-I, C-IV, D-III. Quick Tip: Axoneme forms the core of cilia and flagella, cartwheel pattern is found in centrioles, cristae are folds of mitochondria, and satellite DNA is a part of chromosomes.
Match List I with List II
| List I | List II | ||
|---|---|---|---|
| A. | Typhoid | I. | Bacteria |
| B. | Leishmaniasis | II. | Protozoa |
| C. | Ringworm | III. | Fungus |
| D. | Filariasis | IV. | Nematode |
View Solution
Step 1: Typhoid is caused by the bacterium \textit{Salmonella typhi. (Matching A \(\to\) IV)
Step 2: Leishmaniasis is caused by the protozoan parasite \textit{Leishmania donovani. (Matching B \(\to\) III)
Step 3: Ringworm is a fungal infection caused by \textit{Trichophyton, \textit{Microsporum, or \textit{Epidermophyton. (Matching C \(\to\) I)
Step 4: Filariasis is caused by nematode worms such as \textit{Wuchereria bancrofti. (Matching D \(\to\) II)
Thus, the correct answer is option (4), as the correct matches are A-IV, B-III, C-I, D-II. Quick Tip: Typhoid is caused by bacteria, Leishmaniasis by protozoa, ringworm by fungi, and filariasis by nematodes.
Given below are two statements:
Statement I: In the nephron, the descending limb of loop of Henle is impermeable to water and permeable to electrolytes.
Statement II: The proximal convoluted tubule is lined by simple columnar brush border epithelium and increases the surface area for reabsorption.
View Solution
Step 1: The descending limb of the loop of Henle is permeable to water but impermeable to electrolytes. Thus, Statement I is false.
Step 2: The proximal convoluted tubule (PCT) is lined by simple cuboidal brush border epithelium, not columnar epithelium. Thus, Statement II is false.
Thus, the correct answer is option (4), as both Statement I and Statement II are false. Quick Tip: The descending limb of the loop of Henle is water-permeable, while the proximal convoluted tubule has simple cuboidal brush border epithelium for increased absorption.
Match List I with List II
| List I | List II | ||
|---|---|---|---|
| A. | α-1 antitrypsin | I. | Emphysema |
| B. | Cry IAb | II. | Corn borer |
| C. | Cry IAc | III. | Cotton bollworm |
| D. | Enzyme replacement therapy | IV. | ADA deficiency |
View Solution
Step 1: \(\alpha\)-1 antitrypsin is used in the treatment of emphysema. (Matching A \(\to\) III)
Step 2: Cry IAb is a Bt toxin that is effective against corn borers. (Matching B \(\to\) IV)
Step 3: Cry IAc is another Bt toxin, specifically targeting cotton bollworm. (Matching C \(\to\) I)
Step 4: Enzyme replacement therapy is used for ADA deficiency, a genetic disorder affecting the immune system. (Matching D \(\to\) II)
Thus, the correct answer is option (1), as the correct matches are A-III, B-IV, C-I, D-II. Quick Tip: \(\alpha\)-1 antitrypsin treats emphysema, Cry IAb targets corn borers, Cry IAc affects cotton bollworms, and enzyme replacement therapy helps in ADA deficiency.
Match List I with List II
| List I | List II | ||
|---|---|---|---|
| A. | Non-medicated IUD | I. | Lippes loop |
| B. | Copper releasing IUD | II. | Multiload 375 |
| C. | Hormone releasing IUD | III. | LNG-20 |
| D. | Implants | IV. | Progestogens |
View Solution
Step 1: Non-medicated IUDs include Lippes loop. (Matching A \(\to\) III)
Step 2: Copper-releasing IUDs include Multiload 375. (Matching B \(\to\) I)
Step 3: Hormone-releasing IUDs include LNG-20. (Matching C \(\to\) IV)
Step 4: Implants release progestogens, acting as long-term contraceptives. (Matching D \(\to\) II)
Thus, the correct answer is option (2), as the correct matches are A-III, B-I, C-IV, D-II. Quick Tip: Non-medicated IUDs include Lippes loop, Copper IUDs include Multiload 375, hormone-releasing IUDs include LNG-20, and implants release progestogens.
Consider the following statements:
A. Annelids are true coelomates
B. Poriferans are pseudocoelomates
C. Aschelminthes are acoelomates
D. Platyhelminthes are pseudocoelomates
Choose the correct answer from the options given below:
View Solution
Step 1: Annelids (e.g., earthworms) have a true coelom, which is mesodermally derived. Statement A is correct.
Step 2: Poriferans (sponges) lack body cavities altogether, making Statement B incorrect.
Step 3: Aschelminthes (Nematodes) are pseudocoelomates, not acoelomates, making Statement C incorrect.
Step 4: Platyhelminthes (Flatworms) are acoelomates, not pseudocoelomates, making Statement D incorrect.
Thus, the correct answer is option (4), as only Statement A is correct. Quick Tip: Annelids have a true coelom, Poriferans lack body cavities, Aschelminthes are pseudocoelomates, and Platyhelminthes are acoelomates.
Match List I with List II
| List I | List II | ||
|---|---|---|---|
| A. | Down’s syndrome | I. | 21st chromosome |
| B. | α-Thalassemia | II. | 16th chromosome |
| C. | β-Thalassemia | III. | 11th chromosome |
| D. | Klinefelter’s syndrome | IV. | ‘X’ chromosome |
View Solution
Step 1: Down’s syndrome is caused by trisomy of the 21\(^{st}\) chromosome. (Matching A \(\to\) III)
Step 2: \(\alpha\)-Thalassemia is associated with mutations in the 16\(^{th}\) chromosome. (Matching B \(\to\) IV)
Step 3: \(\beta\)-Thalassemia is caused by mutations in the 11\(^{th}\) chromosome. (Matching C \(\to\) I)
Step 4: Klinefelter’s syndrome occurs due to the presence of an extra ‘X’ chromosome (XXY). (Matching D \(\to\) II)
Thus, the correct answer is option (1), as the correct matches are A-III, B-IV, C-I, D-II. Quick Tip: Down’s syndrome is due to trisomy 21, \(\alpha\)-Thalassemia is linked to chromosome 16, \(\beta\)-Thalassemia to chromosome 11, and Klinefelter’s syndrome to the ‘X’ chromosome.
Following are the stages of pathway for conduction of an action potential through the heart
A. AV bundle
B. Purkinje fibres
C. AV node
D. Bundle branches
E. SA node
Choose the correct sequence of pathway from the options given below:
View Solution
Step 1: The conduction pathway of the heart follows this sequence:
- SA node initiates the impulse. (E)
- AV node delays the impulse slightly. (C)
- AV bundle (Bundle of His) carries the signal to the ventricles. (A)
- Bundle branches transmit impulses through the ventricles. (D)
- Purkinje fibers distribute the impulse, leading to contraction. (B)
Thus, the correct answer is option (3), as the correct sequence is E-C-A-D-B. Quick Tip: The conduction pathway of the heart: SA node → AV node → AV bundle → Bundle branches → Purkinje fibers.

View Solution
Step 1: Lipase breaks down lipids by hydrolyzing ester bonds. (Matching A \(\to\) II)
Step 2: Nuclease hydrolyzes nucleic acids by breaking phosphodiester bonds in DNA/RNA. (Matching B \(\to\) IV)
Step 3: Protease digests proteins by hydrolyzing peptide bonds between amino acids. (Matching C \(\to\) I)
Step 4: Amylase breaks down polysaccharides (starch) by hydrolyzing glycosidic bonds. (Matching D \(\to\) III)
Thus, the correct answer is option (1), as the correct matches are A-II, B-IV, C-I, D-III. Quick Tip: Lipase hydrolyzes ester bonds in fats, nucleases break phosphodiester bonds in DNA/RNA, proteases cleave peptide bonds in proteins, and amylase breaks glycosidic bonds in carbohydrates.
The following diagram shows restriction sites in E. coli cloning vector pBR322. Find the role of ‘X’ and ‘Y’ genes:
View Solution
The vector pBR322 contains the X and Y genes.
- Gene X regulates the copy number of the plasmid DNA.
- Gene Y encodes a protein involved in the replication of the plasmid. This ensures that the plasmid can replicate inside the host cell.
Thus, the correct answer is (1) The gene ‘X’ is responsible for controlling the copy number of the linked DNA and ‘Y’ for protein involved in the replication of Plasmid. Quick Tip: In plasmid vectors like pBR322, genes involved in replication control the number of copies of the plasmid within the host cell.
The "Ti plasmid" of Agrobacterium tumefaciens stands for:
View Solution
Step 1: The Ti plasmid is a tumor-inducing plasmid found in Agrobacterium tumefaciens, a plant pathogen.
Step 2: It is responsible for causing crown gall disease in dicot plants.
Step 3: The T-DNA (transfer DNA) segment of the Ti plasmid integrates into the host genome, leading to uncontrolled cell division.
Thus, the correct answer is option (1), as the Ti plasmid stands for Tumor Inducing plasmid. Quick Tip: The Ti plasmid of \textit{Agrobacterium tumefaciens is widely used in plant genetic engineering due to its ability to transfer genes into plants.
Match List I with List II

View Solution
Step 1: Pleurobrachia belongs to Ctenophora (comb jellies). (Matching A \(\to\) II)
Step 2: Radula is a rasping organ found in Mollusca for feeding. (Matching B \(\to\) I)
Step 3: Stomachord is a structure found in Hemichordates. (Matching C \(\to\) IV)
Step 4: Air bladder is found in Osteichthyes (bony fish) and helps in buoyancy. (Matching D \(\to\) III)
Thus, the correct answer is option (4), as the correct matches are A-II, B-I, C-IV, D-III. Quick Tip: Pleurobrachia belongs to Ctenophora, radula is a feeding organ in Mollusca, stomachord is found in Hemichordata, and air bladder helps in buoyancy in bony fishes.
Which of the following statements is incorrect?
View Solution
Step 1: Bio-reactors are used for large-scale production of biological products, not small-scale. Statement (1) is incorrect.
Step 2: They contain an agitator system, oxygen delivery system, and foam control system to maintain optimal growth conditions. Statements (2), (3), and (4) are correct.
Thus, the correct answer is option (1), as bio-reactors are used for large-scale production, not small-scale. Quick Tip: Bio-reactors are used for large-scale production of enzymes, antibiotics, and recombinant proteins, not small-scale bacterial cultures.
Which one is the correct product of DNA dependent RNA polymerase to the given template?
3’ TACATGGCAAATATCCATTCA 5’
View Solution
Step 1: DNA-dependent RNA polymerase synthesizes mRNA complementary to the DNA template.
Step 2: The given template strand (3’→5’ direction) results in an mRNA strand in the 5’→3’ direction.
Thus, the correct answer is option (3), as the correct mRNA sequence is 5’ AUGUACCGUUUUAUAGGAAGU 3’. Quick Tip: DNA-dependent RNA polymerase synthesizes mRNA complementary to the template strand, replacing thymine (T) with uracil (U).
Match List I with List II

View Solution
Step 1: Cocaine is derived from Erythroxylum coca. (Matching A \(\to\) III)
Step 2: Heroin is synthesized from morphine, which is obtained from \textit{Papaver somniferum (opium poppy). (Matching B \(\to\) IV)
Step 3: Morphine is an effective sedative used in surgery. (Matching C \(\to\) I)
Step 4: Marijuana is derived from \textit{Cannabis sativa. (Matching D \(\to\) II)
Thus, the correct answer is option (2), as the correct matches are A-III, B-IV, C-I, D-II. Quick Tip: Cocaine is derived from \textit{Erythroxylum coca, heroin from Papaver somniferum, morphine is a sedative, and marijuana comes from Cannabis sativa.
Match List I with List II
View Solution
Step 1: Diakinesis is the final stage of prophase I, where chiasmata terminalization is completed. (Matching A \(\to\) II)
Step 2: Pachytene is the stage where recombination nodules appear. (Matching B \(\to\) IV)
Step 3: Zygotene is characterized by synaptonemal complex formation. (Matching C \(\to\) I)
Step 4: Leptotene is the first stage where chromosomes appear as thin threads. (Matching D \(\to\) III)
Thus, the correct answer is option (1), as the correct matches are A-II, B-IV, C-I, D-III. Quick Tip: Prophase I stages: Leptotene (thin chromosomes), Zygotene (synaptonemal complex), Pachytene (crossing over), Diakinesis (chiasmata terminalization).
In both sexes of cockroach, a pair of jointed filamentous structures called anal cerci are present on:
View Solution
Step 1: Cockroaches possess anal cerci, which are jointed filamentous structures that function as sensory organs.
Step 2: These cerci are present in both male and female cockroaches on the 10\(^{th}\) abdominal segment.
Step 3: They help in detecting vibrations and movements in the surrounding environment.
Thus, the correct answer is option (4), as anal cerci are present on the 10\(^{th}\) segment of the cockroach. Quick Tip: Anal cerci in cockroaches are located on the 10\(^{th}\) abdominal segment and serve as sensory organs for detecting movement and vibrations.
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R:
Assertion A: Breast-feeding during the initial period of infant growth is recommended by doctors for bringing a healthy baby.
Reason R: Colostrum contains several antibodies absolutely essential to develop resistance for the newborn baby.
In the light of the above statements, choose the most appropriate answer from the options given below:
View Solution
Step 1: Breastfeeding is recommended as it provides essential nutrients and immunity to the infant. Assertion A is correct.
Step 2: Colostrum, the first milk produced after birth, is rich in maternal antibodies (IgA, IgG, IgM), which help develop immunity in the newborn. Reason R is correct.
Step 3: Since colostrum provides antibodies essential for infant immunity, it directly supports the reason given in Assertion A.
Thus, the correct answer is option (3), as both A and R are correct, and R is the correct explanation of A. Quick Tip: Colostrum is rich in maternal antibodies and helps develop passive immunity in newborns, making breastfeeding highly beneficial.
Match List I with List II

View Solution
Step 1: Pterophyllum is commonly known as the Angel fish. (Matching A \(\to\) III)
Step 2: \textit{Myxine is commonly known as the Hag fish, which is a jawless fish. (Matching B \(\to\) I)
Step 3: \textit{Pristis is known as the Saw fish, characterized by its elongated snout with teeth. (Matching C \(\to\) II)
Step 4: \textit{Exocoetus is called the Flying fish, as it can glide over water surfaces. (Matching D \(\to\) IV)
Thus, the correct answer is option (4), as the correct matches are A-III, B-I, C-II, D-IV. Quick Tip: \textit{Pterophyllum (Angel fish), Myxine (Hag fish - jawless), Pristis (Saw fish), and Exocoetus (Flying fish) are well-known aquatic species.
Three types of muscles are given as a, b, and c. Identify the correct matching pair along with their location in the human body:
Name of muscle/location
View Solution
Step 1: The three major types of muscles in the human body are:
- Skeletal muscle (voluntary and striated, found in biceps, triceps, etc.).
- Smooth muscle (involuntary, non-striated, found in internal organs like the stomach and intestines).
- Cardiac muscle (striated, involuntary, found in the heart).
Step 2:
- Image (a) represents skeletal muscle, found in triceps.
- Image (b) represents smooth muscle, found in stomach.
- Image (c) represents cardiac muscle, found in heart.
Thus, the correct answer is option (4), as (a) Skeletal – Triceps, (b) Smooth – Stomach, (c) Cardiac – Heart. Quick Tip: Skeletal muscles are voluntary and striated, smooth muscles are involuntary and non-striated, and cardiac muscles are striated but involuntary.
Which of the following is not a steroid hormone?
View Solution
Step 1: Steroid hormones are derived from cholesterol and include sex hormones and adrenal cortex hormones.
Step 2:
- Progesterone (option 1) is a steroid hormone involved in pregnancy.
- Cortisol (option 3) is a steroid hormone secreted by the adrenal cortex.
- Testosterone (option 4) is a steroid hormone responsible for male secondary sexual characteristics.
- Glucagon (option 2) is a peptide hormone, not a steroid. It is secreted by the pancreas and regulates blood glucose levels.
Thus, the correct answer is option (2), as Glucagon is a peptide hormone, not a steroid hormone. Quick Tip: Steroid hormones are derived from cholesterol (e.g., testosterone, cortisol, progesterone), while peptide hormones (e.g., glucagon, insulin) are made of amino acids.
Match List I with List II

View Solution
Step 1: Fibrous joints (A) are immovable and found in the skull (Matching A \(\to\) III).
Step 2: Cartilaginous joints (B) provide limited movement and are found in the vertebral column (Matching B \(\to\) I).
Step 3: Hinge joints (C) allow movement in one plane and are found in the knee (Matching C \(\to\) IV).
Step 4: Ball and socket joints (D) allow rotational movement and are found in the humerus and pectoral girdle (Matching D \(\to\) II).
Thus, the correct answer is option (2), as the correct matches are A-III, B-I, C-IV, D-II. Quick Tip: Fibrous joints (skull - no movement), Cartilaginous joints (vertebral column - limited movement), Hinge joints (knee - unidirectional), Ball and socket joints (shoulder - rotational).
Given below are two statements: One is labelled as Assertion A and the other as Reason R:
Assertion A: FSH acts upon ovarian follicles in females and Leydig cells in males.
Reason R: Growing ovarian follicles secrete estrogen in females, while interstitial cells secrete androgens in males.
In the light of the above statements, choose the correct answer from the options given below:
View Solution
Step 1: FSH (Follicle Stimulating Hormone) acts on Sertoli cells, not Leydig cells, in males. It regulates spermatogenesis. Thus, Assertion A is false.
Step 2: Leydig cells in males produce androgens (testosterone). In females, growing ovarian follicles secrete estrogen. Thus, Reason R is correct.
Thus, the correct answer is option (2), as Assertion A is incorrect, but Reason R is correct. Quick Tip: FSH acts on Sertoli cells (not Leydig cells), while growing ovarian follicles secrete estrogen, and Leydig cells secrete testosterone.
Match List I with List II
View Solution
Step 1: Expiratory capacity (A) is the sum of tidal volume + expiratory reserve volume. (Matching A \(\to\) II)
Step 2: Functional residual capacity (B) is the sum of expiratory reserve volume + residual volume. (Matching B \(\to\) IV)
Step 3: Vital capacity (C) is the sum of tidal volume + inspiratory reserve volume + expiratory reserve volume. (Matching C \(\to\) I)
Step 4: Inspiratory capacity (D) is the sum of tidal volume + inspiratory reserve volume. (Matching D \(\to\) III)
Thus, the correct answer is option (3), as the correct matches are A-II, B-IV, C-I, D-III. Quick Tip: Expiratory capacity = TV + ERV, Functional residual capacity = ERV + RV, Vital capacity = TV + IRV + ERV, Inspiratory capacity = TV + IRV.
Following are the stages of cell division:
A. Gap 2 phase
B. Cytokinesis
C. Synthesis phase
D. Karyokinesis
E. Gap 1 phase
Choose the correct sequence of stages from the options given below:
View Solution
Step 1: The correct sequence of cell cycle stages is:
- G1 phase (E): Cell growth occurs.
- S phase (C): DNA replication happens.
- G2 phase (A): Preparation for mitosis.
- Karyokinesis (D): Nuclear division occurs.
- Cytokinesis (B): Cytoplasm divides, forming two daughter cells.
Thus, the correct answer is option (2), as the correct order is E-C-A-D-B. Quick Tip: The cell cycle follows the order: G1 → S → G2 → M (Karyokinesis + Cytokinesis).
Which of the following are Autoimmune disorders?
A. Myasthenia gravis
B. Rheumatoid arthritis
C. Gout
D. Muscular dystrophy
E. Systemic Lupus Erythematosus (SLE)
Choose the most appropriate answer from the options given below:
View Solution
Step 1: Autoimmune disorders occur when the immune system attacks its own body tissues.
Step 2:
- Myasthenia gravis (A): Autoimmune attack on neuromuscular junctions.
- Rheumatoid arthritis (B): Autoimmune inflammation of joints.
- SLE (E): A systemic autoimmune disease affecting multiple organs.
- Gout (C) is due to uric acid crystal accumulation (not autoimmune).
- Muscular dystrophy (D) is a genetic disorder, not autoimmune.
Thus, the correct answer is option (4), as A, B, and E are autoimmune disorders. Quick Tip: Autoimmune diseases include Myasthenia gravis, Rheumatoid arthritis, and Systemic Lupus Erythematosus (SLE), but not gout or muscular dystrophy.
Which of the following is not a component of the Fallopian tube?
View Solution
Step 1: The Fallopian tube consists of four parts:
- Infundibulum (1): Funnel-shaped structure with fimbriae.
- Ampulla (2): The widest and longest part, where fertilization occurs.
- Isthmus (4): A narrow region connecting to the uterus.
Step 2: The Uterine fundus (3) is the uppermost part of the uterus, not the Fallopian tube.
Thus, the correct answer is option (3), as the uterine fundus is not part of the Fallopian tube. Quick Tip: The Fallopian tube consists of Infundibulum, Ampulla, Isthmus, and the Uterine part, but not the Uterine fundus.
As per the ABO blood grouping system, the blood group of the father is B\(^+\), mother is A\(^+\), and child is O\(^-\). Their respective genotypes can be:
Options:
A. I\(^B\)i/I\(^B\)i
B. I\(^B\)I\(^A\)/I\(^A\)i
C. I\(^A\)I\(^B\)/I\(^B\)i
D. I\(^A\)i/I\(^B\)i
E. ii/I\(^B\)i/I\(^A\)I\(^B\)
Choose the most appropriate answer from the options given below:
View Solution
Step 1: Blood type inheritance follows the Mendelian principles of codominance. The possible genotypes are:
- Blood group A: I\(^A\)I\(^A\) or I\(^A\)i
- Blood group B: I\(^B\)I\(^B\) or I\(^B\)i
- Blood group O: ii
Step 2: Since the child has O\(^-\) blood, they must have received the ii genotype (one i from each parent).
Step 3:
- The father’s genotype must be I\(^B\)i or I\(^B\)I\(^B\).
- The mother’s genotype must be I\(^A\)i or I\(^A\)I\(^A\).
- The only correct parental genotype pair that produces an O\(^-\) child is A (I\(^B\)i/I\(^B\)i) & D (I\(^A\)i/I\(^B\)i).
Thus, the correct answer is option (3), as A & D represent the correct possible genotypes. Quick Tip: For a child with O blood type, both parents must contribute an "i" allele (i.e., they must be heterozygous A or B).
Match List I with List II

View Solution
Step 1: Exophthalmic goiter is due to hyperthyroidism, leading to protruding eyeballs. (Matching A \(\to\) III)
Step 2: Acromegaly occurs due to excess secretion of growth hormone. (Matching B \(\to\) IV)
Step 3: Cushing’s syndrome is due to excess cortisol secretion, leading to moon face and hyperglycemia. (Matching C \(\to\) I)
Step 4: Cretinism results from thyroid hormone deficiency, leading to stunted growth. (Matching D \(\to\) II)
Thus, the correct answer is option (2), as the correct matches are A-III, B-IV, C-I, D-II. Quick Tip: Exophthalmic goiter (hyperthyroidism), Acromegaly (excess GH), Cushing’s syndrome (excess cortisol), and Cretinism (thyroid hormone deficiency).
Match List I with List II

View Solution
Step 1: Mesozoic Era is known as the "Age of Reptiles and Birds". (Matching A \(\to\) III)
Step 2: Proterozoic Era had the first lower invertebrates. (Matching B \(\to\) I)
Step 3: Cenozoic Era is known as the "Age of Mammals". (Matching C \(\to\) IV)
Step 4: Paleozoic Era was the period of early Fish and Amphibia. (Matching D \(\to\) II)
Thus, the correct answer is option (2), as the correct matches are A-III, B-I, C-IV, D-II. Quick Tip: Mesozoic Era (Reptiles \& Birds), Proterozoic Era (Lower invertebrates), Cenozoic Era (Mammals), Paleozoic Era (Fish \& Amphibians).
The following are the statements about non-chordates:
A. Pharynx is perforated by gill slits.
B. Notochord is absent.
C. Central nervous system is dorsal.
D. Heart is dorsal if present.
E. Post-anal tail is absent.
Choose the most appropriate answer from the options given below:
View Solution
Step 1: Non-chordates lack notochord (B), post-anal tail (E), and have a dorsal heart (D).
Step 2: Chordates have pharyngeal gill slits (A) and a dorsal CNS (C), so these statements do not apply to non-chordates.
Step 3: Since non-chordates lack notochord, post-anal tail, and have a dorsal heart, the correct answer is B, D, and E only.
Thus, the correct answer is option (1), as B, D, and E are true for non-chordates. Quick Tip: Non-chordates lack a notochord and post-anal tail, while their heart (if present) is dorsal. Chordates have pharyngeal gill slits and a dorsal CNS.
Given below are two statements:
Statement I: The cerebral hemispheres are connected by a nerve tract known as corpus callosum.
Statement II: The brain stem consists of the medulla oblongata, pons, and cerebrum.
In the light of the above statements, choose the most appropriate answer from the options given below:
View Solution
Step 1: The corpus callosum is a bundle of nerve fibers that connects the left and right cerebral hemispheres, making Statement I correct.
Step 2: The brainstem consists of the medulla oblongata, pons, and midbrain (not cerebrum), making Statement II incorrect.
Thus, the correct answer is option (1), as Statement I is correct, but Statement II is incorrect. Quick Tip: The brainstem includes the medulla oblongata, pons, and midbrain but not the cerebrum. The corpus callosum connects the cerebral hemispheres.
Match List I with List II

View Solution
Step 1: Unicellular glandular epithelium (A) consists of goblet cells in the alimentary canal. (Matching A \(\to\) III)
Step 2: Compound epithelium (B) lines moist surfaces like the buccal cavity. (Matching B \(\to\) IV)
Step 3: Multicellular glandular epithelium (C) includes salivary glands. (Matching C \(\to\) I)
Step 4: Endocrine glandular epithelium (D) includes the pancreas. (Matching D \(\to\) II)
Thus, the correct answer is option (1), as the correct matches are A-III, B-IV, C-I, D-II. Quick Tip: Unicellular epithelium (goblet cells), compound epithelium (moist surfaces), multicellular epithelium (salivary glands), endocrine epithelium (pancreas).
Match List I with List II

View Solution
Step 1: P wave represents depolarisation of atria. (Matching A \(\to\) III)
Step 2: QRS complex represents depolarisation of ventricles. (Matching B \(\to\) II)
Step 3: T wave represents repolarisation of ventricles. (Matching C \(\to\) IV)
Step 4: T-P gap indicates electrical silence in heart muscles. (Matching D \(\to\) I)
Thus, the correct answer is option (4), as the correct matches are A-III, B-II, C-IV, D-I. Quick Tip: P wave (atrial depolarisation), QRS complex (ventricular depolarisation), T wave (ventricular repolarisation), T-P gap (electrical silence).
Given below are two statements:
Statement I: Mitochondria and chloroplasts are both double-membrane bound organelles.
Statement II: The inner membrane of mitochondria is relatively less permeable compared to the chloroplast.
In the light of the above statements, choose the correct answer from the options given below:
View Solution
Step 1: Both mitochondria and chloroplasts have double membranes. (Statement I is correct)
Step 2: The inner membrane of mitochondria is highly folded but more permeable than the chloroplast. (Statement II is incorrect)
Thus, the correct answer is option (1). Quick Tip: P wave (atrial depolarisation), QRS complex (ventricular depolarisation), T wave (ventricular repolarisation), T-P gap (electrical silence).
Match List I with List II related to digestive system of cockroach:
View Solution
Step 1: Crop (A) is the structure used for storing food in cockroaches. (Matching A \(\to\) IV)
Step 2: Gastric Caeca (B) are the 6-8 blind tubules at the junction of foregut and midgut. (Matching B \(\to\) II)
Step 3: Malpighian tubules (C) are the yellow-colored filaments found at the junction of midgut and hindgut. (Matching C \(\to\) III)
Step 4: Gizzard (D) is used for grinding food. (Matching D \(\to\) I)
Thus, the correct answer is option (3), as the correct matches are A-IV, B-II, C-III, D-I. Quick Tip: The cockroach digestive system has the crop (storage), gastric caeca (digestion), malpighian tubules (excretion), and gizzard (grinding food).
Match List I with List II:

View Solution
Step 1: RNA polymerase III (A) synthesizes SnRNAs and tRNA. (Matching A \(\to\) IV)
Step 2: Termination of transcription (B) is regulated by the Rho factor. (Matching B \(\to\) II)
Step 3: Splicing of Exons (C) is associated with snRNPs. (Matching C \(\to\) I)
Step 4: The TATA box (D) is a promotor region that facilitates transcription. (Matching D \(\to\) III)
Thus, the correct answer is option (2), as the correct matches are A-IV, B-II, C-I, D-III. Quick Tip: RNA polymerase III produces tRNA and snRNAs. The TATA box is a core promoter for transcription initiation.
Identify the correct option (A), (B), (C), (D) with respect to spermatogenesis.
View Solution
Step 1: FSH (Follicle-stimulating hormone) stimulates Sertoli cells, which support spermatogenesis.
Step 2: LH (Luteinizing hormone) stimulates Leydig cells, which secrete androgens (testosterone) necessary for spermiogenesis.
Thus, the correct answer is option (3), as FSH stimulates Sertoli cells and Leydig cells support spermiogenesis. Quick Tip: FSH acts on Sertoli cells (spermatogenesis), and LH acts on Leydig cells (spermiogenesis).
Choose the correct statement given below regarding juxta medullary nephron.
View Solution
Step 1: Juxta medullary nephrons have a longer loop of Henle that extends deep into the renal medulla, aiding in concentration of urine. (Matching Option 1)
Step 2: Cortical nephrons are more numerous than juxta medullary nephrons.
Thus, the correct answer is option (1), as the loop of Henle in juxta medullary nephron extends deep into medulla. Quick Tip: Juxta medullary nephrons play a critical role in water reabsorption and urine concentration due to their long loop of Henle.
Given below are two statements:
Statement I: Gause's competitive exclusion principle states that two closely related species competing for different resources cannot exist indefinitely.
Statement II: According to Gause's principle, during competition, the inferior will be eliminated. This may be true if resources are limiting.
In the light of the above statements, choose the correct answer from the options given below:
View Solution
Step 1: Gause's competitive exclusion principle does not state that species with different resources cannot coexist indefinitely. It specifically states that closely related species competing for the same resources cannot coexist indefinitely. (Statement I is false)
Step 2: Statement II is true because, in cases of competition with limited resources, the inferior species is eliminated.
Thus, the correct answer is option (2), as Statement I is false, but Statement II is true. Quick Tip: Competitive exclusion occurs when two species competing for the same resources cannot coexist, and the inferior species is eliminated.
Given below are two statements:
Statement I: Bone marrow is the main lymphoid organ where all blood cells including lymphocytes are produced.
Statement II: Both bone marrow and thymus provide microenvironments for the development and maturation of T-lymphocytes.
In the light of the above statements, choose the most appropriate answer from the options given below:
View Solution
Step 1: Bone marrow is indeed the primary site for the production of all blood cells, including lymphocytes. (Statement I is correct)
Step 2: Both bone marrow and thymus play crucial roles in the development and maturation of T-lymphocytes. (Statement II is correct)
Thus, the correct answer is option (3), as both Statement I and Statement II are correct. Quick Tip: Bone marrow produces all blood cells, including lymphocytes. The thymus plays a key role in the maturation of T-lymphocytes.
Regarding catalytic cycle of an enzyme action, select the correct sequential steps:
A. Substrate enzyme complex formation.
B. Free enzyme ready to bind with another substrate.
C. Release of products.
D. Chemical bonds of the substrate broken.
E. Substrate binding to active site.
Choose the correct answer from the options given below:
View Solution
Step 1: The enzyme action begins with substrate binding to the active site (E).
Step 2: This leads to substrate-enzyme complex formation (A).
Step 3: The enzyme is then ready to bind with another substrate (B).
Step 4: The chemical bonds of the substrate are broken (D) as the reaction progresses.
Step 5: Finally, the products are released (C).
Thus, the correct answer is option (3), as the correct sequence is E, A, B, C, D. Quick Tip: The enzyme catalytic cycle follows: Substrate binding → Substrate-enzyme complex → Enzyme ready for next substrate → Substrate broken → Product released.



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