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STPM Biology 964 study materials
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STPM Biology 964 practice questions

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Question 1Biological Molecules and Cell Biology

Which bond is primarily responsible for linking two monosaccharides together in a disaccharide such as maltose?

  1. Peptide bond formed by condensation
  2. Ester bond formed by hydrolysis
  3. Glycosidic bond formed by condensation Correct answer.
  4. Phosphodiester bond formed by condensation

Explanation

Two monosaccharides join when a hydroxyl group of one reacts with the anomeric carbon of the other, releasing water and forming a glycosidic bond. This condensation reaction is the standard linkage in disaccharides and polysaccharides. Peptide and phosphodiester bonds belong to proteins and nucleic acids respectively.

Question 2Biological Molecules and Cell Biology

A triglyceride is described as a saturated fat. Which structural feature accounts for this description?

  1. The molecule carries a phosphate group
  2. Glycerol is esterified to three fatty acids
  3. Fatty acid chains contain one cis double bond
  4. Fatty acid chains contain no carbon-carbon double bonds Correct answer.

Explanation

Saturation refers to the hydrocarbon chains being fully saturated with hydrogen, so no C=C double bonds are present. This gives straight chains that pack closely and are solid at room temperature. The presence of a double bond would make the fat unsaturated.

Question 3Biological Molecules and Cell Biology

Which level of protein structure is stabilised mainly by hydrophobic interactions and disulfide bridges between distant parts of a single polypeptide?

  1. Secondary structure
  2. Primary structure
  3. Tertiary structure Correct answer.
  4. Quaternary structure

Explanation

Tertiary structure is the overall three-dimensional folding of one polypeptide chain, held by hydrophobic interactions, hydrogen bonds, ionic bonds and disulfide bridges between residues far apart in the sequence. Secondary structure concerns local alpha helices and beta pleated sheets. Quaternary structure involves more than one polypeptide chain.

Question 4Biological Molecules and Cell Biology

Which property of water allows insects such as pond skaters to move across its surface?

  1. High latent heat of vaporisation
  2. Low density in the solid state
  3. High specific heat capacity
  4. High surface tension from hydrogen bonding Correct answer.

Explanation

At an air-water interface, surface water molecules hydrogen bond strongly to neighbours below and to the sides but not above, producing an inward pull that creates surface tension. This film can support small objects of low mass. The other properties listed are real but do not explain surface support.

Question 5Biological Molecules and Cell Biology

Which organelle is the principal site of ATP synthesis by oxidative phosphorylation in a eukaryotic cell?

  1. Lysosome
  2. Rough endoplasmic reticulum
  3. Mitochondrion Correct answer.
  4. Golgi apparatus

Explanation

The inner mitochondrial membrane carries the electron transport chain and ATP synthase, so oxidative phosphorylation occurs there. The folding into cristae increases the membrane area available. Other organelles listed perform synthesis, modification or hydrolysis rather than bulk ATP production.

Question 6Biological Molecules and Cell Biology

Which structure controls the movement of substances between the nucleus and the cytoplasm?

  1. Nuclear envelope pores Correct answer.
  2. Nucleolus
  3. Nuclear lamina
  4. Chromatin fibres

Explanation

The nuclear envelope is perforated by pores formed from protein complexes that regulate transport of ions, RNA and proteins across the double membrane. The nucleolus is a site of ribosome subunit assembly, and chromatin is the packaged DNA. The lamina provides internal support.

Question 7Biological Molecules and Cell Biology

Which feature distinguishes prokaryotic cells from eukaryotic cells?

  1. Absence 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 nuclear envelope, so their circular DNA lies free in a nucleoid region rather than inside a membrane-bound nucleus. Ribosomes, plasma membranes and DNA are found in both cell types. This makes the absence of a true nucleus the defining difference in this list.

Question 8Biological Molecules and Cell Biology

In the fluid mosaic model, which component is described as forming a bilayer with hydrophilic heads facing the aqueous environment?

  1. Integral proteins
  2. Cholesterol molecules
  3. Phospholipids Correct answer.
  4. Peripheral glycoproteins

Explanation

Phospholipids are amphipathic: their phosphate heads are hydrophilic while their fatty acid tails are hydrophobic. In water they self-assemble into a bilayer with heads outward and tails inward. Proteins and cholesterol are embedded in or attached to this bilayer rather than forming it.

Question 9Biological Molecules and Cell Biology

An enzyme-catalysed reaction is carried out at a substrate concentration far above the Michaelis constant. What happens to the reaction rate when substrate concentration is doubled?

  1. It increases indefinitely in proportion to substrate concentration
  2. It remains essentially unchanged because the enzyme is saturated Correct answer.
  3. It halves because of substrate inhibition
  4. It approximately doubles because substrate is limiting

Explanation

When substrate concentration greatly exceeds the Michaelis constant, most active sites are occupied, so the enzyme is close to saturation and works at or near maximum velocity. Adding more substrate cannot increase the rate much because active site availability, not substrate, is limiting. Only at low substrate concentrations is rate roughly proportional to substrate concentration.

Question 10Biological Molecules and Cell Biology

A competitive inhibitor is added to an enzyme assay. Which change in kinetic parameters is expected?

  1. Maximum velocity is unchanged while the apparent Michaelis constant increases Correct answer.
  2. Both maximum velocity and Michaelis constant rise
  3. Maximum velocity falls while the Michaelis constant is unchanged
  4. Both maximum velocity and Michaelis constant fall

Explanation

A competitive inhibitor binds the free enzyme at the active site, so higher substrate concentrations can displace it and the maximum velocity is still attainable. However, more substrate is needed to reach half maximum velocity, so the apparent Michaelis constant increases. Non-competitive inhibition is the pattern that lowers maximum velocity.

Question 11Biological Molecules and Cell Biology

A cell is placed in a solution with a lower water potential than its cytoplasm. Which event is expected?

  1. Water leaves the cell and it may become plasmolysed Correct answer.
  2. Water enters the cell and it becomes turgid
  3. Solutes move actively out of the cell to equalise water potential
  4. No net water movement occurs because the cell is isotonic

Explanation

Water moves by osmosis from higher (less negative) to lower (more negative) water potential. The external solution has lower water potential, so water leaves the cell, and in a plant cell the protoplast may shrink away from the wall, causing plasmolysis. Solute movement does not normally equalise the situation quickly.

Question 12Biological Molecules and Cell Biology

Which process allows a white blood cell to engulf a bacterium without the membrane passing through the cytoplasm?

  1. Phagocytosis involving membrane invagination Correct answer.
  2. Active transport by a sodium-potassium pump
  3. Simple diffusion through the bilayer
  4. Facilitated diffusion via channel proteins

Explanation

Phagocytosis is a form of endocytosis in which the plasma membrane invaginates around the particle, then pinches off to form a vesicle inside the cell. The membrane forms the vesicle boundary rather than crossing the cytoplasm. The other options move small molecules across the membrane without engulfing particles.

Question 13Biological Molecules and Cell Biology

In an experiment, a plant tissue is immersed in solutions of different sucrose concentrations and the change in mass is measured. Which conclusion is valid when the tissue shows no change in mass?

  1. The cell walls have become impermeable
  2. The tissue has lost all its solutes
  3. The external solution is isotonic to the cell sap Correct answer.
  4. Active transport has stopped

Explanation

No change in mass indicates no net water movement, which occurs when the external solution and the cell sap have the same water potential. This is the isotonic point for the tissue. It does not mean solutes have been lost or transport has stopped.

Question 14Biological Molecules and Cell Biology

Which statement correctly describes the role of NAD in respiration?

  1. It is the final electron acceptor at the end of the electron transport chain
  2. It phosphorylates ADP directly in the Krebs cycle
  3. It hydrolyses ATP to release energy for glycolysis
  4. It accepts electrons and hydrogen at glycolysis and the Krebs cycle and delivers them to the electron transport chain Correct answer.

Explanation

NAD is reduced to NADH when it accepts electrons and hydrogen during glycolysis, the link reaction and the Krebs cycle. The reduced coenzyme then carries these to the electron transport chain, where it is reoxidised. Oxygen, not NAD, is the final electron acceptor.

Question 15Biological Molecules and Cell Biology

Which feature of the chloroplast is the site of the light-independent reactions?

  1. Stroma Correct answer.
  2. Thylakoid membrane
  3. Intermembrane space
  4. Outer chloroplast envelope

Explanation

The light-independent reactions, including the Calvin cycle, occur in the stroma, the fluid surrounding the thylakoids, where the enzymes such as rubisco are located. The thylakoid membranes host the light-dependent reactions with photosystems and ATP synthase. The envelope and intermembrane space are not sites of carbon fixation.

Question 16Biological Molecules and Cell Biology

During the light-dependent reactions, what is the immediate source of the hydrogen used to reduce NADP?

  1. Breakdown of glucose
  2. Reduction of carbon dioxide
  3. Hydrolysis of ATP
  4. Photolysis of water Correct answer.

Explanation

Photolysis of water in the thylakoid lumen splits water into protons, electrons and oxygen. The electrons replace those lost by photosystem II, and the protons help reduce NADP to NADPH. Carbon dioxide reduction occurs later in the Calvin cycle.

Question 17Biological Molecules and Cell Biology

A cell in metaphase of mitosis is examined. Which description is correct?

  1. Chromosomes are aligned at the equator and attached to spindle fibres Correct answer.
  2. Chromatin is condensing but no spindle has formed
  3. The nuclear envelope is re-forming around two nuclei
  4. Sister chromatids have separated and are moving to opposite poles

Explanation

In metaphase, condensed chromosomes line up along the metaphase plate at the cell equator, each attached to spindle fibres from opposite poles. Chromatid separation defines anaphase, and nuclear envelope re-formation occurs in telophase. Condensation without a spindle describes prophase.

Question 18Biological Molecules and Cell Biology

Which event is characteristic of crossing over during meiosis?

  1. Homologous chromosomes fail to separate during anaphase I
  2. Sister chromatids separate during anaphase II
  3. Non-sister chromatids of homologous chromosomes exchange segments at chiasmata Correct answer.
  4. DNA is replicated before meiosis I

Explanation

Crossing over occurs in prophase I when non-sister chromatids of homologous chromosomes break and rejoin at points called chiasmata, exchanging corresponding segments. This produces recombinant chromatids and increases genetic variation. Sister chromatid separation and DNA replication are separate events.

Question 19Biological Molecules and Cell Biology

A mutation changes a single base in a coding sequence so that one amino acid is replaced by another. Which term describes this change?

  1. Silent mutation
  2. Missense mutation Correct answer.
  3. Frameshift mutation
  4. Nonsense mutation

Explanation

A missense mutation alters a codon so that a different amino acid is incorporated into the polypeptide. A silent mutation changes a base but not the amino acid, and a nonsense mutation produces a premature stop codon. A frameshift arises from insertion or deletion of bases not a multiple of three.

Question 20Biological Molecules and Cell Biology

Which statement best explains why an enzyme is described as having a high degree of specificity?

  1. It works only at high temperatures
  2. Its active site is complementary in shape and charge to a particular substrate Correct answer.
  3. It lowers activation energy for any reaction it contacts
  4. It is consumed during the reaction it catalyses

Explanation

Specificity arises because the active site has a shape and chemical character complementary to a particular substrate, so only that substrate or a closely related molecule binds effectively. Lowering activation energy is a general property of catalysis, not specificity. Enzymes are not consumed and are not restricted to high temperatures.

Question 21Biological Molecules and Cell Biology

A polypeptide has 120 amino acids. Assuming no post-translational modification, what is the minimum number of nucleotides in the coding region of its messenger RNA, excluding the stop codon?

  1. 360 Correct answer.
  2. 40
  3. 480
  4. 120

Explanation

Each amino acid is specified by a codon of three nucleotides, so 120 amino acids require 120 multiplied by three, which is 360 nucleotides. This excludes any stop codon and untranslated regions. The figure of 40 would apply if one nucleotide coded for one amino acid, which is not the case.

Question 22Biological Molecules and Cell Biology

In a mesophyll cell, a substance that uncouples the electron transport chain from ATP synthesis is applied. Which immediate effect is expected?

  1. Oxygen consumption stops completely
  2. Carbon dioxide fixation increases sharply
  3. NADPH production increases proportionally
  4. The proton gradient across the thylakoid membrane is dissipated and ATP synthesis falls Correct answer.

Explanation

Uncouplers allow protons to leak across the membrane without passing through ATP synthase, dissipating the proton gradient that drives ATP synthesis. Electron transport may continue, so oxygen consumption need not stop. ATP supply limits the Calvin cycle, so carbon fixation would not rise sharply.

Question 23Biological Molecules and Cell Biology

A protein is synthesised on the rough endoplasmic reticulum and secreted from the cell. Which route describes its movement?

  1. Free ribosome to nucleus to Golgi apparatus to plasma membrane
  2. Rough endoplasmic reticulum to mitochondrion to vesicle to plasma membrane
  3. Rough endoplasmic reticulum to Golgi apparatus to secretory vesicle to plasma membrane Correct answer.
  4. Golgi apparatus to rough endoplasmic reticulum to lysosome to plasma membrane

Explanation

Secretory proteins enter the rough endoplasmic reticulum during synthesis, are transported in vesicles to the Golgi apparatus for modification and sorting, then packaged into secretory vesicles that fuse with the plasma membrane. This is the standard secretory pathway.

Question 24Biological Molecules and Cell Biology

A student compares a transmission electron micrograph with an optical micrograph of the same tissue. Which advantage is specific to the electron micrograph?

  1. It resolves organelles such as ribosomes because of its much shorter wavelength Correct answer.
  2. It can be used to observe living cells in real time
  3. It shows the natural colour of the specimen
  4. It requires only simple staining with iodine solution

Explanation

Electron microscopes use a beam of electrons whose wavelength is far shorter than that of visible light, giving much higher resolution and allowing very small structures such as ribosomes to be distinguished. Optical microscopes are limited by the wavelength of light. Electron microscopy requires fixed, dehydrated specimens and does not show natural colour.

Question 25Biological Molecules and Cell Biology

A length of DNA has the base sequence 5'-ATGCGT-3' on one strand. What is the sequence of the complementary strand read in the same 5' to 3' direction?

  1. ACGCAT Correct answer.
  2. UACGCA
  3. ATGCGT
  4. TACGCA

Explanation

The complementary strand is antiparallel and pairs A with T and G with C. Pairing the given strand 5'-ATGCGT-3' gives 3'-TACGCA-5', which read in the 5' to 3' direction is ACGCAT. Option TACGCA would be the complement read in the wrong direction.

Question 26Biological Molecules and Cell Biology

Which bond is formed between the carboxyl group of one amino acid and the amino group of the next during protein synthesis, and what is released in the process?

  1. A peptide bond, with the release of water Correct answer.
  2. An ionic bond, with the release of ammonia
  3. A glycosidic bond, with the release of water
  4. A hydrogen bond, with the release of carbon dioxide

Explanation

A condensation reaction joins the carboxyl carbon of one amino acid to the amino nitrogen of the next, forming a covalent peptide bond. Each condensation removes the elements of water, so water is the only by-product of this linkage.

Question 27Biological Molecules and Cell Biology

A triglyceride is hydrolysed completely. Which set of products is obtained?

  1. Three glycerol molecules and one fatty acid molecule
  2. One glycerol and three phosphate groups
  3. One glycerol and three fatty acid molecules Correct answer.
  4. One glycerol and three amino acid molecules

Explanation

A triglyceride consists of a glycerol molecule esterified to three fatty acid chains. Complete hydrolysis breaks the three ester bonds, releasing one glycerol and three fatty acids.

Question 28Biological Molecules and Cell Biology

Which property of water directly explains its high latent heat of vaporisation and its value in transpiration cooling?

  1. The low density of water compared with ice
  2. The non-polar nature of the oxygen-hydrogen bond
  3. The small size of the water molecule
  4. Extensive hydrogen bonding between water molecules Correct answer.

Explanation

Each water molecule can hydrogen bond to several neighbours, and many of these bonds must be broken before a molecule escapes as vapour. This makes evaporation highly endothermic, so transpiration removes large amounts of heat from the leaf.

Question 29Biological Molecules and Cell Biology

In a biochemical test, a sample is boiled with Benedict's solution and a brick-red precipitate forms. Which conclusion is justified?

  1. A lipid is present in the sample
  2. A non-reducing sugar such as sucrose is present in the sample
  3. A reducing sugar is present in the sample Correct answer.
  4. A protein is present in the sample

Explanation

Benedict's reagent contains blue copper(II) ions that are reduced to insoluble copper(I) oxide by reducing sugars on heating, giving a brick-red precipitate. A positive result therefore indicates a reducing sugar such as glucose or maltose.

Question 30Biological Molecules and Cell Biology

A solution of glucose is separated from pure water by a membrane permeable only to water. Which change occurs as the system reaches equilibrium?

  1. Both water and glucose move until the volumes are equal
  2. Glucose moves across the membrane until concentrations are equal
  3. Water moves out of the glucose solution, concentrating it further
  4. Water moves into the glucose solution and its hydrostatic pressure rises Correct answer.

Explanation

The glucose solution has a lower water potential than pure water, so water moves by osmosis into the solution. Because the membrane is impermeable to glucose, the incoming water raises the hydrostatic pressure until this opposes further net movement.

Question 31Biological Molecules and Cell Biology

Which combination of features describes facilitated diffusion across a plasma membrane?

  1. Requires ATP and moves solute against its concentration gradient
  2. Uses vesicles and requires ATP hydrolysis
  3. Moves only non-polar molecules directly through the phospholipid bilayer
  4. Uses channel or carrier proteins and moves solute down its concentration gradient Correct answer.

Explanation

Facilitated diffusion uses transmembrane proteins to allow polar or charged solutes to pass, and it is passive, so net movement is down the concentration gradient. It differs from active transport because no metabolic energy is required.

Question 32Biological Molecules and Cell Biology

A polysaccharide is composed of beta-glucose units joined by 1,4-glycosidic bonds with alternate units inverted. Which property results from this structure?

  1. Branching every ten residues to give many free ends
  2. Formation of a compact coiled helix that stains blue-black with iodine
  3. Ready solubility in water and rapid hydrolysis by amylase
  4. Formation of straight chains held by hydrogen bonds into strong microfibrils Correct answer.

Explanation

Inversion of alternate beta-glucose residues produces straight chains that can lie parallel and be cross-linked by many hydrogen bonds, forming microfibrils of high tensile strength. This is the structure of cellulose, suited to a structural role in plant cell walls.

Question 33Biological Molecules and Cell Biology

In an enzyme-catalysed reaction, raising the temperature from 20 degrees Celsius to 30 degrees Celsius increases the rate, but raising it from 60 degrees Celsius to 70 degrees Celsius decreases the rate. Which explanation accounts for the decrease at the higher range?

  1. The active site loses its specific shape as hydrogen and ionic bonds break Correct answer.
  2. The enzyme becomes a competitive inhibitor of itself
  3. The activation energy of the reaction increases with temperature
  4. Substrate molecules lose kinetic energy at high temperature

Explanation

Above the optimum, increased thermal vibration disrupts the hydrogen and ionic bonds maintaining tertiary structure, so the active site is no longer complementary to the substrate. Denaturation reduces the number of functional enzyme-substrate complexes and the rate falls.

Question 34Biological Molecules and Cell Biology

A length of DNA contains 30 percent adenine. What percentage of its bases is guanine?

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

Explanation

By complementary base pairing, the percentage of thymine equals that of adenine, so A plus T is 60 percent. The remaining 40 percent is shared equally by guanine and cytosine, giving 20 percent guanine.

Question 35Biological Molecules and Cell Biology

During the formation of a mature mRNA in a eukaryotic nucleus, which modification occurs before the mRNA leaves for the cytoplasm?

  1. Replication of the mRNA to produce a double-stranded molecule
  2. Attachment of amino acids to form a polypeptide
  3. Conversion of uracil to thymine throughout the molecule
  4. Removal of introns and joining of exons Correct answer.

Explanation

Post-transcriptional modification splices out non-coding introns and joins exons, and a 5-prime cap and poly-A tail are added. Only the processed mature mRNA is exported through the nuclear pores for translation.

Question 36Biological Molecules and Cell Biology

A protein has a quaternary structure. Which statement about this level of organisation is correct?

  1. It describes the linear sequence of amino acids joined by peptide bonds
  2. It consists of two or more polypeptide chains held together by bonds between their side chains Correct answer.
  3. It refers to the folding of a single chain into alpha helices and beta pleated sheets
  4. It is the term for the prosthetic group attached to a single polypeptide

Explanation

Quaternary structure exists when a functional protein comprises more than one polypeptide subunit, such as haemoglobin with four globin chains. The subunits are held together by hydrogen bonds, ionic bonds and hydrophobic interactions between side chains.

Question 37Biological Molecules and Cell Biology

A student views a specimen at a total magnification of 400 times using an eyepiece graticule. The graticule scale has 100 divisions, and 40 divisions correspond to the full diameter of a field of view known to be 0.45 mm. What is the length represented by one graticule division?

  1. 0.045 mm
  2. 0.0113 mm
  3. 0.0045 mm Correct answer.
  4. 0.180 mm

Explanation

Each division equals the field diameter divided by the number of divisions spanning it, that is 0.45 mm divided by 40, which is 0.0113 mm. Care must be taken to check whether the stated divisions span the whole field or only part of it before calibrating.

Question 38Biological Molecules and Cell Biology

A cell is treated with a drug that prevents the assembly of spindle microtubules. Which stage of mitosis is directly prevented from proceeding normally?

  1. Prophase, because chromosomes fail to condense
  2. Telophase, because nuclear envelopes cannot reform
  3. Metaphase, because chromosomes cannot align at the equator Correct answer.
  4. Cytokinesis, because the cell plate cannot form

Explanation

Spindle microtubules attach to kinetochores and move chromosomes to the metaphase plate. Without microtubule assembly, chromosomes cannot align correctly at the equator, so metaphase and the subsequent separation of chromatids are disrupted.

Question 39Biological Molecules and Cell Biology

In an experiment, mitochondria are supplied with an electron transport inhibitor that blocks electron flow from the carriers to oxygen, while substrate-level phosphorylation continues. Which observation is expected?

  1. The proton gradient across the inner membrane increases indefinitely, producing more ATP
  2. ATP synthesis by oxidative phosphorylation stops, but some ATP is still produced by substrate-level phosphorylation Correct answer.
  3. Glycolysis in the cytoplasm is immediately inhibited by the drug
  4. ATP synthesis increases because oxygen is conserved

Explanation

Oxidative phosphorylation depends on electron flow through the chain to pump protons and drive ATP synthase. Blocking electron transfer to oxygen collapses the proton gradient, so ATP synthesis by the chain stops, while substrate-level phosphorylation in glycolysis and the Krebs cycle can continue.

Question 40Biological Molecules and Cell Biology

A solution contains a competitive inhibitor and a non-competitive inhibitor of the same enzyme. Which statement correctly distinguishes their effects when substrate concentration is increased?

  1. The effect of the competitive inhibitor is reduced, while the effect of the non-competitive inhibitor remains Correct answer.
  2. Both inhibitors become more effective as substrate concentration rises
  3. The effect of the non-competitive inhibitor is reduced, while the competitive inhibitor acts permanently
  4. Neither inhibitor is affected by substrate concentration because both bind the active site

Explanation

A competitive inhibitor competes with substrate for the active site, so a high substrate concentration outcompetes it and the inhibition is reduced. A non-competitive inhibitor binds a site other than the active site and changes enzyme shape, so raising substrate concentration does not restore activity.

Question 41Genetics and Molecular Biology

In the lac operon of Escherichia coli, which statement best describes the role of the repressor protein encoded by the lacI gene?

  1. It binds the promoter and recruits RNA polymerase to initiate transcription.
  2. It cleaves lactose into glucose and galactose to relieve repression.
  3. It binds the operator region and blocks transcription when lactose is absent. Correct answer.
  4. It binds the CAP site and enhances transcription in the presence of glucose.

Explanation

The lacI product is a repressor that binds the operator and physically blocks RNA polymerase from transcribing the structural genes when lactose is absent. When lactose is present, its isomer allolactose binds the repressor, reducing its affinity for the operator and allowing transcription.

Question 42Genetics and Molecular Biology

A molecule of DNA contains 22% adenine. What percentage of its bases is guanine?

  1. 56%
  2. 22%
  3. 44%
  4. 28% Correct answer.

Explanation

Chargaff's rules state that adenine equals thymine, so thymine is also 22%, and together A + T = 44%. The remaining 56% is shared equally between guanine and cytosine, giving guanine 28%.

Question 43Genetics and Molecular Biology

Which enzyme relieves the supercoiling ahead of the replication fork during DNA replication in bacteria?

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

Explanation

DNA gyrase is a type II topoisomerase that introduces negative supercoils and relieves the positive supercoiling generated ahead of the replication fork. Without it, torsional stress would halt fork progression.

Question 44Genetics and Molecular Biology

In a monohybrid cross between two heterozygous individuals (Tt × Tt), what is the probability that an offspring shows the recessive phenotype?

  1. 3/4
  2. 1/2
  3. 0
  4. 1/4 Correct answer.

Explanation

The cross Tt × Tt produces genotypes in the ratio 1 TT : 2 Tt : 1 tt. Only the tt genotype expresses the recessive phenotype, giving a probability of 1/4.

Question 45Genetics and Molecular Biology

Which feature distinguishes the structure of RNA from DNA?

  1. RNA has a phosphate backbone, whereas DNA lacks phosphate groups.
  2. RNA contains thymine and deoxyribose, whereas DNA contains uracil and ribose.
  3. RNA is always double-stranded, whereas DNA is always single-stranded.
  4. RNA contains uracil and ribose, whereas DNA contains thymine and deoxyribose. Correct answer.

Explanation

RNA uses the pyrimidine uracil in place of thymine and has a ribose sugar with a 2'-hydroxyl group, whereas DNA contains thymine and deoxyribose. Both molecules have sugar-phosphate backbones.

Question 46Genetics and Molecular Biology

During which stage of meiosis does crossing over between non-sister chromatids occur?

  1. Pachytene of prophase I Correct answer.
  2. Leptotene of prophase I
  3. Anaphase I
  4. Metaphase II

Explanation

Crossing over occurs during pachytene of prophase I, when homologous chromosomes are paired as bivalents and chiasmata form between non-sister chromatids. This recombination generates new allele combinations.

Question 47Genetics and Molecular Biology

Which type of mutation involves the substitution of a single purine by a pyrimidine?

  1. Transversion Correct answer.
  2. Frameshift
  3. Silent mutation
  4. Transition

Explanation

A transversion is a base substitution in which a purine is replaced by a pyrimidine, or vice versa. A transition replaces a purine with a purine or a pyrimidine with a pyrimidine.

Question 48Genetics and Molecular Biology

In the genetic code, what does the term 'degenerate' mean?

  1. Codons are read in overlapping triplets.
  2. The code differs between prokaryotes and eukaryotes.
  3. Each codon codes for more than one amino acid.
  4. Most amino acids are specified by more than one codon. Correct answer.

Explanation

Degeneracy means that most amino acids correspond to several synonymous codons, although each codon still specifies only one amino acid. This redundancy reduces the impact of some point mutations.

Question 49Genetics and Molecular Biology

A man with blood group AB marries a woman with blood group O. What blood groups can their children have?

  1. A, B, or AB
  2. Only A or B Correct answer.
  3. Only AB
  4. A, B, AB, or O

Explanation

The AB father produces gametes carrying either IA or IB, while the O mother produces only i gametes. Offspring are therefore IAi (group A) or IBi (group B), each with a 50% chance.

Question 50Genetics and Molecular Biology

In a dihybrid cross between two individuals heterozygous for both unlinked genes (AaBb × AaBb), what proportion of offspring shows both recessive phenotypes?

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

Explanation

For unlinked genes, the probability of each recessive phenotype is 1/4, so the joint probability is 1/4 × 1/4 = 1/16. This corresponds to the aabb class in the 9:3:3:1 ratio.