1. | C | 11. | B | 21. | C | 31. | D | 41. | C |
2. | B | 12. | D | 22. | A | 32. | B | 42. | D |
3. | B | 13. | A | 23. | A | 33. | B | 43. | D |
4. | A | 14. | D | 24. | A | 34. | D | 44. | D |
5. | D | 15. | C | 25. | B | 35. | A | 45. | D |
6. | A | 16. | D | 26. | C | 36. | C | 46. | A |
7. | A | 17. | B | 27. | B | 37. | C | 47. | D |
8. | D | 18. | D | 28. | B | 38. | A | 48. | A |
9. | C | 19. | B | 29. | C | 39. | D | 49. | C |
10. | C | 20. | D | 30. | B | 40. | C | 50 | D |
Question | Scheme | Mark |
1 (a)(i) 1(a)(ii) | Plant cell Presence of cell wall/ chloroplast/large vacuole | 1 3 |
1(b)(i) 1(b)(ii) | Chloroplast Site of photosynthesis | 1 1 |
1(c)(i) 1(c)(ii) 1(c)(iii) | Mitochondria Generates energy for the cell chemical reaction like respiration All cell activities will stop as there is no energy | 1 1 1 |
1(d) | R- Composed of cellulose, S- Composed of protein & lipid | 1 |
2(a)(i) 2(a)(ii) | Become firmer. Distilled water is hypotonic to sucrose solution. Water molecules move into the visking tubing through osmosis. The increase in content in visking tubing makes it firmer. | 1 2 |
2(b)(i) 2(b)(ii) | Plasma membrane Allows only certain molecules (smaller molecules) to pass through it. | 1 1 |
2(c)(i) 2(c)(ii) | There will be glucose molecules in distilled water. 2 ml distilled water is added with 1 ml of Benedict’s solution and heated. A red precipitate shows the presence of glucose. | 1 3 |
2(d) | Glucose molecules are smaller in size than the pores in visking tubing. Hence glucose molecules diffuse out from visking tubing due to the concentration gradient. | 2 |
3(a) | Graph (axis with correct scale, transfer correct data, correct shape | 3 |
3(b)(i) 3(b)(ii) | 0.25 M Isotonic | 1 1 |
3(c) | More than 6% | 1 |
3(d)(i) 3(d)(ii) | Plasmolysis 0.5 M sucrose solution is hypertonic compared to the cell sap of a potato. Due to this, the water molecules from the potato cell diffuse out by osmosis. | 1 2 |
3(e)(i) 3(e)(ii) | Turgid/ smooth 0.1 M sucrose solution is hypotonic compared to the cell sap of a potato. Due to this, the water molecules diffuse into the potato cell and cause it to become turgid. | 1 2 |
4(a) | K-Enzyme, L-Substrate, M-Enzyme-substrate complex, N-Product | 4 |
4(b)(i) 4(b)(ii) | K L | 1 1 |
4(c) | The substrate is represented by the key while the enzyme is represented by the lock. Only one key will be fit into the lock and open it. | 2 |
4(d) | Excessive heat denatures the protein causing the configuration to change. Therefore, the substrate molecules are not fit into the active site. | 3 |
4(e)(i) 4(e)(ii) | Soap contain enzyme which act on stains. Enzymes are most active in a temperature range 350C-400C. Enzyme is inactive at low temperature whereas they will be denatured at high temperature. Food processing//Production of alcohol//tenderizing meat and vegetables//making bread | 2 2 |
5(a) | Digest protein into pepton and polipeptide | 1 |
5(b) | Hydrolysis of protein by the enzyme pepsin | 1 |
5(c)(i) 5(c)(ii) | The digestion is incomplete Pepsin needs an acidic medium (pH 2- pH 3) for it to act. Content of the test tube Q is in neutral condition. The contents remains murky because there is no/ incomplete digestion. | 2 2 |
5(d) | 370C is the optimum temperature for pepsin to act | 2 |
5(e) | Contents in the test tube S remain murky because there is no digestion. 40C is not suitable temperature for pepsin to act. Contents in test tube R turns to clear because there is digestion. 370C and acidic medium are suitable condition for pepsin to act. | 2 |
SECTION B: 20 Marks
Question | Scheme | Mark | |||||||||||||||||||
1 (a) | The movement of water molecules from high concentration to low concentration through a semi-permeable membrane | 1 | |||||||||||||||||||
1(b) |
| 9 | |||||||||||||||||||
1(c) | Aim : To investigate the effects of isotonic , hypotonic and hypertonic solutions on red blood cells Problem statement: What are the effects of isotonic, hypotonic and hypertonic solutions on red blood cells?// Does isotonic, hypotonic and hypertonic solutions effects the red blood cells? Hypothesis: When red blood cells immersed in a hypotonic solution, the cells will expand and lyses. When red blood cells immersed in a hypertonic solution, the cells become shrink When red blood cells immersed in an isotonic solution, there is no change to the cell Apparatus & materials
1m: any 3A & 3M Procedure 1. Four slides are prepared and labelled A, B, C and D (√) 2. A drop of blood is dropped onto slide A and the shape of the red blood cells is observed under a microscope (√) 3. A drop of distilled water is dropped onto slide B. Cover the slide with cover slip. Add a drop of blood by the side of cover subsequently and the slide is observed under a microscope (√) 4. Step 1 to 3 are repeated for slide C and D by using salt, solution 0.15M and 0.5M (√) 4 (√) = 2m, 3 (√)= 1m, 2(√)= 1m Presenting data/ Observation
Conclusion: Distilled water is hypotonic to red blood cells (slide C) Salt solution 0.15M is hypotonic to cytoplasm of red blood cells. Salt solution 0.5M is hypertonic to cytoplasm of red blood cells. | 2 1 2 2 2 1 | |||||||||||||||||||
2(a)(i) | Correct diagram and label Cell wall -It maintains the shape of the plant cell -It protects and gives mechanical strength to the cell Cytoplasm -It is a jelly-like medium which stores the organelles of the cell -It acts as a medium where chemical processes occur in the cell Vacuole -It is a fluid-filled space which stores sugars, amino acids and mineral salts -It support herbaceous plants by giving turgid pressure Nucleus -It controls and regulates all the activities of the cell -It contains genetic material Plasma membrane -It controls the substances entering or leaving the cell -It separates the contents of the cell from its external environment Chloroplast -It contains the pigment chlorophyll which makes the leaves green -It acts as a site for photosynthesis | 2 Max:10 | |||||||||||||||||||
2(a)(ii) | Chloroplast. It acts as a site for photosynthesis | 2 | |||||||||||||||||||
3(b)(i) | J The fertilizer dissolves and forms ions in the soil water J The concentration of mineral ions is usually higher in the soil than in the cell, so mineral ions diffuse into the root hairs J Sometimes, the concentration of minerals ions is higher in the cell than in the soil water but the cell still needs more mineral salts J The cell thus absorbs the ions through active transport | Max: 3 | |||||||||||||||||||
3(b)(ii) | J When too much fertilizer is applied, the soil water becomes more concentrated than the cell sap J Water diffuses out of the cell by osmosis J Excessive loss of water causes the plant to wilt J If the condition continues, the plant will then die | Max:3 |
SECTION C: 20 Marks
Question | Scheme | Mark | |||||||||||||||||||||
1 (a)(i) 1(a)(ii) | The time taken for the hydrolysis of starch is faster if the volume of saliva used is larger. The more saliva used, the more enzymes there are to act on the starch molecules | 1 1 | |||||||||||||||||||||
1(b)(i)&(ii) |
| 6 | |||||||||||||||||||||
1(c)(i) 1(c)(ii) | Enzyme concentration Time taken for the complete hydrolysis of starch Temperature, pH, substrate concentration | 1 1 1 | |||||||||||||||||||||
1(d) | The higher the enzyme concentration, the faster the rate of reaction. | 2 | |||||||||||||||||||||
1(e) | Graph (axis with correct scale, transfer correct data, correct shape, title) | 4 | |||||||||||||||||||||
1(f) | There is no reaction in each test tube as the enzymes are denatured at high temperature | 3 |
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