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S4 Biology Practice Paper: Cells, Enzymes and Transport (12 questions, 100 marks, 75 minutes)
S4 Biology: Cells, Enzymes and Transport
S4 Biology practice paper: 12 questions on cell structure, enzymes, diffusion, osmosis, active transport, the fluid mosaic model and microscope calculation. Answers with mark points.
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Paper details Level: S4 | Subject: Biology | Time: 75 minutes | Marks: 100 | Questions: 12 | Answers included
Section A: Cell Structure and Enzymes (32 marks)
Answer all questions in Section A.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A student observes onion epidermal cells and human cheek cells with a light microscope. (a) State two structures present in onion epidermal cells but absent in cheek cells (2 marks). (b) State one structure present in cheek cells but absent in mature onion epidermal cells (2 marks). (c) Explain why physiological saline rather than pure water is used when mounting cheek cells (2 marks). (d) Describe the difference in the position of the nucleus in the two cell types (2 marks). | (a) Cell wall and a large vacuole (chloroplast also accepted); the wall is made of cellulose and gives support (2 marks). (b) Centriole (lysosome also accepted) (2 marks). (c) Saline has a water potential similar to that of the cytoplasm, so the cell neither swells and bursts nor shrinks by osmosis and keeps its shape (2 marks). (d) In the onion cell the nucleus is pushed to the edge by the large vacuole; in the cheek cell it lies near the centre (2 marks). |
| 2 | A student mixes amylase with starch solution and keeps the mixture at different temperatures. (a) Explain why enzymes are specific in action (2 marks). (b) State the product formed when amylase digests starch (2 marks). (c) Explain why the reaction rate falls sharply when the temperature rises from 37 degrees C to 70 degrees C (2 marks). (d) Explain how a pH far from the optimum affects the active site (2 marks). | (a) The shape of the active site is complementary only to its specific substrate, so other molecules cannot bind (2 marks). (b) Maltose, which can be further broken down to glucose (2 marks). (c) Heat breaks bonds in the enzyme, the active site changes shape and the enzyme is denatured, so the substrate can no longer bind and the rate falls sharply (2 marks). (d) A pH far from the optimum breaks ionic and hydrogen bonds in the active site, altering its shape and destroying activity (2 marks). |
| 3 | Red blood cells are placed in distilled water and in 0.9 per cent sodium chloride solution, and plant cells are placed in concentrated sucrose solution. (a) Describe and explain the change in red blood cells in distilled water (3 marks). (b) Explain how plasmolysis occurs in a plant cell in concentrated sucrose solution (3 marks). (c) Explain why 0.9 per cent sodium chloride solution is isotonic to red blood cells (2 marks). | (a) The cells swell and finally burst; the water potential of distilled water is higher than that of the cytoplasm, so water enters by osmosis (3 marks). (b) The sucrose solution has a lower water potential than the cell sap, so water leaves the vacuole by osmosis, the vacuole shrinks and the membrane pulls away from the wall (3 marks). (c) Its water potential equals that of the cytoplasm, so net water movement is zero and cell volume is unchanged (2 marks). |
| 4 | Root hair cells absorb mineral ions from soil solution of lower concentration. (a) Define active transport (2 marks). (b) State two differences between active transport and simple diffusion (2 marks). (c) Explain why the rate of ion uptake falls when the root is deprived of oxygen (2 marks). (d) Explain how the specificity of carrier proteins limits the ions a cell can absorb (2 marks). | (a) Movement of a substance against its concentration gradient, from low to high concentration, using metabolic energy and carrier proteins (2 marks). (b) It can move substances against the gradient and it requires energy and carrier proteins (any two, 2 marks). (c) Lack of oxygen lowers aerobic respiration, so less ATP is available, the ion pumps stop and uptake falls (2 marks). (d) A carrier protein binds only its specific ion, so a cell lacking that carrier cannot actively absorb that ion (2 marks). |
Section B: Transport and Experiments (36 marks)
Answer all questions in Section B and show your working.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | A student uses 1 per cent starch solution, 1 per cent amylase solution and iodine solution to measure, at 20, 30, 37, 50 and 60 degrees C, the time taken for the blue-black colour to disappear. (a) State the independent variable and the dependent variable (2 marks). (b) State two variables that must be kept constant (2 marks). (c) Explain why iodine is added just before the enzyme is added (2 marks). (d) State the error caused by using one test tube for all five temperatures and suggest an improvement (3 marks). | (a) Independent variable: temperature; dependent variable: time for the blue-black colour to disappear (2 marks). (b) Concentration and volume of starch and enzyme solutions, and pH (any two, 2 marks). (c) Iodine turns blue-black with starch at once, so adding it before the reaction starts allows the fading time to represent the rate of starch digestion (2 marks). (d) Enzyme left in the tube and the temperature history of the glass affect later readings; separate tubes should be used for each temperature and all tubes brought to the target temperature before the enzyme is added (3 marks). |
| 6 | Four potato strips, each of mass 2.40 g, are kept for 30 minutes in distilled water and in 0.1, 0.2 and 0.4 mol per dm3 sucrose solutions. After blotting, their masses are 2.76 g, 2.58 g, 2.40 g and 2.16 g. (a) Calculate the percentage change in mass for the distilled water strip (2 marks). (b) Calculate the percentage change in mass for the 0.4 mol per dm3 strip (2 marks). (c) Identify the solution whose concentration equals that of the potato cell sap, and explain (2 marks). (d) Explain why the mass increases in the 0.1 mol per dm3 solution (3 marks). | (a) Distilled water: (2.76 minus 2.40) divided by 2.40 times 100 per cent = +15.0 per cent; 0.4 mol per dm3: (2.16 minus 2.40) divided by 2.40 times 100 per cent = minus 10.0 per cent (4 marks). (b) The 0.2 mol per dm3 solution, whose mass change is 0.0 per cent, so its water potential equals that of the cell sap and net water movement is zero (2 marks). (c) The 0.1 mol per dm3 solution has a higher water potential than the cell sap, so water enters by osmosis, the vacuole enlarges, the cell becomes turgid and its mass rises by 7.5 per cent (3 marks). |
| 7 | Spinach leaves and yeast are used as materials. (a) Write the overall word equation for photosynthesis (2 marks). (b) Write the overall word equation for aerobic respiration (2 marks). (c) State the difference in the direction of energy conversion between the two processes (2 marks). (d) Explain why a plant still respires aerobically in daylight (3 marks). | (a) Photosynthesis: carbon dioxide + water, in the presence of light and chlorophyll, gives glucose + oxygen; aerobic respiration: glucose + oxygen gives carbon dioxide + water + energy (4 marks). (b) Photosynthesis stores light energy as chemical energy, while aerobic respiration releases chemical energy as ATP and heat (2 marks). (c) Respiration runs continuously to supply ATP for life processes regardless of light; in daylight the rate of photosynthesis exceeds that of respiration, so the net effect is uptake of carbon dioxide and release of oxygen (3 marks). |
| 8 | A microscope has a x10 eyepiece and x10 and x40 objectives. With the x10 objective the diameter of the field of view is 1.6 mm. (a) Find the total magnification with the x40 objective (2 marks). (b) Find the diameter of the field of view with the x40 objective, in micrometres (3 marks). (c) If a cell is 50 micrometres in diameter, find how many such cells fit across the diameter (2 marks). (d) Explain why the field becomes dimmer at the higher magnification (2 marks). | (a) Total magnification = 10 times 40 = 400 times (2 marks). (b) The diameter of the field is inversely proportional to magnification: 1.6 mm times 10 divided by 40 = 0.4 mm, which is 400 micrometres (3 marks). (c) 400 micrometres divided by 50 micrometres = 8 cells (2 marks). (d) Higher magnification lets less light into the lens system, so the field is dimmer and more illumination or a wider aperture is needed (2 marks). |
Section C: Models and Data Analysis (32 marks)
Answer all questions in Section C.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | The fluid mosaic model explains the structure and function of the cell membrane. (a) Name three main components of the model (3 marks). (b) Explain why phospholipids form a bilayer in water (2 marks). (c) State the function of cholesterol in an animal cell membrane (3 marks). | (a) Phospholipids, proteins and carbohydrates as glycoproteins, together with cholesterol (any three, 3 marks). (b) The hydrophilic heads face the water on both sides while the hydrophobic tails point inwards, giving a stable bilayer (2 marks). (c) Cholesterol sits between the phospholipid tails and regulates fluidity, restricting movement at high temperature and preventing tight packing at low temperature (3 marks). |
| 10 | Two cube models have sides of 2 cm and 4 cm. (a) Calculate the surface area to volume ratio of each model in per cm (4 marks). (b) Use the results to explain why a larger cell volume limits exchange with the surroundings (2 marks). (c) Explain how the shape of a root hair improves absorption (2 marks). | (a) 2 cm cube: surface area 24 cm2, volume 8 cm3, ratio 3.0 per cm; 4 cm cube: surface area 96 cm2, volume 64 cm3, ratio 1.5 per cm (4 marks). (b) As volume grows the surface area to volume ratio falls, so relatively less surface is available for diffusion and supply or removal of substances becomes limiting (2 marks). (c) Root hairs are long, narrow and numerous, greatly increasing the surface area to volume ratio for water and ion uptake (2 marks). |
| 11 | The relative rate of an enzyme is 20 at 20 degrees C, 35 at 30 degrees C, 40 at 37 degrees C, 25 at 50 degrees C and 8 at 60 degrees C. (a) State the optimum temperature of the enzyme (1 mark). (b) Calculate the rate at 60 degrees C as a percentage of the rate at 37 degrees C (3 marks). (c) Explain why the rates at 50 degrees C and at 20 degrees C are both below the rate at 37 degrees C, although the reasons differ (4 marks). | (a) The optimum temperature is 37 degrees C (1 mark). (b) 8 divided by 40 times 100 per cent = 20.0 per cent (3 marks). (c) At 50 degrees C the enzyme is partly denatured: the active site changes shape irreversibly and the rate falls; at 20 degrees C molecules have less kinetic energy, so collisions between enzyme and substrate are less frequent, the rate falls but the enzyme is not denatured and activity returns on warming (4 marks). |
| 12 | A student uses the time for iodine colour to fade as a measure of amylase activity. (a) State two sources of error in judging the end point (2 marks). (b) Suggest one practical improvement that reduces the error (2 marks). (c) Explain why a control containing only starch solution and iodine is needed (2 marks). (d) State the purpose of repeating the experiment three times and taking the mean (2 marks). | (a) Judging the exact moment the colour disappears differs between observers; the thickness of the tube and the background light affect the reading; the reaction continues during sampling (any two, 2 marks). (b) Use a colorimeter or spectrophotometer to read absorbance at fixed intervals and take a set value as the end point instead of visual judgement (2 marks). (c) The control contains no enzyme, showing that the loss of colour is caused by enzyme action rather than by fading of iodine or other factors (2 marks). (d) Repeats reduce random error and the mean improves the reliability and reproducibility of the result (2 marks). |
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