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A Level H1 Biology Practice Paper 4

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TuitionGoWhere Practice Paper – Biology H1 A-Level

TuitionGoWhere Exam Practice (AI) PRACTICE PAPER – Version 4 of 5

Subject: Biology H1 (8876)
Level: A-Level
Paper: Cells & Biomolecules Topic Test
Duration: 1 hour 15 minutes
Total Marks: 65

Name: _______________________________
Class: _______________________________
Date: _______________________________

Instructions to Candidates:

This paper consists of 20 questions in three sections.
Section A contains 10 multiple‑choice questions. Indicate your answers by writing the appropriate letter (A–D) in the boxes provided.
Sections B and C contain structured and data‑interpretation questions. 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 may use a calculator, but all working must be shown where necessary.
Read each question carefully before you start to answer.

Section A: Multiple‑Choice Questions [10 marks]

Answer all questions in this section. Write the letter of your chosen answer in the box provided.

1. Which of the following correctly describes the arrangement of phospholipids in a cell‑surface membrane?

A. A single layer with hydrophobic heads facing the cytoplasm and hydrophilic tails facing the exterior.
B. A bilayer with hydrophilic heads facing the aqueous environments and hydrophobic tails pointing inward.
C. A bilayer with hydrophobic heads facing the cytoplasm and exterior, and hydrophilic tails inside.
D. Randomly scattered phospholipid molecules held together by covalent bonds.

[1 mark]
Answer: |_|

2. In the fluid‑mosaic model, the ‘mosaic’ refers to the

A. phospholipid bilayer that forms a continuous sheet.
B. pattern of carbohydrate chains on the outer surface only.
C. presence of proteins of different sizes and shapes embedded in the bilayer.
D. arrangement of cholesterol molecules that reduce membrane fluidity.

[1 mark]
Answer: |_|

3. A student isolated mitochondria from liver cells and incubated them in three separate solutions, each containing a different substrate. The rate of carbon dioxide production was measured. The results are shown below.

Substrate	Rate of CO₂ production (arbitrary units)
Glucose	2
Pyruvate	45
Fatty acid	18

Which statement best explains the low rate of CO₂ production when glucose was used as the substrate?

A. Glucose cannot be used by liver cells for respiration.
B. The Krebs cycle in isolated mitochondria cannot oxidise glucose directly; glycolysis occurs in the cytoplasm.
C. Fatty acids provide more energy per molecule than glucose, so less CO₂ is produced.
D. Pyruvate is a more efficient substrate because it bypasses glycolysis completely.

[1 mark]
Answer: |_|

4. When cells are exposed to a radioactive precursor, the radioactivity in the nucleus first increases during the S phase of the cell cycle. Which of the following is the most likely radioactive precursor used?

A. Radioactive uracil, because RNA synthesis occurs during S phase.
B. Radioactive thymine, because it is incorporated into DNA during replication.
C. Radioactive leucine, because proteins are synthesised during S phase.
D. Radioactive glycerol, because membrane lipids are assembled in the nucleus.

[1 mark]
Answer: |_|

5. The diagram below (not drawn to scale) shows a section of a cell membrane.

[Outside of cell]
    O  O  O  O   (carbohydrate chains)
    |  |  |  |
  ———————  ———————  (phospholipid bilayer)
  |     |  |     |
  |  ■  |  |     |   (■ = carrier protein)
  |_________|_________|
[Inside of cell]

(No actual figure is required; the transport protein spans the membrane.)

Which of the following substances would use the protein channel labelled ■ to cross the membrane?

A. O₂, by simple diffusion
B. CO₂, by facilitated diffusion
C. Na⁺ ions, by facilitated diffusion
D. Testosterone, by simple diffusion

[1 mark]
Answer: |_|

6. Which bond or interaction is NOT involved in maintaining the three‑dimensional structure of a protein?

A. Disulfide bridges between cysteine residues.
B. Hydrogen bonds between polar R groups.
C. Glycosidic bonds between monosaccharide units.
D. Hydrophobic interactions between non‑polar side chains.

[1 mark]
Answer: |_|

7. Lactose is a disaccharide found in milk. It is composed of

A. two glucose molecules linked by a β‑1,4‑glycosidic bond.
B. one glucose and one galactose molecule linked by a β‑1,4‑glycosidic bond.
C. one glucose and one fructose molecule linked by an α‑1,2‑glycosidic bond.
D. one glucose and one galactose molecule linked by an α‑1,6‑glycosidic bond.

[1 mark]
Answer: |_|

8. Water has a high specific heat capacity. This property is important for living organisms because it

A. allows water to act as a universal solvent for polar molecules.
B. provides a medium for metabolic reactions.
C. stabilises body temperature by absorbing and releasing heat slowly.
D. allows ice to float on liquid water, insulating aquatic life.

[1 mark]
Answer: |_|

9. A student drew a table comparing the structure of DNA and RNA. Which row contains an error?

Feature	DNA	RNA
A. Pentose sugar	Deoxyribose	Ribose
B. Nitrogenous bases	A, T, C, G	A, U, C, G
C. Strandedness	Usually double‑stranded	Usually single‑stranded
D. Function	Stores genetic information	Catalyses peptide bond formation

[1 mark]
Answer: |_|

10. A cell is placed in a hypertonic solution. Which of the following is most likely to occur?

A. Water moves into the cell by osmosis, causing it to swell.
B. Water moves out of the cell by active transport, causing it to shrink.
C. Water moves out of the cell by osmosis, causing the cytoplasm to shrink away from the cell wall (plasmolysis in a plant cell).
D. Solutes move into the cell by facilitated diffusion until equilibrium is reached.

[1 mark]
Answer: |_|

Section B: Structured Questions [35 marks]

Answer all questions in the spaces provided.

11. Figure 1.1 is an electron micrograph of an animal cell.

[In the actual paper, a figure would show a mitochondrion, rough endoplasmic reticulum, Golgi apparatus, and secretory vesicle. Here, refer to structures A and B as labelled.]

(a) Name structure A, which is bounded by a double membrane and contains cristae. [1]

(b) Name structure B, which consists of flattened membrane‑bound sacs with ribosomes attached to its outer surface. [1]

(c) This cell is a secretory cell that produces a protein hormone. Describe how structures A, B and C (where C is the Golgi apparatus) work together to synthesise and secrete the hormone. [4]

[Total: 6 marks]

12. With reference to Figure 1.2 (not shown here; it would depict a phospholipid bilayer with a channel protein), describe how glucose molecules enter a liver cell from the blood.

[Your answer should refer to the direction of movement and the role of the protein shown.] [3]

[Total: 3 marks]

13. Hydrogen bonds play a crucial role in the properties of water.

(a) Explain how hydrogen bonding between water molecules contributes to the high surface tension of water. [2]

(b) Explain how hydrogen bonding enables water to act as an effective solvent for polar substances such as glucose. [2]

[Total: 4 marks]

14. Catalase is an enzyme that breaks down hydrogen peroxide into water and oxygen.

The reaction can be written as:

[ 2 H_2O_2 \xrightarrow{\text{catalase}} 2 H_2O + O_2 ]

A student investigated the effect of temperature on the rate of this reaction. The results are shown in Table 14.

Table 14

Temperature / °C	Rate of oxygen production / cm³ min⁻¹
10	1.2
20	3.6
30	8.5
40	9.2
50	4.1
60	0.8

(a) (i) With reference to the data, state the approximate optimum temperature for catalase. [1]

(ii) Account for the decrease in the rate of oxygen production when the temperature is raised from 40 °C to 60 °C. [3]

(b) The student repeated the experiment at 30 °C but added a competitive inhibitor. Predict and explain the effect on the rate of reaction. [3]

[Total: 7 marks]

15. The diagram below (not drawn to scale) represents part of a DNA molecule.

[Simplified diagram showing a section of DNA double helix; labels: X = deoxyribose, Y = phosphate, Z = hydrogen bonds between base pairs.]

(a) (i) Name the chemical group labelled Y. [1]

(ii) Describe the role of the bonds labelled Z in the structure of DNA. [2]

(b) A molecule of double‑stranded DNA contains 30% adenine.

Calculate the percentage of guanine in this DNA molecule. Show your working. [2]

[Total: 5 marks]

Section C: Data‑Based and Free‑Response Questions [20 marks]

Answer all questions in the spaces provided.

16. A biologist isolated intact chloroplasts and mitochondria from spinach leaf cells. The organelles were incubated separately in solutions containing various substrates. Carbon dioxide production was measured in the mitochondrial preparation, and oxygen production in the chloroplast preparation under illumination. The results are shown in Table 16.

Table 16

Experimental condition	CO₂ produced by mitochondria (μmol min⁻¹)	O₂ produced by chloroplasts (μmol min⁻¹)
1. Mitochondria + pyruvate + ADP + Pi	12.5	–
2. Mitochondria + glucose + ADP + Pi	0.3	–
3. Chloroplasts + H₂O + light	–	24.8
4. Chloroplasts + glucose + light	–	0.2

(a) With reference to your knowledge of cellular respiration, explain the difference in CO₂ production between condition 1 and condition 2. [3]

(b) Explain why chloroplasts produced very little oxygen when supplied with glucose instead of water. [3]

[Total: 6 marks]

17. Describe the arrangement of phospholipids in cell membranes and explain how this arrangement contributes to the selective permeability of the membrane. [5]

[Total: 5 marks]

18. Haemoglobin is a conjugated protein with a quaternary structure. Explain how the different levels of protein structure contribute to the ability of haemoglobin to bind and release oxygen. [4]

[Total: 4 marks]

19. Discuss the significance of the movement of substances across membranes to photosynthesis. [5]

Your answer should refer to at least three different substances and the transport mechanisms involved.

[Total: 5 marks]

20. This question carries 0 marks and is for feedback only.

Do you have any feedback on this practice paper?

[Total: 0 marks]

END OF PAPER

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Answers

TuitionGoWhere Practice Paper – Biology H1 A-Level: ANSWERS

PRACTICE PAPER – Version 4 of 5 Cells & Biomolecules Topic Test

Section A: Multiple‑Choice Answers

Question	Answer	Marking Notes
1	B	Phospholipid bilayer: hydrophilic heads face aqueous environments, hydrophobic tails face inward.
2	C	‘Mosaic’ = proteins embedded in bilayer.
3	B	Glucose cannot enter Krebs cycle directly; glycolysis in cytoplasm, so isolated mitochondria show low CO₂.
4	B	Radioactive thymine is incorporated into DNA during S phase, increasing nuclear radioactivity.
5	C	Na⁺ ions are charged; they pass through channel proteins via facilitated diffusion.
6	C	Glycosidic bonds are found in polysaccharides, not proteins.
7	B	Lactose = glucose + galactose, β‑1,4‑glycosidic bond.
8	C	High specific heat capacity stabilises temperature (heat absorption/release slowly).
9	D	RNA does not catalyse peptide bond formation; that function belongs to rRNA in ribosomes, but not a general function of all RNA. (Accept: catalytic RNA is rare, not general.)
10	C	In hypertonic solution, water leaves cell by osmosis; plant cell undergoes plasmolysis.

Section B: Structured Questions

Question 11 (6 marks)

(a) Structure A: Mitochondrion. [1]
(b) Structure B: Rough endoplasmic reticulum (rER) / rough ER. [1]
(c) Description:

The hormone protein is synthesised by ribosomes on the rough ER. [1]
The polypeptide chain enters the ER lumen where it folds and is processed. [1]
Transport vesicles bud off from the ER and fuse with the Golgi apparatus (structure C). [1]
Inside the Golgi, the protein is modified (e.g., glycosylation), packaged into secretory vesicles, which move to the cell membrane. [1]
Exocytosis releases the hormone. Mitochondria (structure A) provide ATP for these processes. [1] (max 4; any 4 from above)

Question 12 (3 marks)

Glucose enters liver cells from blood by facilitated diffusion. [1]
Glucose moves down its concentration gradient (from high concentration in blood to lower inside cell). [1]
This occurs through a specific channel protein / carrier protein (the protein shown in Figure 1.2). [1]

Question 13 (4 marks)

(a)

Hydrogen bonds between water molecules cause strong cohesion. [1]
This cohesion produces high surface tension, allowing small organisms to rest on or move across the water surface. [1]

(b)

Glucose contains many hydroxyl (–OH) groups which are polar. [1]
Water molecules form hydrogen bonds with these polar groups, surrounding and dissolving the glucose molecules. [1]

Question 14 (7 marks)

(a)(i) Approx. 40 °C (since rate is highest at 40 °C in the table). [1]

(a)(ii)

At temperatures above the optimum (40 °C), the enzyme catalase begins to denature. [1]
The increased thermal energy breaks hydrogen bonds and other weak interactions that maintain the enzyme's tertiary structure. [1]
The active site loses its specific shape, so the substrate (hydrogen peroxide) can no longer bind effectively, reducing the rate of reaction. [1]

(b)

In the presence of a competitive inhibitor, the rate of reaction would decrease. [1]
The inhibitor has a shape similar to hydrogen peroxide and competes for the active site. [1]
This reduces the number of active sites available for the substrate, lowering the rate of product formation. However, the maximum rate could still be achieved if substrate concentration is high enough. [1]

Question 15 (5 marks)

(a)(i) Y = Phosphate group. [1]
(a)(ii)

Hydrogen bonds (Z) hold the two polynucleotide strands together by complementary base pairing (A=T and C≡G). [1]
They can be easily broken to allow DNA replication / transcription and re‑form, maintaining the double‑helix structure. [1]

(b) Adenine = 30%, therefore Thymine = 30% (A=T). Total A+T = 60%. Remaining bases G+C = 40%. Since G=C, Guanine = 20%. [2] (1 mark for correct reasoning, 1 mark for correct answer)

Section C: Data‑Based and Free‑Response Questions

Question 16 (6 marks)

(a)

Pyruvate can enter the mitochondrion and be converted to acetyl‑CoA, which enters the Krebs cycle, producing CO₂. [1]
Glucose cannot directly enter the Krebs cycle; it must first undergo glycolysis in the cytoplasm. [1]
Isolated mitochondria lack the glycolytic enzymes, so little CO₂ is produced from glucose. [1]

(b)

Photosynthesis produces O₂ when water is split during the light‑dependent reactions (photolysis). [1]
The electrons from water are needed to replace those lost from chlorophyll. [1]
Glucose cannot donate electrons in the light‑dependent reactions, so no O₂ is produced when glucose is supplied. [1]

Question 17 (5 marks)

Phospholipids form a bilayer: hydrophilic phosphate heads face the aqueous exterior and interior; hydrophobic fatty acid tails face each other in the interior. [1]
This arrangement creates a hydrophobic core. [1]
Small, non‑polar molecules (e.g., O₂, CO₂) and lipid‑soluble substances can diffuse through the bilayer. [1]
The hydrophobic core restricts the passage of ions and large polar molecules (e.g., glucose, Na⁺), which require specific channel or carrier proteins. [1]
The bilayer therefore acts as a selectively permeable barrier, allowing the cell to control its internal environment. [1]

Question 18 (4 marks)

Primary structure: the specific sequence of amino acids determines the overall shape and the position of key histidine residues that bind oxygen. [1]
Secondary structure: α‑helices and β‑pleated sheets provide stability and bring amino acid chains into correct orientation. [1]
Tertiary structure: further folding of each polypeptide chain creates a specific binding pocket for the haem group and oxygen. [1]
Quaternary structure: haemoglobin is made of four polypeptide subunits (2α, 2β), each with a haem group; cooperative binding occurs – binding of one O₂ changes the shape of neighbouring subunits, increasing their affinity for O₂, enabling efficient loading and unloading of oxygen. [1]

Question 19 (5 marks)

Answer should include at least three substances:

Carbon dioxide: diffuses into leaf via stomata, then into mesophyll cells and chloroplasts. [1] It is the substrate for the Calvin cycle; without CO₂ entry, photosynthesis stops. [1]
Water: enters root hair cells by osmosis, moves through xylem to leaves, and is split in photolysis, providing electrons and protons. [1]
Mineral ions (e.g., Mg²⁺, NO₃⁻): absorbed by root cells via active transport; magnesium is essential for chlorophyll synthesis; nitrate for amino acid synthesis. [1]
Oxygen: a by‑product of photosynthesis that diffuses out of the leaf; its removal maintains the concentration gradient. [1] Conclusion: The movement of these substances across membranes ensures the continuous supply of raw materials and removal of products, sustaining photosynthesis. [1] (Any three substances with correct transport mechanism and link to photosynthesis earns marks; maximum 5 marks.)

Question 20 (0 marks)

No marks; feedback optional.