A Level H1 Biology Practice Paper 5

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TuitionGoWhere Practice Paper 2 – Biology H1 A-Level (Version 5)

TuitionGoWhere Secondary School (AI)

Subject: Biology H1
Level: A-Level
Paper: Practice Paper 2 (Structured & Free-Response)
Duration: 2 hours
Total Marks: 80
Name: _______________________
Class: _______________________
Date: _______________________

Instructions to Candidates

• Write your name, class and date in the spaces provided.
• Answer all questions in Section A and at least one question from Section B.
• Write your answers in the blank spaces provided. You may use separate paper if more space is needed.
• The number of marks for each question or part question is shown in brackets [ ].
• You are advised to spend about 1 hour 30 minutes on Section A and 30 minutes on Section B.
• Read the questions carefully. Where a question refers to a figure, make sure your answer clearly references that figure.
• Use diagrams and labelled sketches where appropriate to support your answers.
• Electronic calculators may be used, but all working must be shown.

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Section A: Structured Questions (60 marks)

Answer all questions in this section.

1. Fig. 5.1 shows a graph of the relative DNA content per nucleus in a culture of actively dividing mammalian cells. The arrow indicates the point at which a short pulse of radioactive thymidine was added to the culture. The radioactive thymidine is incorporated into DNA during replication.

Fig. 5.1: Graph of relative DNA content per nucleus against time. 
- From 0–4 h, DNA content remains at 1 unit.
- From 4–8 h, DNA content rises steadily from 1 to 2 units.
- From 8–10 h, DNA content stays at 2 units.
- At 10 h, a sharp drop to 1 unit per nucleus occurs (cytokinesis), after which the cycle repeats.
Radioactive pulse added at time 2 h.

(a) (i) With reference to Fig. 5.1, identify the phase of the cell cycle during which the radioactive thymidine is initially incorporated. [1]

(ii) Explain your answer in (a)(i). [2]

(b) Explain why the amount of DNA per nucleus remains constant between 8 and 10 hours, despite radioactive thymidine having been incorporated earlier. [2]

(c) Describe one other experimental method that could be used to confirm the timing of DNA replication in this cell culture, apart from radioactive labelling. [2]

[Total: 7 marks]

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2. The cell membrane is a dynamic structure composed mainly of phospholipids and proteins. Fig. 5.2 diagrams the arrangement of phospholipid molecules in a plasma membrane.

Fig. 5.2: A simplified cross-section of the plasma membrane showing a phospholipid bilayer.
- Head groups (circles) are on the outer and inner surfaces.
- Two wavy lines (fatty acid tails) extend inward from each head group, meeting in the centre.

(a) With reference to Fig. 5.2, describe the arrangement of phospholipids in a cell membrane. [2]

(b) Explain why this arrangement is essential for the membrane’s function as a selective barrier. [3]

(c) The membrane also contains cholesterol. Explain the role of cholesterol in animal cell membranes. [2]

[Total: 7 marks]

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3. A student investigated the effect of temperature on the rate of oxygen uptake by germinating seeds. The results are shown in Table 5.1.

Temperature (°C)	Rate of oxygen uptake (cm³ O₂ min⁻¹ g⁻¹ dry mass)
10	0.12
20	0.28
30	0.45
40	0.52
50	0.23
60	0.05

Table 5.1

(a) With reference to Table 5.1 and your knowledge of enzymes, explain why the rate of oxygen uptake increases from 10°C to 40°C. [3]

(b) Suggest why the rate of oxygen uptake decreases above 40°C. [2]

(c) The student concluded that the optimum temperature for respiration in the seeds is 40°C. Evaluate this conclusion. [2]

[Total: 7 marks]

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4. Fig. 5.3 is an electron micrograph of a liver cell.

Fig. 5.3: Electron micrograph showing several organelles.
- Structure A: elongated, with a double membrane and folded inner membrane (cristae).
- Structure B: network of membrane-bound flattened sacs with ribosomes attached to the cytoplasmic surface.
- Structure C: stack of flattened, membrane-bound cisternae with small vesicles budding off.

(a) (i) Name structure A. [1]
(ii) State the function of structure A in a liver cell. [1]

(b) (i) Name structure B. [1]
(ii) Explain why liver cells contain extensive amounts of structure B. [2]

(c) Describe the role of structure C in processing proteins that are destined for secretion from the liver cell. [3]

[Total: 8 marks]

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5. Isolated mitochondria were incubated with either pyruvate or glucose in separate experiments. The production of carbon dioxide was measured. The results are summarised in Table 5.2.

Substrate incubated with mitochondria	CO₂ produced after 30 minutes
Pyruvate	Yes
Glucose	No

Table 5.2

(a) Explain why carbon dioxide is produced when mitochondria are incubated with pyruvate but not when incubated with glucose. [3]

(b) Name the metabolic pathway in which the carbon dioxide is produced, and state its location within the mitochondrion. [2]

(c) Predict the effect on CO₂ production if the mitochondria were incubated with pyruvate and an inhibitor of the enzyme pyruvate dehydrogenase. Justify your prediction. [3]

[Total: 8 marks]

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6. Fig. 5.4 shows a pedigree of a family with a rare inherited disorder that results in progressive muscle weakness. The disorder is controlled by a single autosomal recessive gene. Squares represent males, circles represent females. Filled symbols are affected individuals.

Fig. 5.4: Pedigree chart (three generations).
Generation I:  1 (male, unaffected) mated to 2 (female, unaffected).
               They have two children: 
               Generation II: 3 (male, affected), 4 (female, unaffected).
               4 mates with 5 (male, unaffected from outside family).
               They have three children:
               Generation III: 6 (female, unaffected), 7 (male, affected), 8 (male, unaffected).

Use the symbols D for the normal allele and d for the disease-causing allele.

(a) State the genotypes of the following individuals:
(i) I‑2 [1]
(ii) II‑4 [1]
(iii) III‑8 [1]

(b) With reference to Fig. 5.4 and your answers in (a), explain how you deduced the genotype of individual II‑4. [2]

(c) If individual III‑6 marries a man who is a carrier of the disease allele, what is the probability that their first child will be affected? Construct a genetic diagram to support your answer. [3]

[Total: 8 marks]

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7. In poultry, a gene for feather colour is carried on an autosome. This gene has two codominant alleles: C^B produces black feathers and C^W produces splashed-white feathers. Heterozygous birds have blue feathers. Barring (a pattern of white bars on feathers) is controlled by a sex-linked dominant allele B, and is absent in b homozygous recessive birds (non-barred). Females are the heterogametic sex (ZW), males are homogametic (ZZ).

(a) State the genotype of a blue-feathered, non-barred male. [1]

(b) A homozygous black-feathered, barred female is crossed with a splashed-white, non-barred male.
(i) Determine the genotypes of the parents. [2]
(ii) State the expected phenotype of the F₁ offspring. [1]
(iii) If the F₁ male offspring are mated with the F₁ female offspring, what is the expected phenotypic ratio in the F₂ generation for feather colour (ignoring barring)? Show your working. [3]

[Total: 7 marks]

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8. Fig. 5.5 shows the changes in DNA content per cell during two types of nuclear division in a diploid organism (2n = 4). Curve X represents mitosis; Curve Y represents meiosis.

Fig. 5.5: Graph of relative DNA content per cell (4C scale) against stage.
Curve X: rises from 2C to 4C in S phase, then after mitosis returns to 2C.
Curve Y: rises from 2C to 4C in S phase, then after meiosis I falls to 2C, and after meiosis II falls to 1C.

(a) With reference to Fig. 5.5, explain why the DNA content per cell returns to 2C after mitosis. [2]

(b) Explain the changes in DNA content per cell from the start of meiosis I to the end of meiosis II. [3]

(c) Discuss the significance of the DNA content changes shown in Curve Y for the production of genetically varied gametes. [3]

[Total: 8 marks]

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Section B: Free-Response Questions (20 marks)

Answer one question from this section. Write your answer in the spaces provided. Where appropriate, use diagrams and examples to illustrate your points.

9. (a) Discuss the significance of the movement of substances across cell membranes to the process of photosynthesis. [10]
(b) Explain how the structure of mitochondria is adapted for the production of ATP during aerobic respiration. [10]

OR

10. Describe the role of the cytoskeleton in maintaining cell shape and facilitating intracellular transport. Use specific examples to illustrate your answer. [20]

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END OF PAPER

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TuitionGoWhere Practice Paper 2 – Biology H1 A-Level (Version 5)

MARKING SCHEME

Subject: Biology H1
Level: A-Level
Paper: Practice Paper 2
Version: 5
Total Marks: 80

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Section A: Structured Questions (60 marks)

Question 1 (7 marks)

(a) (i) S phase (Synthesis phase) [1]
(a) (ii) Thymidine is a nucleoside that is specifically incorporated into DNA. In the cell cycle, DNA replication occurs during S phase; therefore, radioactive thymidine will be incorporated into newly synthesised DNA during S phase. Reference to Fig. 5.1: the steep rise in DNA content from 1 to 2 units between 4 and 8 h corresponds to S phase, so radioactive thymidine added at 2 h would be taken up once cells reach S phase. [2] – 1 mark for linking thymidine to DNA replication, 1 mark for referencing the relevant rise in DNA content in the graph.
(b) Between 8 and 10 hours the DNA content remains at 2 units because DNA replication is complete; the cell is in G2 phase where no net synthesis of DNA occurs. The chromosomes consist of two sister chromatids held together at the centromere, so DNA mass is twice the normal amount, but no further replication happens until after mitosis. [2] – 1 mark for identifying G2, 1 mark for sister chromatid explanation.
(c) Possible methods:

• Use of a DNA-specific fluorescent dye (e.g., DAPI) combined with flow cytometry to measure DNA content per cell at different time points; cells in S phase will show an intermediate DNA content between 2C and 4C. [2]
  OR
• Autoradiography of cells after pulsing with radioactive thymidine; silver grains will be localised over nuclei of cells that were in S phase during the pulse. [2]
  (Any one reasonable alternative with brief description; award 2 marks for clear method linking to timing.)

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Question 2 (7 marks)

(a) The phospholipids form a bilayer. Hydrophilic (polar) phosphate heads face outward on both the extracellular and cytoplasmic sides, while hydrophobic (non‑polar) fatty acid tails point inward, away from water, making up the interior of the membrane. [2] – 1 mark for bilayer, 1 mark for orientation of heads and tails with reference to figure.
(b) The phospholipid bilayer is selectively permeable because the hydrophobic core prevents free passage of most water‑soluble (polar or charged) solutes, such as ions and large polar molecules. Only small, non‑polar molecules (e.g., O₂, CO₂) and very small polar molecules (e.g., water) can diffuse through the bilayer. This allows the cell to control internal composition and maintain gradients. [3] – 1 mark for hydrophobic barrier, 1 mark for restriction of polar/charged substances, 1 mark for implication of selective permeability.
(c) Cholesterol is inserted between phospholipid molecules. It modulates membrane fluidity: at low temperatures it prevents the fatty acid tails from packing too tightly (maintaining fluidity), and at high temperatures it restricts excessive movement (stabilising the membrane). [2] – 1 mark for each role (temperature‑dependent fluidity modulation).

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Question 3 (7 marks)

(a) Rate increases because increasing temperature increases the kinetic energy of enzyme and substrate molecules. This results in more frequent successful collisions between enzymes and substrates, and more molecules have the activation energy required. Thus, the rate of respiration (enzyme‑catalysed) rises. At higher temperatures (up to 40°C), enzyme molecules and substrates move faster, increasing the proportion of enzyme‑substrate complexes formed per unit time. [3] – 1 mark for kinetic energy / collision frequency, 1 mark for enzyme‑substrate complex formation, 1 mark for linking to respiration rate increase.
(b) Above 40°C, the enzymes involved in respiration begin to denature. The high temperature disrupts the hydrogen bonds, ionic bonds and hydrophobic interactions that maintain the tertiary structure of the enzymes. The active site loses its specific shape, so substrate binding decreases dramatically, reducing the rate of oxygen uptake. [2] – 1 mark for denaturation, 1 mark for loss of tertiary structure / active‑site shape.
(c) The data show the highest rate at 40°C among the tested temperatures; however, without testing temperatures between 30°C and 40°C (e.g., 35°C, 38°C) or slightly above 40°C (e.g., 42°C), we cannot be certain that 40°C is the true optimum. The conclusion is valid only for the temperature range tested; the actual optimum might be slightly higher or lower. [2] – 1 mark for need of intermediate temperatures, 1 mark for limited range.

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Question 4 (8 marks)

(a) (i) Mitochondrion [1] (accept ‘mitochondria’, but singular is preferred for a single structure);
(a) (ii) Site of aerobic respiration (Krebs cycle, electron transport chain, oxidative phosphorylation) to produce ATP for energy‑consuming metabolic processes in liver cells, such as protein synthesis, detoxification and glycogen metabolism. [1]
(b) (i) Rough endoplasmic reticulum (RER) [1]
(b) (ii) Liver cells are highly active in protein synthesis. They produce many proteins such as albumins, clotting factors and detoxification enzymes. Proteins are synthesised by ribosomes attached to the RER and translocated into the lumen for folding and transport. Therefore, a large surface area of RER is needed to support this heavy secretory activity. [2] – 1 mark for high protein synthesis role, 1 mark for RER’s function in processing secretory proteins.
(c) Structure C is the Golgi apparatus. It receives proteins from the RER in transport vesicles. Inside the Golgi, proteins undergo post‑translational modifications (e.g., glycosylation, phosphorylation). The Golgi then sorts the modified proteins and packages them into secretory vesicles that pinch off and fuse with the plasma membrane, releasing the proteins outside the cell by exocytosis. [3] – 1 mark for receiving and modifying, 1 mark for sorting/packaging, 1 mark for exocytosis.

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Question 5 (8 marks)

(a) Pyruvate can enter the mitochondrion and is converted to acetyl‑CoA by pyruvate dehydrogenase, after which acetyl‑CoA enters the Krebs cycle. During the Krebs cycle, carbon dioxide is released as a waste product. Glucose cannot directly enter the mitochondrion or the Krebs cycle because the enzymes for glycolysis are located in the cytoplasm, not inside mitochondria. Since isolated mitochondria lack the glycolytic enzymes, glucose is not metabolised and no CO₂ is produced. [3] – 1 mark for pyruvate → acetyl‑CoA → Krebs cycle, 1 mark for CO₂ production, 1 mark for glycolysis location.
(b) The pathway is the Krebs cycle (citric acid cycle), which occurs in the mitochondrial matrix. [2] – 1 mark for naming, 1 mark for location.
(c) If pyruvate dehydrogenase is inhibited, pyruvate cannot be converted to acetyl‑CoA, so the Krebs cycle cannot proceed. Therefore, CO₂ production would cease or be greatly reduced. This is because the inhibitor blocks the entry of pyruvate into the cycle, so no substrate is available for the decarboxylation steps that release CO₂. [3] – 1 mark for linking inhibition to blocked acetyl‑CoA formation, 1 mark for Krebs cycle stopping, 1 mark for explaining loss of CO₂.

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Question 6 (8 marks)

(a) Allele key: D = normal, d = disease allele (autosomal recessive).
(i) I‑2: Dd [1]
(ii) II‑4: Dd [1]
(iii) III‑8: D_ (Dd or DD; cannot be homozygous recessive because he is unaffected). Accept DD or Dd with explanation. [1]
Allow 1 mark for stating that III‑8 could be heterozygous or homozygous dominant; exact genotype Dd if deduced from later cross data? No data, so accept D-.

(b) II‑4 is unaffected, so she must have at least one D allele. Her father (I‑1) is unaffected, and her mother (I‑2) must be a carrier (Dd) because they have an affected son (II‑3, dd). I‑2 could pass on the d allele to II‑4; since II‑4 has a child (III‑7) who is affected and therefore dd, II‑4 must have contributed one d allele. Therefore II‑4 is Dd. [2] – 1 mark for inferring from affected child, 1 mark for using pedigree evidence linking to mother’s carrier status.

(c) III‑6 is unaffected; her brother is affected (dd), and both parents are Dd. Thus, III‑6 has a 2/3 chance of being a carrier (Dd) and a 1/3 chance of being DD (since the Punnett cross of two Dd individuals gives 1 DD : 2 Dd : 1 dd, and affected dd is excluded). She marries a carrier male (Dd).
Probability that their first child is affected = (probability III‑6 is carrier) × (probability of passing d allele from carrier mother) × (probability father passes d).
= (\frac{2}{3} \times \frac{1}{2} \times \frac{1}{2} = \frac{2}{3} \times \frac{1}{4} = \frac{2}{12} = \frac{1}{6}).
Alternatively, genetic diagram:

	D	d
D	DD (normal)	Dd (carrier)
d	Dd (carrier)	dd (affected)

Chance III‑6 is Dd = 2/3.
P(child affected | mother Dd) = 1/4.
So overall probability = ( \frac{2}{3} \times \frac{1}{4} = \frac{1}{6} ).
[3] – 1 mark for correct probability that III‑6 is carrier (2/3), 1 mark for Punnett square and probability of affected child from Dd × Dd = 1/4, 1 mark for final product 1/6.

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Question 7 (7 marks)

(a) For blue feathers, genotype is C^B C^W (heterozygous autosomal). Non‑barred: male must be homozygous recessive on Z chromosome because barring is dominant on Z; Z^b Z^b. So genotype: C^B C^W, Z^b Z^b. [1] – 1 mark for both loci correct.

(b) (i) Homozygous black female: C^B C^B, and barred female (ZW) must have B on Z and W is non‑homologous, so female genotype is Z^B W. Splashed‑white, non‑barred male: C^W C^W, Z^b Z^b. [2] – 1 mark for autosomal genotypes, 1 mark for sex‑linked genotypes.
(ii) Cross female C^B C^B Z^B W × male C^W C^W Z^b Z^b:

• Offspring autosomal: all receive C^B from mother and C^W from father → all C^B C^W (blue feathers).
• Sex‑linked: male offspring (ZZ) receive Z^b from father and Z^B from mother → Z^B Z^b (barred males). Female offspring (ZW) receive Z^b from father and W from mother → Z^b W (non‑barred females). F₁: all blue feathers; males barred, females non‑barred. [1] for stating all blue, with barring pattern.
  (iii) F₁ male C^B C^W Z^B Z^b mate with F₁ female C^B C^W Z^b W. For feather colour only (ignoring barring): cross autosomal C^B C^W × C^B C^W → 1 : 2 : 1 ratio, i.e., 1 black : 2 blue : 1 splashed white. [3] – 1 mark for parental genotype (C^B C^W each), 1 mark for Punnett square or gamete method, 1 mark for correct ratio.

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Question 8 (8 marks)

(a) In mitosis, the 4C state is reached after DNA replication (S phase). During mitosis, sister chromatids separate at anaphase, and each daughter cell receives a set of single chromatids (now chromosomes) that each contain 1C of DNA. Since there are 2n = 4 chromosomes, total DNA per daughter cell is 2C. Thus, the DNA content returns to the diploid amount. [2] – 1 mark for anaphase separation, 1 mark for halving to 2C.

(b) At the start of meiosis I, DNA has already been replicated (4C). During anaphase I, homologous chromosomes separate, each still consisting of two sister chromatids; the DNA content per cell halves to 2C at telophase I. In meiosis II, sister chromatids separate, reducing the DNA content per cell from 2C (after meiosis I) to 1C at telophase II. Therefore, the net change is from 4C → 2C → 1C. [3] – 1 mark for homologous chromosome separation halving to 2C, 1 mark for sister chromatid separation reducing 2C to 1C, 1 mark for linking to stages.

(c) The reduction from 2C to 1C during meiosis ensures that gametes are haploid (n). This, combined with genetic recombination (crossing over and independent assortment), creates gametes with unique combinations of alleles. The halving of DNA content allows the fusion of gametes at fertilisation to restore the diploid number, maintaining species chromosome number and promoting genetic variation in offspring. [3] – 1 mark for haploid gamete production, 1 mark for recombination/crossing over contribution, 1 mark for restoration of diploid at fertilisation and variation.

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Section B: Free-Response Questions (20 marks)

Question 9

(a) Significance of membrane transport to photosynthesis (10 marks)

Marking guidance:

• 1 mark for introducing the role of membranes in controlling movement of substances.
• 2 marks for CO₂ diffusion: CO₂ enters leaf through stomata (diffusion across plasma membrane of guard cells into air spaces), then diffuses across cell wall and plasma membrane of mesophyll cells into chloroplasts. Down concentration gradient; essential for Calvin cycle.
• 2 marks for water uptake: water moves into root cells by osmosis across selectively permeable membranes; the water potential gradient drives water movement to leaves for photolysis in the light-dependent reactions.
• 2 marks for ion transport: active transport of mineral ions (e.g., K⁺, Mg²⁺) across root cell membranes; Mg²⁺ needed for chlorophyll synthesis, K⁺ for stomatal opening; these processes are linked to membrane proteins (carriers, pumps).
• 2 marks for product export: glucose/photosynthate must cross chloroplast envelope, plasma membrane of mesophyll cells, and enter phloem for translocation; transport proteins (e.g., sucrose‑H⁺ symporters) facilitate this active loading.
• 1 mark for conclusion: membrane transport is integral to supplying raw materials, maintaining ion gradients, and exporting products, thereby directly influencing the rate and efficiency of photosynthesis.

(b) Adaptation of mitochondrion for ATP production (10 marks)

Marking guidance:

• 1 mark: mitochondria are the site of aerobic respiration; ATP is produced by oxidative phosphorylation.
• 2 marks: inner membrane highly folded into cristae → large surface area for electron transport chain (ETC) components and ATP synthase complexes.
• 2 marks: matrix contains enzymes of the Krebs cycle and fatty acid oxidation; pyruvate dehydrogenase complex is located here, ensuring acetyl‑CoA production directly feeds the cycle.
• 2 marks: intermembrane space is small → allows proton gradient to be established quickly; protons pumped from matrix to intermembrane space by ETC create a steep electrochemical gradient.
• 1 mark: outer membrane is permeable to small molecules (porins) allowing entry of pyruvate, O₂, etc.; inner membrane is impermeable to most ions, maintaining the proton gradient.
• 1 mark: ATP synthase complexes embedded in inner membrane; protons flow back through them, driving ATP synthesis (chemiosmosis).
• 1 mark: cristae increase the number of ETC and ATP synthase complexes, maximising ATP output per mitochondrion.

Question 10 (alternative to Question 9)

Role of the cytoskeleton in cell shape and intracellular transport (20 marks)
Marking guidance: A full essay should include:

• Introduction: cytoskeleton composed of microfilaments (actin), intermediate filaments, and microtubules.
• Cell shape:
  • Actin filaments form a cortical network beneath plasma membrane, supporting cell shape and enabling changes (e.g., pseudopodia formation).
  • Intermediate filaments provide mechanical strength and anchor organelles (e.g., lamins in nuclear envelope).
  • Microtubules resist compression and help maintain cell asymmetry (e.g., in neurons).
• Intracellular transport:
  • Motor proteins (kinesin and dynein) move along microtubules, transporting vesicles, organelles, and macromolecules. Kinesin moves cargo towards the plus end (often towards the cell periphery); dynein moves towards the minus end (towards the nucleus).
  • Actin filaments serve as tracks for myosin motors, e.g., in cytoplasmic streaming and in muscle contraction (myosin II).
  • Mitotic spindle composed of microtubules segregates chromosomes during cell division.
• Examples: axonal transport in nerve cells (kinesin‑driven vesicle movement), movement of melanosomes in fish pigment cells, organelle positioning in epithelial cells.
• Interactive role: cytoskeleton works with motor proteins and adaptor complexes to ensure targeted delivery.

Award marks for accurate description and suitable examples, with diagrams if used. Up to 20 marks distributed as: 5 for structural components and functions, 5 for cell shape mechanisms, 5 for transport mechanisms and motor proteins, 5 for examples and integration.

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END OF MARKING SCHEME