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A Level H1 Biology Practice Paper 1
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Questions
TuitionGoWhere Practice Paper – Biology H1 A‑Level
TuitionGoWhere Exam Practice (AI)
PRACTICE PAPER – Version 1
Subject: H1 Biology (8876)
Level: A‑Level
Paper: Paper 2 (Practice)
Duration: 2 hours
Total Marks: 80
Name: _______________
Class: _______________
Date: _______________
INSTRUCTIONS
- Answer all questions in Section A.
- Section B contains one essay question. Answer this question.
- 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 1 hour 30 minutes on Section A and 30 minutes on Section B.
SECTION A (60 marks)
Answer all questions in the spaces provided.
1 Fig. 1.1 represents the fluid‑mosaic model of a cell membrane.
(a) Name structures P, Q and R. [3]
P = __________________________
Q = __________________________
R = __________________________
(b) Describe how a small, non‑polar molecule such as oxygen moves across the membrane. [2]
(c) With reference to Fig. 1.1, explain how glucose molecules can be transported into a cell against a concentration gradient. [3]
(d) State one function of cholesterol in animal cell membranes. [2]
2 Fig. 2.1 is an electron micrograph of a liver cell.
(a) Name structures X, Y and Z. [3]
X = __________________________
Y = __________________________
Z = __________________________
(b) Outline the role of structure X in cellular respiration. [2]
(c) Explain why liver cells have extensive networks of structure Y. [3]
3 Biomolecules.
(a) Compare the structure and function of amylose and cellulose. [4]
(b) Describe the structure of a phospholipid and explain how this structure allows it to form a bilayer in an aqueous environment. [4]
4 Fig. 4.1 shows the activity of an enzyme at different temperatures.
(a) With reference to Fig. 4.1, explain the effect of temperature on enzyme activity. [4]
(b) Describe how a non‑competitive inhibitor affects enzyme activity. [3]
(c) Explain why a competitive inhibitor can be overcome by increasing the substrate concentration. [3]
5 Fig. 5.1 shows stages of mitosis in an animal cell.
(a) Identify stages A, B and C, and for each stage state one event that occurs. [6]
Stage A: __________________________
Event: ______________________________
Stage B: __________________________
Event: ______________________________
Stage C: __________________________
Event: ______________________________
(b) Explain the significance of mitosis in growth and repair. [2]
6 DNA and the genetic code.
(a) Name the three components of a DNA nucleotide. [2]
(b) Explain how the base‑pairing rule ensures that semi‑conservative replication produces two identical DNA molecules. [3]
(c) A DNA strand has the base sequence ATGC. Determine the sequence of the complementary strand. [1]
(d) State one structural difference between DNA and mRNA. [2]
7 Photosynthesis.
(a) Describe the role of light in the light‑dependent reactions of photosynthesis. [4]
(b) Explain how the products of the light‑dependent reactions are used in the Calvin cycle. [4]
SECTION B (20 marks)
Answer the following essay question.
8
Describe the structure of a eukaryotic cell membrane and explain how the arrangement of phospholipids, proteins, and cholesterol contributes to its function as a selectively permeable barrier. [20]
[END OF PAPER]
Answers
TuitionGoWhere Practice Paper – Biology H1 A‑Level
Answer Key and Marking Scheme – Version 1
Subject: H1 Biology (8876)
Paper: Paper 2 (Practice)
SECTION A (60 marks)
Question 1
(a)
- P: Phospholipid bilayer [1]
- Q: Channel protein / transport protein (any appropriate integral protein) [1]
- R: Cholesterol [1]
(b) Oxygen is small and non‑polar, so it can dissolve in the hydrophobic core of the phospholipid bilayer and diffuse directly across the membrane down its concentration gradient (simple diffusion). [2]
Award 1 mark for reference to the hydrophobic core/fatty acid tails and 1 mark for movement down a concentration gradient without energy.
(c) Glucose moves against its concentration gradient; this requires energy in the form of ATP. Carrier proteins (proteins embedded in the membrane, such as the sodium‑glucose symporter) use the energy from ATP to undergo a conformational change, allowing glucose to be pumped against its concentration gradient (active transport). [3]
Marking points: energy/ATP (1); involvement of carrier protein (1); conformational change and transport against gradient (1).
(d) Cholesterol increases the packing of phospholipid tails, reducing membrane fluidity at high temperatures, and prevents tight packing at low temperatures, thus maintaining membrane fluidity – it acts as a “fluidity buffer”. [2]
Accept any one clear function: e.g., stabilises membrane / maintains fluidity / prevents crystallisation at low temperature.
Question 2
(a)
- X: Mitochondrion [1]
- Y: Rough endoplasmic reticulum (rough ER) [1]
- Z: Golgi apparatus [1]
(b) Mitochondrion is the site of the Krebs cycle and oxidative phosphorylation. In the matrix, acetyl‑CoA enters the Krebs cycle, producing NADH and FADH₂; on the inner membrane, electron transport chain and ATP synthase generate most of the cell’s ATP. [2]
1 mark for mention of Krebs cycle / link reaction, 1 mark for electron transport chain / ATP production.
(c) Liver cells manufacture and secrete many proteins, such as plasma proteins. Rough ER has ribosomes on its surface where protein synthesis occurs. The extensive network provides a large surface area for protein synthesis and allows the proteins to be transported through the ER lumen for modification and packaging in the Golgi. [3]
Points: protein synthesis role (1); large surface area (1); transport/modification link (1).
Question 3
(a)
- Amylose: unbranched polymer of α‑glucose joined by α‑1,4‑glycosidic bonds; coils into a helix; function – energy storage (starch) in plants. [2]
- Cellulose: unbranched polymer of β‑glucose joined by β‑1,4‑glycosidic bonds, forming long straight chains that hydrogen‑bond to each other to form microfibrils; function – structural component of plant cell walls. [2]
Award marks for correct bonding and shape/function distinction.
(b) A phospholipid has a glycerol backbone linked to two fatty acid tails (hydrophobic) and a phosphate‑containing head group (hydrophilic). In an aqueous environment, the hydrophobic tails face inward to avoid water, while the hydrophilic heads face outward, forming a bilayer spontaneously. This arrangement is the foundation of the cell membrane’s structure. [4]
2 marks for description of structure; 2 marks for bilayer formation explanation: hydrophobic tails inward, hydrophilic heads outward, stabilised by hydrophobic interactions.
Question 4
(a) As temperature increases, enzyme activity initially rises because more molecules have kinetic energy, increasing the frequency of effective collisions between enzyme and substrate. Activity reaches a maximum at the optimum temperature. Beyond this, the enzyme loses activity rapidly because the weak bonds maintaining the tertiary structure (hydrogen bonds, ionic bonds) break, causing denaturation and loss of active site shape. [4]
Marking: 1 for initial increase explanation; 1 for optimum; 1 for denaturation; 1 for linking denaturation to bonds/tertiary structure.
(b) A non‑competitive inhibitor binds to a site on the enzyme other than the active site (allosteric site), changing the shape of the active site so that the substrate can no longer bind. Increasing substrate concentration cannot overcome this because the inhibitor does not compete for the active site. [3]
Points: binding away from active site (1); active site shape change (1); cannot be overcome by extra substrate (1).
(c) A competitive inhibitor has a shape similar to the substrate and competes for the active site. By adding more substrate, the chance of a substrate molecule occupying the active site rather than the inhibitor increases, so the maximum rate can still be achieved. [3]
Points: competition at active site (1); more substrate increases probability of substrate binding (1); max rate eventually achieved (1).
Question 5
(a) Accept correct identification of any three stages that are plausible from a typical diagram.
Example answer:
- A: Prophase – chromosomes condense and become visible; nuclear envelope breaks down.
- B: Metaphase – chromosomes align along the equator of the spindle.
- C: Anaphase – sister chromatids separate and move to opposite poles.
Each stage correctly named = 1 mark; one relevant event = 1 mark. Total 6.
(b) Mitosis produces two genetically identical daughter cells. This allows an organism to increase its cell number for growth (e.g., increase in tissue size) and to replace damaged or worn‑out cells, maintaining tissue integrity. [2]
1 mark for genetic identity; 1 mark for growth/repair context.
Question 6
(a) A deoxyribose sugar, a phosphate group, and a nitrogenous base (adenine, thymine, cytosine, or guanine). [2]
All three components needed.
(b) During semi‑conservative replication, each parental strand acts as a template. Adenine pairs with thymine and cytosine pairs with guanine. This complementary base pairing ensures that each new strand is assembled with bases opposite the template, producing two daughter molecules that are identical to each other and to the original molecule. [3]
1 for template strand concept; 1 for specific base pairing rule; 1 for resulting identity.
(c) TACG. [1]
(d) DNA is double‑stranded (forms a double helix) while mRNA is single‑stranded. (Or DNA contains thymine, mRNA contains uracil; DNA has deoxyribose, mRNA has ribose.) [2]
Award 1 mark for any valid structural difference and 1 mark for explanation.
Question 7
(a) Light energy is absorbed by chlorophyll in photosystem II, exciting electrons to a higher energy level. These electrons are passed along an electron transport chain, and water is split (photolysis) to replace them, releasing oxygen. The energy released from electron transfer is used to pump protons across the thylakoid membrane, creating a proton gradient that drives ATP synthesis via ATP synthase. Light energy also excites electrons in photosystem I, leading to reduction of NADP⁺ to NADPH. [4]
Key points: light absorption/excitation (1); photolysis of water (1); electron transport and chemiosmosis for ATP (1); production of NADPH (1).
(b) The ATP and NADPH produced in the light‑dependent reactions are the essential energy sources for the Calvin cycle. ATP provides the energy for the reduction of 3‑phosphoglycerate to glyceraldehyde‑3‑phosphate and for the regeneration of ribulose bisphosphate. NADPH supplies the reducing power (hydrogen atoms) for the reduction step. Without these products, the Calvin cycle would stop. [4]
Points: ATP used for energy (1); NADPH used for reduction (1); both needed for G3P formation and ribulose bisphosphate regeneration (2).
SECTION B (20 marks)
Question 8 – Essay
Describe the structure of a eukaryotic cell membrane and explain how the arrangement of phospholipids, proteins, and cholesterol contributes to its function as a selectively permeable barrier.
Marking scheme: This 20‑mark essay should be assessed holistically, but the following points indicate the expected content coverage.
Knowledge and understanding (up to 14 marks):
- Phospholipid bilayer: amphipathic nature – hydrophilic phosphate heads face aqueous environment, hydrophobic fatty acid tails face inwards. This arrangement creates a hydrophobic core that is impermeable to most water‑soluble substances.
- Proteins: integral proteins (e.g., carrier, channel) span the membrane, enabling faciliated diffusion and active transport of specific substances. Peripheral proteins may be involved in signalling or structural support.
- Cholesterol: intercalates between phospholipid tails, restricting excessive fluidity at high temperatures and preventing tight packing at low temperatures; thus maintains membrane fluidity and reduces permeability to very small water‑soluble molecules.
- Fluid‑mosaic model: the components are not static; the membrane is fluid, allowing lateral movement of lipids and proteins, which is essential for function.
- Selective permeability based on size, polarity, and charge: small non‑polar molecules (O₂, CO₂) diffuse freely; water passes slowly, often through aquaporins; ions and large polar molecules require specific transport proteins.
Explanation of function (up to 6 marks):
- The hydrophobic core excludes ions and polar molecules, so the cell controls entry/exit via transport proteins.
- Channel proteins form pores for specific ions; carrier proteins undergo conformational changes to move molecules (facilitated diffusion or active transport).
- Cholesterol modulates permeability by reducing the likelihood of cracks or gaps in the bilayer, thus maintaining barrier integrity.
- The membrane’s fluidity allows proteins to diffuse laterally to regions where they are needed, for example, in endocytosis.
- Overall, the composite structure creates a barrier that is both dynamic and highly regulated, essential for maintaining intracellular conditions and homeostasis.
Award marks for:
- Clear description of phospholipid arrangement (0‑4 marks)
- Description of proteins and their transport roles (0‑4 marks)
- Role of cholesterol in fluidity/permeability (0‑3 marks)
- Link between structure and selectivity (0‑4 marks)
- Writing as a coherent essay with correct biological terminology (0‑2 marks)
- Critical integration of all components (0‑3 marks)
A typical full‑mark answer would cover all bullet points with precise terminology and logical flow.
[END OF ANSWERS]