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A Level H2 Biology Practice Paper 1

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Questions

TuitionGoWhere Exam Practice (AI) — Biology H2 A-Level

Practice Paper: Cells & Biomolecules (Version 1 of 5)

School: TuitionGoWhere Exam Practice (AI)
Subject: Biology H2
Level: A-Level
Paper: Practice Paper 1 (Cells & Biomolecules)
Version: 1 of 5
Duration: 75 minutes
Total Marks: 60
Name: ___________________________
Class: ___________________________
Date: ___________________________

Instructions:

Answer all questions in the spaces provided.
Use blue or black pen.
Marks allocated are shown at the end of each question or part.
Show all working where calculations are required.
Diagrams are provided as placeholders to be rendered in a later stage.

Section A: Short Answer and Recall (Questions 1–5) [10 marks]

1. State the three components of the cell theory. [3]

2. Name the bond formed between two glucose monomers in starch and glycogen. [1]

3. Give one structural difference between a typical bacterial cell and a eukaryotic animal cell. [1]

4. Define "induced-fit model" of enzyme action in one sentence. [2]

5. State the role of cholesterol in the fluid mosaic model of the cell membrane. [3]

Section B: Structured Response and Diagram Interpretation (Questions 6–13) [28 marks]

6. With reference to Fig. 1, describe how the rough endoplasmic reticulum (RER) and Golgi body work together to secrete a glycoprotein. [4]

Diagram for Q6 (ALEVEL Biology H2)

Generated diagram for Q6.

7. The figure below shows an enveloped virus. With reference to Fig. 2, explain how this virus challenges the cell theory. [3]

Diagram for Q7 (ALEVEL Biology H2)

Generated diagram for Q7.

8. (a) Draw a simple labelled diagram of a phospholipid molecule. [2]
(b) State two roles of the phospholipid bilayer in membrane transport. [2]

9. A student observed a bacterial cell under an electron microscope. List two features they would see that are absent in a plant cell. [2]

10. Haemoglobin is a globular protein. Outline the four levels of protein structure with one example of a bond involved at each level. [4]

11. With reference to Fig. 3, explain how active transport differs from facilitated diffusion. [3]

Graph for Q11 (ALEVEL Biology H2)

Generated graph for Q11.

12. Describe the structure of cellulose and explain how it provides tensile strength in plant cell walls. [4]

13. Stem cells from bone marrow can differentiate into multiple blood cell types. State the term for this property and name two specific cell lineages produced. [2]

Section C: Data-Based and Applied Questions (Questions 14–20) [22 marks]

14. The table below shows oxygen consumption by isolated mitochondria under three conditions.

Condition	Substrate	Inhibitor	O₂ used (nmol min⁻¹)
A	Pyruvate	None	45
B	Pyruvate	Sodium azide	5
C	None	None	2

(a) Calculate the rate of oxygen consumption in Condition A. [1]
(b) Explain why Condition B shows low O₂ use. [2]
(c) Suggest why Condition C has minimal O₂ use. [1]

15. An enzyme-catalysed reaction was tested at different pH values. The graph (Fig. 4) shows product formed after 5 min.

Graph for Q15 (ALEVEL Biology H2)

Generated graph for Q15.

(a) State the optimum pH. [1]
(b) Explain the shape of the curve using protein structure. [3]

16. Compare the genome organisation of a bacterium and a eukaryotic cell under four headings: DNA shape, presence of introns, ribosome size, membrane-bound nucleus. [4]

17. A virus has a capsid but no envelope. State whether it is more likely a bacteriophage or an enveloped influenza virus, and give one reason. [2]

18. Glycogen and starch are both polysaccharides of glucose. Explain why glycogen is more branched than starch and how this relates to its function in animals. [3]

19. With reference to Fig. 5, describe how a non-competitive inhibitor reduces enzyme activity. [3]

Diagram for Q19 (ALEVEL Biology H2)

Generated diagram for Q19.

20. The fluid mosaic model includes glycolipids and glycoproteins. State two functions of these molecules in the membrane. [2]

End of Paper

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Answers

TuitionGoWhere Exam Practice (AI) — Biology H2 A-Level

Practice Paper: Cells & Biomolecules (Version 1 of 5) — Answer Key

Total Marks: 60

Section A Answers (10 marks)

Q1 [3 marks]

All living organisms are composed of cells. (1)
The cell is the smallest unit of life. (1)
All cells arise from pre-existing cells. (1)
Teaching note: These three statements are the core of cell theory. Do not include "cells contain DNA" as that is not part of the classical definition.

Q2 [1 mark]

Glycosidic bond.
Teaching note: Specifically an α-1,4 glycosidic bond in starch/glycogen; α-1,6 for branches.

Q3 [1 mark]

Bacterial cell has no membrane-bound nucleus (or has peptidoglycan wall / circular DNA / 70S ribosomes). Any one valid.
Teaching note: Must be a structural difference; "bacteria are smaller" alone is not awarded.

Q4 [2 marks]

The enzyme's active site changes shape slightly to better fit the substrate upon binding. (2)
Teaching note: Contrast with lock-and-key (rigid). Award both marks for clear induced-fit definition.

Q5 [3 marks]

Reduces membrane fluidity at high temperature. (1)
Prevents crystallisation of membrane at low temperature. (1)
Maintains membrane stability. (1)
Teaching note: Cholesterol wedges between phospholipids; dual role in temp buffering.

Section B Answers (28 marks)

Q6 [4 marks]

Ribosomes on RER synthesise the protein (1).
Protein enters RER lumen, glycosylation begins (1).
Transport vesicle carries glycoprotein to Golgi (1).
Golgi modifies/add sugars and packages into secretory vesicle to membrane (1).
Teaching note: Must reference figure labels (RER, vesicle, Golgi).

Q7 [3 marks]

Virus has genetic material but no cells/organelles (1).
Cannot reproduce independently without host cell (1).
Challenges "cells are basic unit of life" as virus is acellular yet shows heredity (1).
Teaching note: Link to cell theory explicitly.

Q8 [4 marks]
(a) [2] Draw: glycerol backbone, 2 fatty acid tails, phosphate head. Label all three. (2)
(b) [2] Roles: barrier to hydrophilic substances (1); matrix for embedded proteins (1).
Teaching note: Bilayer hydrophobic core blocks ions.

Q9 [2 marks]

Peptidoglycan wall (1); circular DNA (1) [or 70S ribosomes].
Teaching note: Plant cell has cellulose wall and linear DNA in nucleus.

Q10 [4 marks]

Primary: amino acid sequence; peptide bonds (1).
Secondary: α-helix/β-sheet; H-bonds (1).
Tertiary: 3D folding; disulfide/ionic/hydrophobic (1).
Quaternary: multiple subunits; same bonds as tertiary (1).
Teaching note: Haemoglobin = 4 subunits (α₂β₂).

Q11 [3 marks]

Facilitated diffusion plateaus at low rate, no energy (1).
Active transport needs ATP, higher max rate (1).
Active transport can move against gradient; facilitated only down (1).
Teaching note: Graph shows ATP marker on active transport.

Q12 [4 marks]

β-glucose chains, β-1,4 glycosidic bonds (1).
Chains straight, parallel (1).
H-bonds cross-link chains into microfibrils (1).
Microfibrils give tensile strength (1).
Teaching note: Contrast with starch α-1,4 helical.

Q13 [2 marks]

Multipotency (1).
Lymphoid and myeloid lineages (1).
Teaching note: Not totipotent (that is zygote).

Section C Answers (22 marks)

Q14 [4 marks]
(a) [1] 45 nmol min⁻¹ (from table).
(b) [2] Sodium azide inhibits cytochrome c oxidase / ETC; O₂ not final acceptor used.
(c) [1] No substrate = no NADH produced, no ETC drive.
Teaching note: O₂ is final e⁻ acceptor; rate = table value.

Q15 [4 marks]
(a) [1] pH 7.
(b) [3] Low pH: H⁺ disrupts H-bonds/ionic bonds, denature (1). Optimum: active site intact (1). High pH: same disruption opposite charge (1).
Teaching note: Use protein structure concept.

Q16 [4 marks]

DNA shape: bacterial circular, eukaryotic linear (1).
Introns: bacterial absent, eukaryotic present (1).
Ribosome: 70S vs 80S (1).
Nucleus: none vs membrane-bound (1).

Q17 [2 marks]

Bacteriophage (1). Reason: no envelope, capsid only (1).
Teaching note: Influenza is enveloped.

Q18 [3 marks]

Glycogen more α-1,6 branches (1).
Branches allow rapid glucose release from many ends (1).
Animals need quick energy mobilisation (1).

Q19 [3 marks]

Inhibitor binds allosteric site (1).
Active site shape altered (1).
Substrate cannot bind / lower Vmax (1).
Teaching note: Non-competitive lowers Vmax not Km.

Q20 [2 marks]

Cell recognition (1); receptor/site for signalling (1).
Teaching note: Glycocalyx role.

End of Answer Key