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RBSE SS Agricultural Biology study materials
RBSE SS Agricultural Biology study resources

RBSE SS Agricultural Biology practice questions

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Question 1Plant Physiology and Biochemistry

Which pigment absorbs light most strongly in the blue-violet and red regions while reflecting green light, making it the principal photosynthetic pigment of green plants?

  1. Anthocyanin
  2. Chlorophyll a Correct answer.
  3. Phycoerythrin
  4. β-carotene

Explanation

Chlorophyll a has absorption maxima near 430 nm and 660 nm, so it absorbs blue-violet and red light and reflects green wavelengths. It sits at the reaction centres of both photosystems and is therefore the principal photosynthetic pigment. The other pigments listed are accessory or non-photosynthetic pigments with different absorption profiles.

Question 2Plant Physiology and Biochemistry

In the light reactions of photosynthesis, water is split at which complex, and what is released?

  1. At the cytochrome b6f complex, releasing ATP and oxygen
  2. At photosystem I, releasing carbon dioxide and protons
  3. At ATP synthase, releasing oxygen and NADPH
  4. At photosystem II, releasing oxygen, protons and electrons Correct answer.

Explanation

Photolysis of water occurs at the oxygen-evolving complex associated with photosystem II. The splitting yields electrons that replace those lost by P680, protons that contribute to the thylakoid proton gradient, and molecular oxygen as a by-product.

Question 3Plant Physiology and Biochemistry

Which statement correctly describes the role of Rubisco in the Calvin cycle?

  1. It carboxylates ribulose-1,5-bisphosphate to give two molecules of 3-phosphoglycerate Correct answer.
  2. It regenerates ribulose-1,5-bisphosphate from triose phosphates
  3. It reduces 1,3-bisphosphoglycerate to glyceraldehyde-3-phosphate
  4. It phosphorylates ribulose-5-phosphate using ATP

Explanation

Rubisco catalyses the carboxylation of the five-carbon acceptor ribulose-1,5-bisphosphate, and the unstable six-carbon intermediate immediately splits into two molecules of 3-phosphoglycerate. This is the carbon-fixing step of the Calvin cycle. Reduction and regeneration are carried out by other enzymes.

Question 4Plant Physiology and Biochemistry

A plant is kept in a sealed chamber and illuminated with light of wavelength above 680 nm only. Carbon dioxide uptake falls sharply even though light is present. Which explanation fits best?

  1. Photosystem I absorbs this light too strongly, over-reducing NADP+
  2. Photosystem II is poorly excited, so linear electron flow and NADPH supply decline Correct answer.
  3. Rubisco is denatured by far-red light and loses affinity for carbon dioxide
  4. Stomata close in response to far-red light, blocking all diffusion

Explanation

Photosystem II absorbs maximally near 680 nm, so illumination at longer wavelengths excites photosystem I preferentially. With photosystem II under-excited, water splitting and linear electron flow slow down, reducing NADPH production and hence carbon dioxide fixation. This is the classic red-drop effect.

Question 5Plant Physiology and Biochemistry

Which product of the light reactions is consumed directly in the reduction phase of the Calvin cycle?

  1. NADPH and ATP Correct answer.
  2. Oxygen and ATP
  3. NADH and FADH2
  4. NADPH and oxygen

Explanation

The reduction of 1,3-bisphosphoglycerate to glyceraldehyde-3-phosphate uses NADPH as the reducing agent, and ATP is consumed in phosphorylation steps of the cycle. Both are generated by the thylakoid electron transport chain. Oxygen and NADH are not used in the Calvin cycle.

Question 6Plant Physiology and Biochemistry

In C4 plants such as maize, the first stable product of carbon dioxide fixation is which compound, and in which cell type does it form?

  1. Ribulose-1,5-bisphosphate, formed in mesophyll cells
  2. Malate, formed in guard cells
  3. Oxaloacetate, formed in mesophyll cells Correct answer.
  4. 3-phosphoglycerate, formed in bundle-sheath cells

Explanation

In C4 plants, PEP carboxylase in mesophyll cells fixes bicarbonate onto phosphoenolpyruvate to yield the four-carbon acid oxaloacetate. This acid is converted to malate or aspartate, transported to bundle-sheath cells and decarboxylated there to feed the Calvin cycle.

Question 7Plant Physiology and Biochemistry

Why does the C4 pathway reduce photorespiration in hot, dry conditions?

  1. It keeps stomata fully open at midday to supply carbon dioxide
  2. It concentrates carbon dioxide around Rubisco in bundle-sheath cells Correct answer.
  3. It uses oxygen instead of carbon dioxide as the terminal electron acceptor
  4. It eliminates Rubisco and fixes carbon dioxide with PEP carboxylase only

Explanation

The C4 shuttle pumps carbon dioxide into bundle-sheath cells, raising its concentration around Rubisco. Because Rubisco's oxygenase activity is competitively suppressed at high carbon dioxide to oxygen ratios, photorespiratory losses are small even when stomata are partly closed.

Question 8Plant Physiology and Biochemistry

Which enzyme catalyses the primary carboxylation reaction in CAM plants during the night?

  1. Pyruvate kinase
  2. PEP carboxylase Correct answer.
  3. Carbonic anhydrase
  4. Rubisco

Explanation

CAM plants open their stomata at night and fix carbon dioxide onto phosphoenolpyruvate using PEP carboxylase, storing the product as malate in the vacuole. Rubisco acts later in the day when malate is decarboxylated behind closed stomata.

Question 9Plant Physiology and Biochemistry

Which set of conditions best describes the wavelength dependence of photosynthesis, as shown by an action spectrum?

  1. Highest in green light, lowest in blue light
  2. Highest in blue and red light, lowest in green light Correct answer.
  3. Highest in ultraviolet light, absent in red light
  4. Uniform across all visible wavelengths

Explanation

Action spectra of photosynthesis follow the absorption of chlorophyll, with peaks in the blue-violet and red regions. Green light is largely reflected or transmitted, so photosynthetic rate is lowest there. Ultraviolet is not used and can be damaging.

Question 10Plant Physiology and Biochemistry

Which statement correctly distinguishes cyclic from non-cyclic photophosphorylation?

  1. Cyclic flow involves only photosystem II and produces NADPH without ATP
  2. Cyclic flow involves only photosystem I and produces ATP without NADPH Correct answer.
  3. Non-cyclic flow produces ATP without any electron transport
  4. Non-cyclic flow involves only photosystem I and releases oxygen

Explanation

In cyclic photophosphorylation, electrons leave photosystem I, pass through ferredoxin and the cytochrome b6f complex and return to P700, generating a proton gradient and ATP but no NADPH. Non-cyclic flow uses both photosystems, splits water and yields ATP, NADPH and oxygen.

Question 11Plant Physiology and Biochemistry

The proton gradient used by chloroplast ATP synthase is built up mainly where?

  1. Across the tonoplast, with the vacuole more acidic
  2. Across the thylakoid membrane, with the lumen more acidic Correct answer.
  3. Across the outer mitochondrial membrane, with the matrix more acidic
  4. Across the chloroplast envelope, with the stroma more acidic

Explanation

Electron transport and water splitting release protons into the thylakoid lumen, while protons are consumed in the stroma during NADPH formation. The resulting proton-motive force across the thylakoid membrane drives ATP synthesis as protons return to the stroma.

Question 12Plant Physiology and Biochemistry

Which statement about the oxygen-evolving complex is correct?

  1. It contains manganese and calcium and oxidises water to supply electrons to P680 Correct answer.
  2. It is located on the stromal side of photosystem I
  3. It contains iron and copper and reduces oxygen to water
  4. It transfers electrons directly to NADP+

Explanation

The oxygen-evolving complex is a manganese-calcium cluster bound to the lumenal side of photosystem II. It accumulates oxidising equivalents and splits water, releasing electrons that refill P680, along with protons and oxygen.

Question 13Plant Physiology and Biochemistry

A leaf is exposed to a strong light intensity while carbon dioxide concentration is kept very low. What is the most likely consequence for the Calvin cycle?

  1. The proton gradient across the thylakoid collapses completely
  2. ATP and NADPH accumulate and the acceptor ribulose-1,5-bisphosphate becomes depleted Correct answer.
  3. ATP and NADPH are depleted and ribulose-1,5-bisphosphate accumulates
  4. Oxygen evolution stops immediately because water cannot be split

Explanation

When carbon dioxide is limiting, the Calvin cycle cannot consume ATP and NADPH at the rate the light reactions produce them. The acceptor ribulose-1,5-bisphosphate is not regenerated because its carboxylation is blocked, so it becomes depleted while reduced cofactors build up.

Question 14Plant Physiology and Biochemistry

Which process converts the energy of light into chemical energy stored in ATP and NADPH?

  1. Glycolysis in the cytosol
  2. Photorespiration in the peroxisomes
  3. Oxidative phosphorylation in the mitochondria
  4. Photophosphorylation in the thylakoid membranes Correct answer.

Explanation

Photophosphorylation couples light-driven electron transport across the thylakoid membrane to ATP synthesis, while NADP+ is reduced to NADPH. These products carry the energy and reducing power used in carbon fixation.

Question 15Plant Physiology and Biochemistry

Which statement about the light compensation point of a leaf is correct?

  1. It is the wavelength at which chlorophyll absorbs most strongly
  2. It is the light intensity at which photosynthetic carbon dioxide uptake equals respiratory carbon dioxide release Correct answer.
  3. It is the light intensity at which photosynthesis reaches its maximum rate
  4. It is the carbon dioxide concentration at which stomata close

Explanation

At the light compensation point, the rate of photosynthesis exactly balances the rate of respiration, so net carbon dioxide exchange with the atmosphere is zero. Below this intensity the plant loses carbon overall; above it, net gain occurs.

Question 16Plant Physiology and Biochemistry

Why does the rate of photosynthesis plateau at high light intensity?

  1. Chlorophyll is completely destroyed by excess light
  2. Water can no longer be split by photosystem II
  3. The proton gradient reverses direction across the thylakoid
  4. Another factor such as carbon dioxide supply or Rubisco capacity becomes limiting Correct answer.

Explanation

Once light is saturating, the light reactions supply ATP and NADPH faster than the Calvin cycle can use them. The overall rate is then limited by factors such as carbon dioxide concentration, Rubisco activity or triose phosphate utilisation, producing a plateau.

Question 17Plant Physiology and Biochemistry

Which compound is the primary carbon dioxide acceptor in the Calvin cycle?

  1. Fructose-6-phosphate
  2. Phosphoenolpyruvate
  3. Ribulose-1,5-bisphosphate Correct answer.
  4. Phosphoglycerate

Explanation

Ribulose-1,5-bisphosphate is the five-carbon sugar phosphate that Rubisco carboxylates in the first step of the Calvin cycle. Phosphoenolpyruvate is the acceptor in C4 and CAM initial fixation, not in the Calvin cycle.

Question 18Plant Physiology and Biochemistry

Which statement about chlorophyll biosynthesis is correct?

  1. It requires light in all plants and proceeds from acetyl-CoA via mevalonate
  2. It requires light in angiosperms and proceeds from glutamate via protoporphyrin IX Correct answer.
  3. It proceeds from glucose via glycolysis intermediates without light
  4. It occurs only in roots and is independent of light

Explanation

Chlorophyll is synthesised from glutamate through protoporphyrin IX, and in angiosperms the final step catalysed by protochlorophyllide oxidoreductase requires light. This is why dark-grown seedlings are etiolated and pale.

Question 19Plant Physiology and Biochemistry

Which factor would most directly increase the rate of transpiration from a leaf?

  1. A larger water vapour pressure gradient between leaf interior and outside air Correct answer.
  2. A higher atmospheric humidity around the leaf
  3. Closure of the stomata
  4. A thicker leaf cuticle

Explanation

Transpiration is driven by the vapour pressure gradient between the saturated air spaces inside the leaf and the drier outside air. A steeper gradient increases the rate of water loss through open stomata.

Question 20Plant Physiology and Biochemistry

Which mechanism best explains how water rises in the xylem of tall trees?

  1. Cohesion-tension, driven by transpiration from the leaves Correct answer.
  2. Active pumping by living xylem vessel elements
  3. Capillary action in the phloem
  4. Root pressure alone, pushing water up the stem

Explanation

Evaporation from leaf mesophyll creates tension in the water column, which is transmitted downwards through continuous xylem. Cohesion between water molecules and adhesion to vessel walls allow the column to withstand this tension and move upward.

Question 21Plant Physiology and Biochemistry

Which mineral deficiency symptom is most consistent with the mobility of magnesium in the phloem?

  1. Interveinal chlorosis first appearing on older leaves Correct answer.
  2. Marginal necrosis confined to young leaves
  3. Interveinal chlorosis first appearing on young leaves
  4. Uniform greening of all leaves

Explanation

Magnesium is a mobile nutrient because it is loaded into the phloem and redistributed to growing tissues. Deficiency therefore appears first in older leaves, where chlorophyll breaks down between the veins, producing interveinal chlorosis.

Question 22Plant Physiology and Biochemistry

Which element is a component of the chlorophyll molecule itself?

  1. Iron
  2. Calcium
  3. Magnesium Correct answer.
  4. Manganese

Explanation

Chlorophyll is a magnesium porphyrin, with a magnesium ion held at the centre of the tetrapyrrole ring. Iron and manganese are required for synthesis and for the oxygen-evolving complex but are not part of the chlorophyll molecule.

Question 23Plant Physiology and Biochemistry

Which statement about the stomatal response to blue light is correct?

  1. Blue light inhibits the H+-ATPase, causing K+ efflux and stomatal closure
  2. Blue light causes guard cells to lose turgor and close
  3. Blue light activates the H+-ATPase of guard cells, driving K+ uptake and stomatal opening Correct answer.
  4. Blue light has no effect on guard cells; only carbon dioxide matters

Explanation

Blue light is perceived by phototropins in guard cells, which activate the plasma membrane H+-ATPase. Proton pumping hyperpolarises the membrane, driving potassium uptake through inward-rectifying channels; water follows osmotically and the guard cells become turgid, opening the pore.

Question 24Plant Physiology and Biochemistry

Which statement correctly describes the role of the enzyme nitrate reductase?

  1. It reduces nitrate to nitrite using NADH or NADPH as electron donor Correct answer.
  2. It converts ammonium into glutamate by reductive amination
  3. It fixes atmospheric nitrogen in root nodules
  4. It reduces nitrite to ammonium using ferredoxin

Explanation

Nitrate reductase catalyses the two-electron reduction of nitrate to nitrite, using NADH or NADPH depending on the organism. Nitrite is then further reduced to ammonium by nitrite reductase, and ammonium is assimilated into amino acids.

Question 25Plant Physiology and Biochemistry

A plant is given a substance that uncouples the thylakoid proton gradient from ATP synthase. What effect is expected on photosynthesis?

  1. NADPH production stops immediately while ATP synthesis continues
  2. Oxygen evolution increases because water splitting is stimulated
  3. ATP synthesis falls and carbon dioxide fixation declines despite continued electron transport Correct answer.
  4. Carbon dioxide fixation is unaffected because ATP is not required

Explanation

Uncouplers allow protons to leak across the thylakoid membrane without passing through ATP synthase, so the proton-motive force is dissipated and ATP synthesis falls. Since the Calvin cycle needs ATP as well as NADPH, carbon fixation declines even though electron transport may continue.

Question 26Plant Physiology and Biochemistry

In the Calvin cycle, the regeneration phase produces ribulose-1,5-bisphosphate from which starting intermediate, and how many ATP are consumed per RuBP regenerated?

  1. Ribulose-5-phosphate; three ATP are consumed Correct answer.
  2. Fructose-6-phosphate; two ATP are consumed
  3. Erythrose-4-phosphate; four ATP are consumed
  4. Glyceraldehyde-3-phosphate; five ATP are consumed

Explanation

Regeneration begins with ribulose-5-phosphate, which is phosphorylated by phosphoribulokinase using ATP to form ribulose-1,5-bisphosphate. Three ATP are needed per RuBP regenerated, matching the three ATP per CO2 fixed in the full cycle.

Question 27Plant Physiology and Biochemistry

A leaf is treated with an inhibitor that blocks the reduction of 3-phosphoglycerate to glyceraldehyde-3-phosphate. Which immediate consequence is expected within the Calvin cycle?

  1. Depletion of NADPH and ATP because their consumption stops
  2. Increased oxygenase activity of Rubisco without any change in carbon flux
  3. Immediate cessation of linear electron flow because water splitting stops
  4. Accumulation of 3-phosphoglycerate and depletion of RuBP regeneration Correct answer.

Explanation

Blocking the reduction step stops consumption of 3-phosphoglycerate, so it accumulates while reduced carbon for regeneration falls, depleting RuBP. ATP and NADPH would build up rather than be depleted because their use is blocked.

Question 28Plant Physiology and Biochemistry

Which statement best explains why the net rate of photosynthetic carbon gain in a leaf declines when temperature rises well above the optimum?

  1. Stomata close permanently and never reopen in warm air
  2. Photorespiration and respiratory carbon loss rise faster than carboxylation Correct answer.
  3. The oxygen-evolving complex releases less oxygen per photon absorbed
  4. Light absorption by chlorophyll decreases sharply at high temperature

Explanation

Above the optimum, respiration and photorespiration increase faster than carboxylation, and Rubisco's specificity for CO2 relative to O2 falls, so net carbon gain declines. Chlorophyll light absorption is not the limiting factor at high temperature.

Question 29Plant Physiology and Biochemistry

Which statement correctly describes the role of the enzyme PEP carboxylase in C4 photosynthesis?

  1. It fixes CO2 in bundle-sheath cells using NADPH as the electron donor
  2. It decarboxylates malate in bundle-sheath cells to release CO2
  3. It fixes CO2 in mesophyll cells with high affinity and no oxygenase activity Correct answer.
  4. It regenerates phosphoenolpyruvate from pyruvate in mesophyll cells

Explanation

PEP carboxylase in mesophyll cells has a high affinity for bicarbonate and no oxygenase activity, so it concentrates carbon efficiently. Decarboxylation of malate occurs in bundle-sheath cells by other enzymes, and PEP regeneration is catalysed by pyruvate phosphate dikinase.

Question 30Plant Physiology and Biochemistry

A researcher measures the quantum yield of CO2 fixation in a C3 leaf at 25 degrees Celsius while varying intercellular CO2 from very low to near saturation. Which pattern is expected for the initial slope and plateau of the response curve?

  1. The initial slope is limited by Rubisco carboxylation, and the plateau is limited by electron transport capacity Correct answer.
  2. Both the initial slope and the plateau are limited by stomatal conductance alone
  3. The initial slope is limited by photorespiration, and the plateau is limited by chlorophyll synthesis
  4. The initial slope is limited by electron transport, and the plateau is limited by Rubisco carboxylation

Explanation

At low intercellular CO2, carboxylation by Rubisco limits the rate, giving a steep initial slope. At high CO2, the capacity of electron transport to regenerate RuBP becomes limiting, so the curve plateaus.

Question 31Genetics and Plant Breeding

In a monohybrid cross between two heterozygous tall pea plants (Tt × Tt), what phenotypic ratio is expected in the offspring?

  1. 1 tall : 1 dwarf
  2. 9 tall : 3 dwarf
  3. 3 tall : 1 dwarf Correct answer.
  4. 1 tall : 2 intermediate : 1 dwarf

Explanation

A cross of two heterozygotes yields genotypes in a 1 TT : 2 Tt : 1 tt ratio. Because T is completely dominant over t, both TT and Tt appear tall, giving three tall plants for every dwarf. The 1:1 ratio arises from a test cross, not from heterozygote selfing.

Question 32Genetics and Plant Breeding

Which term describes the genetic constitution of an organism, such as TT or Tt, as distinct from its observable appearance?

  1. Phenotype
  2. Genotype Correct answer.
  3. Pleitropy
  4. Karyotype

Explanation

The genotype is the specific allelic makeup of an organism at a locus or across its genome. The phenotype is the visible or measurable trait produced by the interaction of genotype and environment. Keeping these two concepts separate is essential when predicting segregation ratios.

Question 33Genetics and Plant Breeding

During meiosis, homologous chromosomes separate and move to opposite poles. Which meiotic stage is primarily responsible for Mendel's law of segregation?

  1. Telophase II
  2. Metaphase II
  3. Prophase I
  4. Anaphase I Correct answer.

Explanation

The law of segregation states that the two alleles of a gene separate during gamete formation. This separation is physically achieved when homologous chromosomes are pulled to opposite poles in anaphase I of meiosis. Anaphase II separates sister chromatids, which carry identical alleles rather than the two different alleles of a heterozygote.

Question 34Genetics and Plant Breeding

A cross between a homozygous dominant red-flowered plant and a homozygous recessive white-flowered plant produces an F1 generation. What is the genotype of all F1 plants?

  1. Heterozygous Correct answer.
  2. Hemizygous
  3. Homozygous dominant
  4. Homozygous recessive

Explanation

Each parent contributes one allele, so an RR parent gives R and an rr parent gives r. All F1 offspring therefore receive one R and one r, making them heterozygous. A homozygous outcome would require both parents to contribute the same allele.

Question 35Genetics and Plant Breeding

Two genes located close together on the same chromosome are said to be linked. What is the main consequence of linkage for a dihybrid cross?

  1. The two genes must control the same trait.
  2. Parental allele combinations appear more often than recombinant combinations. Correct answer.
  3. The genes always assort independently.
  4. The genes can no longer undergo crossing over.

Explanation

Linked genes tend to be inherited together because they lie on the same chromosome and crossing over between them is relatively rare. As a result, parental combinations of alleles exceed the 1:1:1:1 expectation, and recombinant types are underrepresented. Independent assortment holds only for genes on different chromosomes or far apart on the same one.

Question 36Genetics and Plant Breeding

In a test cross, an individual with an unknown dominant phenotype is crossed with a homozygous recessive individual. What is the purpose of this cross?

  1. To increase the number of dominant offspring
  2. To produce a pure breeding line
  3. To measure the mutation rate
  4. To determine whether the unknown individual is homozygous or heterozygous Correct answer.

Explanation

Because a homozygous recessive parent contributes only recessive alleles, the offspring phenotypes directly reveal the alleles carried by the unknown parent. If any recessive offspring appear, the unknown parent must be heterozygous. A test cross is therefore a diagnostic tool for genotype, not a breeding method for purity.

Question 37Genetics and Plant Breeding

Which of the following is the most accurate description of a pure line in plant breeding?

  1. A population that is heterozygous at most loci
  2. A hybrid produced from two distinct inbred parents
  3. A group of plants that breed true for a trait over successive generations Correct answer.
  4. A population with a high mutation frequency

Explanation

A pure line is homozygous for the traits of interest, so selfing or crossing within the line produces offspring with the same phenotype generation after generation. Pure lines are obtained by repeated self-pollination or inbreeding with selection. Their uniformity makes them valuable as parents in hybrid breeding programmes.

Question 38Genetics and Plant Breeding

In a dihybrid cross between two plants heterozygous for both genes (RrYy × RrYy), assuming independent assortment, what proportion of offspring is expected to be homozygous recessive for both traits?

  1. 1/16 Correct answer.
  2. 1/4
  3. 9/16
  4. 1/8

Explanation

Each gene pair segregates independently, so the probability of being rr is 1/4 and of being yy is also 1/4. Multiplying these independent probabilities gives 1/16 for the double recessive genotype. This is the basis of the 9:3:3:1 phenotypic ratio in a dihybrid cross.

Question 39Genetics and Plant Breeding

Which of the following best describes the role of the cytoplasm in cytoplasmic inheritance?

  1. It contains genes that are inherited equally from both parents
  2. It carries genes located on organelle DNA that are transmitted mainly through the maternal parent Correct answer.
  3. It determines the segregation of nuclear alleles during meiosis
  4. It prevents the expression of nuclear genes

Explanation

Cytoplasmic inheritance involves genes located in mitochondria and chloroplasts, which are inherited through the cytoplasm of the egg. Because the zygote receives most of its cytoplasm from the female gamete, these traits usually show maternal transmission. Nuclear genes and cytoplasmic genes are distinct systems, although they interact.

Question 40Genetics and Plant Breeding

A plant breeder wants to transfer a single dominant disease-resistance gene from a wild relative into a susceptible cultivated variety. Which breeding method is most direct?

  1. Clonal propagation
  2. Backcross breeding Correct answer.
  3. Mass selection
  4. Mutation breeding

Explanation

Backcross breeding repeatedly crosses the hybrid back to the susceptible cultivated parent while selecting for the resistance gene. This transfers the single dominant gene while progressively recovering the cultivated genotype. It is the standard method for introgressing a specific trait from a wild relative.

Question 41Genetics and Plant Breeding

Which of the following statements about the law of independent assortment is correct?

  1. It applies to genes located on the same chromosome only.
  2. It applies only to genes controlling the same trait.
  3. It states that dominant alleles always appear together in gametes.
  4. It states that alleles of different genes assort independently during gamete formation. Correct answer.

Explanation

Mendel's law of independent assortment states that during gamete formation, the alleles of one gene segregate independently of the alleles of another gene. This holds for genes on different chromosomes or far apart on the same chromosome. It explains the generation of new combinations of traits in the offspring.

Question 42Genetics and Plant Breeding

A plant species has a diploid chromosome number of 24. How many chromosomes are present in a gamete produced by this plant?

  1. 12 Correct answer.
  2. 6
  3. 48
  4. 24

Explanation

Gametes are haploid and carry half the diploid chromosome number because meiosis reduces the chromosome number by half. For a diploid number of 24, each gamete receives 12 chromosomes. Doubling to 48 would occur only after fertilization or during mitotic DNA replication.

Question 43Genetics and Plant Breeding

Which of the following is a key difference between mitosis and meiosis?

  1. Mitosis produces four haploid cells, while meiosis produces two diploid cells.
  2. Mitosis produces two genetically identical diploid cells, while meiosis produces four genetically variable haploid cells. Correct answer.
  3. Mitosis involves crossing over, while meiosis does not.
  4. Mitosis halves the chromosome number, while meiosis maintains it.

Explanation

Mitosis produces two daughter cells that are genetically identical to the parent cell and retain the diploid chromosome number. Meiosis involves two divisions and produces four haploid cells that are genetically variable due to crossing over and independent assortment. These contrasting outcomes explain growth versus gamete formation.

Question 44Genetics and Plant Breeding

A breeder observes that a trait is controlled by two alleles showing incomplete dominance. What phenotypic ratio is expected from a cross between two heterozygous individuals?

  1. 9:3:3:1
  2. 1:2:1 Correct answer.
  3. 3:1
  4. 1:1

Explanation

With incomplete dominance, the heterozygote has an intermediate phenotype distinct from either homozygote. A cross of two heterozygotes therefore yields three phenotypic classes in a 1:2:1 ratio, matching the genotypic ratio. The classic 3:1 ratio requires complete dominance, where heterozygotes resemble one homozygote.

Question 45Genetics and Plant Breeding

What is the primary purpose of emasculation in controlled hybridization of crop plants?

  1. To stimulate root formation in cuttings
  2. To remove the anthers of the female parent to prevent self-pollination Correct answer.
  3. To induce mutations in the pollen grains
  4. To increase the number of chromosomes in the offspring

Explanation

Emasculation involves removing the anthers from a flower before they shed pollen, ensuring that the flower cannot self-pollinate. The emasculated flower is then hand-pollinated with pollen from the chosen male parent. This gives the breeder control over the parentage of the resulting seed.

Question 46Genetics and Plant Breeding

Which of the following is an example of a quantitative trait in crop plants?

  1. Presence of a dominant resistance gene
  2. Flower colour controlled by a single gene
  3. Grain yield per plant Correct answer.
  4. Seed shape following Mendelian inheritance

Explanation

Quantitative traits are controlled by many genes and are strongly influenced by the environment, producing continuous variation. Grain yield per plant is a classic example because it varies continuously and is measured rather than classified. Single-gene traits such as flower colour or seed shape produce discrete classes.

Question 47Genetics and Plant Breeding

Which of the following best explains the genetic basis of heterosis, or hybrid vigour?

  1. The F1 hybrid is homozygous at all loci.
  2. The F1 hybrid has a reduced chromosome number.
  3. The F1 hybrid shows an increased mutation rate.
  4. The F1 hybrid combines favourable dominant alleles from both parents, masking deleterious recessives. Correct answer.

Explanation

Heterosis refers to the superior performance of F1 hybrids compared with their parents. One explanation is the dominance hypothesis, in which the hybrid carries favourable dominant alleles from both parents, masking harmful recessive alleles. The hybrid is highly heterozygous rather than homozygous, and its chromosome number is unchanged.

Question 48Genetics and Plant Breeding

Which term refers to the loss or gain of one or more whole chromosomes, producing an abnormal chromosome number?

  1. Translocation
  2. Polyploidy
  3. Aneuploidy Correct answer.
  4. Inversion

Explanation

Aneuploidy is a numerical chromosome abnormality in which one or a few chromosomes are added or lost, as in trisomy or monosomy. Polyploidy, by contrast, involves whole sets of chromosomes. Inversions and translocations are structural rearrangements, not changes in chromosome number.

Question 49Genetics and Plant Breeding

A plant with the genotype AaBb is selfed, and the two genes are completely linked with no crossing over. What phenotypic ratio is expected in the offspring?

  1. 3:1 Correct answer.
  2. 1:1
  3. 1:2:1
  4. 9:3:3:1

Explanation

With complete linkage and no recombination, the AaBb parent produces only two gamete types, AB and ab, in equal proportions. Selfing gives AABB, AaBb and aabb in a 1:2:1 genotypic ratio, but the two dominant phenotypes are indistinguishable, so the phenotypic ratio is 3 dominant : 1 recessive. Independent assortment would instead give 9:3:3:1.

Question 50Genetics and Plant Breeding

Which of the following is the main advantage of using hybrid varieties in commercial crop production?

  1. Their seed can be saved and replanted without loss of yield.
  2. They breed true in subsequent generations.
  3. They are always resistant to all diseases.
  4. They show uniform, high performance in the first generation and exploit heterosis. Correct answer.

Explanation

Hybrid varieties are produced by crossing selected inbred parents, and the F1 generation shows uniform vigour and high yield due to heterosis. However, seed from the F1 cannot be saved for the next crop because segregation in later generations causes loss of uniformity and vigour. Farmers therefore must buy fresh hybrid seed each season.