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1 2 6 A scientist put samples of animal cells into sucrose solutions with concentrations ranging from 00 mmol dm 3 to 300 mmol dm 3. Before putting them into the sucrose solutions, she found the mean volume of the cells in each sample. She called this the mean initial volume. After the cells had been in the sucrose solutions for 20 minutes, the scientist removed them and found their new mean volume. She called this the mean final volume. The scientist then calculated the ratio of mean final volume to mean initial volume for each sample of cells. She plotted this value against the concentration of sucrose solution in which the cells had been placed. Her graph is shown..4.2 Mean final volume Mean initial volume (a) Describe and explain these results Concentration of sucrose solution / mmol dm 3 (4 marks) (Extra space)... (2) APW/Jun09/BYB
2 3 6 (b) (i) In another experiment, the scientist put a sample of the same animal cells into a 50 mmol dm 3 sucrose solution. After 20 minutes, there were no intact cells, only cell fragments. Explain why. (3 marks) (Extra space)... 6 (b) (ii) Bacterial cells were placed in a 50 mmol dm 3 sucrose solution for the same length of time as the animal cells. These cells remained intact. Explain why. (Extra space)... 9 Turn over (3) APW/Jun09/BYB
3 Biology - AQA GCE Mark Scheme 2009 June series Question Part Sub Part Marking Guidance Mark Comments 6 (a) As the sucrose concentration increases, the ratio decreases; Ratio decreases more slowly after 200 mol dm -3 ; Osmosis takes place; Net diffusion/movement; Water moves to a lower/more negative water potential/down a water potential gradient; 4 max To gain full marks at least one mark must come from the top 2 points. Last four points 3 max. 6 (b) (i) Water enters; Ratio =, solution is isotonic/no net gain or loss of water/same water potential/same concentration; 3 Pressure inside (cell) increases; Membrane ruptures; 6 (b) (ii) Bacteria have a cell wall; Supports cell membrane/maintains shape/structure (of cytoplasm). Reject turgid. 2 Ignore turgid Reject bursting/rupture. 8
4 2 Answer Question 7 in continuous prose. Quality of Written Communication will be assessed in these answers. 7 The diagram shows a cross-section of a red blood cell. 7 (a) The shape of a red blood cell allows it to take up a large amount of oxygen in a short time. Explain how. (Extra space)... 7 (b) A technician estimated the number of red blood cells in a sample of blood. First she diluted the blood sample with an isotonic solution. 7 (b) (i) What is an isotonic solution? ( mark) 7 (b) (ii) The technician did not use distilled water to dilute the blood sample. Use your knowledge of water potential to explain why. (3 marks) (Extra space)... (2) APW/Jun08/BYB
5 Biology - AQA GCE Mark Scheme 2008 June series Question Part Sub Part Marking Guidance Mark Comments 7 (a) Large surface area to volume ratio / thin in centre / concave; Reference to diffusion; 7 (b) (i) Same concentration / same water potential; Allow answers based on no osmosis/diffusion 7 (b) (ii) Higher/less negative water potential of distilled water; Water moves into cells by diffusion / osmosis into cells; Reject turgid for third marking point Cells burst; 7 (c) Phospholipid bilayer; 6 max Allows movement of lipid soluble / non-polar molecules / water / gases / e.g. O2 and CO2 OR prevents movement of water soluble / (named) polar molecules; Carrier proteins; Allow description of active transport which includes use of ATP/energy and movement against the concentration gradient Channel proteins; Facilitated diffusion; (only awarded if linked to carrier/channel proteins) Active transport; (only awarded if linked to carrier protein) Reference to complementary/tertiary structure of protein; Saturated fatty acids decrease permeability/fluidity / converse for unsaturated fatty acids; Cholesterol decreases permeability/fluidity; 9
6 7 8 Students investigated the effect of sucrose concentration on the lengths of strips cut from daffodil flowers. They measured the strips and put them in different sucrose solutions. After two hours they measured the strips again. The table shows their results. Sucrose Initial length concentration / mol dm 3 Final length (a) The initial length of the strip that was put in the sucrose solution which had a concentration of 0.4 mol dm 3 was 45 mm. Calculate its final length. Answer =... ( mark) 8 (b) You could use a graph to find the concentration of sucrose that had the same water potential as the daffodil strips. Describe how. ( mark) 8 (c) There was no further change in the ratio of initial length to final length in solutions above a concentration of 0.6 mol dm 3. Explain why. Question 8 continues on the next page Turn over (7) APW/Jun08/BYA R
7 8 Some species of blood flukes live inside humans. The adult flukes live in the blood vessels near the bladder. They lay eggs which go through the bladder wall into the urine. These eggs do not hatch in the urine but do hatch when an infected person urinates in fresh water such as a lake or a river. Scientists studied the hatching of fluke eggs. They investigated the effect of solute concentration on the hatching of the eggs. The graph shows their results Percentage of eggs hatching Solute concentration /% 8 (d) During the investigation, the scientists kept the temperature constant. Explain how keeping the temperature constant helps to produce results that are reliable. ( mark) 8 (e) (i) Describe how the solute concentration affects the percentage of eggs hatching. (8) APW/Jun08/BYA R
8 9 8 (e) (ii) When an egg hatches, the egg shell bursts and the young blood fluke escapes. Use your knowledge of water potential to explain why the egg shell bursts. 8 (f) The lungs of a mammal are adapted for gas exchange. Use your knowledge of Fick s law to explain how. (6 marks) (Extra space)... END OF QUESTIONS 5 (9) APW/Jun08/BYA R
9 Biology Specification A - AQA GCE Mark Scheme 2008 June series Question 8 (a) 50; (b) Read off value where initial length divided by final length = ; (c) (d) Strip does not shrink any more; Because of cell walls; 2 Only tests one variable/makes sure only one experimental variable is changed/ temperature may also affect percentage of eggs hatching/the results; (e) (i) Decreases as solute concentration increases; Reference to change in gradient, for example, less steep as solute concentration increases; 2 (ii) (At low solute concentrations) water potential higher/less negative outside than inside egg; Water enters by osmosis; Pressure (bursts) egg shell; 2 max (f) Fick s law as (Rate of) diffusion proportional to SA x diff in conc/conc grad Thickness of (exchange surface/membrane/distance). and 2. Allow 2 marks for correct expression;; One mark only for a single error; Must have SA, diff in conc and thickness in right position to gain any credit. Withold one mark for Fick s Law instead of diffusion Equals instead of proportional to Conc instead of conc dif in conc gradient 3. Many (small) alveoli; 4. Many capillaries; 5. Exchange surface consists of squamous/pavement epithelium/epithelial cells flat; 6. Short diffusion pathway (between lumen and blood)/thin for diffusion; 7. (Concentration difference maintained by) circulating blood; 8. (Concentration difference maintained by) ventilation/breathing; 6 max Total 5 7
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