Surname entre Number andidate Number Other Names 2 GE AS/A level 1071/01 S15-1071-01 BIOLOGY/UMAN BIOLOGY BY1 P.M. TURSDAY, 21 May 2015 1 hour 30 minutes For s use Question Maximum Mark Mark Awarded 1. 10 2. 6 3. 9 1071 010001 4. 11 5. 9 6. 15 7. 10 ADDITIONAL MATERIALS Total 70 In addition to this examination paper, you may need a calculator and a ruler. INSTRUTIONS TO ANDIDATES Use black ink or black ball-point pen. Do not use gel pen. Do not use correction fluid. Write your name, centre number and candidate number in the spaces at the top of this page. Answer all questions. Write your answers in the spaces provided in this booklet. If you run out of space, use the continuation pages at the back of the booklet, taking care to number the question(s) correctly. INFORMATION FOR ANDIDATES The number of marks is given in brackets at the end of each question or part-question. You are reminded of the necessity for good English and orderly presentation in your answers. The quality of written communication will affect the awarding of marks. MAY1510710101 J*(S15-1071-01)
2 Answer all questions. 1. The following diagrams represent the structure of some common organic molecules. A 5 2 O O 4 3 O 1 O 2 O B 4 O 6 2 O 5 O 3 O 2 O 1 O 4 α 1,4 link 6 2 O 5 O 3 O 2 O 1 O D O O N N 2 N OO O O 2 S E O O 4 O 6 2 O 5 O 3 O 2 1 O F O G O O 2 O 02
3 (a) Using letters A-G, complete the table below. You may use each letter once, more than once or not at all. [7] Statement Letter(s) is a monosaccharide is a dipeptide would be found in nucleic acids contains bonds contains a glycosidic bond is a triose sugar (b) (i) Describe a biochemical test for the presence of glucose in a solution. [2] 1071 010003 (ii) Explain why a positive result is seen with glucose but not for sucrose. [1] 10 03 Turn over.
4 2. The diagram below shows the DNA structure proposed by Watson and rick in 1953. 0.34 nm 3.4 nm (a) If the double helix takes 3.4 nm to make one complete turn and base pairs are 0.34 nm apart, how many base pairs would you expect to find in five complete turns? Show your working. [2] Answer =... (b) Give a reason why the diagram shown above must be DNA rather than RNA. [1] 04
(c) 5 The table below shows the bases guanine and cytosine as percentages of the total nucleotides present in three different micro-organisms. Microorganism G Base composition (%) yeast 18.7 17.1 bacteria 36.0 35.7 virus 42.0 13.9 (i) Suggest which type of nucleic acid is present in the virus shown in the table. Explain your answer. [2] 1071 010005 (ii) State the type of base to which cytosine belongs. [1]... 6 05 Turn over.
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7 3. The diagram below shows part of the plasma membrane from an animal cell. A B D (a) omplete the table below to identify the structures A-D shown in the diagram. [4] Structure Name 1071 010007 A B D (b) Describe how structure D would transport a molecule against a concentration gradient. [2] 07 Turn over.
8 (c) In an experiment to determine the structure of the plasma membrane, scientists labelled the membrane proteins from two different cells using coloured dyes. One cell had its membrane proteins labelled with a red dye, whilst a second had its membrane proteins labelled with a green dye. The two cells were then fused to become a hybrid cell. This cell was viewed immediately after fusion and again after one hour. The results are shown below. Before fusion ell 1 ell 2 Immediately after fusion KEY Membrane proteins labelled with red dye ybrid cell Membrane proteins labelled with green dye One hour after fusion ybrid cell 08
9 Use your knowledge of the structure and properties of plasma membranes to explain the results seen one hour after fusion. [3] 1071 010009 9 09 Turn over.
10 4. The diagram below shows the relative lengths of the cell cycle in actively dividing cells taken from the root tip of a garlic plant. Mitosis Prophase Anaphase Metaphase Telophase ytokinesis (a) Describe the changes that occur to a plant cell during prophase. [4] 10
11 (b) The percentage of cells in each stage of the cell cycle is proportional to the length of that stage. Using a microscope, a student observed 100 cells and found 5 undergoing prophase. If the total length of the cell cycle is 24 hours, calculate the length of prophase in minutes. Show your working. [2] Answer =... minutes 1071 010011 11 Turn over.
12 (c) Vincristine is a chemical which inhibits mitosis by preventing the formation of the spindle fibres. Garlic bulbs were grown in a solution of vincristine and the quantity of DNA present in a cell from the root tip was measured over the 24 hour length of the cell cycle. The results are shown below together with the results from garlic bulbs grown in water. Quantity of DNA per cell / arbitrary units 6 Garlic grown in vincristine solution Quantity of DNA per cell / arbitrary units 6 Garlic grown in water 4 4 2 2 0 0 4 8 12 16 20 24 Time / hrs 0 0 4 8 12 16 20 24 Time / hrs Using your knowledge of the cell cycle, explain how the results of this experiment show that vincristine inhibits mitosis. [3] 12
13 (d) Meiosis produces four daughter cells, whereas mitosis produces two daughter cells. Describe two other ways in which the cells produced by meiosis would differ from those produced by mitosis. [2] 11 1071 010013 13 Turn over.
14 5. (a) Define the term water potential. [2] A student prepared a microscope slide containing a piece of red onion epidermis which had been soaked in a 0.4 M sucrose solution for 30 minutes. The slide was viewed under high-power and the following drawing was made. Turgid cell Plasmolysed cell Table to convert molarity to solute potential (kpa). Molarity of sucrose solution (M) Solute potential kpa 0.05 130 0.10 260 0.15 410 0.20 540 0.25 680 0.30 860 0.35 970 0.40 1120 0.45 1280 0.50 1450 0.55 1620 0.60 1800 14
15 (b) The student concluded that the water potential of the epidermal tissue was 1120 kpa. Using all the information provided, explain why. [4] (c) Explain the role of the cell wall in maintaining turgidity. [3] 9 15 Turn over.
16 6. A new biological washing powder has been developed containing three protease enzymes (A, B and ), each removing different protein based stains. The manufacturer claims their revolutionary new washing powder works best at 35, but offers superior removal of tough protein stains at higher temperatures. (a) Enzymes are biological catalysts. Describe what is meant by the term biological catalyst. [2] Graph showing protease activity Enzyme A 8 7 6 Enzyme B Enzyme Activity of enzyme / arbitrary units 5 4 3 2 1 0 10 15 20 25 30 35 40 45 50 Temperature of water / Table showing type of food stain removed Protease enzyme Food stain removed Blood Meat Egg A Yes No No B No Yes No No No Yes 16
(b) 17 Using the graph and table, together with your knowledge of enzymes, answer the following questions. (i) Explain why the washing powder works best at 35. [1] (ii) Which stain would not be removed when washing at 40? [1]... (iii) Suggest why three different enzymes are needed to remove blood, meat and egg stains. [2] (iv) Explain fully why washing at 60 would not be recommended for removing protein stains when using this washing powder. [4] 17 Turn over.
18 (c) Many metabolic pathways are switched off by their end products which act as competitive inhibitors of the first enzyme in the pathway. In the example below, the end product isoleucine inhibits threonine deaminase. 3 O Threonine (initial substrate) N 2 OO Intermediate A enzyme 1 (threonine deaminase) enzyme 2 Intermediate B inhibition enzyme 3 Intermediate enzyme 4 3 2 3 Isoleucine (end product) N 2 OO 18
19 (i) Explain how isoleucine works as a competitive inhibitor. [4] (ii) Suggest a reason why this might be useful to the cell. [1] 15 19 Turn over.
20 7. Answer one of the following questions. Any diagrams included must be fully annotated. Either, (a) Explain how a biosensor may be used to measure blood glucose concentration. Outline the advantages of using a biosensor. [10] Or, (b) Describe and explain how a mitochondrion and a chloroplast are adapted to their functions. Explain how they are similar to a prokaryotic cell. [10] 20
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