1. Who demonstrated through experiments with Cladophora and aerobic bacteria that the first action spectrum of photosynthesis roughly matches the absorption spectra of chlorophyll a and b?
Correct Answer: D (T.W. Engelmann)
Using a prism to split white light and illuminating the green alga Cladophora placed in a suspension of aerobic bacteria, T.W. Engelmann discovered that bacteria accumulated mainly in the blue and red light zones.
Using a prism to split white light and illuminating the green alga Cladophora placed in a suspension of aerobic bacteria, T.W. Engelmann discovered that bacteria accumulated mainly in the blue and red light zones.
2. Cornelius van Niel’s microbiological studies on purple and green sulfur bacteria proved that the oxygen evolved during green plant photosynthesis comes from:
Correct Answer: A (Water (H2O))
Van Niel demonstrated that photosynthesis is a light-dependent reaction where hydrogen from an oxidizable compound reduces CO2. In sulfur bacteria H2S is the donor (yielding sulfur), proving that in green plants H2O is the hydrogen donor and O2 source.
Van Niel demonstrated that photosynthesis is a light-dependent reaction where hydrogen from an oxidizable compound reduces CO2. In sulfur bacteria H2S is the donor (yielding sulfur), proving that in green plants H2O is the hydrogen donor and O2 source.
3. Which pigment forms the catalytic “reaction center” in both Photosystem I (PS I) and Photosystem II (PS II)?
Correct Answer: C (A single molecule of Chlorophyll a)
The Light Harvesting Complex (LHC) consists of hundreds of accessory pigments, but the core reaction center is always a specialized single chlorophyll a molecule (P700 in PS I and P680 in PS II).
The Light Harvesting Complex (LHC) consists of hundreds of accessory pigments, but the core reaction center is always a specialized single chlorophyll a molecule (P700 in PS I and P680 in PS II).
4. The Oxygen Evolving Complex (OEC) responsible for the photolysis of water is physically located on which side of the thylakoid membrane?
Correct Answer: A (Inner surface facing the thylakoid lumen, associated with PS II)
Water-splitting complex is located on the luminal side of the thylakoid membrane attached to PS II, requiring Mn2+, Cl-, and Ca2+ ions to liberate O2 and release protons into the lumen.
Water-splitting complex is located on the luminal side of the thylakoid membrane attached to PS II, requiring Mn2+, Cl-, and Ca2+ ions to liberate O2 and release protons into the lumen.
5. Cyclic photophosphorylation involves only:
Correct Answer: B (PS I and produces ATP only without generating NADPH or O2)
Cyclic electron flow occurs predominantly in stroma lamellae (which lack PS II and NADP reductase). Excited electrons from P700 loop back via the ETS, synthesizing ATP via a proton gradient without water splitting or NADPH synthesis.
Cyclic electron flow occurs predominantly in stroma lamellae (which lack PS II and NADP reductase). Excited electrons from P700 loop back via the ETS, synthesizing ATP via a proton gradient without water splitting or NADPH synthesis.
6. According to Peter Mitchell’s Chemiosmotic Hypothesis, ATP synthesis across the thylakoid membrane is driven by:
Correct Answer: B (High proton (H+) accumulation in the thylakoid lumen relative to the stroma)
Water photolysis inside the lumen, cytochrome b6f proton pumping, and stromal NADP+ reduction drop the lumen pH (high H+), creating a proton motive force across the thylakoid membrane.
Water photolysis inside the lumen, cytochrome b6f proton pumping, and stromal NADP+ reduction drop the lumen pH (high H+), creating a proton motive force across the thylakoid membrane.
7. In the chloroplast ATP synthase complex, the CF0 and CF1 components are situated as:
Correct Answer: A (CF0 is a transmembrane proton channel; CF1 protrudes into the stroma)
CF0 forms a channel across the thylakoid membrane through which protons diffuse down their gradient to the stroma, driving conformational changes in the catalytic CF1 headpiece to produce ATP.
CF0 forms a channel across the thylakoid membrane through which protons diffuse down their gradient to the stroma, driving conformational changes in the catalytic CF1 headpiece to produce ATP.
8. What is the primary CO2 acceptor and the first stable product in the C3 pathway (Calvin Cycle)?
Correct Answer: B (Ribulose-1,5-bisphosphate (RuBP) and 3-Phosphoglyceric acid (3-PGA))
In C3 plants, 5-carbon RuBP accepts CO2 in a reaction catalyzed by RuBisCO to form an unstable 6-carbon intermediate that splits into two molecules of 3-carbon 3-PGA.
In C3 plants, 5-carbon RuBP accepts CO2 in a reaction catalyzed by RuBisCO to form an unstable 6-carbon intermediate that splits into two molecules of 3-carbon 3-PGA.
9. To synthesize one net molecule of Glucose via the Calvin cycle, how many molecules of CO2, ATP, and NADPH are consumed?
Correct Answer: B (6 CO2, 18 ATP, 12 NADPH)
Fixation of each CO2 requires 3 ATP (2 for reduction, 1 for RuBP regeneration) and 2 NADPH. For 1 glucose (6 turns): 6 CO2 + 18 ATP + 12 NADPH → 1 Glucose + 18 ADP + 12 NADP+.
Fixation of each CO2 requires 3 ATP (2 for reduction, 1 for RuBP regeneration) and 2 NADPH. For 1 glucose (6 turns): 6 CO2 + 18 ATP + 12 NADPH → 1 Glucose + 18 ADP + 12 NADP+.
10. “Kranz anatomy” with large bundle sheath cells containing agranal chloroplasts and thick suberized walls is characteristic of:
Correct Answer: B (C4 plants (e.g., Maize and Sugarcane))
Kranz (“wreath”) anatomy in C4 plants features concentric rings of bundle sheath cells around vascular bundles that lack intercellular spaces and protect RuBisCO from oxygen exposure.
Kranz (“wreath”) anatomy in C4 plants features concentric rings of bundle sheath cells around vascular bundles that lack intercellular spaces and protect RuBisCO from oxygen exposure.
11. In C4 plants, initial carbon fixation occurs in mesophyll cells catalyzed by which enzyme?
Correct Answer: B (PEP Carboxylase (PEPcase))
In C4 mesophyll cells, PEPcase fixes HCO3- onto 3-carbon Phosphoenolpyruvate (PEP) to form 4-carbon Oxaloacetic acid (OAA). Mesophyll cells completely lack RuBisCO.
In C4 mesophyll cells, PEPcase fixes HCO3- onto 3-carbon Phosphoenolpyruvate (PEP) to form 4-carbon Oxaloacetic acid (OAA). Mesophyll cells completely lack RuBisCO.
12. Photorespiration (C2 cycle) occurs when RuBisCO binds oxygen instead of CO2, leading to the formation of:
Correct Answer: B (One molecule of 3-PGA and one molecule of 2-Phosphoglycolate)
When high O2/CO2 ratios favor RuBisCO’s oxygenase activity, RuBP is oxygenated into one 3-carbon 3-PGA and one 2-carbon phosphoglycolate, which is routed through chloroplasts, peroxisomes, and mitochondria with ATP loss.
When high O2/CO2 ratios favor RuBisCO’s oxygenase activity, RuBP is oxygenated into one 3-carbon 3-PGA and one 2-carbon phosphoglycolate, which is routed through chloroplasts, peroxisomes, and mitochondria with ATP loss.
13. Why is photorespiration absent in C4 plants?
Correct Answer: B (A C4 malate decarboxylation pump maintains high CO2 concentration around RuBisCO in bundle sheath cells)
Decarboxylation of 4C acids in bundle sheath cells elevates intracellular CO2, suppressing RuBisCO’s oxygenase function and preventing photorespiratory loss.
Decarboxylation of 4C acids in bundle sheath cells elevates intracellular CO2, suppressing RuBisCO’s oxygenase function and preventing photorespiratory loss.
14. Blackman’s Law of Limiting Factors (1905) states that if a chemical process is affected by more than one factor, its rate is determined by:
Correct Answer: B (The factor that is nearest to its minimal value)
Blackman’s law states that when a physiological process is conditioned by multiple separate factors, the rate is pace-limited by the single factor nearest its minimum threshold.
Blackman’s law states that when a physiological process is conditioned by multiple separate factors, the rate is pace-limited by the single factor nearest its minimum threshold.
15. What are the atmospheric CO2 saturation levels for C3 and C4 plants respectively?
Correct Answer: B (450 µL/L for C3 and 360 µL/L for C4)
C4 plants saturate at lower CO2 concentrations (~360 ppm) because of PEPcase affinity. C3 plants saturate around 450 ppm, showing increased productivity when enriched with CO2 in greenhouses (greenhouse crops: tomatoes, bell peppers).
C4 plants saturate at lower CO2 concentrations (~360 ppm) because of PEPcase affinity. C3 plants saturate around 450 ppm, showing increased productivity when enriched with CO2 in greenhouses (greenhouse crops: tomatoes, bell peppers).
16. The universal pathway for cellular glucose breakdown (Glycolysis / EMP pathway) occurs in the:
Correct Answer: B (Cytoplasm / Cytosol)
Glycolysis (discovered by Embden, Meyerhof, and Parnas) is an anaerobic sequence of 10 enzyme-catalyzed reactions localized in the cytoplasm of all living cells.
Glycolysis (discovered by Embden, Meyerhof, and Parnas) is an anaerobic sequence of 10 enzyme-catalyzed reactions localized in the cytoplasm of all living cells.
17. What is the net yield of ATP and NADH produced per molecule of Glucose during Glycolysis alone?
Correct Answer: B (2 ATP and 2 NADH)
Glycolysis consumes 2 ATP in the preparatory phase (Hexokinase, PFK-1) and produces 4 ATP + 2 NADH in the payoff phase, yielding a net gain of 2 ATP and 2 NADH per glucose.
Glycolysis consumes 2 ATP in the preparatory phase (Hexokinase, PFK-1) and produces 4 ATP + 2 NADH in the payoff phase, yielding a net gain of 2 ATP and 2 NADH per glucose.
18. In anaerobic alcoholic fermentation by yeast, pyruvate is decarboxylated and reduced into ethanol and CO2 by which enzymes?
Correct Answer: B (Pyruvate decarboxylase and Alcohol dehydrogenase)
Pyruvate decarboxylase converts pyruvate to acetaldehyde releasing CO2, and alcohol dehydrogenase reduces acetaldehyde to ethanol while re-oxidizing NADH to NAD+.
Pyruvate decarboxylase converts pyruvate to acetaldehyde releasing CO2, and alcohol dehydrogenase reduces acetaldehyde to ethanol while re-oxidizing NADH to NAD+.
19. Yeasts poison themselves to death when the natural alcohol concentration reaches approximately:
Correct Answer: B (13%)
Accumulation of ethanol above roughly 13% disrupts membrane integrity and enzyme functions in yeast cells, setting a natural upper limit for unfortified fermented beverages.
Accumulation of ethanol above roughly 13% disrupts membrane integrity and enzyme functions in yeast cells, setting a natural upper limit for unfortified fermented beverages.
20. The “Link Reaction” (oxidative decarboxylation of pyruvate to acetyl-CoA) takes place in the:
Correct Answer: B (Mitochondrial matrix)
Pyruvate enters the mitochondrial matrix where the multi-enzyme pyruvate dehydrogenase complex catalyzes its oxidative decarboxylation into Acetyl-CoA, generating NADH and CO2.
Pyruvate enters the mitochondrial matrix where the multi-enzyme pyruvate dehydrogenase complex catalyzes its oxidative decarboxylation into Acetyl-CoA, generating NADH and CO2.
21. In the first step of the Tricarboxylic Acid (TCA / Krebs) cycle, acetyl-CoA condenses with oxaloacetate (OAA) to form:
Correct Answer: A (Citric acid (6C))
Citrate synthase condenses the 2-carbon acetyl group of Acetyl-CoA with 4-carbon Oxaloacetate and water to produce 6-carbon Citric acid.
Citrate synthase condenses the 2-carbon acetyl group of Acetyl-CoA with 4-carbon Oxaloacetate and water to produce 6-carbon Citric acid.
22. During one complete turn of the Krebs cycle (per single Acetyl-CoA entering), the yield of reduced coenzymes and high-energy phosphates is:
Correct Answer: B (3 NADH, 1 FADH2, 1 GTP/ATP)
Each turn yields 3 NADH (isocitrate → $\alpha$-KG, $\alpha$-KG → Succinyl-CoA, malate → OAA), 1 FADH2 (succinate → fumarate), and 1 GTP/ATP (succinyl-CoA → succinate). For one full glucose (2 acetyl-CoA), values double.
Each turn yields 3 NADH (isocitrate → $\alpha$-KG, $\alpha$-KG → Succinyl-CoA, malate → OAA), 1 FADH2 (succinate → fumarate), and 1 GTP/ATP (succinyl-CoA → succinate). For one full glucose (2 acetyl-CoA), values double.
23. Which enzyme of the TCA cycle is embedded in the inner mitochondrial membrane rather than existing dissolved in the matrix?
Correct Answer: C (Succinate dehydrogenase (Complex II))
Succinate dehydrogenase is an integral inner mitochondrial membrane protein that simultaneously oxidizes succinate to fumarate and transfers electrons to FAD/Ubiquinone as Complex II of the ETS.
Succinate dehydrogenase is an integral inner mitochondrial membrane protein that simultaneously oxidizes succinate to fumarate and transfers electrons to FAD/Ubiquinone as Complex II of the ETS.
24. In the mitochondrial Electron Transport System (ETS), Complex IV is also known as:
Correct Answer: C (Cytochrome c oxidase complex)
Complex IV contains cytochromes a and a3 along with two copper centers (CuA and CuB). It transfers electrons from reduced cytochrome c to final acceptor oxygen, forming water.
Complex IV contains cytochromes a and a3 along with two copper centers (CuA and CuB). It transfers electrons from reduced cytochrome c to final acceptor oxygen, forming water.
25. What acts as the terminal electron acceptor during aerobic cellular respiration?
Correct Answer: C (Molecular oxygen (O2))
Oxygen acts as the ultimate hydrogen/electron sink at Complex IV, combining with electrons and matrix protons to form metabolic water ($H_2O$).
Oxygen acts as the ultimate hydrogen/electron sink at Complex IV, combining with electrons and matrix protons to form metabolic water ($H_2O$).
26. Oxidation of one molecule of NADH and one molecule of FADH2 through the mitochondrial ETS produces how many ATP molecules respectively?
Correct Answer: A (3 ATP and 2 ATP)
According to standard NCERT values, electrons from 1 NADH pump enough protons to generate 3 ATP, while electrons entering downstream via FADH2 generate 2 ATP.
According to standard NCERT values, electrons from 1 NADH pump enough protons to generate 3 ATP, while electrons entering downstream via FADH2 generate 2 ATP.
27. Cellular respiration is correctly termed an “Amphibolic Pathway” because:
Correct Answer: B (It involves both catabolic breakdown of substrates and anabolic withdrawal of carbon skeletons for biosynthesis)
Respiratory intermediates are used both for fuel oxidation (catabolism) and as precursor building blocks for synthesizing amino acids, fatty acids, and chlorophylls (anabolism).
Respiratory intermediates are used both for fuel oxidation (catabolism) and as precursor building blocks for synthesizing amino acids, fatty acids, and chlorophylls (anabolism).
28. What is the Respiratory Quotient (RQ = CO2 evolved / O2 consumed) when pure carbohydrates (glucose) are respired aerobically?
Correct Answer: A (1.0)
For carbohydrates: C6H12O6 + 6 O2 → 6 CO2 + 6 H2O. RQ = 6 CO2 / 6 O2 = 1.0.
For carbohydrates: C6H12O6 + 6 O2 → 6 CO2 + 6 H2O. RQ = 6 CO2 / 6 O2 = 1.0.
29. The Respiratory Quotient (RQ) for the oxidation of Tripalmitin (fat) is:
Correct Answer: B (0.7)
2 (C51H98O6) + 145 O2 → 102 CO2 + 98 H2O. RQ = 102 CO2 / 145 O2 = 0.7. Fats contain less oxygen and require more external O2 for combustion.
2 (C51H98O6) + 145 O2 → 102 CO2 + 98 H2O. RQ = 102 CO2 / 145 O2 = 0.7. Fats contain less oxygen and require more external O2 for combustion.
30. During anaerobic respiration where no oxygen is consumed, the theoretical Respiratory Quotient (RQ) is:
Correct Answer: D (Infinity (∞))
Because no O2 is absorbed (O2 = 0) while CO2 is still evolved, RQ = CO2 / 0 = ∞ (infinity).
Because no O2 is absorbed (O2 = 0) while CO2 is still evolved, RQ = CO2 / 0 = ∞ (infinity).
31. Arithmetic plant growth (where one daughter cell continues dividing while the other differentiates) is expressed mathematically as:
Correct Answer: A (Lt = L0 + rt)
Arithmetic growth results in a straight linear graph where length at time $t$ ($L_t$) equals initial length ($L_0$) plus growth rate ($r$) multiplied by time ($t$).
Arithmetic growth results in a straight linear graph where length at time $t$ ($L_t$) equals initial length ($L_0$) plus growth rate ($r$) multiplied by time ($t$).
32. In geometric growth (sigmoidal curve), the initial slow growth phase followed by rapid exponential growth and finally a stationary phase are termed:
Correct Answer: A (Lag phase, Log (exponential) phase, and Stationary phase)
A typical S-shaped sigmoid curve exhibits an initial lag phase, followed by rapid log/exponential expansion, and plateaus into a stationary phase as nutrients become limiting.
A typical S-shaped sigmoid curve exhibits an initial lag phase, followed by rapid log/exponential expansion, and plateaus into a stationary phase as nutrients become limiting.
33. The phenomenon where differentiated living parenchymal cells regain the capacity to divide under specific conditions is termed:
Correct Answer: B (Dedifferentiation)
Dedifferentiation occurs when mature differentiated cells revert to a meristematic state (e.g., formation of interfascicular cambium and cork cambium from fully differentiated medullary ray and cortical cells).
Dedifferentiation occurs when mature differentiated cells revert to a meristematic state (e.g., formation of interfascicular cambium and cork cambium from fully differentiated medullary ray and cortical cells).
34. Secondary xylem, secondary phloem, and cork produced by secondary meristems undergo maturation to lose dividing capacity again. This process is:
Correct Answer: C (Redifferentiation)
When cells derived from dedifferentiated meristems lose their capacity to divide again and mature to perform specific functions, it is called redifferentiation.
When cells derived from dedifferentiated meristems lose their capacity to divide again and mature to perform specific functions, it is called redifferentiation.
35. Plants follow different developmental pathways in response to environmental conditions or life phases, producing different leaf structures (heterophylly). This capability is called:
Correct Answer: A (Plasticity)
Plasticity is the capacity to alter morphology in response to environment (e.g., aerial vs aquatic leaves in buttercup / Ranunculus) or developmental phase (cotton, coriander, larkspur).
Plasticity is the capacity to alter morphology in response to environment (e.g., aerial vs aquatic leaves in buttercup / Ranunculus) or developmental phase (cotton, coriander, larkspur).
36. Charles Darwin and his son Francis Darwin discovered phototropism by observing coleoptile bending towards unilateral light in:
Correct Answer: B (Phalaris canariensis (Canary grass))
The Darwins observed that the canary grass coleoptile tip perceives unilateral light and transmits a mobile stimulus downwards, causing bending towards light.
The Darwins observed that the canary grass coleoptile tip perceives unilateral light and transmits a mobile stimulus downwards, causing bending towards light.
37. Who first isolated Auxin (Indole-3-acetic acid / IAA) from the coleoptile tips of oat seedlings (Avena curvature test)?
Correct Answer: B (F.W. Went)
F.W. Went (1928) isolated auxin from oat (Avena sativa) coleoptile tips by diffusing the growth chemical into agar blocks.
F.W. Went (1928) isolated auxin from oat (Avena sativa) coleoptile tips by diffusing the growth chemical into agar blocks.
38. Decapitation (removal of the shoot apex) promotes branching into bushy shrubs in tea plantations and hedge-making because it eliminates:
Correct Answer: A (Apical dominance mediated by Auxin)
Apical buds produce auxin which suppresses the outgrowth of lateral axillary buds. Decapitation releases lateral buds from apical dominance, promoting bushy vegetative growth.
Apical buds produce auxin which suppresses the outgrowth of lateral axillary buds. Decapitation releases lateral buds from apical dominance, promoting bushy vegetative growth.
39. Which synthetic auxin is widely used as a selective herbicide to kill broad-leaved dicot weeds in cereal (monocot) lawns?
Correct Answer: C (2,4-Dichlorophenoxyacetic acid (2,4-D))
2,4-D is a synthetic auxin used widely to eradicate broad-leaved weeds in crop fields without harming mature monocotyledonous grasses.
2,4-D is a synthetic auxin used widely to eradicate broad-leaved weeds in crop fields without harming mature monocotyledonous grasses.
40. The “foolish seedling” (bakanae) disease of rice seedlings led to the discovery of which plant hormone?
Correct Answer: B (Gibberellin)
E. Kurosawa (1926) found that sterile filtrates of the fungus Gibberella fujikuroi caused excessive internodal elongation in uninfected rice seedlings, leading to the isolation of gibberellic acid (GA).
E. Kurosawa (1926) found that sterile filtrates of the fungus Gibberella fujikuroi caused excessive internodal elongation in uninfected rice seedlings, leading to the isolation of gibberellic acid (GA).
41. Spraying sugarcane crops with which plant growth regulator increases internode length and boosts yield by up to 20 tonnes per acre?
Correct Answer: A (Gibberellins (GA))
Sugarcane stores carbohydrates in its stems; applying gibberellins elongates stem internodes, significantly increasing cane mass and sugar yield.
Sugarcane stores carbohydrates in its stems; applying gibberellins elongates stem internodes, significantly increasing cane mass and sugar yield.
42. “Bolting”—the rapid internode elongation immediately prior to flowering in rosette plants like beet and cabbage—is induced by:
Correct Answer: B (Gibberellins)
Treatment with gibberellins triggers rapid axis elongation (bolting) in rosette plants that normally require cold exposure or long days to flower.
Treatment with gibberellins triggers rapid axis elongation (bolting) in rosette plants that normally require cold exposure or long days to flower.
43. Kinetin (a modified adenine purine) was discovered by Skoog and Miller from:
Correct Answer: B (Autoclaved herring sperm DNA)
Skoog and Miller crystallized the cytokinesis-promoting active substance kinetin from degraded autoclaved herring sperm DNA. Zeatin was later isolated naturally from corn kernels and coconut milk.
Skoog and Miller crystallized the cytokinesis-promoting active substance kinetin from degraded autoclaved herring sperm DNA. Zeatin was later isolated naturally from corn kernels and coconut milk.
44. Which hormone promotes cell division, helps overcome apical dominance, and delays leaf senescence (Richmond-Lang effect) by mobilizing nutrients?
Correct Answer: C (Cytokinin)
Cytokinins promote cytokinesis, lateral shoot growth, chloroplast development in leaves, and delay senescence by nutrient mobilization.
Cytokinins promote cytokinesis, lateral shoot growth, chloroplast development in leaves, and delay senescence by nutrient mobilization.
45. Which of the following is the only naturally occurring gaseous plant growth regulator?
Correct Answer: C (Ethylene)
Ethylene (C2H4) is a volatile gas synthesized in tissues undergoing senescence and ripening fruits.
Ethylene (C2H4) is a volatile gas synthesized in tissues undergoing senescence and ripening fruits.
46. The sharp respiratory climacteric (increase in respiration rate during fruit ripening) is triggered by:
Correct Answer: C (Ethylene)
Ethylene enhances the respiration rate during fruit ripening, an event known as respiratory climacteric, which accelerates ripening in apples, bananas, and tomatoes.
Ethylene enhances the respiration rate during fruit ripening, an event known as respiratory climacteric, which accelerates ripening in apples, bananas, and tomatoes.
47. Ethephon is widely used in agriculture because it is an aqueous formulation that readily breaks down to release:
Correct Answer: A (Ethylene gas)
Ethephon in aqueous solution is absorbed by plants and slowly releases ethylene, speeding up fruit coloration in tomatoes/apples and promoting female flowers in cucumbers.
Ethephon in aqueous solution is absorbed by plants and slowly releases ethylene, speeding up fruit coloration in tomatoes/apples and promoting female flowers in cucumbers.
48. Which plant hormone is known as the “Stress Hormone” because it stimulates rapid stomatal closure during water deficit?
Correct Answer: C (Abscisic acid (ABA))
Under drought or desiccation stress, ABA synthesizes rapidly in roots and mesophyll cells, triggering K+ efflux from guard cells to close stomata and conserve moisture.
Under drought or desiccation stress, ABA synthesizes rapidly in roots and mesophyll cells, triggering K+ efflux from guard cells to close stomata and conserve moisture.
49. Abscisic Acid (ABA) acts as a direct physiological antagonist to which plant growth regulator regarding seed dormancy and germination?
Correct Answer: B (Gibberellin (GA))
ABA enforces seed dormancy and inhibits germination, whereas gibberellins break seed dormancy by activating hydrolytic enzymes like $\alpha$-amylase during imbibition.
ABA enforces seed dormancy and inhibits germination, whereas gibberellins break seed dormancy by activating hydrolytic enzymes like $\alpha$-amylase during imbibition.
50. Horizontal growth of seedlings, swelling of the axis, and apical hook formation in dicot seedlings (triple response) are induced by:
Correct Answer: B (Ethylene)
The classic “triple response” of etiolated dicot seedlings consists of inhibition of stem elongation, radial swelling of the hypocotyl, and exaggerated apical hook formation mediated by ethylene.
The classic “triple response” of etiolated dicot seedlings consists of inhibition of stem elongation, radial swelling of the hypocotyl, and exaggerated apical hook formation mediated by ethylene.