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Cameroon GCE A-Level Biology study materials
Cameroon GCE A-Level Biology study resources

Cameroon GCE A-Level Biology practice questions

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Question 1Biological inquiry and interpretation of evidence

A researcher predicts that tomato plants given a nitrogen-enriched fertilizer will produce more fruit than plants given a standard fertilizer. In this study, the amount of fruit produced is best described as which of the following?

  1. The hypothesis
  2. A controlled variable
  3. The dependent variable Correct answer.
  4. The independent variable

Explanation

The dependent variable is the measured outcome that responds to the manipulation, which here is fruit yield. The fertilizer type is the factor the researcher deliberately changes, so it is the independent variable rather than the measured result.

Question 2Biological inquiry and interpretation of evidence

A student measures the mass of the same sample five times and obtains nearly identical values, but the balance was never zeroed and reads 2.0 g too high. How should these measurements be described?

  1. Both accurate and precise
  2. Precise but not accurate Correct answer.
  3. Accurate but not precise
  4. Neither accurate nor precise

Explanation

Precision refers to how closely repeated measurements agree with one another, while accuracy refers to how close they are to the true value. Because the readings cluster tightly but are systematically offset by the unzeroed balance, they are precise but inaccurate.

Question 3Biological inquiry and interpretation of evidence

Which of the following best distinguishes a scientific hypothesis from a scientific theory?

  1. A hypothesis is a testable proposed explanation, while a theory is a well-supported explanatory framework built from extensive evidence Correct answer.
  2. A hypothesis is always correct once tested, while a theory remains an untested guess
  3. A hypothesis is a broad framework, while a theory is a single testable prediction
  4. A hypothesis applies only to biology, while a theory applies only to physics

Explanation

In science, a hypothesis is a testable proposed explanation for a specific observation, whereas a theory is a broad, well-substantiated explanation supported by many lines of evidence. The common usage of theory as a mere guess is incorrect in scientific contexts.

Question 4Biological inquiry and interpretation of evidence

A scientist repeats an experiment several times and obtains consistent results. Which conclusion is most justified?

  1. The findings are reliable and reproducible under the tested conditions Correct answer.
  2. The results now apply to all organisms everywhere
  3. The hypothesis has been proven true beyond any doubt
  4. The experiment has no remaining sources of error

Explanation

Reproducible results indicate reliability, meaning the procedure yields consistent outcomes under the tested conditions. Science does not establish absolute proof, and replication does not eliminate error or guarantee universal applicability.

Question 5Biological inquiry and interpretation of evidence

A graph plots plant height on the y-axis against days of growth on the x-axis. Which statement correctly identifies the placement of these variables?

  1. Days of growth is the independent variable on the x-axis, and height is the dependent variable on the y-axis Correct answer.
  2. Both variables are dependent and should share the x-axis
  3. Days of growth is the dependent variable because it is measured over time
  4. Height is the independent variable on the y-axis, and days of growth is the dependent variable on the x-axis

Explanation

By convention, the independent variable is placed on the x-axis and the dependent variable on the y-axis. Height responds to elapsed time, so time is independent and height is dependent.

Question 6Biological inquiry and interpretation of evidence

A student notices that mold grows on bread left in a warm, humid place but not on bread stored in a dry, cool place. Which of the following is the most appropriate next step in the scientific process?

  1. Formulate a testable hypothesis and design a controlled experiment to test it Correct answer.
  2. Conclude immediately that warmth and moisture cause mold growth
  3. Publish the observation as a confirmed scientific law
  4. Discard the observation because it was not made in a laboratory

Explanation

An observation generates questions that should be framed as a testable hypothesis and examined through a controlled experiment. Reaching a firm causal conclusion from a single uncontrolled observation is premature.

Question 7Biochemistry

Which property of water is most directly responsible for its ability to moderate temperature in living organisms and large bodies of water?

  1. Its low specific heat
  2. Its inability to form hydrogen bonds
  3. Its nonpolar covalent bonds
  4. Its high specific heat Correct answer.

Explanation

Water has a high specific heat because hydrogen bonds must absorb or release substantial energy before water molecules change temperature. This allows water and water-rich organisms to resist large temperature swings, helping maintain homeostasis.

Question 8Cell structure, membranes and division

Which structure is found in a typical plant cell but NOT in a typical animal cell?

  1. Mitochondrion
  2. Nucleus
  3. Ribosome
  4. Cell wall made of cellulose Correct answer.

Explanation

Plant cells have a rigid cell wall composed primarily of cellulose that surrounds the plasma membrane, providing structural support. Animal cells lack a cell wall, relying on the cytoskeleton and extracellular matrix instead.

Question 9Genetics, inheritance and molecular information

Meiosis differs from mitosis in that meiosis produces

  1. Two genetically identical diploid cells
  2. One diploid cell with double the chromosomes
  3. Two haploid cells identical to the parent
  4. Four genetically varied haploid cells Correct answer.

Explanation

Meiosis consists of two divisions that produce four haploid gametes, each with half the chromosome number of the parent. Crossing over and independent assortment generate genetic variation, distinguishing meiosis from the identical-diploid products of mitosis.

Question 10Metabolism

An enzyme lowers the activation energy of a reaction. Which statement best describes how this affects the reaction?

  1. It increases the reaction rate without being consumed in the reaction Correct answer.
  2. It changes the overall free-energy difference between reactants and products
  3. It supplies the energy needed to make an endergonic reaction exergonic
  4. It permanently binds to the product to stop the reaction

Explanation

Enzymes are biological catalysts that lower activation energy, speeding up reactions without being consumed. They do not change the net free-energy difference (delta G) between reactants and products; they only make the reaction reach equilibrium faster.

Question 11Biotechnology

The polymerase chain reaction, or PCR, is mainly used to do what?

  1. Amplify a selected DNA sequence Correct answer.
  2. Destroy all proteins in a sample
  3. Make plant roots bend downward
  4. Measure animal breathing rate directly

Explanation

PCR uses primers, DNA polymerase, nucleotides, and temperature cycling to make many copies of a target DNA region. It is widely used in testing and research.

Question 12Evolution

Which of the following is a necessary condition for natural selection to occur in a population?

  1. All individuals are genetically identical
  2. Heritable variation exists and affects survival or reproduction Correct answer.
  3. The environment never changes
  4. Every individual produces the same number of offspring

Explanation

Natural selection requires heritable variation among individuals and differences in reproductive success linked to that variation. Individuals with advantageous heritable traits leave more offspring, shifting the population's traits over generations.

Question 13Plant Physiology

In a transverse section of a dicotyledonous root, which tissue is responsible for transporting water and dissolved mineral ions upward?

  1. Phloem
  2. Xylem Correct answer.
  3. Parenchyma
  4. Epidermis

Explanation

Xylem consists of dead, lignified vessels that conduct water and dissolved mineral ions from the root upward through the plant. Phloem transports organic solutes, and parenchyma is general packing tissue.

Question 14Neurobiology

What ionic mechanism is directly responsible for the rapid depolarizing phase of a neuronal action potential once the threshold potential (~ -55 mV) is reached?

  1. Opening of voltage-gated Na+ channels, causing rapid inward influx of Na+ down its electrochemical gradient Correct answer.
  2. Opening of voltage-gated K+ channels, causing rapid outward efflux of K+
  3. Active extrusion of 3 Na+ ions in exchange for 2 K+ ions by the Na+/K+ pump
  4. Opening of ligand-gated Cl- channels, causing hyperpolarizing influx of Cl-

Explanation

When depolarization reaches the threshold potential (~ -55 mV), voltage-gated sodium channels rapidly open. Sodium ions (Na+) rush into the axon down their steep electrical and chemical gradients, depolarizing the membrane potential to approximately +35 to +40 mV.

Question 15Physiology

Which feature allows a steroid hormone to pass directly through the plasma membrane?

  1. Its lipid (hydrophobic) nature lets it dissolve through the phospholipid bilayer Correct answer.
  2. Its large size lets it push proteins aside
  3. Its negative charge attracts it to the membrane interior
  4. Its attachment to a carbohydrate tag opens a channel

Explanation

Steroid hormones are lipid-soluble (hydrophobic), so they can dissolve through the hydrophobic interior of the phospholipid bilayer without a transport protein. Once inside, they typically bind intracellular receptors that act on the DNA to regulate gene expression.

Question 16Microbiology Immunology

Gram-positive bacteria differ structurally from Gram-negative bacteria in their cell wall composition. Which characteristic feature distinguishes Gram-positive bacterial cell walls upon Gram staining?

  1. Thick multi-layered peptidoglycan meshwork retaining crystal violet-iodine complex (purple) Correct answer.
  2. Thin single layer of peptidoglycan enclosed by an outer membrane containing lipopolysaccharide (LPS)
  3. Absence of peptidoglycan, composed entirely of chitin
  4. Lipid envelope derived from host plasma membrane

Explanation

Gram-positive bacteria possess a thick, multi-layered peptidoglycan (murein) cell wall rich in teichoic and lipoteichoic acids. This dense structure dehydrates during alcohol decolorization, trapping the crystal violet-iodine complex and staining purple. Gram-negative cell walls have thin peptidoglycan and an outer LPS membrane (staining pink with safranin).

Question 17Ecology

In an energy pyramid, only about 10 percent of the energy at one trophic level is transferred to the next. Most of the remaining energy is

  1. Converted directly into new biomass at the next level
  2. Lost as heat, largely through cellular respiration and metabolic processes Correct answer.
  3. Stored permanently in the soil
  4. Recycled back to producers as light energy

Explanation

At each trophic level, organisms use most of the energy they consume for metabolism and lose much of it as heat through cellular respiration. Only about 10 percent is stored as biomass available to the next level, which limits food chain length.

Question 18Cell structure, membranes and division

A student using a light microscope with a 10x eyepiece lens and a 40x objective lens measures an onion epidermal cell to be 48 mm long on a calibrated photomicrograph with a total magnification of 400x. What is the actual length of the cell in micrometres (µm)?

  1. 120 µm Correct answer.
  2. 12 µm
  3. 1,200 µm
  4. 0.12 µm

Explanation

Actual size is calculated using the formula: Actual size = Image size / Total Magnification. Converting 48 mm to micrometres gives 48,000 µm; dividing 48,000 µm by 400 yields exactly 120 µm.

Question 19Cell structure, membranes and division

Which organelle is bounded by a double membrane with pores, contains a nucleolus, and is the primary site of ribosome subunit assembly in eukaryotic cells?

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

Explanation

The nucleus is enclosed by a double membrane known as the nuclear envelope, which contains nuclear pores for macromolecular exchange. The nucleolus inside the nucleus synthesizes ribosomal RNA (rRNA) and combines it with proteins to assemble ribosomal subunits.

Question 20Cell structure, membranes and division

According to the fluid mosaic model proposed by Singer and Nicolson, what maintains the structural fluidity of the eukaryotic plasma membrane at low temperatures?

  1. Cholesterol molecules preventing close packing of phospholipid fatty acid tails Correct answer.
  2. Integral transmembrane glycoproteins forming rigid covalent bridges
  3. High proportions of fully saturated fatty acid chains within phospholipids
  4. Peripheral proteins anchoring cytoskeletal microfilaments to extracellular carbohydrates

Explanation

Cholesterol acts as a bidirectional membrane fluidity buffer. At low temperatures, it intercalates between phospholipid hydrocarbon tails, disrupting regular crystalline packing and preventing membrane solidification.

Question 21Cell structure, membranes and division

Which of the following processes requires direct hydrolysis of ATP to transport ions against their electrochemical gradient across a cell membrane?

  1. Primary active transport by the Na+/K+-ATPase pump Correct answer.
  2. Facilitated diffusion of glucose through GLUT transporters
  3. Osmotic movement of water molecules through aquaporin channels
  4. Voltage-gated potassium ion efflux during action potential repolarization

Explanation

Primary active transport directly couples the hydrolysis of ATP to the uphill movement of ions. The Na+/K+-ATPase hydrolyzes one ATP molecule to pump 3 Na+ ions out of the cell and 2 K+ ions into the cell against their respective concentration gradients.

Question 22Cell structure, membranes and division

Which of the following cellular components is found in both prokaryotic (bacterial) and eukaryotic cells?

  1. Ribosomes Correct answer.
  2. Endoplasmic reticulum
  3. Nuclear envelope
  4. Mitochondria

Explanation

Ribosomes are universal non-membrane-bound ribonucleoprotein complexes essential for protein synthesis found in all living cells (70S in prokaryotes, 80S in eukaryotic cytoplasm).

Question 23Cell structure, membranes and division

A plant cell with a solute potential (Ψs) of -900 kPa and a pressure potential (Ψp) of +400 kPa is placed in a sucrose solution having a water potential (Ψ) of -700 kPa. In which direction will net water movement occur, and what is the initial water potential of the cell?

  1. Cell Ψ = -500 kPa; net water moves out of the cell into the sucrose solution Correct answer.
  2. Cell Ψ = -500 kPa; net water moves into the cell down the water potential gradient
  3. Cell Ψ = -1300 kPa; net water moves into the cell from the sucrose solution
  4. Cell Ψ = -500 kPa; no net water movement occurs because the cell is in dynamic equilibrium

Explanation

Cell water potential is Ψ_cell = Ψs + Ψp = -900 kPa + 400 kPa = -500 kPa. When placed in a solution of Ψ_sol = -700 kPa, water moves from higher (less negative) water potential (-500 kPa in the cell) to lower (more negative) water potential (-700 kPa in the solution), meaning net water moves OUT of the cell down the water potential gradient.

Question 24Cell structure, membranes and division

Which organelle contains acid hydrolases, maintains an internal acidic pH of ~4.5 to 5.0 via vacuolar H+-ATPase pumps, and is responsible for autophagy and heterophagy?

  1. Lysosome Correct answer.
  2. Peroxisome
  3. Centrosome
  4. Glyoxysome

Explanation

Lysosomes are membrane-bound digestive organelles containing over 50 acid hydrolases (proteases, nucleases, lipases). Their luminal acidic pH is maintained by active proton pumping (V-type H+-ATPase), enabling intracellular digestion of organelles (autophagy) and phagocytosed pathogens (heterophagy).

Question 25Cell structure, membranes and division

During which phase of the eukaryotic cell cycle does semi-conservative DNA replication occur?

  1. S phase of interphase Correct answer.
  2. G1 phase of interphase
  3. G2 phase of interphase
  4. Prophase of mitosis

Explanation

DNA synthesis and chromosome duplication occur exclusively during the S (Synthesis) phase of interphase, resulting in two identical sister chromatids per chromosome.

Question 26Cell structure, membranes and division

What is the structural role of the centrioles within the centrosome of animal cells during mitosis?

  1. Organizing the nucleation and polymerization of tubulin dimers into the mitotic spindle Correct answer.
  2. Synthesizing kinetochore proteins that attach chromosomes to centromeres
  3. Contracting the actin-myosin ring during telophase cytokinesis
  4. Dissolving the nuclear envelope via phosphorylation of lamin filaments

Explanation

Centrioles, arranged in pairs at right angles within the centrosome, act as the primary microtubule organizing centres (MTOCs) in animal cells, directing the assembly of the mitotic spindle apparatus that segregates sister chromatids.

Question 27Cell structure, membranes and division

When a red blood cell (erythrocyte) is placed in a 0.1% NaCl solution (distilled water/hypotonic medium), what physiological phenomenon occurs?

  1. Endosmosis occurs, causing cell swelling and osmotic lysis (haemolysis) Correct answer.
  2. Exosmosis occurs, causing cell shrinkage and crenation
  3. Plasmolysis occurs, pulling the cytoplasm away from the cell wall
  4. Equilibrium is maintained because the membrane is impermeable to water

Explanation

A 0.1% NaCl solution is strongly hypotonic relative to human erythrocyte cytoplasm (~0.9% NaCl equivalent). Water enters the erythrocyte rapidly by endosmosis; lacking a rigid cell wall to generate turgor/pressure potential, the membrane ruptures in a process called haemolysis.

Question 28Cell structure, membranes and division

What is the primary function of the Golgi apparatus in eukaryotic secretory cells?

  1. Post-translational glycosylation, sorting, and packaging of proteins into secretory vesicles Correct answer.
  2. Translating mRNA transcripts into nascent polypeptide chains on attached ribosomes
  3. Oxidizing glucose to pyruvate via substrate-level phosphorylation
  4. Synthesizing steroid hormones and detoxifying lipid-soluble metabolic byproducts

Explanation

The Golgi apparatus modifies proteins received from the rough ER (including glycosylation, sulfation, and phosphorylation), sorts them according to their molecular destinations, and packages them into secretory vesicles or lysosomes.

Question 29Biochemistry

What type of chemical bond links the monosaccharide monomers in maltose, and what specific reagent test confirms its reducing property?

  1. α-1,4-glycosidic bond; Benedict's test yielding a brick-red precipitate on heating Correct answer.
  2. β-1,4-glycosidic bond; Biuret test yielding a purple coloration
  3. α-1,6-glycosidic bond; Iodine test yielding a blue-black complex
  4. Ester bond; Emulsion test yielding a cloudy white suspension

Explanation

Maltose is a disaccharide composed of two α-D-glucose residues joined by an α-1,4-glycosidic bond. Because one anomeric carbon retains a free hemiacetal group capable of reducing cupric ions (Cu2+) to cuprous oxide (Cu2O), heating with Benedict's reagent produces a brick-red precipitate.

Question 30Biochemistry

Which level of protein structure is primarily stabilized by hydrogen bonds between the carbonyl oxygen (C=O) of one peptide bond and the amino hydrogen (N-H) of another peptide bond four residues apart along the polypeptide backbone?

  1. Secondary structure (α-helix) Correct answer.
  2. Primary structure (amino acid sequence)
  3. Tertiary structure (3D globular folding)
  4. Quaternary structure (multi-subunit association)

Explanation

The α-helix is a regular secondary structure stabilized by intrachain hydrogen bonds between the C=O of amino acid residue n and the N-H of residue n+4. This regular hydrogen bonding along the polypeptide backbone occurs independently of R-group interactions.

Question 31Biochemistry

In an enzyme kinetics experiment, the maximum velocity (Vmax) of an enzymatic reaction is 80 µmol min⁻¹ mg⁻¹. What is the Michaelis constant (Km) if the reaction rate is 40 µmol min⁻¹ mg⁻¹ when the substrate concentration is 2.5 mmol dm⁻³?

  1. 2.5 mmol dm⁻³ Correct answer.
  2. 5.0 mmol dm⁻³
  3. 1.25 mmol dm⁻³
  4. 10.0 mmol dm⁻³

Explanation

By definition, the Michaelis constant (Km) is the substrate concentration at which the reaction velocity is exactly half of the maximum velocity (1/2 Vmax). Since 40 µmol min⁻¹ mg⁻¹ is exactly half of Vmax (80 µmol min⁻¹ mg⁻¹), the substrate concentration of 2.5 mmol dm⁻³ is the Km.

Question 32Biochemistry

How does a competitive enzyme inhibitor affect the kinetic parameters Vmax and Km of an enzyme-catalysed reaction?

  1. Km increases while Vmax remains unchanged Correct answer.
  2. Km remains unchanged while Vmax decreases
  3. Both Km and Vmax decrease proportionally
  4. Both Km and Vmax increase significantly

Explanation

A competitive inhibitor structurally resembles the substrate and reversibly binds to the enzyme active site. It increases the apparent Km (reflecting lower affinity) because higher substrate concentrations are required to outcompete the inhibitor, but Vmax remains unchanged because saturating substrate concentrations displace all inhibitor molecules.

Question 33Biochemistry

Which property of water allows tall trees like Entandrophragma cylindricum (Sapele wood) in the Cameroon rainforest to transport continuous water columns from roots to leaves over 50 metres above the ground?

  1. High tensile strength and strong cohesive forces due to intermolecular hydrogen bonding Correct answer.
  2. High specific heat capacity preventing rapid temperature fluctuations in xylem vessels
  3. Maximum density at 4 °C causing upward buoyant convective currents in the stem
  4. High latent heat of vaporization generating positive hydrostatic pressure in the xylem

Explanation

Extensive hydrogen bonding between water molecules produces high cohesion, creating high tensile strength that allows sap to withstand strong negative pressures (tension) generated by transpiration without cavitation (column breakage). Adhesion to hydrophilic xylem walls further stabilizes the column.

Question 34Biochemistry

What is the structural basis for the high tensile strength of cellulose microfibrils in plant cell walls?

  1. Linear unbranched chains of β-1,4-glucose stabilized by extensive inter-chain hydrogen bonds Correct answer.
  2. Coiled helical chains of α-1,4-glucose reinforced by periodic α-1,6 branch points
  3. Cross-linked polymers of N-acetylglucosamine linked by covalent peptide bridges
  4. Branching networks of β-1,3-glucan cross-linked with lignin via ether linkages

Explanation

Cellulose consists of unbranched linear chains of β-1,4-linked D-glucose residues where each alternate glucose unit is rotated 180°. This straight geometry allows adjacent parallel chains to form thousands of inter-molecular and intra-molecular hydrogen bonds, aggregating into microfibrils with immense tensile strength.

Question 35Biochemistry

Which of the following lipid classes contains a glycerol molecule esterified to two fatty acid chains and a phosphate group bonded to a polar choline head?

  1. Phosphatidylcholine (phospholipid) Correct answer.
  2. Triacylglycerol (triglyceride)
  3. Cholesterol ester
  4. Sphingomyelin

Explanation

Phosphatidylcholine is a classic phosphoglyceride consisting of a glycerol backbone ester-linked to two hydrophobic fatty acid tails at C-1 and C-2, with the C-3 hydroxyl esterified to a hydrophilic phosphate group attached to choline, giving it an amphipathic nature.

Question 36Biochemistry

Which of the following statements correctly distinguishes purines from pyrimidines in nucleic acid biochemistry?

  1. Purines (adenine, guanine) have a double-ring structure; pyrimidines (cytosine, thymine, uracil) have a single-ring structure Correct answer.
  2. Purines (cytosine, thymine) have a single-ring structure; pyrimidines (adenine, guanine) have a double-ring structure
  3. Purines contain three heterocyclic nitrogen atoms; pyrimidines contain only one nitrogen atom
  4. Purines form three hydrogen bonds in Watson-Crick base pairing; pyrimidines form only two hydrogen bonds

Explanation

Purines (Adenine and Guanine) consist of a fused 6-membered pyrimidine ring and a 5-membered imidazole ring (a 9-membered bicyclic structure). Pyrimidines (Cytosine, Thymine, Uracil) consist of a single 6-membered aromatic heterocyclic ring.

Question 37Biochemistry

Which of the following bonds stabilizes the tertiary structure of proteins and represents a covalent linkage formed between the sulfhydryl (-SH) groups of two cysteine residues?

  1. Disulfide bridge Correct answer.
  2. Peptide bond
  3. Ionic bond (salt bridge)
  4. Hydrophobic interaction

Explanation

A disulfide bridge (-S-S-) is a strong covalent bond formed by the oxidation of two cysteine thiol (-SH) groups. It provides exceptional mechanical and chemical stability to the tertiary and quaternary structures of secreted and extracellular proteins.

Question 38Metabolism

In glycolysis, which enzyme catalyses the phosphorylation of fructose-6-phosphate to fructose-1,6-bisphosphate utilizing ATP, acting as the primary committed and rate-limiting regulatory step?

  1. Phosphofructokinase-1 (PFK-1) Correct answer.
  2. Hexokinase
  3. Pyruvate kinase
  4. Triose phosphate isomerase

Explanation

Phosphofructokinase-1 (PFK-1) catalyses the irreversible phosphorylation of fructose-6-phosphate to fructose-1,6-bisphosphate using one ATP. PFK-1 is the major allosteric control point of glycolysis, inhibited by high ATP and citrate, and activated by AMP and fructose-2,6-bisphosphate.

Question 39Metabolism

During the link reaction occurring in the mitochondrial matrix, what are the products generated per molecule of pyruvate entering from the cytosol?

  1. 1 Acetyl-CoA, 1 NADH + H+, and 1 CO2 Correct answer.
  2. 2 Acetyl-CoA, 2 NADH + H+, and 2 CO2
  3. 1 Oxaloacetate, 1 FADH2, and 1 ATP
  4. 1 Citrate, 1 NADPH + H+, and 1 H2O

Explanation

In the link reaction (oxidative decarboxylation), pyruvate dehydrogenase converts one 3-carbon pyruvate molecule into one 2-carbon Acetyl-CoA, producing one molecule of CO2 and reducing one NAD+ to NADH + H+.

Question 40Metabolism

In the Krebs cycle, how many molecules of NADH, FADH2, and ATP (or GTP by substrate-level phosphorylation) are generated per turn of the cycle from one molecule of Acetyl-CoA?

  1. 3 NADH, 1 FADH2, 1 ATP (or GTP) Correct answer.
  2. 2 NADH, 2 FADH2, 2 ATP (or GTP)
  3. 4 NADH, 0 FADH2, 1 ATP (or GTP)
  4. 6 NADH, 2 FADH2, 2 ATP (or GTP)

Explanation

One turn of the citric acid cycle oxidizes one Acetyl-CoA, yielding 3 NADH (at isocitrate dehydrogenase, α-ketoglutarate dehydrogenase, and malate dehydrogenase), 1 FADH2 (at succinate dehydrogenase), 1 ATP/GTP (at succinyl-CoA synthetase), and releasing 2 CO2.

Question 41Metabolism

What is the immediate source of energy directly driving ATP synthesis by the F0F1-ATP synthase complex during oxidative phosphorylation in mitochondria?

  1. The proton motive force generated by the electrochemical gradient across the inner mitochondrial membrane Correct answer.
  2. Direct covalent transfer of a high-energy phosphate group from 1,3-bisphosphoglycerate
  3. The free energy released from the direct reduction of molecular oxygen to water by cytochrome c
  4. Hydrolysis of GTP in the mitochondrial matrix

Explanation

According to Mitchell's chemiosmotic hypothesis, electron transport along Complexes I-IV pumps protons (H+) from the matrix into the intermembrane space. The resulting proton-motive force (voltage and pH gradient) drives protons through the F0 channel of ATP synthase, causing conformational rotation of the F1 head that phosphorylates ADP to ATP.

Question 42Metabolism

A respiratory chamber contains germinating castor oil seeds (Ricinus communis) oxidizing tripalmitin (C51H98O6). The overall chemical equation is: 2 C51H98O6 + 145 O2 → 102 CO2 + 98 H2O + Energy. What is the Respiratory Quotient (RQ) for the complete aerobic respiration of this substrate?

  1. 0.70 Correct answer.
  2. 1.00
  3. 0.80
  4. 1.33

Explanation

The Respiratory Quotient is calculated as RQ = (Volume or moles of CO2 produced) / (Volume or moles of O2 consumed). For tripalmitin: RQ = 102 CO2 / 145 O2 = 0.7034 ≈ 0.70. This value is typical for lipid respiration because fats contain less oxygen relative to carbon and hydrogen.

Question 43Metabolism

In human skeletal muscle during vigorous anaerobic sprinting, what enzyme catalyses the reduction of pyruvate to lactate, and what is the vital physiological purpose of this reaction?

  1. Lactate dehydrogenase; re-oxidizing NADH to NAD+ to allow glycolysis to continue Correct answer.
  2. Pyruvate decarboxylase; regenerating ATP directly via substrate-level phosphorylation
  3. Alcohol dehydrogenase; converting toxic pyruvate into neutral lactate
  4. Lactate synthetase; pumping accumulated protons out of fatigued myofibrils

Explanation

Lactate dehydrogenase (LDH) reduces pyruvate to lactate using NADH + H+. This regenerates cytosolic NAD+, which is essential for glyceraldehyde-3-phosphate dehydrogenase so that glycolysis can continue generating 2 ATP per glucose under anaerobic conditions.

Question 44Metabolism

Where do the light-dependent reactions and the light-independent Calvin cycle reactions take place within the chloroplast, respectively?

  1. Thylakoid membranes (grana) and Stroma Correct answer.
  2. Stroma and Thylakoid lumen
  3. Outer chloroplast membrane and Inner chloroplast membrane
  4. Thylakoid lumen and Intermembrane space

Explanation

The light-dependent reactions occur on the thylakoid membranes (and within the thylakoid lumen), where photosystems, electron transport chains, and ATP synthase are embedded. The light-independent Calvin cycle takes place in the fluid stroma where soluble enzymes (like RuBisCO) reside.

Question 45Metabolism

What is the ultimate primary electron donor for non-cyclic photophosphorylation in oxygenic photosynthesis, and what complex catalyses its oxidation?

  1. Water (H2O); Oxygen-evolving complex (OEC) associated with Photosystem II (P680) Correct answer.
  2. NADPH; Ferredoxin-NADP+ reductase associated with Photosystem I (P700)
  3. Plastoquinone (PQ); Cytochrome b6f complex
  4. Carbon dioxide (CO2); RuBisCO in the stroma

Explanation

In non-cyclic photophosphorylation, photolysis of water (2 H2O → O2 + 4 H+ + 4 e-) by the oxygen-evolving manganese complex on the luminal side of Photosystem II (P680) provides replacement electrons for photo-oxidized P680 reaction centres, releasing molecular oxygen as a byproduct.

Question 46Metabolism

Which of the following products from the light-dependent reactions are required by the Calvin cycle to reduce 3-phosphoglycerate (PGA) to glyceraldehyde-3-phosphate (GALP/G3P)?

  1. ATP and NADPH Correct answer.
  2. ATP and FADH2
  3. ADP, Pi, and NADP+
  4. Oxygen and Glucose

Explanation

Reduction of 3-phosphoglycerate (PGA) to glyceraldehyde-3-phosphate (GALP/G3P) requires phosphorylation of PGA by ATP (producing 1,3-bisphosphoglycerate) followed by reduction using NADPH + H+ catalysed by glyceraldehyde-3-phosphate dehydrogenase.

Question 47Metabolism

How does cyclic photophosphorylation differ from non-cyclic photophosphorylation?

  1. Cyclic photophosphorylation involves only Photosystem I (P700) and produces only ATP without generating NADPH or O2 Correct answer.
  2. Cyclic photophosphorylation involves only Photosystem II (P680) and produces both ATP and O2
  3. Cyclic photophosphorylation requires photolysis of water to continuously regenerate reduced ferredoxin
  4. Cyclic photophosphorylation fixes carbon dioxide directly into phosphoenolpyruvate

Explanation

In cyclic photophosphorylation, excited electrons from Photosystem I (P700) are transferred to ferredoxin and then cycled back via cytochrome b6f and plastocyanin to P700. This proton pumping drives ATP synthesis without involving PSII, photolysis of water, or the generation of NADPH and oxygen.

Question 48Metabolism

In C4 plants such as maize (Zea mays) and sugarcane, what initial anatomical and biochemical adaptation suppresses photorespiration?

  1. Spatial separation of initial CO2 fixation by PEP carboxylase in mesophyll cells from the Calvin cycle in bundle sheath cells (Kranz anatomy) Correct answer.
  2. Temporal separation of CO2 uptake at night via open stomata from daytime Calvin cycle reactions (CAM)
  3. High concentration of glycolate oxidase in mesophyll chloroplasts that directly consumes O2
  4. Exclusively using cyclic photophosphorylation to eliminate oxygen production in all photosynthetic cells

Explanation

C4 plants exhibit Kranz anatomy, spatially separating initial CO2 capture by PEP carboxylase (which has no affinity for O2) in mesophyll cells into 4-carbon oxaloacetate/malate. Malate is transported to bundle sheath cells and decarboxylated, maintaining high stromal CO2 around RuBisCO and eliminating oxygenase activity (photorespiration).

Question 49Metabolism

According to Blackman's principle of limiting factors, when a photosynthetic leaf is exposed to bright sunshine at 25 °C in normal atmospheric air (0.04% CO2), which factor is the primary limiting factor for the rate of photosynthesis?

  1. Carbon dioxide concentration Correct answer.
  2. Light intensity
  3. Temperature
  4. Water availability

Explanation

Blackman's Law of Limiting Factors states that when a physiological process is governed by several factors, its rate is limited by the factor that is nearest its minimal value. Under bright light and warm temperature (25 °C), atmospheric CO2 concentration (0.04% or 400 ppm) is sub-saturating and acts as the primary limiting factor.

Question 50Metabolism

What is the net gain of ATP molecules produced per molecule of glucose solely through substrate-level phosphorylation during glycolysis?

  1. 2 ATP Correct answer.
  2. 4 ATP
  3. 32 ATP
  4. 36 ATP

Explanation

Glycolysis consumes 2 ATP molecules during the investment phase (hexokinase and PFK-1) and produces 4 ATP molecules during the payoff phase (phosphoglycerate kinase and pyruvate kinase). The NET gain is 4 - 2 = 2 ATP per glucose.