A Level H1 Biology Practice Paper 1

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TuitionGoWhere Practice Paper - Biology H1 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Biology H1
Level: A-Level
Paper: Topic Quiz – Cells and Biomolecules
Version: 1
Duration: 50 minutes
Total Marks: 40

Name: ______________________
Class: ______________________
Date: ______________________

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Instructions to Candidates

• 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 about 1 minute per mark.

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1. The cell membrane performs several vital functions.

(a) State two functions of the cell membrane. [2]

(b) Name the component of the cell membrane that is primarily responsible for cell-to-cell recognition. [1]

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2. A student examined a transmission electron micrograph of a cell from a muscle. The micrograph contained a structure with a double membrane and many internal folds.

(a) Name this structure. [1]

(b) Explain why muscle cells contain a large number of this structure. [2]

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3. (a) Describe the arrangement of phospholipids in a cell membrane. [2]

(b) State one property of the phospholipid bilayer that allows lipid‑soluble substances to pass through easily. [1]

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4. (a) State what is meant by the fluid mosaic model of membrane structure. [2]

(b) Identify one piece of experimental evidence that supports this model. [1]

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5. (a) Distinguish between simple diffusion and facilitated diffusion across a cell membrane. [2]

(b) Give an example of a molecule that enters cells by facilitated diffusion. [1]

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6. (a) Name the process by which sodium ions move out of a cell against their concentration gradient. [1]

(b) Outline the role of the carrier protein and ATP in this process. [2]

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7. A red blood cell is placed in a beaker of distilled water.

Predict what will happen to the cell and explain your answer. [2]

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8. (a) Name the type of covalent bond that links two glucose molecules in the disaccharide maltose. [1]

(b) State one structural difference between amylose and cellulose. [1]

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9. Triglycerides are formed by condensation reactions.

Draw a simple diagram to show the formation of an ester bond between a glycerol molecule and a fatty acid. Label the ester bond.

[3]

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10. The primary structure of a polypeptide determines its three‑dimensional shape.

Explain why changing a single amino acid in the primary structure can make the protein non‑functional. [2]

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11. Water is an essential molecule for life.

State two properties of water and, for each property, give one biological importance. [2]

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12. A student investigated the effect of pH on the activity of the enzyme catalase. The results are shown in Table 1.

Table 1: Catalase activity at different pH values

pH	Rate of oxygen production / cm³ min⁻¹
3.0	2.5
5.0	8.2
7.0	14.0
9.0	6.1
11.0	0.8

(a) Plot a line graph of the results. Label the axes clearly. [2]

(b) Explain why the enzyme shows negligible activity at pH 11.0. [2]

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13. (a) On the sketch of an enzyme molecule below, label the active site. [1]

(b) Explain the role of the active site in enzyme catalysis. [2]

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14. A competitive inhibitor is a molecule that resembles the substrate.

Explain how a competitive inhibitor reduces the rate of an enzyme‑catalysed reaction. [2]

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15. Fig. 2.1 shows the effect of increasing substrate concentration on the rate of an enzyme‑catalysed reaction in the absence and presence of an inhibitor.

(No figure is reproduced here; the graph shows two curves. Curve A (no inhibitor) reaches a high, flat plateau. Curve B (with inhibitor) also plateaus but at a lower maximum rate, although at very high substrate concentrations the two curves appear to approach similar rates.)

(a) Identify whether the inhibitor is competitive or non‑competitive. [1]

(b) Explain your choice. [2]

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16. A typical animal cell and a typical plant cell are each placed in a hypotonic solution.

Describe what happens to each cell and explain the reason for any difference. [2]

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17. (a) Name the two types of nucleic acid found in cells. [1]

(b) State two differences between the structures of DNA and RNA. [2]

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18. Explain why DNA is classified as a polymer. [1]

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19. Draw and label a single DNA nucleotide. Name each component. [2]

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20. The fatty acid chains of phospholipids from a cold‑adapted fish contain a higher proportion of unsaturated fatty acids than those from a warm‑water fish.

Explain the advantage of this difference to the cold‑adapted fish. [2]

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End of Paper

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TuitionGoWhere Practice Paper – Answers and Mark Scheme

Biology H1 A-Level – Cells and Biomolecules (Version 1)

Question 1

(a) [2 marks]
Any two of:

• Forms a selectively permeable barrier, controlling the passage of substances into/out of the cell [1]
• Provides mechanical support to the cell / maintains cell shape [1]
• Contains receptors for cell signalling and communication [1]
• Compartmentalises organelles (for eukaryotic cells) [1]
  (Award 1 mark per correct function, max 2.)

(b) [1 mark]
Glycocalyx / glycoprotein / glycolipid [1]

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Question 2

(a) [1 mark]
Mitochondrion [1] (accept mitochondria)

(b) [2 marks]

• Mitochondria are the sites of aerobic respiration / ATP production [1]
• Muscle cells require a large amount of ATP to power contraction; more mitochondria increase the cell’s capacity for ATP synthesis [1]

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Question 3

(a) [2 marks]

• Phospholipids form a bilayer [1]
• Hydrophilic phosphate heads face outward (towards the aqueous cytoplasm and extracellular fluid); hydrophobic fatty acid tails point inwards, away from water [1]

(b) [1 mark]
The hydrophobic interior / non‑polar core allows lipid‑soluble (non‑polar) substances to dissolve through the bilayer and diffuse across [1]

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Question 4

(a) [2 marks]

• ‘Fluid’ refers to the ability of phospholipid molecules and many proteins to move laterally within the bilayer [1]
• ‘Mosaic’ describes the pattern of different proteins (integral and peripheral) embedded in / attached to the phospholipid bilayer [1]

(b) [1 mark]
Any one of:

• Freeze‑fracture electron micrographs show proteins scattered throughout the membrane [1]
• Fluorescent antibody tagging experiments demonstrate that membrane proteins can move laterally [1]
• Fusion of mouse and human cells shows mixing of membrane proteins [1]

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Question 5

(a) [2 marks]

• Simple diffusion occurs directly through the phospholipid bilayer (mainly for small, non‑polar molecules) and does not require a transport protein; facilitated diffusion uses specific channel or carrier proteins [1]
• Both are passive processes (driven by a concentration gradient), but the rate of facilitated diffusion can become saturated when all transport proteins are occupied [1]

(b) [1 mark]
Any one: glucose (via GLUT transporter) / amino acids / ions (e.g., Na⁺, K⁺, Cl⁻ through channel proteins) / fructose [1]

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Question 6

(a) [1 mark]
Active transport [1]

(b) [2 marks]

• The carrier protein binds sodium ions on the cytoplasmic side; ATP is hydrolysed, donating a phosphate group that causes the protein to change shape [1]
• This conformational change releases the ions on the exterior side, moving them against the concentration gradient [1]

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Question 7

[2 marks]

• Water enters the red blood cell by osmosis because the cell’s interior has a lower water potential (more negative solute potential) than the distilled water [1]
• The cell swells and eventually bursts (haemolysis) because animal cells lack a cell wall to resist osmotic pressure [1]

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Question 8

(a) [1 mark]
Glycosidic bond / 1,4‑glycosidic bond [1]

(b) [1 mark]
Any one:

• Amylose is unbranched (straight chain) while cellulose forms straight, parallel chains with hydrogen bonds between them / amylose is a polymer of α‑glucose, cellulose is a polymer of β‑glucose [1]
• Amylose coils into a helix; cellulose chains are straight and form microfibrils [1]
• Amylose contains α‑1,4 glycosidic bonds; cellulose contains β‑1,4 glycosidic bonds [1]

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Question 9

[3 marks]

• Diagram showing a glycerol backbone and one fatty acid chain, with the hydroxyl group (–OH) of glycerol and the carboxyl group (–COOH) of the fatty acid reacting [1]
• Water molecule eliminated (condensation reaction) [1]
• The resulting –COO– (ester) bond labelled clearly [1]

(Accept any clear diagram showing the ester bond formation; marks for correct reactants, water removal and ester linkage.)

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Question 10

[2 marks]

• The primary structure is the sequence of amino acids; this sequence determines the folding (secondary and tertiary structure) through hydrogen, ionic, hydrophobic and disulphide bonds [1]
• A single amino acid substitution can alter these bonds, changing the overall 3‑D shape (conformation) of the protein so that the active site / binding site is distorted and can no longer function [1]

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Question 11

[2 marks]
Students must give two properties, each with a biological importance, for 2 marks (1 per pair). Examples:

• High specific heat capacity → buffers temperature changes in organisms / aquatic environments [1]
• Cohesion and adhesion → enables capillary action in plants / transpiration stream [1]
• Good solvent (polar) → allows transport of dissolved substances in blood / cytoplasm [1]
• High latent heat of vaporisation → effective cooling via sweating / transpiration [1]
  (Accept any valid property and linked importance.)

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Question 12

(a) [2 marks]

• Graph with correct axes: x‑axis ‘pH’, y‑axis ‘Rate of oxygen production / cm³ min⁻¹’, both labelled with units [1]
• Points plotted accurately and joined with a smooth curve showing a peak at pH 7.0, falling on both sides [1]

(b) [2 marks]

• At pH 11.0, the enzyme is denatured [1]
• Very high pH disrupts the hydrogen and ionic bonds that maintain the enzyme’s tertiary structure; the active site loses its specific shape, so the substrate no longer fits and no enzyme‑substrate complexes form [1]

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Question 13

(a) [1 mark]
Active site correctly indicated on the enzyme diagram (a depression or cleft on the surface of the enzyme).

(b) [2 marks]

• The active site has a specific shape complementary to the substrate [1]
• It binds the substrate, forming an enzyme‑substrate complex, lowering the activation energy of the reaction and enabling catalysis to take place [1]

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Question 14

[2 marks]

• The competitive inhibitor has a shape similar to the substrate and binds to the active site, blocking the substrate from binding [1]
• This reduces the number of enzyme‑substrate complexes formed; the inhibition can be overcome by increasing substrate concentration [1]

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Question 15

(a) [1 mark]
Competitive inhibitor [1]

(b) [2 marks]

• At low substrate concentrations, the inhibitor‑bound enzyme reduces the reaction rate (curve B is lower) [1]
• At very high substrate concentrations, the substrate out‑competes the inhibitor for the active site, so the maximum rate achieved by curve B approaches that of the uninhibited enzyme – a hallmark of competitive inhibition [1]

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Question 16

[2 marks]

• The animal cell swells and may burst (lyse) because water enters by osmosis down a water‑potential gradient; the cell has no cell wall [1]
• The plant cell also takes in water and becomes turgid, but it does not burst because the rigid cellulose cell wall resists expansion and exerts inward pressure, stopping further water entry [1]

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Question 17

(a) [1 mark]
DNA (deoxyribonucleic acid) and RNA (ribonucleic acid) [1]

(b) [2 marks]
Any two differences (1 mark each):

• DNA contains deoxyribose sugar; RNA contains ribose [1]
• DNA is double‑stranded (double helix); RNA is usually single‑stranded [1]
• DNA has the nitrogenous base thymine; RNA has uracil instead [1]
• DNA is larger / longer than most RNA molecules [1]

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Question 18

[1 mark]
DNA is a polymer because it is a long chain made of many repeating monomer units – nucleotides – linked by phosphodiester bonds [1]

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Question 19

[2 marks]
Diagram should show:

• A phosphate group, a deoxyribose sugar (pentose ring), and a nitrogenous base (any one of A, T, C, G) correctly labelled [1]
• The three components connected correctly: phosphate attached to the 5′ carbon, base to the 1′ carbon, and the 3′ hydroxyl shown [1]

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Question 20

[2 marks]

• Unsaturated fatty acids contain double bonds, which introduce kinks in the hydrocarbon tails, preventing tight packing of phospholipids [1]
• This increases membrane fluidity at low temperatures, allowing the cell membrane to remain functional (e.g., for transport) in cold environments [1]

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End of Mark Scheme