- A. 4√2 Ω
- B. 5√2 Ω
- C. 4 Ω
- D. 5 Ω
Correct answer & explanation hidden
Log in to view solution →- A. towards the left as its potential energy will increase.
- B. towards the right as its potential energy will decrease.
- C. towards the left as its potential energy will decrease.
- D. towards the right as its potential energy will increase.
Correct answer & explanation hidden
Log in to view solution →- A. I_d = V_0 ωC cosωt
- B. I_d = (V_0 / ωC) cosωt
- C. I_d = (V_0 / ωC) sinωt
- D. I_d = V_0 ωC sinωt
Correct answer & explanation hidden
Log in to view solution →- A. 25
- B. 15
- C. 50
- D. 30
Correct answer & explanation hidden
Log in to view solution →- A. (A)-(R), (B)-(S), (C)-(P), (D)-(Q)
- B. (A)-(R), (B)-(S), (C)-(Q), (D)-(P)
- C. (A)-(R), (B)-(P), (C)-(S), (D)-(Q)
- D. (A)-(R), (B)-(Q), (C)-(S), (D)-(P)
Correct answer & explanation hidden
Log in to view solution →- A. 0.25 Ω
- B. 0.5 Ω
- C. 1 Ω
- D. 4 Ω
Correct answer & explanation hidden
Log in to view solution →- A. 1/2
- B. 1/(2√2)
- C. 2/3
- D. 2/(3√2)
Correct answer & explanation hidden
Log in to view solution →- A. 10¹⁸
- B. 10¹⁷
- C. 10¹⁶
- D. 10¹⁵
Correct answer & explanation hidden
Log in to view solution →- A. 10^18
- B. 10^17
- C. 10^16
- D. 10^15
Correct answer & explanation hidden
Log in to view solution →- A. n
- B. 2n
- C. 3n
- D. 4n
Correct answer & explanation hidden
Log in to view solution →- A. n
- B. 2n
- C. 3n
- D. 4n
Correct answer & explanation hidden
Log in to view solution →- A. 4×10⁻²⁰ N
- B. 8π×10⁻²⁰ N
- C. 4π×10⁻²⁰ N
- D. 8×10⁻²⁰ N
Correct answer & explanation hidden
Log in to view solution →- A. [F] [A] [T]
- B. [F] [A] [T²]
- C. [F] [A] [T⁻¹]
- D. [F] [A⁻¹] [T]
Correct answer & explanation hidden
Log in to view solution →- A. [F] [A] [T]
- B. [F] [A] [T^2]
- C. [F] [A] [T^-1]
- D. [F] [A^-1] [T]
Correct answer & explanation hidden
Log in to view solution →- A. (A) - (R), (B) - (P), (C) - (S), (D) - (Q)
- B. (A) - (Q), (B) - (R), (C) - (S), (D) - (P)
- C. (A) - (Q), (B) - (P), (C) - (S), (D) - (R)
- D. (A) - (R), (B) - (Q), (C) - (P), (D) - (S)
Correct answer & explanation hidden
Log in to view solution →- A. 0.9 MeV
- B. 9.4 MeV
- C. 804 MeV
- D. 216 MeV
Correct answer & explanation hidden
Log in to view solution →- A. 0.52 cm
- B. 0.026 cm
- C. 0.26 cm
- D. 0.052 cm
Correct answer & explanation hidden
Log in to view solution →- A. a large aperture contributes to the quality and visibility of the images.
- B. a large area of the objective ensures better light gathering power.
- C. a large aperture provides a better resolution.
- D. all of the above.
Correct answer & explanation hidden
Log in to view solution →- A. R1/R2
- B. R2/R1
- C. (R1/R2)^2
- D. R2^2/R1^2
Correct answer & explanation hidden
Log in to view solution →- A. 0.0628 s
- B. 6.28 s
- C. 3.14 s
- D. 0.628 s
Correct answer & explanation hidden
Log in to view solution →- A. v
- B. 2v
- C. 3v
- D. 4v
Correct answer & explanation hidden
Log in to view solution →- A. 60 cm
- B. 21.6 cm
- C. 64 cm
- D. 62 cm
Correct answer & explanation hidden
Log in to view solution →- A. Mg/2
- B. Mg
- C. 3Mg/2
- D. 2Mg
Correct answer & explanation hidden
Log in to view solution →- A. Mg/2
- B. Mg
- C. 3Mg/2
- D. 2Mg
Correct answer & explanation hidden
Log in to view solution →- A. 1/2 * ε0 * E^2
- B. ε0 * E * A * d
- C. 1/2 * ε0 * E^2 * A * d
- D. ε0 * E^2 * A * d
Correct answer & explanation hidden
Log in to view solution →- A. 1/2 ε0 E^2
- B. ε0 EAd
- C. 1/2 ε0 E^2 Ad
- D. ε0 E Ad / 2
Correct answer & explanation hidden
Log in to view solution →- A. current in n-type = current in p-type.
- B. current in p-type > current in n-type.
- C. current in n-type > current in p-type.
- D. No current will flow in p-type, current will only flow in n-type.
Correct answer & explanation hidden
Log in to view solution →- A. (A) and (B) both are correct.
- B. (A) and (B) both are incorrect.
- C. (A) is correct and (B) is incorrect.
- D. (A) is incorrect but (B) is correct.
Correct answer & explanation hidden
Log in to view solution →- A. having zero dipole moment.
- B. acquire a dipole moment only in the presence of electric field due to displacement of charges.
- C. acquire a dipole moment only when magnetic field is absent.
- D. having a permanent electric dipole moment.
Correct answer & explanation hidden
Log in to view solution →- A. [M^2][L^-1][T^0]
- B. [M][L^-1][T^-1]
- C. [M][L^0][T^0]
- D. [M^2][L^-2][T^-1]
Correct answer & explanation hidden
Log in to view solution →- A. 10.2 kW
- B. 8.1 kW
- C. 12.3 kW
- D. 7.0 kW
Correct answer & explanation hidden
Log in to view solution →- A. 20 m/s, 5 m/s^2
- B. 20 m/s, 0
- C. 20√2 m/s, 0
- D. 20√2 m/s, 10 m/s^2
Correct answer & explanation hidden
Log in to view solution →- A. 0 kg m/s
- B. 4.2 kg m/s
- C. 2.1 kg m/s
- D. 1.4 kg m/s
Correct answer & explanation hidden
Log in to view solution →- A. 20 cm from the lens, it would be a real image.
- B. 30 cm from the lens, it would be a real image.
- C. 30 cm from the plane mirror, it would be a virtual image.
- D. 20 cm from the plane mirror, it would be a virtual image.
Correct answer & explanation hidden
Log in to view solution →- A. 0.2 A
- B. 0.4 A
- C. 2 A
- D. 4 A
Correct answer & explanation hidden
Log in to view solution →- A. 0.2 A
- B. 0.4 A
- C. 2 A
- D. 4 A
Correct answer & explanation hidden
Log in to view solution →- A. (k^2 / (1-k^2)) * R
- B. (k^2 / (1+k^2)) * R
- C. (R * k) / (1+k)
- D. (R*k^2) / (1-k^2)
Correct answer & explanation hidden
Log in to view solution →- A. R * k^2 / (1 - k^2)
- B. R * k^2 / (1 + k^2)
- C. R * k / (1 + k)
- D. R * k^2 / (1 - k)
Correct answer & explanation hidden
Log in to view solution →- A. 660 V
- B. 1320 V
- C. 1520 V
- D. 1980 V
Correct answer & explanation hidden
Log in to view solution →- A. 25 rad/s and 75 rad/s
- B. 50 rad/s and 25 rad/s
- C. 46 rad/s and 54 rad/s
- D. 42 rad/s and 58 rad/s
Correct answer & explanation hidden
Log in to view solution →- A. Both Statement I and Statement II are true.
- B. Both Statement I and Statement II are false.
- C. Statement I is correct but Statement II is false.
- D. Statement I is incorrect but Statement II is true.
Correct answer & explanation hidden
Log in to view solution →- A. sp3 and 4
- B. sp3 and 6
- C. sp2 and 6
- D. sp2 and 8
Correct answer & explanation hidden
Log in to view solution →- A. Noble gases are sparingly soluble in water.
- B. Noble gases have very high melting and boiling points.
- C. Noble gases have weak dispersion forces.
- D. Noble gases have large positive values of electron gain enthalpy.
Correct answer & explanation hidden
Log in to view solution →- A. Noble gases are sparingly soluble in water.
- B. Noble gases have very high melting and boiling points.
- C. Noble gases have weak dispersion forces.
- D. Noble gases have large positive values of electron gain enthalpy.
Correct answer & explanation hidden
Log in to view solution →- A. 201.28 S cm2 mol−1
- B. 390.71 S cm2 mol−1
- C. 698.28 S cm2 mol−1
- D. 540.48 S cm2 mol−1
Correct answer & explanation hidden
Log in to view solution →- A. 201.28 S cm^2 mol^-1
- B. 390.71 S cm^2 mol^-1
- C. 698.28 S cm^2 mol^-1
- D. 540.48 S cm^2 mol^-1
Correct answer & explanation hidden
Log in to view solution →- A. NaCl solution
- B. Glucose solution
- C. Starch solution
- D. Urea solution
Correct answer & explanation hidden
Log in to view solution →- A. NaCl solution
- B. Glucose solution
- C. Starch solution
- D. Urea solution
Correct answer & explanation hidden
Log in to view solution →- A. Vitamin B12
- B. Vitamin B6
- C. Vitamin B1
- D. Vitamin B2
Correct answer & explanation hidden
Log in to view solution →- A. 120°
- B. 180°
- C. 60°
- D. 0°
Correct answer & explanation hidden
Log in to view solution →- A. 120°
- B. 180°
- C. 60°
- D. 0°
Correct answer & explanation hidden
Log in to view solution →- A. Actinoid contraction is greater for element to element than Lanthanoid contraction.
- B. Most of the trivalent Lanthanoid ions are colorless in the solid state.
- C. Lanthanoids are good conductors of heat and electricity.
- D. Actinoids are highly reactive metals, especially when finely divided.
Correct answer & explanation hidden
Log in to view solution →- A. Saytzeff’s Rule
- B. Hund’s Rule
- C. Hofmann Rule
- D. Huckel’s Rule
Correct answer & explanation hidden
Log in to view solution →- A. CP/CV = R
- B. CP - CV = R
- C. CP = RCV
- D. CV = RCP
Correct answer & explanation hidden
Log in to view solution →- A. Teflon
- B. Nylon-66
- C. Novolac
- D. Dacron
Correct answer & explanation hidden
Log in to view solution →- A. (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
- B. (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
- C. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
- D. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
Correct answer & explanation hidden
Log in to view solution →- A. Electrolysis
- B. Chromatography
- C. Distillation
- D. Zone refining
Correct answer & explanation hidden
Log in to view solution →- A. Beta (β-)
- B. Alpha (α)
- C. Gamma (γ)
- D. Neutron (n)
Correct answer & explanation hidden
Log in to view solution →- A. CH3–F < CH3–Cl < CH3–Br < CH3–I
- B. CH3–F > CH3–Cl > CH3–Br > CH3–I
- C. CH3–F < CH3–Cl > CH3–Br > CH3–I
- D. CH3–Cl > CH3–F > CH3–Br > CH3–I
Correct answer & explanation hidden
Log in to view solution →- A. 8, 4
- B. 6, 12
- C. 2, 1
- D. 12, 6
Correct answer & explanation hidden
Log in to view solution →- A. 2KClO3 → 2KCl + 3O2
- B. Cr2O3 + 2Al → Al2O3 + 2Cr
- C. Fe + 2HCl → FeCl2 + H2
- D. 2Pb(NO3)2 → 2PbO + 4NO2 + O2
Correct answer & explanation hidden
Log in to view solution →- A. 8.50
- B. 5.50
- C. 7.75
- D. 6.25
Correct answer & explanation hidden
Log in to view solution →- A. Calcium chloride
- B. Strontium chloride
- C. Magnesium chloride
- D. Beryllium chloride
Correct answer & explanation hidden
Log in to view solution →- A. Calcium chloride
- B. Strontium chloride
- C. Magnesium chloride
- D. Beryllium chloride
Correct answer & explanation hidden
Log in to view solution →- A. upto 1200 K
- B. upto 2200 K
- C. upto 1900 K
- D. upto 5000 K
Correct answer & explanation hidden
Log in to view solution →- A. Hexadentate ligand with four “O” and two “N” donor atoms
- B. Unidentate ligand
- C. Bidentate ligand with two “N” donor atoms
- D. Tridentate ligand with three “N” donor atoms
Correct answer & explanation hidden
Log in to view solution →- A. Hexadentate ligand with four “O” and two “N” donor atoms
- B. Unidentate ligand
- C. Bidentate ligand with two “N” donor atoms
- D. Tridentate ligand with three “N” donor atoms
Correct answer & explanation hidden
Log in to view solution →- A. P2 > P1 > P3
- B. P1 > P2 > P3
- C. P2 > P3 > P1
- D. P3 > P1 > P2
Correct answer & explanation hidden
Log in to view solution →- A. P2 > P1 > P3
- B. P1 > P2 > P3
- C. P2 > P3 > P1
- D. P3 > P1 > P2
Correct answer & explanation hidden
Log in to view solution →- A. Both Statement I and Statement II are true.
- B. Both Statement I and Statement II are false.
- C. Statement I is correct but Statement II is false.
- D. Statement I is incorrect but Statement II is true.
Correct answer & explanation hidden
Log in to view solution →- A. Chain and dimer, respectively
- B. Linear in both
- C. Dimer and Linear, respectively
- D. Chain in both
Correct answer & explanation hidden
Log in to view solution →- A. belonging to same group
- B. diagonal relationship
- C. lanthanoid contraction
- D. having similar chemical properties
Correct answer & explanation hidden
Log in to view solution →- A. 219.3 m
- B. 219.2 m
- C. 2192 m
- D. 21.92 cm
Correct answer & explanation hidden
Log in to view solution →- A. CH
- B. CH2
- C. CH3
- D. CH4
Correct answer & explanation hidden
Log in to view solution →- A. C5H12
- B. C3H8O
- C. C3H6O
- D. C4H10O
Correct answer & explanation hidden
Log in to view solution →- A. 7
- B. 5
- C. 2
- D. 3
Correct answer & explanation hidden
Log in to view solution →- A. 2.518
- B. 2.602
- C. 25.18
- D. 26.02
Correct answer & explanation hidden
Log in to view solution →- A. ΔU=0, ΔStotal=0
- B. ΔU ≠ 0, ΔStotal ≠ 0
- C. ΔU=0, ΔStotal ≠ 0
- D. ΔU ≠ 0, ΔStotal=0
Correct answer & explanation hidden
Log in to view solution →- A. HF < HCl < HBr < HI : Increasing acidic strength
- B. H2O < H2S < H2Se < H2Te : Increasing pKa values
- C. NH3 < PH3 < AsH3 < SbH3 : Increasing acidic character
- D. CO2 < SiO2 < SnO2 < PbO2 : Increasing oxidizing power
Correct answer & explanation hidden
Log in to view solution →- A. 1.75×10^-4 mol L^-1
- B. 2.50×10^-4 mol L^-1
- C. 1.75×10^-5 mol L^-1
- D. 2.50×10^-5 mol L^-1
Correct answer & explanation hidden
Log in to view solution →- A. 41.5 kJ mol^-1
- B. 83.0 kJ mol^-1
- C. 166 kJ mol^-1
- D. -83 kJ mol^-1
Correct answer & explanation hidden
Log in to view solution →- A. POCl3
- B. CH2O
- C. SbCl5
- D. NO2
Correct answer & explanation hidden
Log in to view solution →- A. O^2-, F-
- B. Na+, Mg^2+
- C. Mn^2+, Fe^3+
- D. Fe^2+, Mn^2+
Correct answer & explanation hidden
Log in to view solution →- A. 160 mm of Hg
- B. 168 mm of Hg
- C. 336 mm of Hg
- D. 350 mm of Hg
Correct answer & explanation hidden
Log in to view solution →- A. Resource partitioning
- B. Competitive release
- C. Mutualism
- D. Predation
Correct answer & explanation hidden
Log in to view solution →- A. RNA
- B. DNA
- C. Histones
- D. Polysaccharides
Correct answer & explanation hidden
Log in to view solution →- A. RNA
- B. DNA
- C. Histones
- D. Polysaccharides
Correct answer & explanation hidden
Log in to view solution →- A. Cohesion (i) More attraction in liquid phase; (b) Adhesion (ii) Mutual attraction among water molecules; (c) Surface tension (iii) Water loss in liquid phase; (d) Guttation (iv) Attraction towards polar surfaces
- B. (1)(ii)(iv)(i)(iii); (2)(iv)(iii)(ii)(i); (3)(iii)(i)(iv)(ii); (4)(ii)(i)(iv)(iii)
- C. Wait for mapping
- D. Wait for mapping
Correct answer & explanation hidden
Log in to view solution →- A. Xenogamy
- B. Geitonogamy
- C. Chasmogamy
- D. Cleistogamy
Correct answer & explanation hidden
Log in to view solution →- A. Xenogamy
- B. Geitonogamy
- C. Chasmogamy
- D. Cleistogamy
Correct answer & explanation hidden
Log in to view solution →- A. Metaphase I
- B. Metaphase II
- C. Anaphase II
- D. Telophase II
Correct answer & explanation hidden
Log in to view solution →- A. Metaphase I
- B. Metaphase II
- C. Anaphase II
- D. Telophase II
Correct answer & explanation hidden
Log in to view solution →- A. Denaturation, Annealing, Extension
- B. Denaturation, Extension, Annealing
- C. Extension, Denaturation, Annealing
- D. Annealing, Denaturation, Extension
Correct answer & explanation hidden
Log in to view solution →- A. Mosses
- B. Pteridophytes
- C. Some Gymnosperms
- D. Some Liverworts
Correct answer & explanation hidden
Log in to view solution →- A. Bullet square
- B. Punch square
- C. Punnett square
- D. Net square
Correct answer & explanation hidden
Log in to view solution →- A. Natural selection
- B. Genetic recombination
- C. Mutation
- D. Genetic drift
Correct answer & explanation hidden
Log in to view solution →- A. Homosorus
- B. Heterosorus
- C. Homosporous
- D. Heterosporous
Correct answer & explanation hidden
Log in to view solution →- A. Elasticity
- B. Flexibility
- C. Plasticity
- D. Maturity
Correct answer & explanation hidden
Log in to view solution →- A. Morphine, codeine
- B. Amino acids, glucose
- C. Vinblastin, curcumin
- D. Rubber, gums
Correct answer & explanation hidden
Log in to view solution →- A. (a)-Replication; (b)-Transcription; (c)-Transduction; (d)-Protein
- B. (a)-Translation; (b)-Replication; (c)-Transcription; (d)-Transduction
- C. (a)-Replication; (b)-Transcription; (c)-Translation; (d)-Protein
- D. (a)-Transduction; (b)-Translation; (c)-Replication; (d)-Protein
Correct answer & explanation hidden
Log in to view solution →- A. Metacentric
- B. Telocentric
- C. Sub-metacentric
- D. Acrocentric
Correct answer & explanation hidden
Log in to view solution →- A. Yellow bands
- B. Bright orange bands
- C. Dark red bands
- D. Bright blue bands
Correct answer & explanation hidden
Log in to view solution →- A. Shoot apex
- B. Stem
- C. Axillary bud
- D. Leaf
Correct answer & explanation hidden
Log in to view solution →- A. Biopiracy
- B. Gene therapy
- C. Molecular diagnosis
- D. Safety testing
Correct answer & explanation hidden
Log in to view solution →- A. Carica papaya
- B. Chara
- C. Marchantia polymorpha
- D. Cycas circinalis
Correct answer & explanation hidden
Log in to view solution →- A. Molecular diagnosis
- B. Gene amplification
- C. Purification of isolated protein
- D. Detection of gene mutation
Correct answer & explanation hidden
Log in to view solution →- A. (a) Cristae (i) Primary constriction in chromosome; (b) Thylakoids (ii) Disc-shaped sacs in Golgi apparatus; (c) Centromere (iii) Infoldings in mitochondria; (d) Cisternae (iv) Flattened membranous sacs in stroma of plastids
- B. (1)(iv)(iii)(ii)(i); (2)(i)(iv)(iii)(ii); (3)(iii)(iv)(i)(ii); (4)(ii)(iii)(iv)(i)
- C. Wait for mapping
- D. Wait for mapping
Correct answer & explanation hidden
Log in to view solution →- A. China rose
- B. Citrus
- C. Pea
- D. China rose and citrus
Correct answer & explanation hidden
Log in to view solution →- A. China rose
- B. Citrus
- C. Pea
- D. China rose and citrus
Correct answer & explanation hidden
Log in to view solution →- A. (1)(iii)(iv)(ii)(i)
- B. (2)(ii)(i)(iv)(iii)
- C. (3)(iii)(iv)(i)(ii)
- D. (4)(iv)(iii)(ii)(i)
Correct answer & explanation hidden
Log in to view solution →- A. (a) Protoplast fusion (i) Totipotency; (b) Plant tissue culture (ii) Pomato; (c) Meristem culture (iii) Somaclones; (d) Micropropagation (iv) Virus free plants
- B. (1)(iii)(iv)(ii)(i); (2)(ii)(i)(iv)(iii); (3)(iii)(iv)(i)(ii); (4)(iv)(iii)(ii)(i)
- C. Wait for mapping
- D. Wait for mapping
Correct answer & explanation hidden
Log in to view solution →- A. Species A (−) ; Species B (0)
- B. Species A (1) ; Species B (1)
- C. Species A (−) ; Species B (−)
- D. Species A (1) ; Species B (0)
Correct answer & explanation hidden
Log in to view solution →- A. Species A (−) ; Species B (0)
- B. Species A (1) ; Species B (1)
- C. Species A (−) ; Species (−)
- D. Species A (1) ; Species B (0)
Correct answer & explanation hidden
Log in to view solution →- A. Mature sieve tube elements possess a conspicuous nucleus and usual cytoplasmic organelles.
- B. Microbodies are present both in plant and animal cells.
- C. The perinuclear space forms a barrier between the materials present inside the nucleus and that of the cytoplasm.
- D. Nuclear pores act as passages for proteins and RNA molecules in both directions between nucleus and cytoplasm.
Correct answer & explanation hidden
Log in to view solution →- A. Mature sieve tube elements possess a conspicuous nucleus and usual cytoplasmic organelles.
- B. Microbodies are present both in plant and animal cells.
- C. The perinuclear space forms a barrier between the materials present inside the nucleus and that of the cytoplasm.
- D. Nuclear pores act as passages for proteins.
Correct answer & explanation hidden
Log in to view solution →- A. 8-nucleate and 7-celled
- B. 7-nucleate and 8-celled
- C. 7-nucleate and 7-celled
- D. 8-nucleate and 8-celled
Correct answer & explanation hidden
Log in to view solution →- A. Ectocarpus
- B. Gracilaria
- C. Volvox
- D. Ulothrix
Correct answer & explanation hidden
Log in to view solution →- A. IAA
- B. NAA
- C. 2, 4-D
- D. IBA
Correct answer & explanation hidden
Log in to view solution →- A. Climax
- B. Climax community
- C. Standing state
- D. Standing crop
Correct answer & explanation hidden
Log in to view solution →- A. Kinetin
- B. Infrared rays
- C. Gamma rays
- D. Zeatin
Correct answer & explanation hidden
Log in to view solution →- A. Pyramid of biomass in sea is generally inverted.
- B. Pyramid of biomass in sea is generally upright.
- C. Pyramid of energy is always upright.
- D. Pyramid of numbers in a grassland ecosystem is upright.
Correct answer & explanation hidden
Log in to view solution →- A. Radiant energy
- B. Retardation factor
- C. Environment factor
- D. Respiration losses
Correct answer & explanation hidden
Log in to view solution →- A. Green algae
- B. Brown algae
- C. Red algae
- D. Blue-green algae
Correct answer & explanation hidden
Log in to view solution →- A. Pyruvic acid
- B. Oxaloacetic acid
- C. Succinic acid
- D. Phosphoglyceric acid
Correct answer & explanation hidden
Log in to view solution →- A. (1)(iv)(i)(iii)(ii)
- B. (2)(iii)(i)(iv)(ii)
- C. (3)(ii)(iii)(iv)(i)
- D. (4)(iv)(ii)(i)(iii)
Correct answer & explanation hidden
Log in to view solution →- A. During aerobic respiration, role of oxygen is limited to the terminal stage.
- B. In ETC (Electron Transport Chain), one molecule of NADH+H+ gives rise to 2 ATP molecules, and one FADH2 gives rise to 3 ATP molecules.
- C. ATP is synthesized through complex V.
- D. Oxidation-reduction reactions produce proton gradient in respiration.
Correct answer & explanation hidden
Log in to view solution →- A. (1)(iii)(ii)(i)(iv)
- B. (2)(iv)(ii)(iii)(i)
- C. (3)(iv)(i)(ii)(iii)
- D. (4)(ii)(iv)(iii)(i)
Correct answer & explanation hidden
Log in to view solution →- A. it will not be able to confer ampicillin resistance to the host cell.
- B. the transformed cells will have the ability to resist ampicillin as well as produce β-galactoside.
- C. it will lead to lysis of host cell.
- D. it will be able to produce a novel protein with dual ability.
Correct answer & explanation hidden
Log in to view solution →- A. it will not be able to confer ampicillin resistance to the host cell.
- B. the transformed cells will have the ability to resist ampicillin as well as produce β-galactoside.
- C. it will lead to lysis of host cell.
- D. it will be able to produce a novel protein with dual ability.
Correct answer & explanation hidden
Log in to view solution →- A. In capping, methyl guanosine triphosphate is added to the 39 end of hnRNA.
- B. RNA polymerase binds with Rho factor to terminate the process of transcription in bacteria.
- C. The coding strand in a transcription unit is copied to an mRNA.
- D. Split gene arrangement is characteristic of prokaryotes.
Correct answer & explanation hidden
Log in to view solution →- A. In capping, methyl guanosine triphosphate is added to the 3' end of hnRNA.
- B. RNA polymerase binds with Rho factor to terminate the process of transcription in bacteria.
- C. The coding strand in a transcription unit is copied to an mRNA.
- D. Split gene arrangement is characteristic of prokaryotes.
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Log in to view solution →- A. mutated gene partially appears on a photographic film.
- B. mutated gene completely and clearly appears on a photographic film.
- C. mutated gene does not appear on a photographic film as the probe has no complimentarity with it.
- D. mutated gene does not appear on photographic film as the probe has complimentarity with it.
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Log in to view solution →- A. The base of number logarithms
- B. The base of exponential logarithms
- C. The base of natural logarithms
- D. The base of geometric logarithms
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Log in to view solution →- A. Large colorless empty cells in the epidermis of grass leaves - Subsidiary cells
- B. In dicot leaves, vascular bundles are surrounded by large thick-walled cells - Conjunctive tissue
- C. Cells of medullary rays that form part of cambial ring - Interfascicular cambium
- D. Loose parenchyma cells rupturing the epidermis and forming a lens-shaped opening in bark - Spongy parenchyma
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Log in to view solution →- A. Poaceae ; Rosaceae
- B. Poaceae ; Leguminosae
- C. Poaceae ; Solanaceae
- D. Rosaceae ; Leguminosae
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Log in to view solution →- A. Transcribes rRNAs (28S, 18S and 5.8S)
- B. Transcribes tRNA, 5s rRNA and snRNA
- C. Transcribes precursor of mRNA
- D. Transcribes only snRNAs
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Log in to view solution →- A. Both ATP and NADPH + H+ are synthesized during non-cyclic photophosphorylation.
- B. Stroma lamellae have PS I only and lack NADP reductase.
- C. Grana lamellae have both PS I and PS II.
- D. Cyclic photophosphorylation involves both PS I and PS II.
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Log in to view solution →- A. Fusion of two cells is called Karyogamy.
- B. Fusion of protoplasms between two motile or non-motile gametes is called plasmogamy.
- C. Organisms that depend on living plants are called saprophytes.
- D. Some of the organisms can fix atmospheric nitrogen in specialized cells called sheath cells.
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Log in to view solution →- A. (1)(iv)(i)(ii)(iii)
- B. (2)(i)(iv)(iii)(ii)
- C. (3)(ii)(i)(iv)(iii)
- D. (4)(iv)(iii)(i)(ii)
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Log in to view solution →- A. Satellite DNA
- B. Repetitive DNA
- C. Single nucleotides
- D. Polymorphic DNA
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Log in to view solution →- A. (1)(ii)(iv)(i)(iii)
- B. (2)(i)(ii)(iii)(iv)
- C. (3)(iii)(i)(iv)(ii)
- D. (4)(iv)(iii)(ii)(i)
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Log in to view solution →- A. Degenerate primer sequence
- B. Okazaki sequences
- C. Palindromic Nucleotide sequences
- D. Poly(A) tail sequences
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Log in to view solution →- A. Degenerate primer sequence
- B. Okazaki sequences
- C. Palindromic Nucleotide sequences
- D. Poly(A) tail sequences
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Log in to view solution →- A. 8
- B. 16
- C. 4
- D. 32
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Log in to view solution →- A. 8
- B. 16
- C. 4
- D. 32
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Log in to view solution →- A. Fire fly
- B. Grasshopper
- C. Cockroach
- D. House fly
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Log in to view solution →- A. Fire fly
- B. Grasshopper
- C. Cockroach
- D. House fly
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Log in to view solution →- A. Pancreatic juice
- B. Intestinal juice
- C. Gastric juice
- D. Chyme
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Log in to view solution →- A. (b) and (d) only
- B. (b) and (c) only
- C. (a), (c) and (d) only
- D. (a) and (d) only
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Log in to view solution →- A. Absence of antigens A and B on the surface of RBCs
- B. Absence of antigens A and B in plasma
- C. Presence of antibodies, anti-A and anti-B, on RBCs
- D. Absence of antibodies, anti-A and anti-B, in plasma
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Log in to view solution →- A. 50%
- B. 75%
- C. 25%
- D. 100%
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Log in to view solution →- A. Thrombin
- B. Renin
- C. Epinephrine
- D. Thrombokinase
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Log in to view solution →- A. pO2=104 and pCO2=40
- B. pO2=40 and pCO2=45
- C. pO2=95 and pCO2=40
- D. pO2=159 and pCO2=0.3
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Log in to view solution →- A. Arthritis
- B. Muscular dystrophy
- C. Myasthenia gravis
- D. Gout
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Log in to view solution →- A. DNA dependent DNA polymerase
- B. DNA dependent RNA polymerase
- C. DNA Ligase
- D. DNase
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Log in to view solution →- A. mRNA
- B. tRNA
- C. rRNA
- D. siRNA
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Log in to view solution →- A. CuT
- B. LNG 20
- C. Cu 7
- D. Multiload 375
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Log in to view solution →- A. T : 20 ; G : 30 ; C : 20
- B. T : 20 ; G : 20 ; C : 30
- C. T : 30 ; G : 20 ; C : 20
- D. T : 20 ; G : 25 ; C : 25
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Log in to view solution →- A. (1) (iii)(i)(iv)(ii)
- B. (2) (i)(ii)(iii)(iv)
- C. (3) (ii)(iii)(i)(iv)
- D. (4) (iv)(ii)(i)(iii)
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Log in to view solution →- A. (c), (d) and (e) are correct
- B. (a), (b) and (c) are correct
- C. (a), (d) and (e) are correct
- D. (b), (c) and (e) are correct
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Log in to view solution →- A. Corona radiata
- B. Vitelline membrane
- C. Perivitelline space
- D. Zona pellucida
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Log in to view solution →- A. Corona radiata
- B. Vitelline membrane
- C. Perivitelline space
- D. Zona pellucida
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Log in to view solution →- A. (1) (iv)(iii)(i)(ii)
- B. (2) (iii)(iv)(i)(ii)
- C. (3) (iii)(iv)(ii)(i)
- D. (4) (iv)(i)(ii)(iii)
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Log in to view solution →- A. (1)(iv)(iii)(i)(ii)
- B. (2)(iii)(iv)(i)(ii)
- C. (3)(iii)(iv)(ii)(i)
- D. (4)(iv)(i)(ii)(iii)
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Log in to view solution →- A. Alpha cells of pancreas
- B. The cells of rostral adenohypophysis
- C. The cells of bone marrow
- D. Juxtaglomerular cells of the kidney
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Log in to view solution →- A. Alpha cells of pancreas
- B. The cells of rostral adenohypophysis
- C. The cells of bone marrow
- D. Juxtaglomerular cells of the kidney
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Log in to view solution →- A. (1)(a), (b) and (c) only
- B. (2)(b), (c) and (d) only
- C. (3)(b) and (c) only
- D. (4)(a) and (c) only
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Log in to view solution →- A. A ring of gastric caeca is present at the junction of midgut and hind gut.
- B. Hypopharynx lies within the cavity enclosed by the mouth parts.
- C. In females, 7th-9th sterna together form a genital pouch.
- D. 10th abdominal segment in both sexes, bears a pair of anal cerci.
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Log in to view solution →- A. (1)(ii)(iii)(i)(iv)
- B. (2)(iv)(i)(iii)(ii)
- C. (3)(ii)(iii)(iv)(i)
- D. (4)(i)(iv)(iii)(ii)
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Log in to view solution →- A. Neophron
- B. Hemidactylus
- C. Macropus
- D. Ornithorhynchus
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Log in to view solution →- A. S-phase
- B. Prophase
- C. Metaphase
- D. G2 phase
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Log in to view solution →- A. Annealing
- B. Extension
- C. Denaturation
- D. Ligation
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Log in to view solution →- A. Improve protein content
- B. Improve resistance to diseases
- C. Improve vitamin content
- D. Improve micronutrient and mineral content
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Log in to view solution →- A. CFCs
- B. Stratosphere
- C. Ozone
- D. Troposphere
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Log in to view solution →- A. Ileo-caecal junction
- B. Junction of hepato-pancreatic duct and duodenum
- C. Gastro-oesophageal junction
- D. Junction of jejunum and duodenum
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Log in to view solution →- A. High pO2, low pCO2, less H+, lower temperature
- B. Low pO2, high pCO2, more H+, higher temperature
- C. High pO2, high pCO2, less H+, higher temperature
- D. Low pO2, low pCO2, more H+, higher temperature
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Log in to view solution →- A. Alkaloids- Codeine
- B. Toxin - Abrin
- C. Lectins - Concanavalin A
- D. Drugs - Ricin
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Log in to view solution →- A. These muscle have no striations
- B. They are involuntary muscles
- C. Communication among the cells is performed by intercalated discs
- D. These muscles are present in the wall of blood vessels
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Log in to view solution →- A. Western Blotting Technique
- B. Southern Blotting Technique
- C. ELISA Technique
- D. Hybridization Technique
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Log in to view solution →- A. Western Blotting Technique
- B. Southern Blotting Technique
- C. ELISA Technique
- D. Hybridization Technique
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