Chapter 7
Plant Physiology and Developmental Botany
MBBS and BDS entrance past exam questions
Past questions and answers
45 questions set from this chapter. Most repeated first.
- 2016 · NEET-UG (India) · 4 marks
Emerson's enhancement effect and Red drop have been instrumental in the discovery of:
(a) Photophosphorylation and non-cyclic electron transport
(b) Two photosystems operating simultaneously
(c) Photophosphorylation and cyclic electron transport
(d) Oxidative phosphorylation
Answer
(b) Two photosystems operating simultaneously
The red drop (quantum yield falls beyond about 680 nm) is removed by adding shorter-wavelength light, which is Emerson's enhancement effect. This showed that two light reactions, PS I and PS II, work together in photosynthesis.
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- 2016 · NEET-UG (India) · 4 marks
In a chloroplast the highest number of protons are found in:
(a) Stroma
(b) Lumen of thylakoids
(c) Inter membrane space
(d) Antennae complex
Answer
(b) Lumen of thylakoids
Water splitting and plastoquinone proton pumping release protons into the thylakoid lumen, creating the proton gradient. ATP synthase uses this gradient to make ATP on the stroma side, so the lumen has the highest proton concentration.
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- 2016 · NEET-UG (India) · 4 marks
The Avena curvature is used for bioassay of:
(a) ABA
(b)
(c) IAA
(d) Ethylene
Answer
(c) IAA
The Avena (oat) coleoptile curvature test, by Went, measures auxin: agar blocks containing IAA placed asymmetrically cause bending in decapitated coleoptiles. The bending angle is proportional to IAA concentration. GA3 is assayed by dwarf pea or maize tests, ABA and ethylene by other tests.
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- 2016 · NEET-UG (India) · 4 marks
A plant in your garden avoids photorespiratory losses, has improved water use efficiency, shows high rates of photosynthesis at high temperatures and has improved efficiency of nitrogen utilisation. In which of the following physiological groups would you assign this plant?
(a)
(b)
(c) CAM
(d) Nitrogen fixer
Answer
(b)
The listed features are the hallmarks of C plants: Kranz anatomy and PEP carboxylase concentrate CO around RuBisCO, so photorespiration is avoided. They also have better water use efficiency and keep high photosynthetic rates at high temperature. C plants lose carbon through photorespiration, and CAM plants are adapted to arid conditions with slow growth.
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- 2016 · NEET-UG (India) · 4 marks
Water vapour comes out from the plant leaf through the stomatal opening. Through the same stomatal opening carbon dioxide diffuses into the plant during photosynthesis. Reason out the above statements using one of following options.
(a) Both processes can not happen simultaneously.
(b) Both processes can happen together because the diffusion coefficient of water and is different.
(c) The above processes happen only during nighttime.
(d) One process occurs during day time, and the other at night.
Answer
(b) Both processes can happen together because the diffusion coefficient of water and is different.
Water vapour moves out and CO moves in along their own concentration gradients through the same open stomata, and the two gases diffuse independently. They differ in diffusion coefficient and gradient, so both go on simultaneously during the day. Stomata are open mainly in the day, so it is wrong to say they happen only at night or at separate times.
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- 2015 · NEET-UG (India) · 4 marks
Root pressure develops due to:
(a) Increase in transpiration
(b) Active absorption
(c) Low osmotic potential in soil
(d) Passive absorption
Answer
(b) Active absorption
Root pressure develops from active uptake of ions by root cells into the xylem, which lowers its water potential so water enters by osmosis and pushes up. Increased transpiration causes suction (negative pressure), not root pressure, and passive absorption is driven by transpiration pull.
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- 2015 · NEET-UG (India) · 4 marks
The oxygen evolved during photosynthesis comes from water molecules. Which one of the following pairs of elements is involved in this reaction?
(a) Magnesium and Chlorine
(b) Manganese and Chlorine
(c) Manganese and Potassium
(d) Magnesium and Molybdenum
Answer
(b) Manganese and Chlorine
Photolysis of water at photosystem II needs Mn (in the oxygen-evolving complex) and Cl as cofactors to release O, protons and electrons. Mg is in chlorophyll, and K and Mo are not directly involved in water splitting.
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- 2015 · NEET-UG (India) · 4 marks
During biological nitrogen fixation, inactivation of nitrogenase by oxygen poisoning is prevented by:
(a) Cytochrome
(b) Leghaemoglobin
(c) Xanthophyll
(d) Carotene
Answer
(b) Leghaemoglobin
Nitrogenase is highly oxygen-sensitive. Leghaemoglobin in root nodules is an oxygen scavenger that keeps the free oxygen level low. Cytochromes, xanthophyll and carotene are not oxygen scavengers in nodules.
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- 2015 · NEET-UG (India) · 4 marks
Auxin can be bioassayed by:
(a) Lettuce hypocotyl elongation
(b) Avena coleoptile curvature
(c) Hydroponics
(d) Potometer
Answer
(b) Avena coleoptile curvature
The Avena coleoptile curvature test is the classic bioassay for auxin (IAA). Lettuce hypocotyl elongation is used for gibberellins, while hydroponics and the potometer are not bioassays.
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- 2015 · NEET-UG (India) · 4 marks
Roots play insignificant role in absorption of water in:
(a) Wheat
(b) Sunflower
(c) Pistia
(d) Pea
Answer
(c) Pistia
Pistia is a floating aquatic plant that absorbs water through its whole surface, so roots play only an insignificant role in water absorption. Wheat, sunflower and pea are terrestrial plants that depend on roots for water.
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- 2015 · NEET-UG (India) · 4 marks
A column of water within xylem vessels of tall trees does not break under its weight because of:
(a) Positive root pressure
(b) Dissolved sugars in water
(c) Tensile strength of water
(d) Lignification of xylem vessels
Answer
(c) Tensile strength of water
Water columns in xylem do not break because of cohesion and adhesion, which give water high tensile strength. Root pressure is too weak for tall trees, dissolved sugars do not hold the column together, and lignification gives wall strength but not column strength.
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- 2015 · NEET-UG (India) · 4 marks
In photosynthesis, the light-independent reaction takes place at:
(a) Stromal matrix
(b) Thylakoid lumen
(c) Photosystem I
(d) Photosystem II
Answer
(a) Stromal matrix
The light-independent reactions (Calvin cycle) occur in the stroma of the chloroplast, using ATP and NADPH made in the light reactions. The thylakoid lumen and photosystems I and II are on the thylakoid membrane and are sites of the light reactions.
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- 2014 · NEET-UG (India) · 4 marks
A few normal seedlings of tomato were kept in a dark room. After a few days they were found to have become white-coloured like albinos. Which of the following terms will you use to describe them?
(a) Embolised
(b) Etiolated
(c) Defoliated
(d) Mutated
Answer
(b) Etiolated
Seedlings grown in the dark become pale and spindly because there is no chlorophyll or light-regulated growth. This is etiolation. Embolism is an air bubble in xylem, defoliation is leaf loss, and mutation is a genetic change.
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- 2014 · NEET-UG (India) · 4 marks
Which one of the following growth regulators is known as 'stress hormone'?
(a) Ethylene
(b)
(c) Indole acetic acid
(d) Abscissic acid
Answer
(d) Abscisic acid
ABA is called the stress hormone because its level rises under drought, salinity and cold, closing stomata and helping the plant tolerate stress. Ethylene is the fruit-ripening hormone, GA promotes stem elongation and IAA is the main auxin for growth.
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- 2014 · NEET-UG (India) · 4 marks
Dr. F. Went noted that if coleoptile tips were removed and placed on agar for one hour, the agar would produce a bending when placed on one side of freshly-cut coleoptile stumps. Of what significance is this experiment?
(a) It is the basis for quantitative determination of small amounts of growth-promoting substances.
(b) It supports the hypothesis that IAA is auxin.
(c) It demonstrated polar movement of auxins.
(d) It made possible the isolation and exact identification of auxin.
Answer
(a) It is the basis for quantitative determination of small amounts of growth-promoting substances.
Went's agar block experiment (the Avena curvature test) showed that the degree of bending of the stump is proportional to the amount of auxin in the agar, giving a bioassay for tiny quantities of auxin. It did not identify IAA chemically or prove polar transport, so (b), (c) and (d) are not what the experiment is significant for.
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- 2013 · NEET-UG (India) · 4 marks
During seed germination its stored food is mobilized by:
(a) Ethylene
(b) Cytokinin
(c) ABA
(d) Gibberellin
Answer
(d) Gibberellin
During germination gibberellins from the embryo induce alpha-amylase and other hydrolases in the aleurone layer, which mobilise stored food in the endosperm. ABA and ethylene do not do this, and ABA inhibits germination.
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- 2013 · NEET-UG (India) · 4 marks
Which of the following criteria does not pertain to facilitated transport?
(a) Requirement of special membrane proteins
(b) High selectivity
(c) Transport saturation
(d) Uphill transport
Answer
(d) Uphill transport
Facilitated diffusion needs membrane proteins, is selective and saturates, but it moves molecules down their concentration gradient without energy. Uphill transport needs ATP, so it is active transport.
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- 2013 · NEET-UG (India) · 4 marks
The first stable product of fixation of atmospheric nitrogen in leguminous plants is:
(a)
(b) Ammonia
(c)
(d) Glutamate
Answer
(b) Ammonia
Nitrogenase reduces to , which is the first stable product in root nodules (). Glutamate forms later by reductive amination of ammonia with alpha-ketoglutarate.
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- 2017 · NEET-UG (India) · 4 marks
Which of the following facilitates opening of stomatal aperture?
(a) Contraction of outer wall of guard cells
(b) Decrease in turgidity of guard cells
(c) Radial orientation of cellulose microfibrils in the cell wall of guard cells
(d) Longitudinal orientation of cellulose microfibrils in the cell wall of guard cells
Answer
(c) Radial orientation of cellulose microfibrils in the cell wall of guard cells
Cellulose microfibrils in guard cell walls are arranged radially, so the cells cannot widen. When they become turgid they lengthen and bend outward, pulling the stomatal pore open. Decreased turgidity closes the stoma. Longitudinal microfibrils would not give this response.
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- 2017 · NEET-UG (India) · 4 marks
With reference to factors affecting the rate of photosynthesis, which of the following statements is not correct?
(a) Light saturation for fixation occurs at 10% of full sunlight
(b) Increasing atmospheric concentration upto 0.05% can enhance fixation rate
(c) plants responds to higher temperatures with enhanced photosynthesis while plants have much lower temperature optimum
(d) Tomato is a greenhouse crop which can be grown in - enriched atmosphere for higher yield
Answer
(c) C plants respond to higher temperatures with enhanced photosynthesis while C plants have much lower temperature optimum
This statement is not correct because it is reversed. C plants respond to higher temperatures (optimum about 30-45 °C), while C plants have a lower optimum (about 20-25 °C). The other statements are correct: light saturates at about 10% of full sunlight, CO up to 0.05% increases fixation, and tomato is grown in CO-enriched greenhouses.
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- 2017 · NEET-UG (India) · 4 marks
Fruit and leaf drop at early stages can be prevented by the application of
(a) Cytokinins
(b) Ethylene
(c) Auxins
(d) Gibberellic acid
Answer
(c) Auxins
Auxins delay abscission of young leaves and fruits and so prevent early drop. Ethylene promotes abscission. Cytokinins delay senescence and gibberellins promote elongation, but neither is the standard answer for preventing fruit and leaf drop.
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- 2017 · NEET-UG (India) · 4 marks
Phosphoenol pyruvate (PEP) is the primary acceptor in :
(a) plants
(b) plants
(c) plants
(d) and plants
Answer
(b) C plants
In C plants the primary CO acceptor is phosphoenolpyruvate (PEP) in the mesophyll cells, with PEP carboxylase fixing CO into oxaloacetate. In C plants the acceptor is RuBP, fixed by RuBisCO.
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- 2017 · NEET-UG (India) · 4 marks
The water potential of pure water is
(a) Zero
(b) Less than zero
(c) More than zero but less than one
(d) More than one
Answer
(a) Zero
Water potential of pure water at standard temperature and atmospheric pressure is taken as zero. Adding solute lowers it, so the water potential of solutions is negative.
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- 2018 · NEET-UG (India) · 4 marks
Stomatal movement is not affected by
(a) Temperature
(b) Light
(c) concentration
(d) concentration
Answer
(c) concentration
Guard cells respond to light, temperature and level inside the leaf (low opens stomata, high closes them), as well as water status and ABA. Ambient concentration has no significant role in stomatal movement.
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- 2018 · NEET-UG (India) · 4 marks
Which of the following is not a product of light reaction of photosynthesis?
(a) ATP
(b) NADH
(c) NADPH
(d) Oxygen
Answer
(b) NADH
The light reaction produces ATP (photophosphorylation), NADPH (reduction of at PS I) and (photolysis of water at PS II). NADH is a product of respiration, not photosynthesis; the light reaction makes NADPH.
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- 2018 · NEET-UG (India) · 4 marks
Oxygen is not produced during photosynthesis by
(a) Green sulphur bacteria
(b) Nostoc
(c) Cycas
(d) Chara
Answer
(a) Green sulphur bacteria
Green sulphur bacteria carry out anoxygenic photosynthesis, using rather than water as the electron donor, so no is released. Nostoc (cyanobacterium), Cycas and Chara (green alga) all use water and release oxygen.
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- 2019 · NEET-UG (India) · 4 marks
It takes very long time for pineapple plants to produce flowers. Which combination of hormones can be applied to artificially induce flowering in pineapple plants throughout the year to increase yield?
(a) Gibberellin and Abscisic acid
(b) Cytokinin and Abscisic acid
(c) Auxin and Ethylene
(d) Gibberellin and Cytokinin
Answer
(c) Auxin and Ethylene
Auxins (e.g. NAA) and ethylene promote flowering in pineapple and make flowering synchronous, so it can be induced throughout the year. Gibberellin promotes bolting and cytokinin promotes cell division, while ABA is inhibitory.
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- 2019 · NEET-UG (India) · 4 marks
What is the site of perception of photoperiod necessary for induction of flowering in plants?
(a) Shoot apex
(b) Leaves
(c) Lateral buds
(d) Pulvinus
Answer
(b) Leaves
Photoperiod is perceived by the leaves, which then send a floral stimulus (florigen) through the phloem to the shoot apex. The shoot apex is where flowering takes place, not where the light signal is sensed. Experiments in which only a leaf is exposed to the right photoperiod still induce flowering.
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- 2019 · NEET-UG (India) · 4 marks
Xylem translocates
(a) Water, mineral salts and some organic nitrogen only
(b) Water, mineral salts, some organic nitrogen and hormones
(c) Water only
(d) Water and mineral salts only
Answer
(b) Water, mineral salts, some organic nitrogen and hormones
Xylem mainly carries water and mineral salts. It also carries some organic nitrogen (amino acids, amides) and hormones such as cytokinins made in the roots. Options (c) and (d) leave these out. Option (a) leaves out the hormones.
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- 2019 · NEET-UG (India) · 4 marks
What is the direction of movement of sugars in phloem?
(a) Downward
(b) Bi-directional
(c) Non-multidirectional
(d) Upward
Answer
(b) Bi-directional
Sugars move in phloem from source to sink, and the direction depends on where the source and sink are. Leaves can send sugar down to roots, or up to growing buds and fruits, and a sink can be above or below the source. Movement in phloem is therefore bi-directional, unlike the one-way upward flow in xylem.
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- 2020 · NEET-UG (India) · 4 marks
In light reaction, plastoquinone facilitates the transfer of electrons from
(a) PS-I to
(b) PS-I to ATP synthase
(c) PS-II to Cytb6f complex
(d) Cytb6f complex to PS-I
Answer
(c) PS-II to Cytb6f complex
Plastoquinone carries electrons from PS-II to the Cytb6f complex, and it also carries protons into the lumen. Plastocyanin then carries electrons from Cytb6f to PS-I. Ferredoxin and NADP reductase take them from PS-I to , so options (a), (b) and (d) are wrong.
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- 2020 · NEET-UG (India) · 4 marks
Name the plant growth regulator which upon spraying on sugarcane crop increases the length of stem, thus increasing the yield of sugarcane crop
(a) Ethylene
(b) Abscisic acid
(c) Cytokinin
(d) Gibberellin
Answer
(d) Gibberellin
Gibberellins cause elongation of internodes. Spraying sugarcane with gibberellin increases stem length and so the yield of sugar. Ethylene promotes ripening, abscisic acid is a growth inhibitor, and cytokinin promotes cell division.
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- 2020 · NEET-UG (India) · 4 marks
The process of growth is maximum during
(a) Senescence
(b) Dormancy
(c) Log phase
(d) Lag phase
Answer
(c) Log phase
The log (exponential) phase is the period of fastest growth, when the growth rate is greatest. The lag phase is slow growth, and in senescence growth ends. Dormancy is a resting state with no growth.
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- 2020 · NEET-UG (India) · 4 marks
The oxygenation activity of RuBisCO enzyme in photorespiration leads to the formation of:
(a) 1 molecule of 6-C compound
(b) 1 molecule of 4-C compound and 1 molecule of 2-C compound
(c) 2 molecules of 3-C compound
(d) 1 molecule of 3-C compound
Answer
(d) 1 molecule of 3-C compound
When RuBisCO acts as an oxygenase, RuBP (5-C) is split into one 3-phosphoglycerate (3-C) and one phosphoglycolate (2-C). NCERT-style keys count the 3-C PGA, which enters the Calvin cycle, as the answer. Option (c), two 3-C molecules, is the carboxylation product, and no 6-C compound forms in oxygenation.
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- 2020 · NEET-UG (India) · 4 marks
Which of the following is not an inhibitory substance governing seed dormancy?
(a) Phenolic acid
(b) Para-ascorbic acid
(c) Gibberellic acid
(d) Abscisic acid
Answer
(c) Gibberellic acid
Gibberellic acid promotes germination and breaks seed dormancy, so it is not an inhibitor. Abscisic acid, phenolic acids and para-ascorbic acid are inhibitory substances that maintain dormancy.
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- 2020 · NEET-UG (India) · 4 marks
The process responsible for facilitating loss of water in liquid form from the tip of grass blades at night and in early morning is:
(a) Imbibition
(b) Plasmolysis
(c) Transpiration
(d) Root pressure
Answer
(d) Root pressure
Loss of water as liquid droplets from grass blade tips at night and early morning is guttation, caused by positive root pressure when transpiration is low and soil moisture is high. Transpiration is loss as vapour, while imbibition and plasmolysis do not move water out as droplets.
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- 2023 · NEET-UG (India) · 4 marks
Which hormone promotes internode/petiole elongation in deep water rice?
(a) Kinetin
(b) Ethylene
(c) 2, 4-D
(d) GA3
Answer
(b) Ethylene
Deep water rice responds to submergence by accumulating ethylene, which promotes rapid elongation of internodes and petioles so the leaves reach the surface for air. Kinetin and 2,4-D do not drive this response, and GA3 acts downstream in the stem elongation but ethylene is the primary trigger.
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- 2023 · NEET-UG (India) · 4 marks
Movement and accumulation of ions across a membrane against their concentration gradient can be explained by
(a) Facilitated Diffusion
(b) Passive Transport
(c) Active Transport
(d) Osmosis
Answer
(c) Active Transport
Moving ions against their concentration gradient needs energy (ATP) and carrier proteins, which is active transport. Facilitated diffusion, passive transport and osmosis all move substances down a gradient without energy expenditure.
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- 2023 · NEET-UG (India) · 4 marks
In tissue culture experiments, leaf mesophyll cells are put in a culture medium to form callus. This phenomenon may be called as
(a) Dedifferentiation
(b) Development
(c) Senescence
(d) Differentiation
Answer
(a) Dedifferentiation
Mature, differentiated mesophyll cells regain the capacity to divide and form an undifferentiated mass of cells (callus). This reversal is dedifferentiation; later, callus cells mature again by redifferentiation.
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- 2023 · NEET-UG (India) · 4 marks
How many ATP and NADPH2 are required for the synthesis of one molecule of Glucose during Calvin cycle?
(a) 18 ATP and 12 NADPH2
(b) 12 ATP and 16 NADPH2
(c) 18 ATP and 16 NADPH2
(d) 12 ATP and 12 NADPH2
Answer
(a) 18 ATP and 12 NADPH2
One CO2 fixed needs 3 ATP and 2 NADPH. Glucose has 6 carbons, so 6 CO2 are fixed: ATP and NADPH.
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- 2023 · NEET-UG (India) · 4 marks
Given below are two statements :
Statement I: The forces generated by transpiration can lift a xylem-sized column of water over 130 meters height.
Statement II: Transpiration cools leaf surfaces sometimes 10 to 15 degree evaporative cooling.
In the light of the above statements, choose the most appropriate answer from the options given below :
(a) Both Statement I and Statement II are incorrect
(b) Statement I is correct but Statement II is incorrect
(c) Statement I is incorrect but Statement II is correct
(d) Both Statement I and Statement II are correct
Answer
(d) Both Statement I and Statement II are correct
Transpiration pull generates forces able to lift a xylem water column over 130 m, and evaporative cooling can lower leaf temperature by 10 to 15 degrees. Both statements are standard textbook facts.
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- 2023 · NEET-UG (India) · 4 marks
Spraying of which of the following phytohormone on juvenile conifers helps in hastening the maturity period, that leads to early seed production?
(a) Gibberellic Acid
(b) Zeatin
(c) Abscisic Acid
(d) Indole-3-butyric Acid
Answer
(a) Gibberellic Acid
Spraying GA (gibberellins) on juvenile conifers hastens maturity and leads to early seed production. Zeatin promotes cell division, ABA is a stress hormone, and IBA is an auxin used for rooting.
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- 2023 · NEET-UG (India) · 4 marks
The reaction centre PS II has an absorption maxima at
(a) 700 nm
(b) 660 nm
(c) 780 nm
(d) 680 nm
Answer
(d) 680 nm
The reaction centre of PS II (P680) absorbs at 680 nm, whereas PS I (P700) absorbs at 700 nm. The other values are not reaction centre maxima.
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- 2023 · NEET-UG (India) · 4 marks
Which micronutrient is required for splitting of water molecule during photosynthesis?
(a) Molybdenum
(b) Magnesium
(c) Copper
(d) Manganese
Answer
(d) Manganese
Manganese is required for the water-splitting (oxygen-evolving) complex of PS II. Magnesium is part of chlorophyll but is a macronutrient, and molybdenum and copper act elsewhere (nitrogen metabolism, plastocyanin).
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- 2023 · NEET-UG (India) · 4 marks
Match List I with List II:
List I List II 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
Choose the correct answer from the options given below:
(a) A-IV, B-III, C-II, D-I
(b) A-III, B-I, C-IV, D-II
(c) A-II, B-I, C-IV, D-III
(d) A-II, B-IV, C-I, D-III
Answer
(d) A-II, B-IV, C-I, D-III
Cohesion is mutual attraction among water molecules, adhesion is attraction towards polar surfaces, surface tension arises from more attraction in the liquid phase, and guttation is water loss in liquid phase.
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Questions from NEET-UG 2013 (India), question paper (NTA NEET-UG 2013 question paper; only questions on topics in Nepal's MBBS entrance syllabus), AIPMT 2014 (NEET-UG, India), Code Q (NTA NEET-UG 2014 question paper; only questions on topics in Nepal's MBBS entrance syllabus), NEET-UG 2015 (India), Re-exam Code A (NTA NEET-UG 2015 question paper; only questions on topics in Nepal's MBBS entrance syllabus), NEET-UG 2016 (India), Phase 1 Code A (NTA NEET-UG 2016 question paper; only questions on topics in Nepal's MBBS entrance syllabus), NEET-UG 2017 (India), Code A (NTA NEET-UG 2017 question paper; only questions on topics in Nepal's MBBS entrance syllabus), NEET-UG 2018 (India), Code PP (NTA NEET-UG 2018 question paper; only questions on topics in Nepal's MBBS entrance syllabus), NEET-UG 2019 (India), Code S2 (NTA NEET-UG 2019 question paper; only questions on topics in Nepal's MBBS entrance syllabus), NEET-UG 2020 (India), Code F1 (September 2020) (NTA NEET-UG 2020 question paper; only questions on topics in Nepal's MBBS entrance syllabus) and NEET-UG 2023 (India), Code F1 Botany (NTA NEET-UG 2023 question paper; only questions on topics in Nepal's MBBS entrance syllabus). Answers are written for this site; check them against your class notes.