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A Level H1 Biology Practice Paper 5

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TuitionGoWhere Exam Practice (AI) - Biology H1 A-Level

Practice Paper - Version 5 of 5

Subject: Biology
Level: H1 (8876)
Paper: Practice Assessment (Topic: Cells & Biomolecules)
Duration: 1 hour 15 minutes
Total Marks: 60

Name: __________________________
Class: __________________________
Date: __________________________

Instructions to Candidates

Write your name, class, and date in the spaces provided.
Answer all questions.
Write your answers in the spaces provided in this booklet.
The number of marks is given in brackets [ ] at the end of each question or part question.
You are advised to spend approximately 45 minutes on Section A and 30 minutes on Section B.

Section A: Structured Questions

Answer all questions in this section.

1. Fig. 1.1 shows a diagram of a phospholipid molecule and its arrangement in a cell membrane.

(a) With reference to Fig. 1.1, describe the arrangement of phospholipids in the cell membrane. [2]

(b) Explain why this arrangement makes the membrane selectively permeable to water-soluble molecules. [2]

2. A student investigated the effect of temperature on the activity of the enzyme amylase. The results are shown in Table 2.1.

Temperature / °C	Time taken for starch to disappear / s	Rate of reaction / s⁻¹
20	120	0.0083
30	60	0.0167
40	30	0.0333
50	45	0.0222
60	>300	~0

(a) Calculate the rate of reaction at 50°C. Show your working. [1]

Answer space

(b) With reference to Table 2.1 and your knowledge of enzymes, explain the change in the rate of reaction between 40°C and 60°C. [3]

3. Fig. 3.1 shows an electron micrograph of a liver cell. Structure A is identified as the rough endoplasmic reticulum (RER).

(a) State one structural feature visible in Fig. 3.1 that identifies structure A as the RER. [1]

(b) Liver cells synthesize many plasma proteins. Explain the role of the RER in this process. [2]

4. Mitochondria were isolated from liver cells and incubated in a solution containing pyruvate. Oxygen consumption was measured. In a second experiment, mitochondria were incubated with glucose instead of pyruvate. No oxygen consumption was observed.

(a) Explain why oxygen was consumed when pyruvate was present. [2]

(b) Explain why no oxygen was consumed when glucose was present. [2]

5. Fig. 5.1 shows the structure of a triglyceride molecule.

(a) Name the type of reaction that joins the fatty acids to the glycerol molecule. [1]

(b) Triglycerides are used as energy storage molecules in animals. Explain two properties of triglycerides that make them suitable for this function. [2]

6. Water is essential for life.

(a) Explain how the polarity of water molecules contributes to its effectiveness as a solvent. [2]

(b) State one other biological importance of water’s high specific heat capacity. [1]

7. Fig. 7.1 shows a model of the cell membrane including a channel protein.

(a) Describe how potassium ions (K⁺) move through the channel protein shown in Fig. 7.1. [2]

(b) State whether this movement requires ATP. Give a reason for your answer. [2]

8. A student observed plant cells under a microscope after placing them in a concentrated sucrose solution. The cytoplasm shrank away from the cell wall.

(a) Name this process. [1]

(b) Explain why the cytoplasm shrank. [2]

9. DNA and RNA are nucleic acids.

(a) State two structural differences between DNA and RNA. [2]

(b) Explain the significance of the complementary base pairing in DNA structure. [2]

10. Fig. 10.1 shows the structure of an amino acid.

(a) Identify the group labeled X. [1]

(b) Amino acids join to form polypeptides. Name the bond formed between two amino acids. [1]

Section B: Data Interpretation and Extended Response

Answer all questions in this section.

11. Fig. 11.1 shows the results of an experiment investigating the uptake of substance X by cells at different external concentrations.

Curve A: Uptake in the presence of a respiratory inhibitor (stops ATP production).
Curve B: Uptake in the absence of a respiratory inhibitor.

(a) With reference to Fig. 11.1, describe the difference in uptake between Curve A and Curve B at high concentrations of substance X. [2]

(b) Identify the transport mechanism responsible for Curve B. Justify your answer using evidence from the graph. [3]

12. Haemoglobin is a globular protein found in red blood cells.

(a) Describe the levels of protein structure present in haemoglobin. [4]

(b) Explain how a change in the primary structure of haemoglobin (as seen in sickle cell anaemia) can affect its function. [3]

13. Enzymes are biological catalysts.

(a) Define the term 'active site'. [1]

(b) Explain the 'induced fit' model of enzyme action. [3]

14. Cell membranes contain cholesterol.

(a) State the effect of cholesterol on membrane fluidity at high temperatures. [1]

(b) Explain why maintaining membrane fluidity is important for cell function. [2]

15. Fig. 15.1 shows a diagram of a nucleotide.

(a) Name the components labeled P, Q, and R. [3] P: ______________________ Q: ______________________ R: ______________________

(b) Describe how nucleotides are linked together to form a single strand of DNA. [2]

16. Glycogen is a polysaccharide found in animal cells.

(a) Describe the structure of glycogen. [2]

(b) Explain why glycogen is a suitable storage molecule in animals. [2]

17. Fig. 17.1 shows the fluid mosaic model of the cell membrane.

(a) Explain why the membrane is described as 'fluid'. [2]

(b) Explain why the membrane is described as a 'mosaic'. [2]

18. Compare and contrast the structure of a prokaryotic cell and a eukaryotic cell.

(a) State one structural feature present in eukaryotic cells but absent in prokaryotic cells. [1]

(b) State one structural feature present in both prokaryotic and eukaryotic cells. [1]

19. Water potential is a key concept in understanding transport in cells.

(a) Define water potential. [1]

(b) Pure water has a water potential of 0 kPa. Explain why the water potential of a cell cytoplasm is always negative. [2]

20. Discuss the significance of the movement of substances across membranes to the process of photosynthesis. [6]

*** End of Paper ***

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Answers

TuitionGoWhere Exam Practice (AI) - Biology H1 A-Level

Answer Key & Marking Scheme - Version 5

Topic: Cells & Biomolecules
Total Marks: 60

Section A: Structured Questions

1. Phospholipids and Membrane Structure (a) [2 marks]

Phospholipids form a bilayer [1].
Hydrophilic (phosphate) heads face outward towards the aqueous environment, and hydrophobic (fatty acid) tails face inward, away from water [1].

(b) [2 marks]

The hydrophobic interior/tails repel water-soluble (polar/charged) molecules [1].
Therefore, these molecules cannot pass freely through the lipid bilayer and require transport proteins [1].

2. Enzyme Kinetics (a) [1 mark]

Rate = 1 / time = 1 / 45 = 0.0222 s⁻¹ (Accept 0.022).

(b) [3 marks]

At 40°C, the enzyme is near its optimum temperature, so kinetic energy is high and frequent successful collisions occur [1].
As temperature increases to 60°C, the heat energy breaks the hydrogen bonds holding the tertiary structure [1].
The active site changes shape (denaturation), so the substrate no longer fits, and the rate decreases to zero [1].

3. Organelles (a) [1 mark]

Presence of ribosomes on the outer surface / membrane appears rough/studded [1].

(b) [2 marks]

Ribosomes on the RER synthesize proteins (polypeptides) [1].
These proteins are processed/modified/folded within the lumen of the RER before transport to the Golgi [1].

4. Mitochondrial Substrates (a) [2 marks]

Pyruvate enters the mitochondrial matrix and is converted to Acetyl-CoA, entering the Krebs cycle [1].
The Krebs cycle and subsequent oxidative phosphorylation use oxygen as the final electron acceptor [1].

(b) [2 marks]

Glucose cannot enter the mitochondria directly; it must first undergo glycolysis in the cytoplasm [1].
Isolated mitochondria lack the enzymes for glycolysis, so glucose cannot be broken down to pyruvate to fuel respiration [1].

5. Lipids (a) [1 mark]

Condensation / Esterification [1].

(b) [2 marks] (Any two)

High energy content per gram (more than carbohydrates) [1].
Hydrophobic/insoluble in water, so does not affect cell water potential/osmosis [1].
Compact storage volume [1].

6. Water Properties (a) [2 marks]

Water molecules are polar (dipole), with partial positive and negative charges [1].
They form hydrogen bonds with other polar/ionic solutes, surrounding them and keeping them in solution [1].

(b) [1 mark]

Buffers temperature changes in organisms/environment, preventing enzyme denaturation due to rapid temperature fluctuation [1].

7. Membrane Transport (a) [2 marks]

Potassium ions bind to specific sites on the channel protein [1].
The channel opens (or provides a hydrophilic pore), allowing ions to move down their concentration gradient [1].

(b) [2 marks]

No, ATP is not required [1].
This is facilitated diffusion, which is passive movement down a concentration gradient [1].

8. Osmosis in Plant Cells (a) [1 mark]

Plasmolysis [1].

(b) [2 marks]

The external sucrose solution has a lower water potential (more negative) than the cell cytoplasm [1].
Water leaves the cell by osmosis, causing the vacuole/cytoplasm to shrink [1].

9. Nucleic Acids (a) [2 marks] (Any two)

DNA contains deoxyribose sugar; RNA contains ribose sugar [1].
DNA contains Thymine; RNA contains Uracil [1].
DNA is double-stranded; RNA is typically single-stranded [1].

(b) [2 marks]

Ensures accurate replication of genetic information [1].
Allows for the formation of the stable double helix structure via hydrogen bonds [1].

10. Amino Acids (a) [1 mark]

R-group / Side chain / Variable group [1].

(b) [1 mark]

Peptide bond [1].

Section B: Data Interpretation and Extended Response

11. Transport Mechanisms (a) [2 marks]

Curve B shows a higher rate of uptake than Curve A at high concentrations [1].
Curve A levels off (saturates) at a lower rate, while Curve B continues to increase or saturates at a higher level [1].

(b) [3 marks]

Mechanism: Active Transport [1].
Justification 1: Curve B (normal conditions) shows uptake against a concentration gradient or at a higher rate than passive diffusion [1].
Justification 2: Curve A (with respiratory inhibitor) shows reduced uptake because ATP production is stopped, indicating the process requires energy [1].

12. Protein Structure (a) [4 marks]

Primary: Sequence of amino acids held by peptide bonds [1].
Secondary: Folding into alpha-helices or beta-pleated sheets held by hydrogen bonds [1].
Tertiary: 3D folding of the polypeptide chain held by ionic, hydrogen, disulfide bonds, and hydrophobic interactions [1].
Quaternary: Association of multiple polypeptide subunits (haemoglobin has 4) [1].

(b) [3 marks]

Change in primary structure (amino acid sequence) alters the R-groups involved in bonding [1].
This changes the tertiary structure/shape of the protein, specifically the active site or binding pocket [1].
Haemoglobin may lose its ability to bind oxygen effectively or may polymerize (clump), changing cell shape [1].

13. Enzyme Action (a) [1 mark]

The region on the enzyme surface where the substrate binds and the reaction occurs [1].

(b) [3 marks]

The substrate enters the active site [1].
The enzyme changes shape slightly to fit the substrate more closely (mold around it) [1].
This puts strain on the substrate bonds, lowering the activation energy and facilitating the reaction [1].

14. Cholesterol (a) [1 mark]

Reduces fluidity / Restricts movement of phospholipids [1].

(b) [2 marks]

Maintains membrane integrity and prevents it from becoming too fluid/leaky at high temperatures [1].
Ensures membrane proteins remain in correct orientation/function and prevents the membrane from breaking down [1].

15. Nucleotides (a) [3 marks]

P: Phosphate group [1]
Q: Deoxyribose (pentose) sugar [1]
R: Nitrogenous base [1]

(b) [2 marks]

Nucleotides are linked by phosphodiester bonds [1].
Formed between the phosphate of one nucleotide and the sugar of the next (condensation reaction) [1].

16. Glycogen (a) [2 marks]

Polymer of alpha-glucose [1].
Highly branched structure with 1,4 and 1,6 glycosidic bonds [1].

(b) [2 marks]

Compact due to coiling/branching, allowing large amounts of energy to be stored in small space [1].
Insoluble, so does not affect water potential of the cell [1].
Many ends for rapid hydrolysis/release of glucose when needed [1] (Any two).

17. Fluid Mosaic Model (a) [2 marks]

Phospholipids and proteins can move laterally within the layer [1].
Due to weak hydrophobic interactions between tails, not rigid bonds [1].

(b) [2 marks]

Proteins are embedded in the lipid bilayer like tiles in a mosaic [1].
Proteins vary in size, shape, and distribution [1].

18. Prokaryotic vs Eukaryotic (a) [1 mark]

Membrane-bound nucleus / Membrane-bound organelles (e.g., mitochondria, Golgi) [1].

(b) [1 mark]

Cell membrane / Ribosomes / Cytoplasm / DNA [1].

19. Water Potential (a) [1 mark]

The measure of the tendency of water molecules to move from one area to another (kinetic energy of water molecules) [1].

(b) [2 marks]

Cytoplasm contains dissolved solutes (salts, sugars, proteins) [1].
Solutes lower the free energy of water, making the potential negative relative to pure water [1].

20. Extended Response: Membrane Transport and Photosynthesis [6 marks]

Indicative Content:

CO₂ Uptake: CO₂ enters leaf mesophyll cells via diffusion across the plasma membrane. It moves down a concentration gradient maintained by its consumption in the Calvin cycle. Without this transport, the substrate for carbon fixation would be limited [2].
Water Uptake: Water enters root hair cells via osmosis and is transported to leaves. In chloroplasts, water is split (photolysis) in the thylakoid lumen. Membrane integrity is required to maintain the proton gradient for ATP synthesis [2].
Ion Transport: Minerals like Magnesium (for chlorophyll) and Nitrates (for enzymes/proteins) are taken up by root cells via active transport against concentration gradients. This requires ATP and specific carrier proteins. Deficiency limits photosynthetic capacity [1].
Product Export: Triose phosphate/glucose produced in photosynthesis must be transported out of the chloroplast and then out of the cell via transport proteins to be used or stored. Accumulation of products would inhibit photosynthesis via feedback inhibition [1].

Marking Guidance:

5-6 marks: Detailed explanation of at least 3 distinct transport processes linked clearly to photosynthetic stages.
3-4 marks: Description of 2 processes with some link to photosynthesis.
1-2 marks: General statement about membranes or listing substances without explaining the transport mechanism or significance.