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WBCHSE HS Biological Science study materials
WBCHSE HS Biological Science study resources

WBCHSE HS Biological Science practice questions

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Question 1Cell Biology and Cell Division

Which cell organelle is described as the 'powerhouse of the cell' because it generates most of the ATP through oxidative phosphorylation?

  1. Golgi apparatus
  2. Ribosome
  3. Lysosome
  4. Mitochondrion Correct answer.

Explanation

The mitochondrion carries the electron transport chain and ATP synthase on its inner membrane, so oxidative phosphorylation and most cellular ATP production occur there. Ribosomes synthesise protein, the Golgi modifies and packages secretions, and lysosomes carry hydrolytic enzymes. Hence only the mitochondrion matches the described function.

Question 2Cell Biology and Cell Division

During which stage of mitosis do sister chromatids separate and move towards opposite poles of the spindle?

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

Explanation

In anaphase the centromeres split and the two sister chromatids of each chromosome are pulled to opposite poles by shortening spindle fibres. Prophase involves chromosome condensation, metaphase alignment at the equator, and telophase nuclear reformation. Separation of chromatids is therefore the defining event of anaphase.

Question 3Cell Biology and Cell Division

A cell is placed in a hypotonic solution. What is the most likely immediate consequence for an animal cell?

  1. It shows no net movement of water
  2. It actively pumps solutes out to prevent change
  3. It gains water and may burst Correct answer.
  4. It loses water and shrinks

Explanation

A hypotonic solution has a lower solute concentration than the cytoplasm, so water enters the cell by osmosis down its water-potential gradient. An animal cell lacks a rigid wall and may swell and lyse. Loss of water would occur in a hypertonic solution, not a hypotonic one.

Question 4Cell Biology and Cell Division

Which of the following is the correct sequence of phases in the cell cycle of a typical eukaryotic cell?

  1. M → S → G1 → G2
  2. G2 → G1 → S → M
  3. G1 → S → G2 → M Correct answer.
  4. S → G1 → M → G2

Explanation

Interphase proceeds as G1 (cell growth), S (DNA replication) and G2 (further growth and preparation), after which mitosis (M phase) occurs. DNA must be replicated in S before the mitotic division in M. Any order placing S after G2 or M before interphase is inconsistent with the cycle.

Question 5Cell Biology and Cell Division

Which structural feature distinguishes a prokaryotic cell from a eukaryotic cell?

  1. Presence of a membrane-bound nucleus Correct answer.
  2. Presence of ribosomes
  3. Presence of DNA
  4. Presence of a plasma membrane

Explanation

Prokaryotes lack a membrane-bound nucleus; their genetic material lies in a nucleoid region. Eukaryotes enclose their chromosomes within a nuclear envelope. Ribosomes, a plasma membrane and DNA are present in both cell types, so they cannot be used to distinguish them.

Question 6Cell Biology and Cell Division

Which molecule is the principal structural component of the plasma membrane according to the fluid mosaic model?

  1. Chitin
  2. Cellulose
  3. Phospholipid bilayer Correct answer.
  4. Starch

Explanation

The fluid mosaic model describes the membrane as a phospholipid bilayer in which proteins are embedded or attached. The amphipathic phospholipids form the continuous barrier with hydrophilic heads outward and hydrophobic tails inward. Cellulose, chitin and starch are polysaccharides not forming this bilayer.

Question 7Cell Biology and Cell Division

A cell in G2 phase is treated with a drug that blocks synthesis of tubulin. Which process will be most directly prevented?

  1. Replication of DNA
  2. Formation of the mitotic spindle Correct answer.
  3. Formation of the nuclear envelope
  4. Synthesis of ribosomal RNA

Explanation

Tubulin is the protein subunit of microtubules, which assemble into the mitotic spindle. Blocking tubulin synthesis therefore prevents spindle formation and chromosome segregation. DNA replication, rRNA synthesis and nuclear envelope assembly depend on other molecules, not on tubulin.

Question 8Cell Biology and Cell Division

In a dihybrid cross between two heterozygotes (AaBb × AaBb) with independent assortment, what fraction of the offspring is expected to show both recessive phenotypes?

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

Explanation

Each parent produces four gamete types in equal proportion, and independent assortment gives 16 equally likely combinations. The double recessive phenotype requires the homozygous recessive genotype at both loci, which occurs in only one of the 16 combinations. Hence the expected fraction is 1/16.

Question 9Cell Biology and Cell Division

Which observation about the behaviour of chromosomes during meiosis best supports the law of independent assortment?

  1. Separation of sister chromatids in anaphase II
  2. Crossing over between non-sister chromatids
  3. Random orientation of bivalents at metaphase I Correct answer.
  4. Homologous chromosomes pair during prophase I

Explanation

At metaphase I the bivalents align independently of one another, so the maternal and paternal chromosomes of different pairs are distributed to poles in random combinations. This random orientation underlies independent assortment of unlinked genes. Pairing and crossing over relate to synapsis and recombination, while anaphase II separation concerns sister chromatids.

Question 10Cell Biology and Cell Division

Which of the following correctly describes the role of the Golgi apparatus in a secretory cell?

  1. It modifies, sorts and packages proteins for secretion Correct answer.
  2. It generates ATP for vesicle transport
  3. It synthesises proteins directly from mRNA
  4. It degrades worn-out organelles

Explanation

Proteins made on the rough endoplasmic reticulum pass to the Golgi, where they are glycosylated, sorted and packed into vesicles for secretion or delivery. Protein synthesis itself occurs on ribosomes, and organelle degradation is a lysosomal function. The Golgi is therefore a processing and sorting station, not a synthetic or energy-generating organelle.

Question 11Cell Biology and Cell Division

A student observes a cell with chromosomes aligned at the equatorial plate and distinct spindle fibres attached to each centromere. Which stage is being observed?

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

Explanation

Alignment of condensed chromosomes along the metaphase plate with spindle fibres attached to kinetochores is the hallmark of metaphase. In prophase chromosomes are still condensing and scattered, in anaphase they are moving apart, and in telophase nuclei are reforming. The described configuration therefore identifies metaphase.

Question 12Cell Biology and Cell Division

Which property of the plasma membrane is best explained by the presence of cholesterol in animal cells?

  1. It makes the membrane freely permeable to ions
  2. It modulates membrane fluidity across temperature changes Correct answer.
  3. It provides the main site of ATP synthesis
  4. It acts as the genetic material of the cell

Explanation

Cholesterol intercalates between phospholipid tails and buffers membrane fluidity, preventing rigid packing at low temperature and excessive fluidity at high temperature. It does not create ion channels, generate ATP or carry genetic information. Its role is therefore modulation of fluidity and stability.

Question 13Cell Biology and Cell Division

During meiosis, crossing over occurs between which structures?

  1. Two different bivalents
  2. Sister chromatids of the same chromosome
  3. Non-sister chromatids of homologous chromosomes Correct answer.
  4. Chromatids of non-homologous chromosomes

Explanation

Crossing over takes place at chiasmata between non-sister chromatids belonging to the two homologues of a bivalent during prophase I. Exchange between identical sister chromatids would not produce new allele combinations. Exchange between non-homologous chromosomes is a translocation, not meiotic crossing over.

Question 14Cell Biology and Cell Division

A cell has completed S phase but is prevented from entering mitosis. Which checkpoint has most likely failed to give clearance?

  1. G1/S checkpoint
  2. Spindle assembly checkpoint
  3. G2/M checkpoint Correct answer.
  4. Restriction point

Explanation

The G2/M checkpoint verifies that DNA replication is complete and undamaged before mitosis begins. Since S phase is already finished, the block on entry to mitosis implicates this checkpoint. The G1/S checkpoint and restriction point act before replication, and the spindle assembly checkpoint operates during mitosis itself.

Question 15Cell Biology and Cell Division

Which statement about the nuclear envelope is correct?

  1. It is a protein coat surrounding the nucleolus
  2. It is a double membrane perforated by nuclear pores Correct answer.
  3. It is a single membrane continuous with the endoplasmic reticulum
  4. It is absent in all metabolically active cells

Explanation

The nuclear envelope consists of two membranes separated by a perinuclear space, and it is interrupted by nuclear pores that allow exchange of molecules with the cytoplasm. The outer membrane is continuous with the rough endoplasmic reticulum. A single membrane, a protein coat, or complete absence would misdescribe this structure.

Question 16Cell Biology and Cell Division

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

  1. To produce offspring with a new combination of traits
  2. To determine whether the dominant phenotype is homozygous or heterozygous Correct answer.
  3. To induce mutations in the offspring
  4. To increase the number of dominant offspring

Explanation

Crossing with a homozygous recessive parent reveals the genotype of the dominant individual, because recessive alleles from the unknown parent cannot mask anything. A 1:1 ratio of dominant to recessive offspring indicates heterozygosity, while all dominant offspring indicate homozygosity. The cross is diagnostic rather than aimed at producing a particular phenotype.

Question 17Cell Biology and Cell Division

A cell is observed with a cell plate forming at its equator during cytokinesis. Which type of cell is this most likely to be?

  1. A fungal spore
  2. A plant cell Correct answer.
  3. An animal cell
  4. A bacterial cell

Explanation

Plant cells possess a rigid cellulose wall, so cytokinesis proceeds by the assembly of a cell plate from Golgi-derived vesicles at the equator, which later becomes the middle lamella. Animal cells divide by cleavage furrow formation. Bacteria divide by binary fission with septum formation, not a cell plate.

Question 18Cell Biology and Cell Division

A mutation prevents the formation of the synaptonemal complex during prophase I. Which outcome is most directly expected?

  1. DNA replication will be blocked in S phase
  2. Homologous chromosomes will fail to pair and recombine normally Correct answer.
  3. The mitotic spindle will fail to assemble
  4. Sister chromatids will fail to separate in anaphase II

Explanation

The synaptonemal complex is a protein scaffold that holds homologous chromosomes together during synapsis, allowing crossing over at chiasmata. Without it, homologues fail to pair and recombine properly, which can lead to mis-segregation. It has no direct role in sister chromatid separation, spindle assembly or DNA replication.

Question 19Cell Biology and Cell Division

Which experimental observation provided the strongest evidence that DNA, rather than protein, is the genetic material?

  1. The helical structure revealed by X-ray diffraction
  2. The transformation of non-virulent bacteria by heat-killed virulent bacteria Correct answer.
  3. The presence of DNA in the nucleus of eukaryotic cells
  4. The equal ratio of purines to pyrimidines in DNA

Explanation

The bacterial transformation experiments showed that a heat-stable substance from virulent cells could permanently change the heritable traits of non-virulent cells, and this substance was later identified as DNA. Base ratios, nuclear location and helical structure are consistent with DNA being genetic material but do not by themselves exclude protein. Transformation therefore provided the decisive functional evidence.

Question 20Cell Biology and Cell Division

A cell with a diploid number of 2n = 24 undergoes meiosis. How many chromosomes are present in each cell at metaphase II?

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

Explanation

Meiosis I reduces the chromosome number from 24 to 12 per cell, so each cell entering meiosis II has 12 chromosomes, each consisting of two sister chromatids. At metaphase II these 12 chromosomes align at the equator. The number 24 is the diploid count, 48 would be the replicated count immediately after S phase, and 6 is not related to this sequence.

Question 21Biomolecules and Enzymes

Which pair of monomers correctly matches the polymer it builds during condensation polymerisation in living cells?

  1. Fatty acids and starch
  2. Glucose and glycogen
  3. Nucleotides and cellulose
  4. Amino acids and proteins Correct answer.

Explanation

Proteins are polymers of amino acids joined by peptide bonds formed through condensation reactions with the loss of water. Each polymer must be matched with its true monomer, so the correct pairing is amino acids with proteins.

Question 22Biomolecules and Enzymes

A Benedict's test is performed on a fruit extract and a brick-red precipitate appears after heating. Which conclusion is justified?

  1. The extract contains only non-reducing polysaccharide
  2. The extract contains protein with peptide bonds
  3. The extract contains a reducing sugar such as glucose Correct answer.
  4. The extract contains sucrose but no monosaccharide

Explanation

Benedict's reagent contains copper(II) ions that are reduced to red copper(I) oxide by reducing sugars when heated. Glucose and other reducing monosaccharides give this brick-red precipitate, so the result indicates a reducing sugar.

Question 23Biomolecules and Enzymes

Which statement about the induced-fit model of enzyme action is correct?

  1. Enzymes lower activation energy by permanently altering the substrate
  2. The substrate changes shape to fit a fixed active site
  3. The active site moulds itself around the substrate as it binds Correct answer.
  4. The active site is a rigid lock that fits only one substrate shape

Explanation

In the induced-fit model the active site is flexible and adjusts its shape as the substrate binds, forming a complementary enzyme-substrate complex. This conformational change brings catalytic groups into position and strains bonds in the substrate.

Question 24Biomolecules and Enzymes

Which property of water arises mainly because water molecules form hydrogen bonds with one another?

  1. Its inability to act as a solvent for ions
  2. Its low boiling point compared with similar-sized molecules
  3. Its high specific heat capacity and high latent heat of vaporisation Correct answer.
  4. Its ability to dissolve non-polar lipids readily

Explanation

Extensive hydrogen bonding between water molecules requires a large amount of energy to break, giving water a high specific heat capacity and a high latent heat of vaporisation. These properties help organisms buffer temperature changes and cool by evaporation.

Question 25Biomolecules and Enzymes

Which bond links two adjacent amino acids in a polypeptide chain?

  1. Glycosidic bond
  2. Ester bond
  3. Phosphodiester bond
  4. Peptide bond Correct answer.

Explanation

A peptide bond forms between the carboxyl group of one amino acid and the amino group of the next, releasing water in a condensation reaction. Repeated peptide bonds create the primary structure of a polypeptide.

Question 26Biomolecules and Enzymes

An enzyme-catalysed reaction is studied while substrate concentration is increased with enzyme concentration kept constant. Which change is expected as substrate concentration continues to rise?

  1. Rate increases and then levels off as active sites become saturated Correct answer.
  2. Rate remains unchanged from the start because enzyme concentration is fixed
  3. Rate falls steadily because excess substrate blocks the active sites
  4. Rate rises indefinitely in direct proportion to substrate concentration

Explanation

At first the rate rises with substrate concentration because more active sites are occupied. Once every active site is saturated, the maximum rate is reached and further substrate gives no further increase.

Question 27Biomolecules and Enzymes

A competitive inhibitor is added to an enzyme-catalysed reaction. Which effect is observed when substrate concentration is very high?

  1. The inhibitor raises the activation energy by binding only to the product
  2. The inhibitor resembles the substrate and competes for the active site, so inhibition is reduced by excess substrate Correct answer.
  3. The inhibitor binds the enzyme-substrate complex and lowers the maximum rate
  4. The inhibitor binds the allosteric site and permanently denatures the enzyme

Explanation

A competitive inhibitor has a shape similar to the substrate and binds reversibly at the active site. Raising substrate concentration outcompetes the inhibitor, so the maximum rate can still be reached.

Question 28Biomolecules and Enzymes

Which statement best explains why an enzyme is denatured by heating far above its optimum temperature?

  1. Hydrogen and ionic bonds maintaining the tertiary structure break, altering the active site Correct answer.
  2. Peptide bonds are hydrolysed, breaking the chain into amino acids
  3. The enzyme is converted into a competitive inhibitor of itself
  4. The substrate gains kinetic energy and no longer collides with the enzyme

Explanation

High temperature disrupts the weak hydrogen and ionic interactions that hold the tertiary structure, so the active site loses its complementary shape. The substrate can no longer bind, and the loss of activity is usually irreversible.

Question 29Biomolecules and Enzymes

In a DNA molecule, if adenine makes up 30% of the bases, what percentage of the bases is guanine?

  1. 70%
  2. 40%
  3. 30%
  4. 20% Correct answer.

Explanation

By complementary base pairing, adenine equals thymine, so thymine is also 30% and together they make 60% of the bases. The remaining 40% is shared equally by guanine and cytosine, giving guanine 20%.

Question 30Biomolecules and Enzymes

Which structural feature distinguishes a phospholipid from a triglyceride?

  1. Phospholipids contain amino acids joined by peptide bonds
  2. Phospholipids lack glycerol and contain only a phosphate group
  3. Phospholipids have a phosphate-containing polar head group and two fatty acid tails Correct answer.
  4. Phospholipids contain three fatty acid chains attached to glycerol

Explanation

A phospholipid has glycerol linked to two fatty acid tails and a phosphate-containing group, giving it a hydrophilic head and hydrophobic tails. A triglyceride instead carries three fatty acid chains and is entirely non-polar.

Question 31Biomolecules and Enzymes

A solution of starch is incubated with salivary amylase at 37 degrees Celsius. Which observation after testing with iodine solution indicates that hydrolysis is complete?

  1. A brick-red precipitate forms, showing reducing sugar
  2. A blue-black colour persists, showing starch remains
  3. A violet colour appears, showing peptide bonds have formed
  4. No blue-black colour appears, showing starch has been broken down Correct answer.

Explanation

Iodine gives a blue-black colour with intact starch because of the helical amylose-iodine complex. When amylase hydrolyses the starch to maltose, no such complex forms, so the absence of blue-black colour indicates completion.

Question 32Biomolecules and Enzymes

Which level of protein structure is directly stabilised by disulphide bridges between cysteine residues?

  1. Primary structure
  2. Secondary structure
  3. Tertiary structure Correct answer.
  4. The sequence of nucleotides

Explanation

Disulphide bridges are covalent cross-links between the sulphur-containing groups of cysteine residues that help hold the folded three-dimensional shape. This folded shape is the tertiary structure of the protein.

Question 33Biomolecules and Enzymes

A student finds that an enzyme works fastest at pH 2 and loses activity sharply at pH 7. Which interpretation is most reasonable?

  1. The enzyme is inhibited competitively by hydrogen ions at pH 2
  2. The enzyme is adapted to a strongly acidic environment such as the stomach Correct answer.
  3. The enzyme is a typical intracellular enzyme with an optimum near neutral pH
  4. The enzyme has no ionisable groups in its active site

Explanation

An optimum near pH 2 suggests the enzyme normally functions in a strongly acidic compartment, such as gastric juice. At pH 7 the change in ionisation disrupts ionic bonds and the active site, sharply reducing activity.

Question 34Biomolecules and Enzymes

A polysaccharide is found to be unbranched, insoluble in water and composed of beta-glucose units joined by 1,4-glycosidic bonds. Which polymer is described?

  1. Sucrose
  2. Glycogen
  3. Starch
  4. Cellulose Correct answer.

Explanation

Cellulose is an unbranched polymer of beta-glucose linked by 1,4-glycosidic bonds, and its straight chains form hydrogen-bonded microfibrils that make it insoluble and strong. This suits its structural role in plant cell walls.

Question 35Biomolecules and Enzymes

An enzyme follows Michaelis-Menten behaviour. When substrate concentration equals the Michaelis constant, Km, which statement about the reaction is correct?

  1. The reaction rate equals half of the maximum rate Correct answer.
  2. The inhibitor has displaced all substrate from the active site
  3. The reaction rate equals the maximum rate
  4. The enzyme is fully saturated with substrate

Explanation

By definition, Km is the substrate concentration at which the reaction proceeds at half of its maximum velocity. At this point half of the active sites are occupied on average, so the rate is Vmax divided by two.

Question 36Biomolecules and Enzymes

Which combination correctly pairs a biological molecule with the element that it always contains in addition to carbon, hydrogen and oxygen?

  1. Carbohydrate with nitrogen
  2. Nucleic acid with nitrogen and phosphorus Correct answer.
  3. Protein with phosphorus
  4. Triglyceride with sulphur

Explanation

Nucleic acids are built from nucleotides that contain a nitrogenous base and a phosphate group, so they always contain nitrogen and phosphorus as well as carbon, hydrogen and oxygen. Proteins contain nitrogen and sometimes sulphur, while pure carbohydrates and triglycerides contain only carbon, hydrogen and oxygen.

Question 37Genetics and Molecular Basis of Inheritance

In the classical monohybrid cross of Mendel using pure tall (TT) and pure dwarf (tt) pea plants, what phenotypic ratio appears in the F2 generation?

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

Explanation

The F1 (Tt) is all tall because tallness is dominant. Selfing F1 gives genotypic ratio 1 TT : 2 Tt : 1 tt, so three-quarters of the plants show the dominant tall phenotype and one-quarter show dwarf. Hence the classic 3:1 phenotypic ratio.

Question 38Genetics and Molecular Basis of Inheritance

A cross between a heterozygous tall pea plant and a dwarf pea plant produces progeny in which the expected proportion of dwarf individuals is

  1. three quarters
  2. zero
  3. one half Correct answer.
  4. one quarter

Explanation

The heterozygote Tt forms T and t gametes in equal numbers, while the dwarf tt forms only t gametes. Fertilisation gives 1 Tt : 1 tt, so half the offspring are dwarf. This is the diagnostic 1:1 outcome of a test cross.

Question 39Genetics and Molecular Basis of Inheritance

Which term describes an organism carrying two identical alleles of a gene at a given locus?

  1. Hemizygous
  2. Heterozygous
  3. Codominant
  4. Homozygous Correct answer.

Explanation

Homozygosity means both homologous chromosomes carry the same allele, whether dominant (TT) or recessive (tt). Such individuals breed true for that trait. The other terms describe different allele constitutions or expression patterns.

Question 40Genetics and Molecular Basis of Inheritance

In a dihybrid cross between plants heterozygous for two independently assorting gene pairs (RrYy × RrYy), the expected number of F2 phenotypes is

  1. two
  2. sixteen
  3. four Correct answer.
  4. eight

Explanation

Each gene pair contributes two phenotypes, and independent assortment multiplies the possibilities, giving 2 × 2 = 4 phenotypic classes in the ratio 9:3:3:1. Sixteen refers to the number of genotypic combinations, not phenotypes.

Question 41Genetics and Molecular Basis of Inheritance

Which nitrogenous base is found in RNA but not in DNA?

  1. Thymine
  2. Uracil Correct answer.
  3. Adenine
  4. Guanine

Explanation

RNA contains uracil in place of the thymine found in DNA; both molecules share adenine, guanine and cytosine. Uracil pairs with adenine during transcription and translation. Thymine is characteristic of DNA.

Question 42Genetics and Molecular Basis of Inheritance

According to the Watson–Crick model, the two strands of a DNA double helix are held together mainly by

  1. peptide bonds between sugar residues
  2. hydrogen bonds between complementary bases Correct answer.
  3. phosphodiester bonds between adjacent bases
  4. ionic bonds between phosphate groups

Explanation

Adenine pairs with thymine by two hydrogen bonds and guanine with cytosine by three, holding the two antiparallel strands together. The sugar-phosphate backbone is joined by phosphodiester bonds along each strand. Hydrogen bonding between bases is the key stabilising interaction across the helix.

Question 43Genetics and Molecular Basis of Inheritance

A DNA sample contains 30% adenine. What percentage of its bases is guanine?

  1. 30%
  2. 70%
  3. 40%
  4. 20% Correct answer.

Explanation

By Chargaff's rule A equals T, so thymine is also 30%, making A + T = 60%. The remaining 40% is shared equally by G and C, so guanine is 20%. Base pairing therefore fixes the composition once one base percentage is known.

Question 44Genetics and Molecular Basis of Inheritance

In the semi-conservative replication of DNA, each daughter molecule contains

  1. two parental strands
  2. fragments of both parental strands only
  3. one parental strand and one newly synthesised strand Correct answer.
  4. two newly synthesised strands

Explanation

Meselson and Stahl's experiment showed that each new DNA duplex retains one intact parental strand paired with a newly made complementary strand. This is the meaning of semi-conservative replication. It explains why the parental label is diluted by half in each generation.

Question 45Genetics and Molecular Basis of Inheritance

The enzyme that relieves the supercoiling ahead of the replication fork during DNA replication is

  1. Primase
  2. DNA topoisomerase Correct answer.
  3. DNA polymerase III
  4. DNA ligase

Explanation

As the fork advances, torsional stress builds up ahead of it, and topoisomerase nicks and reseals the backbone to relieve this strain. Ligase instead joins Okazaki fragments, polymerase extends the chain and primase lays down RNA primers. Each enzyme has a distinct role at the fork.

Question 46Genetics and Molecular Basis of Inheritance

During transcription, the DNA strand that is copied into mRNA is called the

  1. leading strand
  2. template strand Correct answer.
  3. coding strand
  4. sense strand

Explanation

RNA polymerase reads the template strand in the 3′ to 5′ direction and builds mRNA complementary to it. The coding or sense strand has the same sequence as the mRNA except that T replaces U. Leading strand refers to replication, not transcription.

Question 47Genetics and Molecular Basis of Inheritance

A DNA triplet reads 5′-TAC-3′ on the template strand. Which mRNA codon is transcribed from it?

  1. 5′-ATG-3′
  2. 5′-GUA-3′
  3. 5′-UAC-3′
  4. 5′-AUG-3′ Correct answer.

Explanation

Transcription follows complementary base pairing, so T pairs with A, A with U and C with G, giving AUG. This codon also serves as the start codon in translation. Remembering that RNA uses uracil in place of thymine is essential here.

Question 48Genetics and Molecular Basis of Inheritance

Which feature of the genetic code explains why a point mutation in the third base of a codon often leaves the amino acid unchanged?

  1. Non-overlapping nature of codons
  2. Degeneracy of the code Correct answer.
  3. Comma-less reading of codons
  4. Universality of the code

Explanation

Degeneracy means several codons specify the same amino acid, so a change in the third base frequently still codes for the same residue. This is called a silent mutation. The other options describe how codons are arranged and read, not redundancy.

Question 49Genetics and Molecular Basis of Inheritance

A frameshift mutation is most likely produced by

  1. insertion of a single nucleotide Correct answer.
  2. deletion of three consecutive nucleotides
  3. substitution of a pyrimidine by a purine
  4. substitution of one purine by another

Explanation

Adding one base shifts the reading frame of every downstream codon, altering the whole protein. Transitions and transversions change only a single codon, and a three-base deletion removes one whole codon without shifting the frame. Hence only single-base indels cause frameshifts.

Question 50Genetics and Molecular Basis of Inheritance

In the lac operon of E. coli, the repressor protein is encoded by

  1. the lacZ gene
  2. the promoter region
  3. the lacI gene Correct answer.
  4. the operator region

Explanation

The lacI gene lies upstream of the operon and produces the repressor that binds the operator in the absence of lactose. The lacZ gene codes for β-galactosidase, while the operator and promoter are regulatory DNA sequences rather than protein-coding genes. This distinction is central to negative control of the operon.