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A Level H2 Biology Practice Paper 2

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Questions

TuitionGoWhere Practice Paper - Biology H2 A-Level

PRACTICE PAPER 2 (Version 2 of 5)

TuitionGoWhere Exam Practice (AI)

Subject: Biology H2 (9477) Level: A-Level Paper: 2 - Structured Questions Duration: 2 hours Total Marks: 75

Name: _________________________ Class: _________________________ Date: _________________________

INSTRUCTIONS TO CANDIDATES

This paper consists of THREE sections: Section A, Section B, and Section C.
Answer ALL questions in Section A.
Answer ONE question from Section B.
Answer ONE question from Section C.
Write your answers in the spaces provided.
The number of marks is given in brackets [ ] at the end of each question or part question.
You are advised to spend no more than 50 minutes on Section A, 35 minutes on Section B, and 35 minutes on Section C.

SECTION A: Structured Questions (45 marks)

Answer ALL questions in this section.

Question 1: Membrane Transport Mechanisms

Fig. 1.1 shows the structure of a cell surface membrane with three transport proteins labelled P, Q, and R.

(a) With reference to Fig. 1.1, identify the type of transport mediated by protein P and explain how the structure of protein P facilitates this function. [3]

(b) Protein Q is a carrier protein that transports glucose into the cell against its concentration gradient.

(i) Name the specific transport process mediated by protein Q. [1]

(ii) Explain how protein Q uses energy to transport glucose against its concentration gradient. [3]

(c) Protein R is an aquaporin. Describe how the structure of aquaporins enables rapid water transport across membranes while preventing the passage of ions. [3]

[Total: 10 marks]

Question 2: Enzyme Kinetics and Inhibition

A student investigated the effect of an inhibitor on the activity of the enzyme succinate dehydrogenase. The enzyme catalyses the conversion of succinate to fumarate. The student measured the initial rate of reaction at different substrate concentrations, both in the absence and presence of the inhibitor. The results are shown in Fig. 2.1.

(a) Describe the effect of increasing substrate concentration on the rate of reaction in the absence of the inhibitor. [2]

(b) With reference to Fig. 2.1, explain the effect of the inhibitor on the maximum rate of reaction (Vmax) and the Michaelis constant (Km) of the enzyme. [3]

(c) Based on your answer in (b), deduce the type of inhibition shown by this inhibitor. Explain your reasoning. [2]

(d) Malonate is a competitive inhibitor of succinate dehydrogenase. Explain, at the molecular level, how malonate inhibits the enzyme. [3]

[Total: 10 marks]

Question 3: DNA Replication

Fig. 3.1 shows a replication fork during DNA replication in a prokaryotic cell.

(a) Identify the enzyme labelled X in Fig. 3.1 and state its function. [2]

(b) Explain why DNA replication on the lagging strand is discontinuous. [3]

(c) DNA polymerase III has proofreading activity. Explain how this activity contributes to the accuracy of DNA replication and state the consequence if this activity is defective. [3]

(d) The antibiotic ciprofloxacin inhibits DNA gyrase in prokaryotes. Suggest and explain why ciprofloxacin is effective against bacterial infections but has minimal effect on human cells. [2]

[Total: 10 marks]

Question 4: Cell Cycle and Mitosis

Fig. 4.1 shows the changes in the mass of DNA per cell during the cell cycle in a population of actively dividing human cells.

(a) Identify the phases of the cell cycle represented by the regions labelled A, B, and C in Fig. 4.1. [3]

(b) Colchicine is a drug that prevents the polymerisation of tubulin into microtubules. Explain how colchicine affects mitosis and state the stage at which mitosis would be arrested. [3]

(c) A student prepared a root tip squash to observe cells undergoing mitosis. The student counted 200 cells and recorded the number of cells in each stage of mitosis. The results are shown in Table 4.1.

Stage	Number of cells
Prophase	42
Metaphase	18
Anaphase	10
Telophase	30
Interphase	100

(i) Calculate the mitotic index for this root tip. Show your working. [2]

(ii) The total duration of the cell cycle in this root tip is 20 hours. Calculate the duration of metaphase. Show your working. [2]

[Total: 10 marks]

Question 5: Protein Structure and Function

Haemoglobin is a globular protein that transports oxygen in red blood cells. Fig. 5.1 shows the oxygen dissociation curves for adult haemoglobin (HbA) and foetal haemoglobin (HbF).

(a) Compare the oxygen dissociation curves of HbA and HbF. [2]

(b) Explain the physiological significance of the difference in oxygen affinity between HbA and HbF. [3]

(c) Sickle cell anaemia is caused by a single base substitution in the gene coding for the β-globin chain of haemoglobin. This results in the substitution of glutamic acid with valine at position 6 of the polypeptide.

(i) Explain how this single amino acid substitution leads to the sickling of red blood cells under low oxygen tension. [3]

(ii) Suggest why individuals who are heterozygous for the sickle cell allele (HbA/HbS) have a selective advantage in regions where malaria is endemic. [2]

[Total: 10 marks]

SECTION B: Long Structured Questions (15 marks)

Answer ONE question from this section.

Question 6: Cellular Respiration

(a) Outline the process of glycolysis, stating the location, the initial substrate, and the net products. [4]

(b) Describe the role of coenzymes in the link reaction and the Krebs cycle. [3]

(c) Fig. 6.1 shows the electron transport chain and chemiosmosis in the inner mitochondrial membrane.

With reference to Fig. 6.1, explain how the transfer of electrons along the electron transport chain leads to the synthesis of ATP. [5]

(d) Cyanide is a potent inhibitor of cytochrome c oxidase (Complex IV) in the electron transport chain. Explain why cyanide poisoning is rapidly fatal. [3]

[Total: 15 marks]

Question 7: Gene Expression and Regulation

(a) Describe the process of transcription in prokaryotes, including the roles of RNA polymerase and the promoter. [4]

(b) The lac operon in E. coli is an example of an inducible operon. Explain how the lac operon is regulated in the presence and absence of lactose. [5]

(c) Fig. 7.1 shows the structure of a eukaryotic gene and its mature mRNA transcript.

With reference to Fig. 7.1, explain the differences between the eukaryotic gene and its mature mRNA transcript. [3]

(d) Suggest why the regulation of gene expression in eukaryotes is more complex than in prokaryotes. [3]

[Total: 15 marks]

SECTION C: Data-Based and Application Questions (15 marks)

Answer ONE question from this section.

Question 8: Gel Electrophoresis and Genetic Disease Diagnosis

Sickle cell anaemia is caused by a mutation in the β-globin gene. The normal allele (HbA) and the sickle cell allele (HbS) can be distinguished by gel electrophoresis of haemoglobin protein.

Fig. 8.1 shows the results of haemoglobin gel electrophoresis for four individuals, W, X, Y, and Z.

(a) Describe and explain how gel electrophoresis separates proteins. [4]

(b) With reference to Fig. 8.1, identify the genotype of each individual (W, X, Y, and Z) with respect to the HbA and HbS alleles. Explain your reasoning for each. [4]

(c) Individual X is heterozygous for the sickle cell allele. Explain why individual X shows two bands on the gel. [2]

(d) A couple, both with the same genotype as individual X, are planning to have a child. Using a genetic diagram, determine the probability that their child will have sickle cell anaemia. [3]

(e) Discuss the ethical considerations of using genetic screening for sickle cell anaemia in pre-natal diagnosis. [2]

[Total: 15 marks]

Question 9: Mitochondrial Function and Metabolic Inhibitors

A researcher investigated the effect of different substrates and inhibitors on oxygen consumption by isolated mitochondria. The mitochondria were suspended in a buffer containing ADP and inorganic phosphate (Pi). Oxygen concentration was monitored using an oxygen electrode.

Fig. 9.1 shows the oxygen electrode trace obtained during the experiment. At the times indicated by the arrows, the following additions were made:

A: Addition of succinate (a substrate for Complex II)
B: Addition of ADP
C: Addition of sodium azide (an inhibitor of cytochrome c oxidase)

(a) Explain why oxygen consumption by the mitochondria is low before the addition of succinate at point A. [2]

(b) With reference to Fig. 9.1, describe and explain the change in the rate of oxygen consumption after the addition of ADP at point B. [3]

(c) Sodium azide inhibits cytochrome c oxidase (Complex IV). Explain the effect of sodium azide on: (i) The rate of oxygen consumption [2]

(ii) The synthesis of ATP [2]

(d) The experiment was repeated using pyruvate instead of succinate as the substrate. Predict and explain how the oxygen electrode trace would differ from Fig. 9.1. [3]

(e) Uncouplers such as 2,4-dinitrophenol (DNP) increase the permeability of the inner mitochondrial membrane to protons. Predict and explain the effect of DNP on: (i) The rate of oxygen consumption [1]

(ii) The rate of ATP synthesis [2]

[Total: 15 marks]

END OF PAPER

Answers

TuitionGoWhere Practice Paper - Biology H2 A-Level

PRACTICE PAPER 2 (Version 2 of 5) - ANSWER KEY AND MARKING SCHEME

TuitionGoWhere Exam Practice (AI)

Subject: Biology H2 (9477) Level: A-Level Paper: 2 - Structured Questions Total Marks: 75

SECTION A: Structured Questions (45 marks)

Question 1: Membrane Transport Mechanisms

(a) With reference to Fig. 1.1, identify the type of transport mediated by protein P and explain how the structure of protein P facilitates this function. [3]

Answer:

Protein P mediates facilitated diffusion [1]
Protein P is a channel protein with a hydrophilic pore/central channel [1]
The hydrophilic pore allows specific ions/polar molecules to pass through the hydrophobic core of the membrane down their concentration gradient [1]

Marking notes:

Accept: "Protein P forms a water-filled pore" for the second mark
Must link structure (hydrophilic pore) to function (passage of polar/charged molecules)
Reject: "Active transport" or "simple diffusion"

(b)(i) Name the specific transport process mediated by protein Q. [1]

Answer:

Active transport / Primary active transport [1]

Marking notes:

Accept: "Sodium-glucose co-transport" or "Secondary active transport" if context provided
Reject: "Facilitated diffusion" or "Simple diffusion"

(b)(ii) Explain how protein Q uses energy to transport glucose against its concentration gradient. [3]

Answer:

ATP is hydrolysed to ADP + Pi, releasing energy [1]
The energy is used to change the conformation/shape of the carrier protein [1]
This conformational change moves glucose across the membrane against its concentration gradient [1]

Marking notes:

Must mention ATP hydrolysis as the energy source
Must describe conformational change
Accept: "Phosphorylation of the carrier protein causes conformational change"
Reject: Vague statements like "energy is used to pump glucose"

(c) Protein R is an aquaporin. Describe how the structure of aquaporins enables rapid water transport across membranes while preventing the passage of ions. [3]

Answer:

Aquaporins have a narrow central pore/channel [1]
The pore is lined with hydrophilic amino acids that allow water molecules to pass through in single file [1]
The narrow pore and specific amino acid residues (asparagine) prevent the passage of ions/hydrated ions which are too large to pass through [1]

Marking notes:

Must mention narrow pore and selective permeability
Accept: "The pore contains positively charged residues that repel protons/cations"
Must explain why ions cannot pass (size exclusion and/or charge repulsion)

[Total: 10 marks]

Question 2: Enzyme Kinetics and Inhibition

(a) Describe the effect of increasing substrate concentration on the rate of reaction in the absence of the inhibitor. [2]

Answer:

At low substrate concentrations, the rate of reaction increases as substrate concentration increases [1]
At high substrate concentrations, the rate of reaction plateaus/levels off as the enzyme becomes saturated / all active sites are occupied [1]

Marking notes:

Must describe both phases (increasing and plateau)
Accept: "The curve follows Michaelis-Menten kinetics"
Reject: "Rate increases indefinitely"

(b) With reference to Fig. 2.1, explain the effect of the inhibitor on the maximum rate of reaction (Vmax) and the Michaelis constant (Km) of the enzyme. [3]

Answer:

Vmax is decreased/reduced in the presence of the inhibitor [1]
Km is unchanged/remains the same [1]
This indicates non-competitive inhibition where the inhibitor binds to an allosteric site, reducing the number of functional enzyme molecules without affecting substrate binding affinity [1]

Marking notes:

Must state both Vmax and Km effects
Must link to non-competitive inhibition mechanism
Accept: "Vmax decreases because inhibitor reduces the number of active enzyme molecules"
Reject: "Km increases" (this would indicate competitive inhibition)

(c) Based on your answer in (b), deduce the type of inhibition shown by this inhibitor. Explain your reasoning. [2]

Answer:

Non-competitive inhibition [1]
Reasoning: Vmax decreases but Km is unchanged, which is characteristic of non-competitive inhibition where the inhibitor does not compete with the substrate for the active site [1]

Marking notes:

Must state "non-competitive" explicitly
Must link to Vmax/Km pattern
Accept: "The inhibitor binds to an allosteric site, not the active site"

(d) Malonate is a competitive inhibitor of succinate dehydrogenase. Explain, at the molecular level, how malonate inhibits the enzyme. [3]

Answer:

Malonate has a similar shape/structure to succinate (the normal substrate) [1]
Malonate binds to the active site of succinate dehydrogenase [1]
This prevents succinate from binding to the active site, so the enzyme cannot catalyse the conversion of succinate to fumarate [1]

Marking notes:

Must mention structural similarity between malonate and succinate
Must state that malonate binds to the active site
Must explain the consequence (substrate cannot bind)
Accept: "Malonate is a structural analogue of succinate"

[Total: 10 marks]

Question 3: DNA Replication

(a) Identify the enzyme labelled X in Fig. 3.1 and state its function. [2]

Answer:

Enzyme X is helicase [1]
Function: Unwinds the DNA double helix by breaking hydrogen bonds between complementary base pairs / separates the two DNA strands to form the replication fork [1]

Marking notes:

Must identify helicase
Must describe unwinding/strand separation
Accept: "Breaks hydrogen bonds between base pairs"

(b) Explain why DNA replication on the lagging strand is discontinuous. [3]

Answer:

DNA polymerase III can only synthesise DNA in the 5' to 3' direction [1]
The lagging strand template runs in the 3' to 5' direction (away from the replication fork) [1]
Therefore, DNA polymerase III must synthesise short fragments (Okazaki fragments) in the opposite direction to fork movement, which are later joined by DNA ligase [1]

Marking notes:

Must state directionality of DNA polymerase III (5' to 3')
Must explain why this necessitates discontinuous synthesis
Accept: "The two template strands are antiparallel"
Must mention Okazaki fragments

(c) DNA polymerase III has proofreading activity. Explain how this activity contributes to the accuracy of DNA replication and state the consequence if this activity is defective. [3]

Answer:

DNA polymerase III has 3' to 5' exonuclease activity that can detect and remove incorrectly paired nucleotides [1]
This reduces the error rate/mutation rate during DNA replication [1]
If proofreading is defective, the mutation rate would increase, leading to more errors in the DNA sequence / increased risk of genetic diseases/cancer [1]

Marking notes:

Must mention 3' to 5' exonuclease activity
Must explain that it removes mismatched nucleotides
Must state consequence (increased mutation rate)
Accept: "Proofreading increases fidelity of replication"

(d) The antibiotic ciprofloxacin inhibits DNA gyrase in prokaryotes. Suggest and explain why ciprofloxacin is effective against bacterial infections but has minimal effect on human cells. [2]

Answer:

DNA gyrase is present in prokaryotes but absent in eukaryotes/human cells [1]
DNA gyrase relieves supercoiling tension ahead of the replication fork in prokaryotes; inhibition prevents DNA replication and bacterial cell division [1]

Marking notes:

Must state that DNA gyrase is specific to prokaryotes
Must explain the consequence of inhibition (prevents replication)
Accept: "Human cells do not have DNA gyrase, so ciprofloxacin has no target"

[Total: 10 marks]

Question 4: Cell Cycle and Mitosis

(a) Identify the phases of the cell cycle represented by the regions labelled A, B, and C in Fig. 4.1. [3]

Answer:

A: G1 phase [1]
B: S phase (DNA synthesis) [1]
C: G2 phase [1]

Marking notes:

Must correctly identify all three phases
Accept: "G1, S, G2" in order
Reject: "Interphase" for any single region (interphase includes G1, S, and G2)

(b) Colchicine is a drug that prevents the polymerisation of tubulin into microtubules. Explain how colchicine affects mitosis and state the stage at which mitosis would be arrested. [3]

Answer:

Colchicine prevents the formation of spindle fibres/microtubules [1]
Without spindle fibres, chromosomes cannot attach to the spindle and cannot be separated/moved to opposite poles [1]
Mitosis would be arrested at metaphase (as chromosomes cannot align at the metaphase plate and anaphase cannot proceed) [1]

Marking notes:

Must link tubulin polymerisation to spindle fibre formation
Must explain the consequence (no chromosome separation)
Must state metaphase arrest
Accept: "Arrested at prometaphase/metaphase transition"

(c)(i) Calculate the mitotic index for this root tip. Show your working. [2]

Answer:

Number of cells in mitosis = 42 + 18 + 10 + 30 = 100 [1]
Mitotic index = (Number of cells in mitosis / Total number of cells) × 100 = (100 / 200) × 100 = 50% [1]

Marking notes:

Must show correct addition of mitotic cells (prophase + metaphase + anaphase + telophase)
Must show correct calculation
Accept: 0.5 or 50%
Award 1 mark for correct method even if final answer is incorrect due to arithmetic error

(c)(ii) The total duration of the cell cycle in this root tip is 20 hours. Calculate the duration of metaphase. Show your working. [2]

Answer:

Proportion of cells in metaphase = 18 / 200 = 0.09 [1]
Duration of metaphase = 0.09 × 20 hours = 1.8 hours (or 108 minutes) [1]

Marking notes:

Must show correct proportion calculation
Must multiply by total cell cycle duration
Accept: 1.8 hours or 108 minutes
Award 1 mark for correct method even if final answer is incorrect

[Total: 10 marks]

Question 5: Protein Structure and Function

(a) Compare the oxygen dissociation curves of HbA and HbF. [2]

Answer:

The HbF curve is shifted to the left of the HbA curve [1]
HbF has a higher affinity for oxygen than HbA at any given partial pressure of oxygen / HbF becomes saturated at a lower pO₂ than HbA [1]

Marking notes:

Must state left shift
Must state higher affinity
Accept: "HbF has a lower P₅₀ than HbA"

(b) Explain the physiological significance of the difference in oxygen affinity between HbA and HbF. [3]

Answer:

Foetal haemoglobin (HbF) must obtain oxygen from maternal haemoglobin (HbA) in the placenta [1]
The higher oxygen affinity of HbF allows it to bind oxygen at the lower pO₂ found in the placenta / allows oxygen to move from maternal blood to foetal blood [1]
This ensures adequate oxygen supply to the developing foetus for aerobic respiration and growth [1]

Marking notes:

Must mention oxygen transfer from mother to foetus
Must link higher affinity to function at placental pO₂
Must state the significance (foetal oxygen supply)
Accept: "HbF can load oxygen at pO₂ where HbA unloads"

(c)(i) Explain how this single amino acid substitution leads to the sickling of red blood cells under low oxygen tension. [3]

Answer:

Glutamic acid is hydrophilic/polar/negatively charged, while valine is hydrophobic/non-polar [1]
The substitution causes a hydrophobic patch on the surface of the β-globin chain [1]
Under low oxygen tension, deoxyhaemoglobin molecules aggregate/polymerise via hydrophobic interactions, forming long fibres that distort the red blood cell into a sickle shape [1]

Marking notes:

Must contrast properties of glutamic acid and valine
Must mention hydrophobic interactions/aggregation
Must link to sickling under low oxygen
Accept: "Hydrophobic valine residues stick together"

(c)(ii) Suggest why individuals who are heterozygous for the sickle cell allele (HbA/HbS) have a selective advantage in regions where malaria is endemic. [2]

Answer:

Heterozygotes produce both normal HbA and sickle HbS; they have some sickle-shaped cells but generally do not suffer from severe sickle cell anaemia [1]
The presence of HbS/sickle trait provides some resistance to malaria because the malaria parasite (Plasmodium) has reduced survival in sickle-shaped cells / infected cells are removed more rapidly by the spleen [1]

Marking notes:

Must state that heterozygotes have malaria resistance
Must explain the mechanism (reduced parasite survival)
Accept: "Heterozygotes have a balanced polymorphism"
Reject: "Heterozygotes are immune to malaria" (they are resistant, not immune)

[Total: 10 marks]

SECTION B: Long Structured Questions (15 marks)

Question 6: Cellular Respiration

(a) Outline the process of glycolysis, stating the location, the initial substrate, and the net products. [4]

Answer:

Location: Cytoplasm/cytosol [1]
Initial substrate: Glucose (6C) [1]
Process: Glucose is phosphorylated (using 2 ATP) and then split into two molecules of triose phosphate (3C); triose phosphate is oxidised to pyruvate (3C) [1]
Net products: 2 ATP (by substrate-level phosphorylation), 2 reduced NAD (NADH), and 2 pyruvate [1]

Marking notes:

Must state location, substrate, and net products
Must mention phosphorylation and splitting
Accept: "Net gain of 2 ATP" (gross 4 ATP minus 2 ATP used)
Reject: "Produces 4 ATP" without stating net gain

(b) Describe the role of coenzymes in the link reaction and the Krebs cycle. [3]

Answer:

NAD⁺ accepts hydrogen atoms/electrons and is reduced to NADH [1]
FAD accepts hydrogen atoms/electrons and is reduced to FADH₂ [1]
NADH and FADH₂ carry electrons/hydrogen to the electron transport chain for ATP synthesis via oxidative phosphorylation [1]

Marking notes:

Must mention both NAD and FAD
Must state they are reduced (accept electrons/hydrogen)
Must state they carry electrons to the ETC
Accept: "Coenzyme A carries acetyl groups into the Krebs cycle" (for link reaction context)

(c) With reference to Fig. 6.1, explain how the transfer of electrons along the electron transport chain leads to the synthesis of ATP. [5]

Answer:

Electrons from NADH and FADH₂ are passed along a series of electron carriers (Complexes I, II, III, IV) in the inner mitochondrial membrane [1]
As electrons are transferred, energy is released [1]
This energy is used to pump protons (H⁺) from the mitochondrial matrix into the intermembrane space [1]
This creates a proton gradient / electrochemical gradient across the inner mitochondrial membrane (higher [H⁺] in intermembrane space) [1]
Protons flow back down their concentration gradient through ATP synthase (chemiosmosis), and this flow drives the synthesis of ATP from ADP and Pi [1]

Marking notes:

Must describe electron transfer, proton pumping, gradient formation, and chemiosmosis
Must mention ATP synthase
Accept: "Proton-motive force drives ATP synthesis"
Award marks for each key step

(d) Cyanide is a potent inhibitor of cytochrome c oxidase (Complex IV) in the electron transport chain. Explain why cyanide poisoning is rapidly fatal. [3]

Answer:

Cyanide inhibits Complex IV, preventing the transfer of electrons to oxygen (the final electron acceptor) [1]
This halts the entire electron transport chain; no proton gradient is generated [1]
Without the proton gradient, ATP synthesis via oxidative phosphorylation stops; cells cannot produce sufficient ATP for vital processes, leading to rapid cell death, particularly in tissues with high energy demand (brain, heart) [1]

Marking notes:

Must state that electron transfer to oxygen is blocked
Must explain that proton gradient cannot form
Must link to ATP depletion and cell death
Accept: "Cyanide causes histotoxic hypoxia"

[Total: 15 marks]

Question 7: Gene Expression and Regulation

(a) Describe the process of transcription in prokaryotes, including the roles of RNA polymerase and the promoter. [4]

Answer:

RNA polymerase binds to the promoter region of the DNA (a specific sequence upstream of the gene) [1]
RNA polymerase unwinds the DNA double helix, exposing the template strand [1]
RNA polymerase synthesises a complementary mRNA strand in the 5' to 3' direction using ribonucleoside triphosphates (ATP, UTP, GTP, CTP) [1]
Transcription continues until RNA polymerase reaches a terminator sequence, where it detaches and the mRNA is released [1]

Marking notes:

Must mention promoter binding, unwinding, synthesis, and termination
Must state direction of synthesis (5' to 3')
Accept: "Sigma factor helps RNA polymerase recognise the promoter" (prokaryotic detail)
Reject: "RNA polymerase binds to the start codon" (start codon is on mRNA, not DNA)

(b) The lac operon in E. coli is an example of an inducible operon. Explain how the lac operon is regulated in the presence and absence of lactose. [5]

Answer:

In the absence of lactose: The lac repressor protein is active and binds to the operator region [1]; this blocks RNA polymerase from binding to the promoter/prevents transcription of the structural genes (lacZ, lacY, lacA) [1]
In the presence of lactose: Lactose (or allolactose, an isomer of lactose) acts as an inducer [1]; the inducer binds to the lac repressor protein, causing a conformational change that inactivates the repressor [1]; the repressor can no longer bind to the operator, allowing RNA polymerase to transcribe the structural genes, producing enzymes for lactose metabolism [1]

Marking notes:

Must describe both absence and presence scenarios
Must mention repressor, operator, and inducer
Must explain the conformational change of the repressor
Accept: "Allolactose is the natural inducer"
Reject: "Lactose binds to the promoter"

(c) With reference to Fig. 7.1, explain the differences between the eukaryotic gene and its mature mRNA transcript. [3]

Answer:

The eukaryotic gene contains introns (non-coding sequences) and exons (coding sequences), while the mature mRNA contains only exons [1]
During RNA processing, introns are removed by splicing (spliceosomes remove introns and join exons together) [1]
The mature mRNA also has a 5' cap (modified guanine nucleotide) and a 3' poly-A tail, which are not present in the original gene sequence [1]

Marking notes:

Must mention intron removal/splicing
Must mention 5' cap and 3' poly-A tail
Accept: "The gene is longer than the mature mRNA due to introns"

(d) Suggest why the regulation of gene expression in eukaryotes is more complex than in prokaryotes. [3]

Answer:

Eukaryotes have more genes and larger genomes that require more sophisticated regulation [1]
Eukaryotic DNA is packaged with histones into chromatin; gene expression requires chromatin remodelling to make genes accessible [1]
Eukaryotes are multicellular with different cell types; differential gene expression allows cell specialisation/differentiation [1]

Marking notes:

Must mention chromatin structure
Must mention cell differentiation/multicellularity
Accept: "Eukaryotes have more levels of regulation (transcriptional, post-transcriptional, translational, post-translational)"
Accept: "Eukaryotic transcription occurs in the nucleus, translation in the cytoplasm, allowing more regulatory steps"

[Total: 15 marks]

SECTION C: Data-Based and Application Questions (15 marks)

Question 8: Gel Electrophoresis and Genetic Disease Diagnosis

(a) Describe and explain how gel electrophoresis separates proteins. [4]

Answer:

Protein samples are loaded into wells in a gel (usually polyacrylamide) [1]
An electric field/potential difference is applied across the gel [1]
Proteins migrate through the gel based on their charge and molecular mass/size [1]
Smaller proteins move faster/through the gel more easily, while larger proteins are retarded by the gel matrix; proteins with different charges migrate at different rates [1]

Marking notes:

Must mention electric field application
Must mention separation by size and charge
Must explain why smaller proteins move faster
Accept: "SDS-PAGE denatures proteins and gives them uniform negative charge, so separation is by size only"

(b) With reference to Fig. 8.1, identify the genotype of each individual (W, X, Y, and Z) with respect to the HbA and HbS alleles. Explain your reasoning for each. [4]

Answer:

W: HbA/HbA (homozygous normal) - shows only one band corresponding to HbA [1]
X: HbA/HbS (heterozygous) - shows two bands, one for HbA and one for HbS [1]
Y: HbS/HbS (homozygous sickle cell) - shows only one band corresponding to HbS [1]
Z: HbA/HbA (homozygous normal) - shows only one band corresponding to HbA [1]

Marking notes:

Must correctly identify all four genotypes
Must explain reasoning for each (number and position of bands)
Award 1 mark per correct genotype with reasoning
Accept: "HbAA, HbAS, HbSS" notation

(c) Individual X is heterozygous for the sickle cell allele. Explain why individual X shows two bands on the gel. [2]

Answer:

Individual X has two different alleles (HbA and HbS), each coding for a different form of haemoglobin [1]
The two haemoglobin variants have different charges/molecular masses, so they migrate to different positions on the gel, producing two distinct bands [1]

Marking notes:

Must state that two different alleles produce two different proteins
Must explain why the proteins separate (different charge/mass)
Accept: "HbS has a different amino acid sequence, altering its charge"

(d) A couple, both with the same genotype as individual X, are planning to have a child. Using a genetic diagram, determine the probability that their child will have sickle cell anaemia. [3]

Answer:

Parental genotypes: HbA/HbS × HbA/HbS [1]
Gametes: HbA and HbS from each parent [1]
Punnett square:
HbA HbS
HbA HbA/HbA HbA/HbS
HbS HbA/HbS HbS/HbS
Probability of child having sickle cell anaemia (HbS/HbS) = 1/4 or 25% [1]

	HbA	HbS
HbA	HbA/HbA	HbA/HbS
HbS	HbA/HbS	HbS/HbS

Marking notes:

Must show correct parental genotypes and gametes
Must show Punnett square or equivalent
Must state correct probability
Award 1 mark for correct genotypes, 1 mark for correct gametes/cross, 1 mark for correct probability

(e) Discuss the ethical considerations of using genetic screening for sickle cell anaemia in pre-natal diagnosis. [2]

Answer:

Genetic screening allows parents to make informed reproductive decisions / prepare for a child with sickle cell anaemia [1]
However, there are ethical concerns including: potential for discrimination against affected individuals, the possibility of termination of affected pregnancies (which some consider unethical), and the psychological impact on parents [1]

Marking notes:

Must present at least one benefit and one concern
Accept: "Issues of consent, confidentiality, and genetic privacy"
Accept: "Screening may lead to stigmatisation of carriers"
Award 1 mark for a valid benefit, 1 mark for a valid ethical concern

[Total: 15 marks]

Question 9: Mitochondrial Function and Metabolic Inhibitors

(a) Explain why oxygen consumption by the mitochondria is low before the addition of succinate at point A. [2]

Answer:

Before succinate addition, there is no substrate available for the electron transport chain / Krebs cycle [1]
Without substrate, there are no electrons to pass along the electron transport chain, so oxygen (the final electron acceptor) is not consumed [1]

Marking notes:

Must state lack of substrate
Must link to absence of electron flow and oxygen consumption
Accept: "Mitochondria require respiratory substrate to generate NADH/FADH₂"

(b) With reference to Fig. 9.1, describe and explain the change in the rate of oxygen consumption after the addition of ADP at point B. [3]

Answer:

The rate of oxygen consumption increases (the slope becomes steeper) [1]
ADP is required for ATP synthesis by ATP synthase [1]
The addition of ADP allows ATP synthesis to proceed; this uses the proton gradient, which increases electron flow along the ETC and thus increases oxygen consumption [1]

Marking notes:

Must describe the increase in rate
Must explain that ADP is needed for ATP synthesis
Must link to increased electron flow (respiratory control)
Accept: "ADP stimulates respiration / respiratory control"

(c)(i) Sodium azide inhibits cytochrome c oxidase (Complex IV). Explain the effect of sodium azide on the rate of oxygen consumption. [2]

Answer:

Sodium azide inhibits Complex IV, preventing the transfer of electrons to oxygen [1]
Therefore, oxygen consumption decreases/stops because oxygen can no longer act as the final electron acceptor [1]

Marking notes:

Must state that electron transfer to oxygen is blocked
Must state that oxygen consumption decreases/stops
Accept: "The ETC is blocked, so oxygen is not reduced to water"

(c)(ii) Explain the effect of sodium azide on the synthesis of ATP. [2]

Answer:

Inhibition of Complex IV halts the entire electron transport chain [1]
No proton gradient is generated, so ATP synthase cannot synthesise ATP; ATP synthesis stops [1]

Marking notes:

Must explain that the ETC is blocked
Must link to absence of proton gradient and ATP synthesis
Accept: "No proton-motive force, so no ATP production"

(d) The experiment was repeated using pyruvate instead of succinate as the substrate. Predict and explain how the oxygen electrode trace would differ from Fig. 9.1. [3]

Answer:

Pyruvate enters the Krebs cycle via the link reaction (producing acetyl-CoA), while succinate enters directly at Complex II [1]
Pyruvate would generate more NADH and FADH₂ per molecule than succinate (pyruvate yields more electrons) [1]
Therefore, the rate of oxygen consumption would be higher / the slope would be steeper after substrate addition compared to succinate [1]

Marking notes:

Must state that pyruvate generates more reducing equivalents
Must predict higher oxygen consumption rate
Accept: "Pyruvate produces NADH via the link reaction and Krebs cycle, while succinate only produces FADH₂"
Reject: "Pyruvate produces less ATP" (it produces more)

(e)(i) Uncouplers such as 2,4-dinitrophenol (DNP) increase the permeability of the inner mitochondrial membrane to protons. Predict the effect of DNP on the rate of oxygen consumption. [1]

Answer:

The rate of oxygen consumption would increase [1]

Marking notes:

Must state increase
Accept: "Oxygen consumption continues/increases even without ADP"

(e)(ii) Predict and explain the effect of DNP on the rate of ATP synthesis. [2]

Answer:

The rate of ATP synthesis would decrease/stop [1]
DNP dissipates the proton gradient by allowing protons to leak back across the inner membrane without passing through ATP synthase; without the proton gradient, ATP synthase cannot synthesise ATP [1]

Marking notes:

Must state decrease in ATP synthesis
Must explain that the proton gradient is dissipated
Accept: "DNP uncouples electron transport from ATP synthesis"
Reject: "DNP inhibits the ETC" (it does not inhibit electron transport, it uncouples it from ATP synthesis)

[Total: 15 marks]

END OF ANSWER KEY