CBSE EXAMINATION PAPER-2022
NEET-ZOOLOGY
(Solved)
General Instructions :
Read the following instructions carefully and follow them :
- This question paper contains 200 questions. All questions are compulsory.
- This question paper is divided into 1 sections.
- Section A – questions number 1 to 200 are multiple choice questions
- There is no overall choice given in the question paper. However, an internal choice has been provided in few questions.
- Use of calculator is NOT allowed.
Section A
Two hollow conducting spheres of radii R₁ and R₂ (R₁ >> R₂) have equal charges. The potential would be :
Potential, = KQ / R ∵ Q = constant ∴ V ∝ 1/R vₛₘₐₗₗₛₚₕₑᵣₑ > v_larger sphere
In the given circuits (1), (2) and (3), the potential drop across the two p-n junctions are equal in:
Left P-N junction diode is forward biased whereas right one is reverse biased so obviously both will have different resistance.
Two objects of mass 10 kg and 20 kg respectively are connected to the two ends of a rigid rod of length 10 m with negligible mass. The distance of the center of mass of the system from the 10 kg mass is:
As, distance of COM from = m₂r / m₁ + m₂ Here, m₁ = 10 kg, m₂ = 20 kg and r = 10 m So, required distance = 20 × 10 / 20 + 10 = 20/3 m
A biconvex lens has radii of curvature, 20 cm each, if the refractive index of the material of the lens is 1.5, the power of the lens is:
We have P = 1/f = (μ − 1) ( 1/R₁ − 1/R₂ ) = (1.5 − 1) ( 1/20 -(1/-20)) × 1/(10⁻²) = 0.5 × 2/20 × 100 = +5 D
A spherical ball is dropped in a long column of a highly viscous liquid. The curve in the graph shown, which represents the speed of the ball (v) as a function of time (t) is:
When ball is dropped in highly viscous fluid, the ball's velocity will increase untill it becomes equal to terminal velocity it moves downward with constant velocity so, graph will be like as shown.
The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the dis about its diameter is:
I_केन्द्र = I₁ = 1/2 MR² ⇒ I_Diameter = I₂ = 1/4 MR² ⇒ I_Center / I_Diameter = (1/2 MR²) / (1/4 MR²) ⇒ MK²_C / MK²_D = 2 ⇒ K_C / K_D = √2
A shell of mass m is at rest initially. It explodes into three fragments having mass in the ratio 2:2: 1. If the fragments having equal mass fly off along mutually perpendicular directions with speed v, the speed of the third (lighter) fragment is
𝐏⃗ᵢ = 0 𝐏⃗_f = 2mvî + 2mvĵ + m𝐯⃗ As, 𝐏⃗ᵢ = 𝐏⃗_f 0 = m𝐯⃗ + 2mvî + 2mvĵ 𝐯⃗ = −2v(î + ĵ) ⇒ |𝐯⃗| = 2v√(1² + 1²) = 2√2 v
A long solenoid of radius 1 mm has 100 turns per mm. If 1A current flows in the solenoid, the magnetic field strength at the centre of the solenoid is
Magnetic field B_centre = μ₀nI = 4π × 10⁻⁷ × (100/10⁻³) × 1 = 4π × 10⁻² = 12.56 × 10⁻² T
Let T₁ and T₂ be the energy of an electron in the first and second excited states of hydrogen atom, respectively. According to the Bohr’s model of an atom, the ratio T₁: T₂ is:
We have, n₁ = 2 and n₂ = 3 And, Tₙ = −13.6/n² So, T₁/T₂ = (n₂/n₁)² = (3/2)² = 9/4
A light ray falls on a glass surface of refractive index √3 at an angle 60°. The angle between the refracted and reflected rays would be:
We have, μ₂/μ₁ = sin i / sin r ⇒ √3 = sin 60° / sin r ⇒ sin r = √3 / (2 × √3) ⇒ sin r = 1/2 ⇒ r = sin⁻¹(1/2) = 30° Now, angle between refracted ray and reflected ray = α = (180 − 60 − 30)° = 90°
NOTES
Snell's law for two media,
₁μ₂ = μ₂/μ₁ = sin i / sin r
or, μ₁ × sin i = μ₂ × sin r
i.e., μ sin θ = constant
Plane angle and solid angle have:
Plane angle unit is radian, whereas solid angle unit is steradian, but they don’t have any dimensions.
When light propagates through a material medium of relative permittivity εᵣ and relative permeability μᵣ, the velocity of light, v is given by: (c-velocity of light in vacuum)
We know that V = 1/√με and C = 1/√μ₀ε₀ So, V/C = (√μ₀ε₀)/(√με) = (√μ₀ε₀)/(√μ₀ε₀ μᵣεᵣ) = 1/√μᵣ εᵣ V = C/√μᵣ εᵣ
Two resistors of resistance, 100 Ω and 200 Ω are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in 100 Ω to that in 200 Ω in a given time is:
As power, P = V²/R P₁/P₂ = R₂/R₁ P₁₀₀_Ω/P₂₀₀_Ω = 200/100 = 2/1
The graph which shows the variation of the de Broglie wavelength (λ) of a particle and its associated momentum (p) is:
-
(A) a
-
(B) d
-
(C) b
-
(D) c
Wavelength, λ = h/P ⇒ λ ∝ 1/P So graph will be rectangular hyperbola as shown above
A square loop of side 1 m and resistance 1 Ω is placed in a magnetic field of 0.5 T. If the plane of loop is perpendicular to the direction of magnetic field, the magnetic flux through the loop is:
Clearly angle between 𝐁⃗ and 𝐀⃗ is 0° so, ϕ (flux) = 𝐁⃗.𝐀⃗ = BA cos 0° = BA = 0.5 × 1² = 0.5 weber
When two monochromatic lights of frequency, υ and υ/2 are incident on a photoelectric metal, their stopping potential becomes Vₛ/2 and Vₛ respectively. The threshold frequency for this metal is:
(None) By Einstein's photoelectric equation, hf = ϕ + eV_stopping when f = ν hν = hν₀ + e Vₛ/2 ..... (i) when f = ν/2 h(ν/2) = hν₀ + eVₛ ..... (ii) ⇒ 1/2 ( hν₀ + e(Vₛ/2) ) = hν₀ + eVₛ [From (i)] ⇒ -hν₀/2 = −3/4 eVₛ ⇒ ϕ = hν₀ = −3/2 eVₛ, which is −ve. So data given is incorrect because ‘ϕ’can never be negative.
In half wave rectification, if the input frequency is 60 Hz, then the output frequency is would be:
We can see from graph, of input and output of Half Wave Rectifier Tᵢₙₚᵤₜ = Tₒᵤₜₚᵤₜ ⇒ fᵢₙₚᵤₜ = fₒᵤₜₚᵤₜ [ ∵ T = 1/f ]
An ideal gas undergoes four different processes from the same initial state as shown in the figure below. Those processes are adiabatic, isothermal, isobaric and isochoric. The curve which represents the adiabatic process among 1, 2, 3 and 4 is:
Curve 1 and curve 4 are isochoric and isobaric, whereas, slope of curve 3 is less than slope of curve 2. So curve 2 should be adiabatic and curve 3 should be isothermal because (slope) isothermal < (slope) adiabatic.
Match List-I with List-II Choose the correct answer from the options given below:
Radio waves has longest wavelength and x-rays have shortest wavelength, so correct A - ii; B - iii: C - iv; D - i.
The displacement-time graphs of two moving particles make angles of 30° and 45° with the x-axis as shown in the figure. The ratio of their respective velocity is:
V = ds/dt = slope of s − t curve = tan θ V_30°/V_45° = tan 30°/tan 45° = 1/√3
InaYoung’s doubleslitexperiment, astudentobserves 8 fringes in a certain segment of screen when amonochromatic light of 600 nm wavelength is used. If the wavelength of light is changed to 400 nm, then the number of fringes he would observe in the same region of the screen is:
No. of fringes (n) = length of segment (x) / fringe width (β) ⇒ n = x/β ⇒ x = nβ = cons. n₁β₁ = n₂β₂ ⇒ n₁ (λ₁D/d) = n₂ (λ₂D/d) ⇒ n₁λ₁ = n₂λ₂ ⇒ 8 × 600 = 400 × n₂ ⇒ n₂ = 12
The peak voltage of the ac source is equal to:
We know that Vᵣₘₛ = V₀/√2 ⇒ V₀ = √2 Vᵣₘₛ
If the initial tension on a stretched string is doubled, then the ratio of the initial and final speeds of a transverse wave along the string is:
Speed V = √(T/μ) ⇒ V α √T So, V₁/V₂ = √(T₁/T₂) = 1/√2
Given below are two statements:
Statement I: Biot-Savart’s law gives us the expression for the magnetic field strength of an infinitesimal current element(Idl) of a current carrying conductor only.
statement II: Biot-Savart’s law is analogous to Coulomb’s inverse square law of charge g, with the former being related to the field produced by a scalar source, Idl while the latter being produced by a vector source q.
In light of above statement choose the most appropriate answer from the options given below:
Coulomb’s law and universal law of gravitation are analogous to each other, whereas Biot-savart’s law is different from the above two laws as it is used to calculate magnetic produced by vector source, Idl. so statement II is incorrect and statement I is correct.
As the temperature increase, the electrical resistance:
For conductors, R α T so, T↑↑ ⇒ R↑↑ For semiconductor, R α 1/T so T↑↑ ⇒ R↓↓
The energy that will be ideally rediated by a 100 kW transmitter in 1 hour is:
Energy radiated = Power × time = 100 kW × 1 hr = 100 kWh = 100 × 3.6 × 10⁶ J = 3.6 × 10⁸ J = 36 × 10⁷ J
A body of mass 60 g experiences a gravitational force of 3.0 N, when placed at a particular point. The magnitude of the gravitational field intensity at that point is:
𝐅⃗ₘ = m 𝐄⃗ |𝐄⃗| = 𝐅⃗ₘ / m = 3/(60 × 10⁻³) = 50 N/kg
In the given nuclear reaction, the element X is : ²²₁₁Na → X + e⁺ + ν
²²₁₁Na → ᴬ_ZX + ⁰₊₁e + ν So, 22 = A + 0 ⇒ A = 22 ,
11 = Z + 1 ⇒ Z = 10 ⇒ Element is Neon Clearly, answer is ²²₁₀Ne
The angle between the electric lines of force and the quintessential surface is:
Electric field vectors are always ⊥r to equipotential surface.
A copper wire of length 10 m and radius (10⁻²/√π) m has electrical resistance of 10 Ω. The current density in the wire for an electric field strength of 10 (V/m) is :
We have J = σE J = E/ρ ⇒ J = El/RA ⇒ J = (10 × 10)/(10 × π) × (10⁻²/√π)² = 10⁵
The ratio of the distances travelled by a freely falling body in the 1ˢᵗ, 2ⁿᵈ, 3ʳᵈ and 4ᵗʰ second:
Sₙ = u + a/2 (2n − 1) S₁ : S₂ : S₃ : S₄ = (2 × 1 − 1) : (2 × 2 − 1) : (2 × 3 − 1) : (2 × 4 − 1) = 1 : 3 : 5 : 7
An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 ms⁻¹. The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is :(g = 10ms⁻²)
Power = F × V = (2000 g + 3000) × 1.5 = 23000 × 1.5 = 34500 watt
The volume occupied by the molecules contained in 4.5 kg water of STP, if the intermolecular forces vanish away is:
Number of moles of water molecules (n) = Mass of water / Molecular weight = 4.5 / (18 × 10⁻³) = 250 ∴ PV = nRT V = nRT / P = (250 × 8.314 × 273) / 10⁵ = 5.6 m³ [at STP, T = 273 K, P = 10⁵ Pa]
The area of a rectangular field (in m²) of length 55.3 m and breadth 25 m after rounding off the value of correct significant digits is:
Area = length × breadth = 55.3 × 25 As, Answer of multiplication is rounded off to the same number of significant figure as present in least precise term. Here, 25 has least significant figure and it is 2 so, Answer will also be 2 SF. Therefore, correct option is (c) i.e. 14 × 10²
A capacitor of capacitance C = 900 pF is charged fully by 100 V battery B as shown in figure (a). Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance C = 900 pF as shown in figure (b). The electrostatic energy stored by the system (b) is
By conservation of charge qᵢ = q_f ⇒ C × 100 = CV′ + CV′ ⇒ V′ = 50 V So, U_f = 1/2 CV′² × 2 = CV'² = 900 × 10⁻¹² × 25 × 10² = 225 × 10⁻⁸ = 2.25 × 10⁻⁶ J
Match List-I with List-II Choose the correct answer from the options given below:
[G] = [Fr² / (m₁m₂)] = (MLT⁻²L²) / M² = M⁻¹L³T²
[U] = [W] = ML²T⁻² [V] = [U/M] = (ML²T⁻²) / M = L²T⁻² [I] = [F/M] = LT⁻²
Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is:
T ∝ √ℓ Tℓ / Tₛ = √121 / √100 ⇒ Tℓ/ Tₛ = 11/10 ⇒ 11Tₛ = 10T_c Hence, smaller pendulum will make 1 oscillation more than longer pendulum.
If longer makes n oscillation, then smaller makes (n + 1) oscillation. For oscillating in same phase (n + 1)T_S = nTₗ ⇒ (n + 1) · 2π √(100/g) = n · 2π √(121/g) ⇒ (n + 1)10 = n × 11
⇒ n = 10
Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): The stretching of a spring is determined by the shear modulus of the material of the spring. Reason (R): A coil spring of copper has more tensile strength than a steel spring of same dimensions.
In the light of the above statements, choose the most appropriate answer from the options given below:
By stretching coil's, shape changes whereas its length of wire remains same. Due to which shear modulus of elasticity is involved, so assertion is correct. Elasticity of steel is more than that of copper and also has more tensile stress. So reason is false.
A ball is projected with a velocity, 10 ms⁻¹, at an angle of 60° with the vertical direction. Its speed at the highest point of its trajectory will be:
V_Horizontal = ucosθ = 10 sin 60° = (10 × √3)/2 = 5√3 As there is zero acceleration along horizontal direction, at highest point we will have same horizontal velocity also there is no vertical velocity at highest point. |𝐕⃗| = |𝐕⃗_horizontal| = 5√3 m/s
The truth table for the given logic circuit is:
-
(A) a
-
(B) b
-
(C) d
-
(D) c
The truth table for this logic gate is given as Let A = 0 and B = 0. Then Similarly,
A = 0, B = 1 ⇒ C = 0
A = 1, B = 0 ⇒ C = 1
A = 1, B = 1 ⇒ C = 0
So, truth table is
A | B | C
0 | 0 | 1
0 | 1 | 0
1 | 0 | 1
1 | 1 | 0
A series LCR circuit with inductance 10 H, capacitance 10 μF, resistance 50 Ω is connected to an ac source of voltage, V = 200 sin (100 t) volt. If the resonant frequency of the LCR circuit is ν₀ and the frequency of the ac source is ν, then
At resonance X_L = X_c ⇒ ωᵣL = 1/(ωᵣC) ⇒ ωᵣ = 1/√LC ⇒ fᵣ = 1/(2π√LC) ⇒ ν₀ = 1/(2π√(10 × 10 × 10⁻⁶)) ⇒ ν₀ = 10³/(10 × 2π) = 50/π So, if frequency of AC source is 50/π i.e. ν = 50/π then there will be only resonance.
A wheatstone bridge is used to determine the value of unknown resistance X by adjusting the variable resistance Y as shown in the figure. For the most precise measurement of X, the resistances P and Q:
For precise measurement, we always put P and Q approx. equal and small.
From Ampere’s circuital law for a long straight wire of circular cross-section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire is:
For long straight wire B = (μ₀ i r)/(2πR²) ⇒ B ∝ r, r < R
and B = (μ₀ I)/(2πr) ⇒ B ∝ 1/r
Two point charges —q and +q are placed at a distance of L, as shown in the figure. The magnitude of electric field intensity at a distance R(R>>L) varies as:
If R >> L ₁, it will behave like a dipole and in dipole we have, E ∝ 1/r³
NOTES The magnitude of electric field intensity due to dipole at large distances varies as 1/r³ and can never be zero unless r → ∞ and p → 0. Here, p is dipole moment.
Given below are two statements:
Statement I: In the coagulation of a negative sol, the flocculating power of the three given ions is in the order: Al³⁺ > Ba²⁺ > Na⁺
Statement II: In the coagulation of a positive sol, the flocculating power of the three given salts is in the order: NaCl > Na₂SO₄ > Na₃PO₄
In the light of above statements, choose the most appropriate answer from the options given below:
According to Hardy Schulze rule statement I is correct. For coagulation of a positive sol the flocculating power is in the order: PO₄³⁻ > SO₄²⁻ > Cl⁻ Hence, statement II is incorrect.
The incorrect statement regarding chirality is:
Enantiomers are non superimposable mirror images of each other.
What is Y in the above reaction:
Match List-I with List-II Choose the correct answer from the options given below:
(A) - (iv), (B) - (i), (C) - (ii), (D) - (iii)
Given below are two statements : -
Statement I : The boiling points of aldehydes and ketones are higher than hydrocarbons of comparable molecular masses because of weak molecular association in aldehydes and ketones due to dipole - dipole interactions.
Statements II : The boiling points aldehydes and ketones are lower than the alcohols of similar molecular masses due to the absence of H-bonding. In the light of the statements, choose the most appropriate answer from the options given below:
Due to hydrogen bonding in alcohol and non-polar nature of alkane, the boiling point of aldehydes and ketones are in- - between of these two. B.P. order:
Alcohol > Aldehyde/Ketone > Alkane
Match List-I with List-II Choose the correct answer from the options given below:
Which one is not correct mathematical equation for Dalton's Law of partial pressure? Here p = total pressure of gaseous mixture
Partial pressure of gas = Mole fraction of gas x Total pressure pᵢ = xᵢ .p
Match List-I with List-II. Choose the correct answer from the options given below:
(A) - (iii), (B) - (iv), (C) - (ii), (D) -(i)
Given below are two statements:
Statement I: The boiling points of the following hydrides of group 16 elements increases in the order —H₂O < H₂S < H₂Se < H₂Te
Statement II: The boiling points of these hydrides increase with increase in molar mass.
In the light of the above statements, choose the most appropriate answer from the options given below:
It is true that generally b.p. increases with increase in ‘molar mass but due to H-bonding in H₂O the correct order of b.p. is: H₂S < H₂Se < H₂Te < H₂O
The IUPAC name of the complex [Ag(H₂O)][Ag(CN)₂] is :
[⁺¹Ag(OH₂)₂][⁺¹Ag(⁻¹CN)₂]
diaquasilver (l) diicyanidoargentate(I)
Which of the following is suitable to synthesize chlorobenzene ?
-
(A) a
-
(B) b
-
(C) d
-
(D) c
Given below are two statements; one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : ICl is more reactive than I₂.
Reason (R) : I-Cl bond is weaker than I-I bond.
In the light of the above statements, choose the most appropriate answer from the options given below:
Due to weak I - Cl bond it is more reactive than I - I. F₂ is more reactive than most of the interhalogen compounds due to weak F - F bond.
The IUPAC name of an element with atomic number 119 is
119 = un - un - enn - ium
Which compound amongst the following is not an aromatic compound?
-
(A) A
-
(B) C
-
(C) B
-
(D) D
The molecule is not conjugated, hence it is not an aromatic compound.
Choose the correct statement:
(a) Diamond and graphite both are covalent. (c) Only graphite can be used as dry Lubricant. (d) Both diamond and graphite have 3-Dimensional structure.
Given below are two statements:
Statement I: Primary aliphatic amines react with HNO₂ to give unstable diazonium salts. Statement II: Primary aromatic amines react with HNO₂ to form diazonium salts which are stable even above 300 K.
In the light of the above statements, choose the most appropriate answer from the options given below:
Aryl diazonium salts are stable at low temperature due to resonating structure
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): In a particular point defect, an ionic solid is electrically neutral, even if few of its cations are missing from its unit cells.
Reason (R): In an ionic solid, Frenkel defect arises due to dislocation of cation from its lattice site to interstitial site, maintaining overall electrical neutrality.
In the light of the above statements, choose the most appropriate answer from the options given below:
Assertion: For any kind of point defect, electrical neutrality is an essential condition, e.g., metal deficiency defect. Reason: This is correct but it does not explain the given assertion.
The Kjeldahl’s method for the estimation of nitrogen can be used to estimate the amount of nitrogen in which one of the following compounds?
-
(A) D
-
(B) C
-
(C) A
-
(D) B
Only the nitrogen of amine from the given options can be changed to ammonium sulphate under Kjeldahl’s method.
Which of the following p-V curve represents maximum work done?
-
(A) d
-
(B) c
-
(C) a
-
(D) b
Area under the p-V curve is maximum in the first option, which is equal to work done.
Which of the following statement is not correct about diborane ?
Boran atoms are sp³ hybridised.
The pH of the solution containing 50 mL each of 0.10 M sodium acetate and 0.01 M acetic acid is [Given pKₐ of CH₃COOH = 4.57]
For acidic buffer pH = p_Kᵃ + log [Salt] / [Acid] = 4.57 + log (0.1 + 0.01) = 5.57
Amongst the following which one will have maximum ‘lone pair-lone pair” electron repulsions?
:XeF₂, due to 3 lone pairs, the lp-lp repulsions are maximum.
What mass of 95% pure CaCO₃ will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction? CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l) [Calculate upto second place of decimal point]
From the given reaction, No. of moles of CaCO₃ = 1 / 2 × HCl of Solution = 1 / 2 × 0.5 × 50 / 1000 = 0.0125 Mass of pure CaCO₃ =0.0125 ×100 =1.25g Mass of impure sample 1.25 × 100 / 95 = 1.32 g
Gadolinium has a low value of third ionisation enthalpy because of
High exchange enthalpy
The given graph is a representation of kinetics of a reaction.
The y and x axes for zero and first order reactions, respectively are
Given below are half cell reactions:
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, E°_Mn²⁺ / mno₄⁻ = -1.510V (1/2)O₂ + 2H⁺ + 2e⁻ → H₂O
E°_o₂ / H₂O = +1.223V
Will the permanganate ion, MnO₄⁻ liberate O₂ from water in the presence of an acid ?
[MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O] × 2
E°_MnO₄⁻/Mn²⁺ = +1.510 V
[H₂O → (½)O₂ + 2H⁺ + 2e⁻] × 5
E°_H₂O/O₂ = −1.223 V
2MnO₄⁻ + 6H⁺ → 2Mn²⁺ + (5/2) O₂ + 3H₂O
E°_cell = +1.510 V − 1.223 V = 0.287 V
Match List-I with List-II.
(a) Li is used in electrochemical cells. (b) Na is used as a coolant in fast breeder reactors because liquid metal atoms are weak neutron moderators. (c) KOH can absorb CO₂ to form K₂CO₃. (d) Cs is used in photoelectric cell due to low ionisation enthalpy.
Given below are two statements:
Statement I: The acidic strength of monosubstituted nitrophenol is higher than phenol because of electron withdrawing nitro group.
Statement II: o-nitrophenol, m-nitrophenol and p-nitophenol will have same acidic strength as they have one nitro group attached to the phenolic ring.
In the light of the above statements, choose the most appropriate answer from the options given below:
Electron withdrawing group increases the acidic character while electron donating group decreases the acidic character, More the number of electron withdrawing group, more will be acidic strength. Order of acidic strength
The pollution due to oxides of sulphur gets enhanced due to the presence of:
(A) particulate matter
(B) ozone
(C) hydrocarbons
(D) hydrogen peroxide Choose the most appropriate answer from the options given below:
The presence of particulate matter in polluted air catalyses the oxidation of sulphurdioxide to sulphur trioxide.
2SO₂(g) + O₂(g) → 2SO₃(g)
The reaction can also be promoted by ozone and hydrogen peroxide. SO₂(g) + O₃(g) → SO₃(g) + O₂(g) SO₂(g) + H₂O₂(l) → H₂SO₄(aq)
The correct IUPAC name of the following compound is:
IUPAC name: 1-Bromo-5-chloro-4-methylhexan-3-ol
3O₂(g) ⇌ 2O₃(g) for the above reaction at 298 K, K_C is found to be 3.0 × 10⁻⁵⁹. If the concentration of O₂ at equilibrium is 0.040 M then concentration of O₃ in M is
he given reaction is: 3O₂(g) ⇌ 2O₃(g) K_c = [O₃]² / [O₂]³ 3 × 10⁻⁵⁹ = [O₃]² / (4 × 10⁻²)³ [O₃]² = 3 × 10⁻⁵⁹ × 64 × 10⁻⁶ [O₃] = √(19.2 × 10⁻⁶⁴) [O₃] = 4.38 × 10⁻³² M
Match List-I with List-II. Choose the correct answer from the options given below:
The chemical composition of Haematite is Fe₂O₃ The chemical composition of Magnetite is Fe₃O₄ The chemical composition of Calamine is ZnCO₃ The chemical composition of Kaolinito is [Al₂(OH)₄Si₂O₅]
Given below are two statements:
Statement I: In Lucas test, primary, secondary and tertiary alcohols are distinguished on the basis of their reactivity with cone. HCI + ZnCl₂, known as Lucas Reagent.
Statement II: Primary alcohols are most reactive and immediately produce turbidity at room temperature on reaction with Lucas Reagent.
In the light of the above statements, choose the most appropriate answer from the options given below:
Lucas test is used to distinguished between 1°, 2° and 3° alcohol on the basis of time taken for turbidity to appear by using a solution of anhydrous ZnCl₂ in cone. HCl.
In the neutral or faintly alkaline medium, KMnO₄ oxidises iodide into iodate. The change in oxidation state of manganese in this reaction is from
The reaction of potassium permanganate in neutral or weak alkaline medium is KMnO₄ oxidises iodide to iodate in neutral or weak alkaline medium so, the O.S. of Mn changes from + 7 to + 4.
For a first order reaction A → Products, initial concentration of A is 0.1 M, which becomes 0.001 M after 5 minutes. Rate constant for the reaction in min⁻¹ is
A → Products Given: Initial conc.
[A₀] =0.1 M
Con. After 5 min [Aₜ] = 0.001 M
t = 5 min. For first order reaction
k = (2.303 / t) log
[A₀] / [Aₜ] = (2.303 / 5)log (0.1 /0.001) k = 0.9212 min⁻¹
Compound X on reaction with O₃ followed by Zn/H₂O gives formaldehyde and 2-methyl propanal as products. The compound X is:
A 10.0 L flask contains 64 g of oxygen at 27°C. (Assume O₃ gas is behaving ideally). The pressure inside the flask in bar is (Given R =0.0831 L bar K⁻¹ mol⁻¹)
Given: V=10.0L W_o₂ = 64g
n_o₂ = 2 mol
L bar K⁻¹mol⁻¹
T = 27°C
R = 0.0831
Ideal gas equation; PV =nRT
P = 2 × 0.0831 × 300 / 10.0
P = 4.9 bar
The order of energy absorbed which is responsible for the colour of complexes
(A) [Ni(H₂O)₂(en)₂]²⁺
(B) [Ni(H₂O)₄(en)]²⁺
(C) [Ni(en)₃]²⁺
(A) [Ni(H₂O)₂(en)₂]²⁺
(B) [Ni(H₂O)₄(en)]²⁺
(C) [Ni(en)₃]²⁺
en behaves as SFL while H₂O behaves as WFL In option (C); number of en (strong ligand) increase, which increases the splitting and absorb maximum eenrgy. So, Δ₀ increases.
Notes SFL increases the splitting and abosrb maximum energy while WFL decreases the splitting energy.
Which one of the following is not formed when acetone reacts with 2-pentanone in the presence of dilute NaOH followed by heating?
-
(A) c
-
(B) d
-
(C) b
-
(D) a
Find the emf of the cell in which the following reaction takes place at 298 K. Ni(s) + 2Ag⁺(0.001M) → Ni²⁺(0.001 M) + 2Ag(s)
(Given that E°_cell = 10.5V, 2.303RT / F = 0.059 at 298 K)
(None) Ni(s) + 2Ag⁺ (0.001 M) → Ni²⁺ (0.001 M) + 2Ag(s)
E_cell = E°_cell − 0.059 / n (log) [Ni²⁺] / [Ag⁺]²
E_cell = 10.5 − 0.059 (log) (10⁻³) / (10⁻³)²
= 10.5 −0.059 log 10⁺³ =
= 10.5 - 0.059/2 × 3 = 10.4115 V
(Calculated answer is not given in options)
If radius of second Bohr orbit of the He⁺ ion is 105.8 pm, what is the radius of third Bohr orbit of Li²⁺ ion?
According to Bohr’s atomic model r ∝ n² / z 3rd orbit of Li²⁺ n₁ = 3
z₁ = 3 ⇒ 2ⁿᵈ orbit of He⁺ n₂ = 2
z₂ = 2
(r₃)_Li²⁺ / (r₂)He⁺ (n₁² / n₂²) × z₂ / z₁
(r₃)_Li²⁺ / 105.8 pm= (3)² / (2)² × 2 / 3
(r₃)_Li²⁺ = 158.7 pm
Copper crystallises in fec unit cell with cell edge length of 3.608 × 10⁻⁸ cm. The density of copperis 8.92 g cm⁻³ Calculate the atomic mass of copper.
Given: Edge Length (a) = 3.608 × 10⁻⁸ cm Density of copper (d) = 8.92
In fcc crystal structure, the value of z is 4. d = z × M / N_A × a³
8.92 = 4 × M / 6.022 × 10²³ × (3.608 × 10⁻⁸)³ M = (8.92 × 6.022 × 10²³ / 4) × 46.96 × 10⁻²⁴ M = 63.1 g/mol (Molar Atomic Mass) M = 63.1 u (Atomic Mass)
The product formed from the following reaction sequence is
-
(A) c
-
(B) a
-
(C) b
-
(D) d
Read the following statements about the vascular bundles:
(i) In roots, xylem and phloem in a vascular bundle are arranged in an altenate manner along the different radii.
(ii) Conjoint closed vascular bundles do not possess cambium
(iii) In open vascular bundles, cambium is present in between xylem and phloem
(iv) The vascular bundles of dicotyledonous stem possess endarch protoxylem
(v) In monocotyledonous root, usually there are more than six xylem bundles present Choose the correct answer from the options given below :
All the statements are correct about vascular bundles.
Identify the correct set of statements:
(i) The leaflets are modified into pointed hard thorns in Citrus and Bougainvillea
(ii) Axillary buds form slender and spirally coiled tendrils in cucumber and pumpkin
(iii) Stem is flattened and fleshy in Opuntia and modified to perform the function of leaves
(iv) Rhizophora shows vertically upward growing roots that help to get oxygen for respiration
(v) . Subaerially growing stems in grasses and strawberry help in vegetative propagation Choose the correct answer from the options given below:
Thorns of citrus and Bougainvillea are modified stem not leaves.
Read the following statements and choose the set of correct statements:
(i) Euchromatin is loosely packed chromatin
(ii) Heterochromatin is transcriptionally active
(iii) Histone octomer is wrapped by negatively charged DNA in nucleosome
(iv) Histones are rich in lysine and arginine
(v) Atypical nucleosome contains 400 by of DNA helix
Choose the correct answer from the options given below:
Heterochromatin is highly condensed, gene-poor, and transcriptionally silent, whereas euchromatin is less condensed, gene-rich, and more easily transcribed. So, option b is not correct. A typical nucleosome contains 200 bp of DNA helix. So, option e is not correct.
Given below are two statements :
Statement I: The primary CO₂ acceptor in C₄ plants is phosphoenolpyruvate and is found in the mesophy!l cells.
Statement II: Mesophyll cells of C₄ plants lack RuBisCo enzyme.
In the light of the above statements, choose the correct answer from the options given below:
The primary CO₂ acceptor is a 3-carbon molecule phosphoenol pyruvate (PEP) and is present in the mesophyll cells. The enzyme responsible for this fixation is PEP carboxylase or PEPcase. It is important to register that the mesophyll cells lack RuBisCO enzyme. The C4 acid OAA is formed in the mesophyll cells. Therefore, both the statements are correct.
Identify the incorrect statement related to Pollination:
Bees are the most common pollination agent. Therefore, Moths and butterflies are the most dominant pollinating agents among insects
Which one of the following statement is not true regarding get electrophoresis technique?
The separated DNA fragments can be visualised only after staining the DNA with a compound known as ethidium bromide followed by exposure to UV radiation. So, The presence of chromogenic substrate gives blue coloured DNA bands on the gel is not correct.
Habitat loss and fragmentation, over exploitation, alien species invasion and co-extinction are causes for:
Habitat loss and fragmentation, over-exploitation are the causes for loss in biodiversity.
Production of Cucumber has increased manifold in recent years. Application of which of the following phytohormones has resulted in this increased yield as the hormone is known to produce female flowers in the plants:
Ethephon is the compound which is the source of Ethylene hormone. The ethehone promotes female flowers in cucumbers thereby increasing their yield. So, ethylene is correct answer.
What is the net gain of ATP when each molecule of glucose is converted to two molecules of pyruvic acid ?
During glycolysis 2 ATP molecules are used up and four ATP molecules are generated. So, there are only 2 ATPs net gain when each molecule of glucose is converted into 2 molecules of plyruvic acid.
Given below are two statements:
Statement I: Cleistogamous flowers are invariably autogamous
Statement II: Cleistogamy is disadvantageous as there is no chance for cross pollination.
In the light of the above statements, choose the correct answer from the options given below:
Cleistogamy is the phenomenon, where flowers never open and in such flowers, only self-pollination occurs within the bud (unopened flower). Bisexual flowers which do not open at all are called cleistogamous. In such flowers, anthers and stigma lie close to each other. When the anthers dehisce in the flower buds, pollen grains come in contact with the stigma and pollination occurs. So, there is no chance of cross-pollination in this type of flower. Therefore, both statements are true.
Hydrocolloid carrageen is obtained from
Hydrocolloid carrageen is obtained from red algae (rhodophyceae).
"Girdling Experiment" was performed by Plant Physiologists to identify the plant tissue through which:
A simple experiment called girdling, was used to identify the tissues through which food is transported.
Which of the following is incorrectly matched?
Ulothrix belong to Chlorophyceae (green algae). Members of this class have starch as reserve food material.
DNA polymorphism forms the basis of:
DNA polymorphism form the basis of genetic mapping and DNA fingerprinting.
Match List-I with List-II.
Manganese help in splitting of water during photosynthesis.
Magnesium activates the enzyme of respiration. Boron is required for pollen germination. Iron activates the catalase enzyme.
The process of translation of mRNA to proteins begins as soon as :
The process of translation of mRNA to protein begins as soon small unit of ribosome encounters Mrna.
The device which can remove particulate matter present in the exhaust from a thermal power plant is:
An electrostatic precipitator (ESP) removes particles from a gas stream by using electrical energy to charge particles either positively or negatively. The charged particles are then attracted to collector plates carrying the opposite charge.
The flowers are Zygomorphic in:
(i) Mustard (ii) Gulmohar
(iii) Cassia (iv) Datura
(v) Chilly
Choose correct answer from the options given below
A zygomorphic flower is one that is capable of dividing into two equal parts (mirror image) by only one longitudinal plane passing through the axis. Example include gulmohar, Cassia.
Given below are two statements; one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Polymerase chain reaction is used in DNA amplification.
Reason (R): The ampicillin resistant gene is used as a selectable marker to check transformation.
In the light of the above statements, choose the correct answer from the options given below:
PCR (polymerase chain reaction) is used in DNA amplification. So, this is true. Ampicillin resistant gene is used as a selectable marker. Therefore, both A and R are correct, but R is not the correct explanation of A.
NOTES
Polymerase chain reaction (abbreviated PCR) is a laboratory technique for rapidly producing (amplifying) millions to billions of copies of a specific segment of DNA, which can then be studied in greater detail. PCR involves using short synthetic DNA fragments called primers to select a segment of the genome to be amplified, and then multiple rounds of DNA synthesis to amplify that segment.
Which one of the following statements cannot be connected to Predation?
Predation is a biological interaction where one organism, the predator, kills and eats another organism, its prey. In predation, one species is benefitted and other is harmed.
Which one of the following never occurs during mitotic cell division?
Mitosis occurs in somatic cells; this means that it takes place in all types of cells that are not involved in the production of gametes. Prior to each mitotic division, a copy of every chromosome is created; thus, following division, a complete set of chromosomes is found in the nucleus of each new cell. So, it do not include pairing of homologous chromosomes.
Which of the following is not a method of ex situ conservation?
Examples of ex-situ conservation of biodiversity include aquariums, botanical gardens, cryopreservation, DNA banks, zoos, etc. National parks is an example of in-situ conservation.
Given below are two statements:
Statement I: Mendel studied seven pairs of contrasting traits in pea plants and proposed the Laws of Inheritance
Statement II: Seven characters examined by Mendel in his experiment on pea plants were seed shape and colour, flower colour, pod shape and colour, flower position and stem height
In the light of the above statements, choose the correct answer from the options given below:
There were 7 characters of the pea plant which were selected by Mendel for the experiments. The characters which were chosen by Mendel for his study were stem height, flower colour, flower position, pod shape, pod colour, seed shape, seed colour. All these characters belong to different chromosomes in the pea plant. Therefore both statements are correct.
Which one of the following plants does not show plasticity?
Plants follow different pathways in response to environment or phases of life to form different kinds of structures. This ability is called plasticity, e.g., heterophylly in cotton, coriander and larkspur. In such plants, the leaves of the juvenile plant are different in shape from those in mature plants. On the other hand, difference in shapes of leaves produced in air and those produced in water in buttercup also represent the heterophyllous development due to environment. This phenomenon of heterophylly is an example of plasticity.
What amount of energy is released from glucose during lactic acid fermentation?
In both lactic acid and alcohol fermentation, less than 7 percent of the energy in glucose is released.
NOTES
Lactic acid fermentation is a metabolic process by which glucose or other six-carbon sugars (also, disaccharides of six-carbon sugars, e.g. sucrose or lactose) are converted into cellular energy and the metabolite lactate, which is lactic acid in solution.
The gaseous plant growth regulator is used in plants to :
Ethylene is the gaseous hormone that promotes root growth and root hair formation.
Which of the following is not observed during apoplastic pathway?
In the apoplast pathway, water is transported from root hair to xylem through the cell wall of intervening cells. Apoplastic water movement beyond the cortex is blocked by the Casparian strip of endodermal cells. Cytoplasmic streaming is involved symplast pathway.
Which one of the following is not true regarding the release of energy during ATP synthesis through chemiosmosis? It involves :
Breakdown of electron gradient does not take place in chemiosmosis.
Which one of the following plants shows vexillary aestivation and diadelphous stamens?
Pisum Sativum show Vexillary aestivation and diadelphous stamens.
Given below are two statements:
Statement I: Decomposition is a process in which the detritus is degraded into simpler substances by microbes.
Statement II: Decomposition is faster if the detritus is rich in lignin and chitin
In the light of the above statements, choose the correct answer from the options given below:
decomposers break down complex organic matter into inorganic substances like carbon dioxide, water and nutrients and the process is called decomposition. In a particular climatic condition, decomposition rate is slower if detritus is rich in lignin and chitin, and quicker, if detritus is rich in nitrogen and water-soluble substances like sugars. So, statement (Statement I is correct but Statement II is incorrect) is not correct.
Which one of the following produces nitrogen fixing nodules on the root of Alnus?
Frankia alni is a Gram-positive species of actinomycete filamentous bacterium that lives in symbiosis with actinorhizal plants in the genus Alnus. It is a nitrogen-fixing bacterium and forms nodules on the roots of alder trees.
Exoskeleton of arthropods is composed of:
Exoskeleton of arthropods is made up of chitin.
XO type of sex determination can be found in:
XO type of sex-determination is found in grass-hoppers.
In old trees the greater part of secondary xylem is dark brown and resistant to insect attack due to:
(i) secretion of secondary metabolites and their deposition in the lumen of vessels.
(ii) deposition of organic compounds like tannins and resins in the central layers of stem.
(iii) deposition of suberin and aromatic substances in the outer layer of stem.
(iv) deposition of tannins, gum, resin and aromatic substances in the peripheral layers of stem.
(v) presence of parenchyma cells, functionally active xylem elements and essential oils
Choose the correct answer from the options given below:
In old trees, greater part of secondary xylem is dark brown and resistant to insect is because of secretion of secondary metabolites and deposition of organic compounds.
Match List-I with List-II
Every chromosome (visible only in dividing cells) essentially has a primary constriction or the centromere on the sides of which disc shaped structures called kinetochores are present (Figure 8.12). Centromere holds two chromatids of a chromosome. Based on the position of the centromere, the chromosomes can be classified into four types (Figure 8.13). The metacentric chromosome has middle centromere forming two equal arms of the chromosome. The sub- metacentric chromosome has centromere slightly away from the middle of the chromosome resulting into one shorter arm and one longer arm. In case of acrocentric chromosome the centromere is situated close to its end forming one extremely short and one very long arm, whereas the telocentric chromosome has a terminal centromere.
The entire fleet of buses in Delhi were converted to CNG from diesel. In reference to this, which one of the following statements is false?
CNG used different engine not diesel engine as it uses expensive technology.
Read the following statements on lipids and find out correct set of statements:
(i) Lecithin found in the plasma membrane is a glycolipid
(ii) Saturated fatty acids possess one or more C = C bonds
(iii) Gingely oil has lower melting point, hence remains as oil in winter
(iv) Lipids are generally insoluble in water but soluble in some organic solvents
(v) When fatty acid is esterified with glycerol, monoglycerides are formed
Choose the correct answer from the options given below:
Lecithins are mixtures of glycerophospholipids including phosphatidylcholine, phosphatidylethanolamine, phosphatidyl-inositol, phosphatidylserine and phosphatidic acid. So, option a is not correct. Saturated fatty acids (SFAs) are made up of a carbon chain with no double bonds. So, option (iii), (iv) and (v) only is not correct.
The anatomy of springwood shows some peculiar features.
Identify the correct set of statements about springwood.
(i) It is also called as the earlywood
(ii) In spring season cambium produces xylem elements with narrow vessels
(iii) It is lighter in colour
(iv) The springwood along with autumnwood shows alternate concentric rings forming annual rings
(v) It has lower density
Choose the correct answer from the options given below:
In spring season, cambium produces xylem elements with wider vessels. So, option b is not correct. Rest all statements are
correct.
Transposons can be used during which one of the following?
Transposon silencing is a form of transcriptional gene silencing targeting transposons. Transcriptional gene silencing is a product of histone modifications that prevent the transcription of a particular area of DNA. Transcriptional silencing of transposons is crucial to the maintenance of a genome.
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Mendel's law of independent assortment does not hold good for the genes that are located closely on the same chromosome.
Reason (R): Closely located genes assort independently.
In the light of the above statements, choose the most appropriate answer from the options given below:
Mendel's law of independent assortment states that the alleles of two (or more) different genes get sorted into gametes independently of one another. In other words, the allele a gamete receives for one gene does not influence the allele received for another gene. This law does not hold good for the genes that are closely located. Therefore, assertion is correct and reason is not correct.
The following palindromic base sequences of DNA, which one can be cut easily by particular restriction enzyme?
Restriction enzymes, also called restriction endonucleases, cut double stranded DNA molecules by cleaving phosphodiester bonds at palindromic sequences. This means that the majority of the restriction enzymes cut the DNA into fragments that are characterized by a twofold rotational symmetry. As in the first option, both the sequence are palindromic, so it will be the correct answer.
Which one of the following will accelerate phosphorus cycle?
Phosphorus moves in a cycle through rocks, water, soil and sediments and organisms. Over time, rain and weathering cause rocks to release phosphate ions and other minerals. This inorganic phosphate is then distributed in soils and water. Plants take up inorganic phosphate from the soil.
Match the plant with the kind of life cycle it exhibits:
Choose the correct answer from the options given below:
Spirogyra is a green algae which have dominant haploid – free living gametophyte. Fern is a pteridophyte having dominant sporophyte alternating with reduced gametophyte.
While explaining interspecific interaction of population, (+) sign is assigned for beneficial interaction, (−) sign is assigned for detrimental interaction and (0) for neutral interaction. Which of the following interactions can be assigned (+) for one species and (−) for another species involved in the interaction?
Predation, a “+/-” interaction, includes predator-prey, herbivore-plant, and parasitoid host interactions. These linkages are the prime movers of energy through food chains and are an important factor in the ecology of populations, determining mortality of prey and birth of new predators.
Addition of more solutes in a given solution will:
Addition of more solutes in a given solution will lower its water potential.
Which part of the fruit, labelled in the given figure makes it a false fruit?
In few species, such as apple, strawberry, cashew, thalamus
contribute to fruit formation and these fruits are known as false
fruit.
Which of the following occurs due to the presence of autosome linked dominant trait?
Myotonic dystrophy is a autosomal dominant trait.
If a geneticist uses the blind approach for sequencing the whole genome of an organism, followed by assignment of function to different segments, the methodology adopted by him is called as:
Sequence annotation is the process of marking specific features in a DNA, RNA or protein sequence with descriptive information about structure or function.
What is the role of large bundle shealth cells found around the vascular bundles in C₄ plants?
Function of large bundle sheath is to increase the number of chloroplast for operation of Calvin cycle.
Nitrogenous waste is excreted in the form of pellet or paste by:
Reptiles, birds, land snails and insects excrete nitrogenous wastes as uric acid in the form of pellet or paste with a minimum loss of water and are called uricotelic animals. Hence correct option is Pavo contains uric acid and excrete uric acid.
NOTES
Of the four major macromolecules in biological systems, both proteins and nucleic acids contain nitrogen. Excess nitrogen is excreted from the body. Nitrogenous wastes tend to form toxic ammonia, which raises the pH of body fluids. The formation of ammonia itself requires energy
in the form of ATP and large quantities of water to dilute it out of a biological system. Animals that live in aquatic environments tend to release ammonia into the water. Animals that excrete ammonia are said to be ammonotelic. Terrestrial organisms have evolved other mechanisms to excrete nitrogenous wastes. The animals must detoxify ammonia by converting it into a relatively nontoxic form such as urea or uric acid. Mammals including humans, produce urea, whereas reptiles and many terrestrial invertebrates produce uric acid. Animals that secrete urea as the primary nitrogenous waste are called ureotelic animals.
Select the incorrect statement with reference to mitosis:
Spindle fibres attach to kinetochores of chromosomes.
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Osteoporosis is characterised by decreased bone mass and increased chances of fractures.
Reason (R): Common cause of osteoporosis is increased levels of estrogen.
In the light of the above statements, choose the most appropriate answer from the options given below:
Osteoporosis is a age-related disorder characterised by decreased bone mass and increased chances of fractures.
Common cause of osteoporosis is due to decreased levels of estrogen.
Under normal physiological conditions in human being every 100 ml of oxygenated blood can deliver ______ ml of O₂ to the tissues.
Every 100 ml of oxygenated blood can deliver around 5 ml of O₂ to the tissues under normal physiological conditions.
A dehydration reaction links two glucose molecules to produce maltose. If the formula for glucose is C₆H₁₂O₆, then what is the formula for maltose?
2 molecule of glucose is considered as maltose i.e. C₁₂H₂₄O₁₂. If water (H₂O) molecule is dehydrated from maltose then molecules of maltose considered as C₁₂H₂₂O₁₁.
In which of the following animals digestive tract has additional chambers like crop and gizzard?
The digestive tract of birds has additional chambers, the crop and gizzard. Example: Corvus (Crow), Columba (Pigeon), Psittacula (Parrot), Struthio (Ostrich), Pavo (Peacock), Aptenodytes (Penguin), Neophron (Vulture).
Given below are two statements:
Statement I: The release of sperms into the seminiferous tubules is called spermiation.
Statement II: Spermiogenesis is the process of formation of sperms from spermatogonia.
In the light of the above statements, choose the most appropriate answer from the options given below:
The spermatids are transformed into spermatozoa (sperms) by the process called spermiogenesis. After spermiogenesis, sperms heads become embedded in the Sertoli cells, and are finally release from the seminiferous tubules by the process called spermiation.
Natural selection where more individuals acquire specific character value other than the mean character value, leads to:
Natural selection can lead to stabilisation (in which more individuals acquire mean character value), directional change (more individuals acquire value other than the mean character value) or disruption (more individuals acquire peripheral character value at both ends of the distribution curve).
Which of the following statements with respect to Endoplasmic Reticulum is incorrect?
The eukaryotic cells reveal the presence of a network or reticulum called the endoplasmic reticulum. The ER often shows ribosomes attached to their outer surface refer to the rough endoplasmic reticulum.
Which of the following is present between the adjacent bones of the vertebral column?
Cartilage is present between the adjacent bones of the vertebral column.
Which of the following functions is not performed by secretions from salivary glands?
About 30 per cent of starch is hydrolysed here by this enzyme (optimum pH 6.8) into a disaccharide (in the form of maltose). Rest of the disaccharides are digested in stomach and small intestine. Lysozyme present in saliva acts as an antibacterial agent that prevents infections.
In an E.coli strain i gene gets mutated and its product can not bind the inducer molecule. If growth medium is provided with lactose, what will be the outcome?
If lactose is provided in the growth medium of the bacteria, the lactose is transported into the cells through the action of permease (Remember, a very low level of expression of lac operon has to be present in the cell all the time, otherwise lactose cannot enter the cells). The lactose then induces the operon in the following manner.
Identify the asexual reproductive structure associated with Penicillium:
The asexual spores are conidia produced exogenously in ascomycetes, on the special mycelium called conidiophores. Hence, correct option is (Conidia) as Penicillium is the example of Ascomycetes.
If the length of a DNA molecule is 1.1 metres, what will be the approximate number of base pairs?
If the length of DNA double helix in a typical mammalian cell is calculated (simply by multiplying the total number of bp with distance between two consecutive bp, that is, 3.3 × 109 bp × 0.34 × 10⁻⁹ m/bp), it comes out to be approximately 1.1 metres. A length that is far greater than the dimensions of a typical nucleus (approximately 10⁻⁶ m).
Which of the following is not a connective tissue?
Neuroglia, are non-neuronal cells in the central nervous system and the peripheral nervous system. They maintain homeostasis, form myelin, and provide support and protection for neurons. Hence, it is neural tissue not a connective tissue.
NOTES
Neuroglia are located throughout the CNS and PNS. In the CNS, the most important glial cells include astrocytes, microglia, oligodendrocytes, and ependymal cells.
Given below are two statements:
Statement I: Restriction endonucleases recognise specific sequence to cut DNA known as palindromic nucleotide sequence.
Statement II: Restriction endonucleases cut the DNA strand a little away from the centre of the palindromic site.
In the light of the above statements, choose the most appropriate answer from the options given below:
Both the statements are correct.
Detritivores breakdown detritus into smaller particles. This process is called:
Detritivores (e.g., earthworm) break down detritus into smaller particles. This process is called fragmentation.
Which of the following statements are true spermatogenesis but do not hold true Oogenesis?
(i) It results in the formation of haploid gametes
(ii) Differentiation of gamete occurs after the completion of meiosis
(iii) Meiosis occurs continuously in a mitotically dividing stem cell population
(iv) It is controlled by the Luteinising hormone (LH) and Follicle Stimulating Hormone (FSH) secreted by the anterior pituitary
(v) It is initiated at puberty
Choose the most appropriate answer from the options given below:
Oocytes start division and enter into prophase-I of the meiotic division and get temporarily arrested at that stage, called primary oocytes.
Oogenesis is initiated during the embryonic development stage when a couple of million gamete mother cells (oogonia) are formed within each fetal ovary; no more oogonia are formed and added after birth.
Given below are two statements:
Statement I: Fatty acids and glycerols cannot be absorbed into the blood.
Statement II: Specialized lymphatic capillaries called lacteals carry chylomicrons into lymphatic vessels and ultimately into the blood.
In the light of the above statements, choose the most appropriate answer from the options given below:
Fatty acids and glycerol being insoluble, cannot be absorbed into the blood. They are first incorporated into small droplets called micelles which move into the intestinal mucosa. They are re-formed into very small protein coated fat globules called chylomicrons which are transported into the lymph vessels (lacteals) in the villi. These lymph vessels ultimately release the absorbed substances into the blood stream.
If ‘8’ Drosophila in a laboratory population of ‘80’ died during a week, the death rate in the population is individuals per Drosophila per week.
If 8 individuals in a laboratory population of 80 (Drosophila melanogaster) fruit flies died during a specified time interval, say a week, the death rate in the population during that period is 8/80 = 0.1 individuals per fruitfly per week.
Given below are two statements:
Statement I: The coagulum is formed of network of threads called thrombins.
Statement II: Spleen is the graveyard of erythrocytes.
In the light of the above statements, choose the most appropriate answer from the options given below:
Coagulum is formed mainly of a network of threads called fibrins in which dead and damaged formed elements of blood are trapped.
Spleen mainly contains lymphocytes and phagocytes. It acts as a filter of the blood by trapping blood-borne microorganisms.
Spleen also has a large reservoir of erythrocytes. Henc, also refer to the graveyard of erythrocytes.
Tegmina in cockroach, arises from:
Forewings (mesothoracic) called tegmina are opaque dark and leathery and cover the hind wings when at rest.
In the taxonomic categories which hierarchial arrangement in ascending order is correct in case of animals?
Taxonomical studies of all known organisms have led to the development of common categories such as kingdom, phylum or division (for plants), class, order, family, genus and species. (provide option are in descending order).
Identify the microorganism which is responsible. for the production of an immunosuppressive molecule cyclosporin A:
Cyclosporin A is used as an immunosuppressive agent in organ-transplant patients, is produced by the fungus Trichoderma polysporum.
Which of the following is not the function of conducting part of respiratory system?
The part starting with the external nostrils up to the terminal bronchioles constitute the conducting part whereas the alveoli and their ducts form the respiratory or exchange part of the respiratory system. The conducting part transports the atmospheric air to the alveoli, clears it from foreign particles, humidifies and also brings the air to body temperature. Exchange
part is the site of actual diffusion of O₂ and CO₂ between blood and atmospheric air.
Lippe’s loop is a type of contraceptive used as:
Intra Uterine Devices are presently available as the non-medicated IUDs (e.g., Lippes loop), copper releasing IUDs (CuT, Cu7, Multiload 375) and the hormone releasing IUDs (Progestasert, LNG-20)
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): All vertebrates are chordates but all chordates are not vertebrates.Reason (R): Notochord is replaced by vertebral column
in the adult vertebrates.
In the light of the above statements, choose the most appropriate answer from the options given below :
The members of subphylum Vertebrata possess notochord during the embryonic period. The notochord is replaced by a cartilaginous or bony vertebral column in the adult. Thus all vertebrates are chordates but all chordates are not vertebrates.
Given below are two statements :
Statement I: Mycoplasma can pass through less than 1 micron filter size.
Statement II: Mycoplasma are bacteria with cell wall.
In the light of the above statements, choose the most appropriate answer from the options given below:
Mycoplasma or PPLO can pass through less than 1 micron filter size. the prokaryotes have a cell wall surrounding the cell membrane except in mycoplasma.
Regarding Meiosis, which of the statements is incorrect?
M phase represent the phase when the actual cell division (Mitosis/ meiosis) occurs. During G1 phase the cell is metabolically active and continuously grows but does not replicate its DNA. Hence, S or synthesis phase marks the period during which DNA synthesis or replication takes place. During this time the amount of DNA per cell doubles.
In-situ conservation refers to:
When we conserve and protect the whole ecosystem, its biodiversity at all levels is protected - we save the entire forest to save the tiger. This approach is called in situ (on site) conservation.
At which stage of life the oogenesis process is initiated?
Oogenesis is initiated during the embryonic development stage when a couple of million gamete mother cells (oogonia) are formed within each fetal ovary; no more oogonia are formed and added after birth.
Which of the following is a correct match for disease and its symptoms?
Arthritis: Inflammation of joints.
Myastheniz gravis: Auto immune disorder affecting neuromuscular junction leading to fatigue, weakening and paralysis of skeletal muscle.
Muscular dystrophy: Progressive degeneration of skeletal muscle mostly due to genetic disorder.
Tetany: Rapid spasms (wild contractions) in muscle due to low Ca⁺⁺ in body fluid.
Given below are two statements:
Statement I: Autoimmune disorder is a condition where body defense mechanism recognizes its own cells as foreign bodies.
Statement II: Rheumatoid arthritis is a condition where body does not attack self cells.
In the light of the above statements, choose the most appropriate answer from the options given below:
two reason for causing auto immune disorder:
First, higher vertebrates can distinguish foreign molecules as well as foreign organisms. Most of the experimental immunology deals with this aspect. Second, sometimes due to genetic and other unknown reasons, the body attack self-cells. This results in damage to the body and is called auto-immune disease.
Rheumatoid arthritis which affects many people in our society is an auto-immune disease.
In gene therapy of Adenosine Deaminase (ADA) deficiency, the patient requires periodic infusion of genetically engineered lymphocytes because:
A functional ADA cDNA (using a retroviral vector) is introduced into the treated lymphocytes, which are subsequently returned to the patient. However, as these cells are not immortal, the patient requires periodic infusion of such genetically engineered lymphocytes.
Breeding crops with higher levels of vitamins and minerals or higher proteins and healthier fats is called:
Biofortification is the process of breeding crops with higher levels of vitamins and minerals, or higher protein and healthier fats — is the most practical means to improve public health.
Which one of the following statements is correct?
During cardiac cycle, As ‘the ventricular pressure declines further, the tricuspid and bicuspid valves are pushed open by the pressure in the atria exerted by the blood which was being emptied into them by the veins. The blood moves freely to the ventricles. The ventricles and atria are again in a relaxed during joint diastole) state, as earlier.
Select the incorrect statement regarding synapses:
Impulse transmission across an electrical synapse is always faster than that across a chemical synapse.
Select the incorrect statement with respect to acquired immunity.
Acquired immunity s pathogen specific. It is characterised by memory. This means when our body encounters a pathogen for the first time it produces a response in a low intensity (primary response). Subsequent encounter with the same pathogen elicits a highly intensified secondary or anamnestic response. This is ascribed to the fact that our body appears to have memory of the first encounter.
Match List-I with List-II.
Choose the correct answer from the options given below:
Glycogen — storage product
Globulin — antibody
Steroids - hormone
Thrombin — thrombin
Match List-I with List-II with respect to methods of Contraception and their respective actions.
Choose the correct answer from the options given
below:
Diaphragms- they cover the cervix blocking the entry of sperms
Contraceptive pills — inhibit ovulation and implantation
Intra uterine devices — increases phagocytosis of sperm within uterus
Lactational amenorrhea — absences of menstrual cycle and ovulation following parturition.
Ten E.coli cells with ‘5N - dsDNA are incubated in medium containing “N nucleotide. After 60 minutes, how many E.coli cells will have DNA totally free from ‘5N?
E.coli cells with ¹⁶N-dsDNA are incubated in medium contain ¹⁵N Nucleotide.
If the culture ran for a generation (i.e. after 20 min.) had 100% hybrid DNA. DNA extracted from the culture after IInd generation (i.e. after 40 min.), then hybrid DNA was composed of equal amounts i.e. 50% light DNA and 50 hybrid. And for III generation (i.e. after 60 min.) amount of DNA is 75% hybrid and 25% light DNA. Therefore, if 10 cells of E.coli is incubated in medium, after 60 min, 80 cells will formed in IIIrd generation, there, 60 cells (75%) will have DNA totally free from ¹⁵N.
The recombination frequency between the genes a & c is 5%, b & c is 15%, b & d is 9%, a & b is 20%, c & d is 24% and a & d is 29%. What will be the sequence of these genes on a linear chromosome?
According to given percentage
So the sequence of genes, a, b, c & d will be a, c, b, d.
Given below are two statements:
Statement I: In a scrubber the exhaust from the thermal plant is passed through the electric wires to charge the dust particles.
Statement II: Particulate matter (PM 2.5) can not be removed by scrubber but can be removed by an electrostatic precipitator.
In the light of the above statements, choose the most appropriate answer from the options given below:
The collecting plates are grounded and attract the charged dust particles.
The dangers of particulate matter that are very very small and are not removed by electrostatic precipitators. According to Central Pollution Control Board (CPCB), particulate size 2.5 micrometers or less in diameter (PM 2.5) are responsible for causing the greatest harm to human health.
Statements related to human Insulin are given below.
Which statement(s) is/are correct about genetically engineered Insulin?
(i) Pro-hormone insulin contain extra stretch of C-peptide
(ii) A-peptide and B-peptide chains of insulin were produced separately in E.coli, extracted and combined by creating disulphide bond between them.
(iii) Insulin used for treating Diabetes was extracted from Cattles and Pigs.
(iv) Pro-hormone Insulin needs to be processed for converting into a mature and functional hormone.
(v) Some patients develop allergic reactions to the foreign insulin.
Choose the most appropriate answer from the options given below:
Chains A and B were produced separately, extracted and combined by creating disulfide bonds to form human insulin.
Insulin used for diabetes was earlier extracted from pancreas of slaughtered cattle and pigs.
Insulin is synthesised as a pro-hormone (like a pro-enzyme, the pro-hormone also needs to be processed before it becomes a fully mature and functional hormone) which contains an extra stretch called the C peptide. Insulin from an animal source, though caused some patients to develop allergy or other types of reactions to the foreign protein,
Which of the following statements is not true?
The eye of the octopus and of mammals or the flippers of Penguins and Dolphins are the example of analogy (analogous organ).
Which of the following is not a desirable feature of a cloning vector?
In the alien DNA, if Presence of more than one recognition sites within the vector will generate several fragments, which will complicate the gene cloning. Hence, the vector needs to have very few, preferably single, recognition sites for the commonly used restriction enzymes.
Which of the following is a correct statement?
The cyanobacteria have chlorophyll a similar to green plants and are photosynthetic autotrophs. The cyanobacteria are unicellular, colonial or filamentous, freshwater/marine or terrestrial algae including in kingdom monera.
Match List-I with List-II
Choose the correct answer from the options given below:
Bronchioles —ciliated epithelium
Goblet cell - glandular tissue
Tendons — dense regular connective tissue
Adipose tissue — loose connective tissue
Which of the following are not the effects of Parathyroid
hormone?
(i) Stimulates the process of bone resorption
(ii) Decreases Ca²⁺ level in blood
(iii) Reabsorption of Ca²⁺ by renal tubules
(iv) Decreases the absorption of Ca²⁺ from digested food
(v) Increases metabolism of carbohydrates
Choose the most appropriate answer from the options
given below:
Parathyroid hormone (PTH) increases the Ca²⁺ levels in the blood. PTH acts on bones and stimulates the process of bone resorption. PTH also stimulates reabsorption of Ca²⁺ by the renal tubules and increases Ca²⁺ absorption from the digested food. PTH plays a significant role in calcium balance not in metabolism of carbohydrates.
If a colour blind female marries a man whose mother was
also colour blind, what are the chances of her progeny
having colour blindness?
There is 100% chance of her progeny having colour blindness.
A nucleus of mass number 189 splits into two nuclei having mass number 125 and 64. The ratio of radius of two daughter nuclei respectively is:
We know that R = R₀A¹ᐟ³ R₁/R₂ = (A₁/A₂)¹ᐟ³ = (125/64)¹ᐟ³ = 5/4
NOTES The volume of a nucleus varies directly with mass number. V = (4/3)πR³ = (4/3)πR₀³A (∵ R = R₀A¹ᐟ³) ∴ V ∝ A Also, density of nucleus is constant, independent of A. Nuclear density (ρ) = Mass of nucleus / Volume of nucleus = mA / [(4/3)π(R₀A¹ᐟ³)³] = 3m / (4πR₀³)= Constant Here, m = Average mass of a nucleon = 1.66 × 10⁻²⁷ kg
R₀ = 1.2 × 10⁻¹⁵ m = 1.2 fm
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