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Secondary 4 Combined Science Biology Cells Biomolecules Quiz
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Questions
Secondary 4 Combined Science Biology Quiz - Cells Biomolecules
Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 40
Duration: 45 minutes Total Marks: 40
Instructions:
- Answer ALL questions in the spaces provided.
- The number of marks is given in brackets [ ] at the end of each question or part question.
- Show your working for calculation questions where appropriate.
- Use scientific terminology accurately.
Section A: Cell Structure and Organelles (10 marks)
Answer all questions in this section.
1. The diagram below shows two cells, A and B, viewed under a light microscope.
[Diagram showing Cell A: rectangular shape with cell wall, large central vacuole, chloroplasts, nucleus]
[Diagram showing Cell B: irregular round shape, no cell wall, small vacuoles, nucleus]
(a) Identify cell A and cell B. [2]
Cell A: _________________________
Cell B: _________________________
(b) State ONE structural feature visible in the diagram that distinguishes cell A from cell B. [1]
(c) Explain why cell A contains chloroplasts while cell B does not. [2]
2. The table below shows the number of mitochondria in three different types of human cells.
| Cell Type | Number of Mitochondria per Cell |
|---|---|
| Skin cell | 200 |
| Liver cell | 1,000 |
| Muscle cell | 2,500 |
(a) Using the data in the table, explain why muscle cells have the highest number of mitochondria. [2]
(b) Predict the relative number of mitochondria you would expect to find in a human red blood cell. Explain your answer. [3]
3. The diagram below shows a typical animal cell as seen under an electron microscope.
[Diagram showing an animal cell with nucleus, rough endoplasmic reticulum, Golgi body, mitochondria, and ribosomes labelled]
(a) State the function of the Golgi body. [1]
(b) Describe the role of ribosomes in the cell. [1]
4. A student prepared a slide of onion epidermal cells and observed it under a microscope. The student could see the cell wall and nucleus clearly but could not see the mitochondria.
Explain why the mitochondria were not visible. [1]
5. Compare the structure of a typical plant cell and a typical animal cell by stating TWO differences. [2]
Difference 1: _________________________________________________________________________
Difference 2: _________________________________________________________________________
Section B: Movement of Substances (10 marks)
Answer all questions in this section.
6. A student carried out an experiment to investigate osmosis using potato strips. Three potato strips of equal mass were placed in three different solutions: distilled water, 5% sucrose solution, and 20% sucrose solution. After 30 minutes, the potato strips were removed, blotted dry, and reweighed.
The results are shown in the table below.
| Solution | Initial Mass (g) | Final Mass (g) | Change in Mass (g) |
|---|---|---|---|
| Distilled water | 5.0 | 5.8 | +0.8 |
| 5% sucrose | 5.0 | 5.1 | +0.1 |
| 20% sucrose | 5.0 | 4.4 | -0.6 |
(a) Explain why the potato strip in distilled water gained mass. [3]
(b) Explain why the potato strip in 20% sucrose solution lost mass. [3]
(c) The potato strip in 5% sucrose solution showed very little change in mass. What does this suggest about the water potential of the 5% sucrose solution compared to the potato cells? [1]
7. The diagram below shows an investigation into the effect of temperature on the rate of diffusion of potassium permanganate crystals in water.
[Diagram showing three beakers of water at 10°C, 30°C, and 60°C, each with a crystal of potassium permanganate at the bottom. The purple colour has spread furthest in the 60°C beaker and least in the 10°C beaker.]
(a) Describe the results shown in the diagram. [1]
(b) Explain why the rate of diffusion is fastest at 60°C. [2]
8. Define the term diffusion. [1]
9. State ONE factor, other than temperature, that affects the rate of diffusion. [1]
10. A red blood cell is placed in a concentrated salt solution. Describe and explain what happens to the cell. [2]
Section C: Enzymes and Biomolecules (10 marks)
Answer all questions in this section.
11. An enzyme extracted from bacteria found in hot springs was investigated. The enzyme activity was measured at different temperatures. The results are shown in the graph below.
[Graph showing enzyme activity on y-axis (arbitrary units) against temperature on x-axis (°C).
Activity increases from 20°C, peaks at 75°C, then drops sharply to near zero at 90°C.]
(a) State the optimum temperature for this enzyme. [1]
(b) Describe and explain the shape of the graph between 20°C and 75°C. [3]
(c) Explain why the enzyme activity decreases sharply after 75°C. [3]
(d) Suggest why this enzyme has a much higher optimum temperature than enzymes found in the human body. [3]
12. State the monomer units that make up proteins. [1]
13. Describe the 'lock and key' hypothesis of enzyme action. [2]
14. Name the type of biomolecule that enzymes belong to. [1]
15. Explain why enzymes are described as biological catalysts. [2]
Section D: Integrated Application (10 marks)
Answer all questions in this section.
16. A student set up an experiment to investigate the effect of pH on the activity of pepsin, an enzyme found in the human stomach. Egg white (a protein) was placed in test tubes containing buffer solutions of different pH values. Pepsin was added to each tube, and the time taken for the egg white to be completely digested was recorded.
| pH | Time for Complete Digestion (minutes) |
|---|---|
| 1 | 4 |
| 2 | 2 |
| 3 | 5 |
| 5 | 30 |
| 7 | No digestion after 60 minutes |
| 9 | No digestion after 60 minutes |
(a) Identify the optimum pH for pepsin based on the data. [1]
(b) Explain why no digestion occurred at pH 7 and pH 9. [3]
(c) The stomach lining secretes hydrochloric acid. Using your knowledge of enzymes, explain why this is important for protein digestion. [2]
(d) State the product(s) formed when pepsin digests egg white. [1]
(e) The student repeated the experiment at pH 2 but first boiled the pepsin solution for 10 minutes before adding it to the egg white. Predict the result and explain your answer. [3]
17. A plant cell was placed in a concentrated sugar solution. The cell became plasmolysed. Explain what is meant by plasmolysis. [2]
18. State the test used to identify the presence of reducing sugars in a food sample and describe the expected positive result. [2]
Test: _________________________________________________________________________
Positive result: _________________________________________________________________________
19. Explain why a person with a high fever (above 40°C) may be at risk of serious health complications, with reference to enzymes. [2]
20. A student investigated the effect of surface area on the rate of diffusion using agar cubes of different sizes containing a pH indicator. The cubes were placed in hydrochloric acid. State the relationship between surface area to volume ratio and the rate of diffusion. [1]
END OF QUIZ
Check your answers carefully before submitting.
Answers
Secondary 4 Combined Science Biology Quiz - Cells Biomolecules
ANSWER KEY AND MARKING SCHEME
Total Marks: 40
Section A: Cell Structure and Organelles (10 marks)
1. (a) Identify cell A and cell B. [2]
- Cell A: Plant cell [1 mark]
- Cell B: Animal cell [1 mark]
Accept: Accept "palisade cell" or "leaf cell" for Cell A; accept "cheek cell" or "human cell" for Cell B.
1. (b) State ONE structural feature visible in the diagram that distinguishes cell A from cell B. [1]
Any ONE of:
- Cell A has a cell wall / Cell B has no cell wall
- Cell A has chloroplasts / Cell B has no chloroplasts
- Cell A has a large central vacuole / Cell B has small vacuoles
- Cell A is regular/rectangular in shape / Cell B is irregular/round in shape
Award 1 mark for any correct structural difference.
1. (c) Explain why cell A contains chloroplasts while cell B does not. [2]
- Cell A is a plant cell and carries out photosynthesis to make its own food [1 mark]
- Chloroplasts contain chlorophyll which absorbs light energy for photosynthesis [1 mark]
- Cell B is an animal cell; animals obtain food by feeding/eating, not by photosynthesis, so chloroplasts are not needed
Award 2 marks for clear explanation linking chloroplast presence to photosynthetic function in plants and absence to heterotrophic nutrition in animals.
2. (a) Using the data in the table, explain why muscle cells have the highest number of mitochondria. [2]
- Muscle cells require more energy/ATP for contraction/movement [1 mark]
- Mitochondria are the site of aerobic respiration, which produces ATP/energy [1 mark]
- Therefore, muscle cells have more mitochondria to meet their higher energy demands
Award 2 marks for linking high mitochondrial count to high energy requirement and mitochondrial function.
2. (b) Predict the relative number of mitochondria you would expect to find in a human red blood cell. Explain your answer. [3]
- Red blood cells would have zero/no mitochondria [1 mark]
- Mature human red blood cells lack a nucleus and most organelles, including mitochondria [1 mark]
- Red blood cells transport oxygen and rely on anaerobic respiration for their energy needs, so mitochondria are not required [1 mark]
Award 3 marks for correct prediction with full explanation including structural adaptation and metabolic reasoning.
3. (a) State the function of the Golgi body. [1]
- The Golgi body modifies, sorts, and packages proteins (and lipids) for secretion or transport to other parts of the cell [1 mark]
Accept any correct description of Golgi body function.
3. (b) Describe the role of ribosomes in the cell. [1]
- Ribosomes are the site of protein synthesis / Ribosomes synthesise proteins [1 mark]
Award 1 mark for correct function.
4. A student prepared a slide of onion epidermal cells and observed it under a microscope. The student could see the cell wall and nucleus clearly but could not see the mitochondria. Explain why the mitochondria were not visible. [1]
- Mitochondria are too small to be seen under a light microscope / A light microscope does not have sufficient magnification/resolution to see mitochondria [1 mark]
Award 1 mark for correct explanation relating to size or microscope limitation.
5. Compare the structure of a typical plant cell and a typical animal cell by stating TWO differences. [2]
Any TWO of:
- Plant cells have a cell wall; animal cells do not [1 mark]
- Plant cells have chloroplasts; animal cells do not [1 mark]
- Plant cells have a large central vacuole; animal cells have small/temporary vacuoles [1 mark]
- Plant cells are regular/rectangular in shape; animal cells are irregular/round [1 mark]
Award 1 mark for each correct structural difference, up to 2 marks.
Section B: Movement of Substances (10 marks)
6. (a) Explain why the potato strip in distilled water gained mass. [3]
- Distilled water has a higher water potential than the potato cell contents/sap [1 mark]
- Water molecules move from the distilled water into the potato cells by osmosis [1 mark]
- Osmosis is the net movement of water molecules from a region of higher water potential to a region of lower water potential through a partially permeable membrane [1 mark]
- The cells become turgid, and the potato strip increases in mass
Award 3 marks for explanation including water potential comparison, osmosis definition, and reference to partially permeable membrane.
6. (b) Explain why the potato strip in 20% sucrose solution lost mass. [3]
- The 20% sucrose solution has a lower water potential than the potato cell contents/sap [1 mark]
- Water molecules move from the potato cells into the sucrose solution by osmosis [1 mark]
- Water moves from a region of higher water potential (inside cells) to a region of lower water potential (outside cells) through the partially permeable cell membrane [1 mark]
- The cells become plasmolysed/flaccid, and the potato strip decreases in mass
Award 3 marks for explanation including water potential comparison, direction of water movement, and reference to partially permeable membrane.
6. (c) The potato strip in 5% sucrose solution showed very little change in mass. What does this suggest about the water potential of the 5% sucrose solution compared to the potato cells? [1]
- The water potential of the 5% sucrose solution is approximately equal/similar to the water potential of the potato cells / The solution is isotonic to the potato cells [1 mark]
Award 1 mark for correct inference about equal water potential or isotonic condition.
7. (a) Describe the results shown in the diagram. [1]
- The purple colour (potassium permanganate) spread furthest in the 60°C beaker and least in the 10°C beaker / The rate of diffusion increases with increasing temperature [1 mark]
Award 1 mark for correct description of the trend.
7. (b) Explain why the rate of diffusion is fastest at 60°C. [2]
- At higher temperatures, particles have more kinetic energy [1 mark]
- Particles move faster and collide more frequently, so they spread out/diffuse more quickly [1 mark]
Award 2 marks for linking temperature to kinetic energy and rate of particle movement.
8. Define the term diffusion. [1]
- Diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration down a concentration gradient [1 mark]
Award 1 mark for correct definition including net movement and concentration gradient.
9. State ONE factor, other than temperature, that affects the rate of diffusion. [1]
Any ONE of:
- Concentration gradient [1 mark]
- Surface area (to volume ratio) [1 mark]
- Size of particles / molecular mass [1 mark]
- Diffusion distance / thickness of membrane [1 mark]
Award 1 mark for any correct factor.
10. A red blood cell is placed in a concentrated salt solution. Describe and explain what happens to the cell. [2]
- The cell shrinks/crinkles (becomes crenated) [1 mark]
- The concentrated salt solution has a lower water potential than the cell contents; water moves out of the cell by osmosis [1 mark]
Award 2 marks for correct description and explanation involving water potential and osmosis.
Section C: Enzymes and Biomolecules (10 marks)
11. (a) State the optimum temperature for this enzyme. [1]
- 75°C [1 mark]
Award 1 mark for correct value.
11. (b) Describe and explain the shape of the graph between 20°C and 75°C. [3]
- The enzyme activity increases as temperature increases from 20°C to 75°C [1 mark]
- Increasing temperature increases the kinetic energy of enzyme and substrate molecules [1 mark]
- This leads to more frequent successful collisions and more enzyme-substrate complexes formed per unit time [1 mark]
- The rate of reaction increases up to the optimum temperature
Award 3 marks for description of increasing trend and explanation involving kinetic energy, collision frequency, and enzyme-substrate complex formation.
11. (c) Explain why the enzyme activity decreases sharply after 75°C. [3]
- Above 75°C, the enzyme is denatured [1 mark]
- The high temperature breaks the hydrogen bonds (and other bonds) that maintain the enzyme's specific three-dimensional shape/tertiary structure [1 mark]
- The active site loses its specific shape, so the substrate can no longer bind/fit into the active site [1 mark]
- The enzyme can no longer catalyse the reaction
Award 3 marks for explanation including denaturation, bond disruption, and active site shape change preventing substrate binding.
11. (d) Suggest why this enzyme has a much higher optimum temperature than enzymes found in the human body. [3]
- The enzyme was extracted from bacteria living in hot springs, which have high environmental temperatures [1 mark]
- The enzyme has evolved/adapted to function optimally at high temperatures [1 mark]
- The enzyme's structure is more stable at high temperatures; it has more bonds (e.g., disulfide bridges, hydrogen bonds) or a more compact structure that resists denaturation [1 mark]
- Human body temperature is ~37°C, so human enzymes have optimum temperatures around 37°C; this bacterial enzyme is adapted to its hot environment
Award 3 marks for linking the enzyme's origin to its thermal adaptation, explaining structural stability, and contrasting with human enzyme temperature requirements.
12. State the monomer units that make up proteins. [1]
- Amino acids [1 mark]
Award 1 mark for correct answer.
13. Describe the 'lock and key' hypothesis of enzyme action. [2]
- The active site of the enzyme has a specific shape that is complementary to the shape of the substrate [1 mark]
- The substrate fits into the active site to form an enzyme-substrate complex, like a key fitting into a lock [1 mark]
Award 2 marks for clear description including complementary shape and enzyme-substrate complex formation.
14. Name the type of biomolecule that enzymes belong to. [1]
- Proteins [1 mark]
Award 1 mark for correct answer.
15. Explain why enzymes are described as biological catalysts. [2]
- Enzymes speed up the rate of chemical reactions in living organisms [1 mark]
- They remain chemically unchanged at the end of the reaction and can be reused [1 mark]
Award 2 marks for explanation covering both speed increase and reusability.
Section D: Integrated Application (10 marks)
16. (a) Identify the optimum pH for pepsin based on the data. [1]
- pH 2 [1 mark]
Award 1 mark for correct value (shortest digestion time indicates optimum pH).
16. (b) Explain why no digestion occurred at pH 7 and pH 9. [3]
- Pepsin is an enzyme with a specific optimum pH of around 2 (acidic conditions) [1 mark]
- At pH 7 (neutral) and pH 9 (alkaline), the pH is far from the enzyme's optimum [1 mark]
- The enzyme is denatured; the change in pH disrupts the ionic and hydrogen bonds that maintain the enzyme's specific three-dimensional shape/active site [1 mark]
- The substrate (egg white protein) can no longer bind to the active site, so no enzyme-substrate complexes form and no digestion occurs
Award 3 marks for explanation linking pH deviation to denaturation and loss of active site function.
16. (c) The stomach lining secretes hydrochloric acid. Using your knowledge of enzymes, explain why this is important for protein digestion. [2]
- Hydrochloric acid provides the optimum pH (acidic conditions) for pepsin to function [1 mark]
- Pepsin requires an acidic pH (around pH 2) to maintain its active site shape and catalyse protein digestion effectively [1 mark]
Award 2 marks for linking HCl secretion to providing optimum pH for pepsin activity.
16. (d) State the product(s) formed when pepsin digests egg white. [1]
- Polypeptides / peptides / amino acids [1 mark]
Accept any correct product of protein digestion by pepsin.
16. (e) The student repeated the experiment at pH 2 but first boiled the pepsin solution for 10 minutes before adding it to the egg white. Predict the result and explain your answer. [3]
- No digestion would occur / The egg white would not be digested [1 mark]
- Boiling causes the enzyme (pepsin) to be denatured [1 mark]
- The high temperature breaks the bonds maintaining the enzyme's three-dimensional shape/active site, so the active site loses its specific shape and the substrate can no longer bind [1 mark]
Award 3 marks for correct prediction and explanation involving denaturation and loss of active site function.
17. A plant cell was placed in a concentrated sugar solution. The cell became plasmolysed. Explain what is meant by plasmolysis. [2]
- Plasmolysis is the shrinkage of the cytoplasm/cell contents away from the cell wall [1 mark]
- It occurs when a plant cell is placed in a solution of lower water potential, causing water to leave the cell by osmosis [1 mark]
Award 2 marks for correct definition and explanation of water movement.
18. State the test used to identify the presence of reducing sugars in a food sample and describe the expected positive result. [2]
- Test: Benedict's test [1 mark]
- Positive result: Brick-red/orange precipitate forms upon heating [1 mark]
Award 1 mark for correct test name and 1 mark for correct positive result.
19. Explain why a person with a high fever (above 40°C) may be at risk of serious health complications, with reference to enzymes. [2]
- High temperatures can cause enzymes in the body to denature [1 mark]
- Denatured enzymes lose their active site shape and can no longer catalyse essential metabolic reactions, leading to cell damage/death [1 mark]
Award 2 marks for explanation linking fever temperature to enzyme denaturation and metabolic disruption.
20. A student investigated the effect of surface area on the rate of diffusion using agar cubes of different sizes containing a pH indicator. The cubes were placed in hydrochloric acid. State the relationship between surface area to volume ratio and the rate of diffusion. [1]
- The larger the surface area to volume ratio, the faster the rate of diffusion / As surface area to volume ratio increases, the rate of diffusion increases [1 mark]
Award 1 mark for correct relationship stated.