AQA A-level BIOLOGY Paper 3, rated A+

Document Content and Description Below

Please write clearly in block capitals. Centre number Candidate number Surname Forename(s) Candidate signature I declare this is my own work. A-level BIOLOGY Paper 3 2 *02* IB/M/Jun22/7402/ ... 3 Do not write outside the There are no questions printed on this page box DO NOT WRITE ON THIS PAGE ANSWER IN THE SPACES PROVIDED 3 *03* Turn over ► IB/M/Jun22/7402/3 Do not write outside the Section A box Answer all questions in this section. You are advised to spend no more than 1 hour and 15 minutes on this section. 0 1 Amino acids are used to make proteins. Table 1 shows the R groups of six different amino acids. Table 1 Amino acid R group Alanine CH3 Asparagine CH2CONH2 Aspartic acid CH2COOH Amino acid R group Glutamic acid CH2CH2COOH Glycine H Serine CH2OH 0 1 . 1 Use Table 1 to identify the three different amino acids used to make the polypeptide shown in Figure 1. [2 marks] Figure 1 Left amino acid Middle amino acid Right amino acid Question 1 continues on the next page 4 *04* IB/M/Jun22/7402/3 Do not write outside the box 0 1 . 2 Table 2 shows three statements and names of four biological molecules. Put a tick () in each box where the statement is true for the biological molecule. [3 marks] Table 2 Statement DNA ATP Reverse transcriptase Phospholipid Contains peptide bonds Is formed using a condensation reaction Is a polymer Figure 2 represents the structure of adult human haemoglobin. Figure 2 5 *05* Turn over ► IB/M/Jun22/7402/3 Do not write outside the box 0 1 . 3 The number of amino acids in the beta chains in Figure 2 is 3.546% greater than in the alpha chains. Each alpha chain contains 141 amino acids. Calculate how many amino acids there are in total in the haemoglobin molecule shown in Figure 2. Give your answer to the nearest whole number. [2 marks] Answer amino acids When a substance called BPG binds to haemoglobin, it reduces the affinity of haemoglobin for oxygen. 0 1 . 4 Figure 3 shows an oxyhaemoglobin dissociation curve for haemoglobin in normal conditions. Sketch a curve on Figure 3 to show the oxyhaemoglobin dissociation curve for haemoglobin when BPG binds to it. [1 mark] Figure 3 Question 1 continues on the next page 6 *06* IB/M/Jun22/7402/3 Do not write outside the box 0 1 . 5 Suggest and explain when it would be an advantage to a human for BPG to bind to haemoglobin. [2 marks] 10 7 *07* Turn over ► IB/M/Jun22/7402/3 Do not write outside the box 0 2 Figure 4 shows a diagram of a Pacinian corpuscle. Figure 4 0 2 . 1 Name the structures labelled P, Q and R shown in Figure 4. [2 marks] P Q R Question 2 continues on the next page 8 *08* IB/M/Jun22/7402/3 Do not write outside the box Two students (A and B) investigated reaction time in response to touch. • Student A sat with her eyes shut and her forearm resting on a worktop so that her hand was over the edge. • Student B held a ruler vertically between student A’s thumb and first finger, with the ruler at 0 mm lightly touching student A’s first finger. • Student B released the ruler. • As soon as student A felt the ruler fall, she closed her thumb and first finger to catch the ruler as shown in Figure 5. • Student B measured the distance the ruler had fallen to the nearest mm Figure 5 The test was repeated three more times using the same hand to catch the ruler. Table 3 shows student A’s results. Table 3 Trial Distance the ruler has fallen / mm 1 79 2 97 3 10 4 94 The student was able to convert these distances into reaction times using Table 4. Table 4 Distance the ruler fell / mm Reaction time / ms 10 45 20 64 30 78 40 90 50 101 60 111 70 120 80 128 90 136 9 *09* Turn over ► IB/M/Jun22/7402/3 Do not write outside the box 0 2 . 2 Calculate the percentage uncertainty in the measurement of Trial 1 in Table 3. Put a tick () in the correct box below. [1 mark] 0.633% 1.27% 2.53% 12.6% 0 2 . 3 In this investigation, it is not possible for a student to react in less than 45 ms Suggest one explanation for the value recorded in Trial 3 in Table 3. [1 mark] 0 2 . 4 Student A estimated that the length of the nerve pathway involved was 175 cm Use Table 3 and Table 4 to calculate the mean speed of nerve impulse transmission. Do not use the value for Trial 3 in your calculation. [2 marks] Answer m s–1 Question 2 continues on the next page 10 *10* IB/M/Jun22/7402/3 Do not write outside the box 0 2 . 5 In response to touch, nerve impulses can be transmitted at speeds of 76.2 m s–1 Suggest three reasons why, in this investigation, the estimated speed of student A’s impulse transmission was less than 76.2 m s–1 [3 marks] 1 2 3 9 11 *11* Turn over ► IB/M/Jun22/7402/3 Do not write outside the box 0 3 A student prepared a stained squash of cells from the root tips of garlic to calculate a mitotic index. He: 1. cut the end 5 mm from 10 garlic roots 2. placed the root tips into a Petri dish containing 5 cm3 of hydrochloric acid for 12 minutes 3. rinsed the root tips in distilled water 4. placed one of the root tips on a microscope slide and added toluidine blue stain 5. placed a coverslip onto the microscope slide, and gently pressed the coverslip downwards on the root tip 6. observed the root tip using an optical microscope. 0 3 . 1 Suggest why the student soaked the root tips in hydrochloric acid in step 2. [2 marks] 0 3 . 2 Pressing the coverslip downwards enabled the student to observe the stages of mitosis clearly. Explain why. [2 marks] Question 3 continues on the next page 12 *12* IB/M/Jun22/7402/3 Do not write outside the Figure 6 shows the student’s drawing of one field of view. box Figure 6 0 3 . 3 Name the stage of mitosis shown in cell G. Explain the appearance of this cell. [2 marks] Stage of mitosis Explanation 0 3 . 4 Use Figure 6 to calculate a mitotic index for the cells in this field of view. [1 mark] Mitotic index 13 *13* Turn over ► IB/M/Jun22/7402/3 Do not write outside the box 0 3 . 5 Other students in the class followed the same method, but calculated different mitotic indices. Apart from student errors, suggest two explanations why. [2 marks] 1 2 Turn over for the next question 9 14 *14* IB/M/Jun22/7402/3 Do not write outside the box 0 4 . 1 Complete the following definitions. [2 marks] The genome is The proteome is Recombinant DNA technology can involve the transfer of fragments of human DNA into bacteria. The bacteria are then used to produce human proteins. 0 4 . 2 Give two reasons why bacteria are able to use human DNA to produce human proteins. [2 marks] 1 2 0 4 . 3 Suggest and explain one reason why bacteria might not be able to produce every human protein. [1 mark] 15 *15* Turn over ► IB/M/Jun22/7402/3 Do not write outside the box Antithrombin is a protein. Antithrombin prevents blood from clotting too much. Some people have a deficiency of antithrombin in their blood, so they need to inject the protein. Genetically modified goats are used to produce this protein. The human antithrombin gene is transferred into goat embryos. The adult goats then make human antithrombin protein. Figure 7 shows an example of a DNA fragment that can be transferred into the cells of goats. Figure 7 0 4 . 4 The enhancer stimulates region M. Name regions M and N shown in Figure 7. [2 marks] Region M Region N 0 4 . 5 Explain the purpose of the marker gene. [1 mark] 0 4 . 6 The enhancer only stimulates region M in the milk-producing glands of a goat. Suggest two explanations for the importance of the enhancer being included in the DNA fragment transferred. [2 marks] 1 2 10 16 *16* IB/M/Jun22/7402/3 Do not write outside the box 0 5 Scientists investigated the effect of full sun and shade on the rate of photosynthesis in a species of shade-tolerant tree. To estimate the rate of photosynthesis, the scientists measured uptake of carbon dioxide by trees in a forest. They measured uptake of carbon dioxide during two parts of the day: • 08.30 – 09.40 hours • 11.40 – 13.15 hours. Figure 8 shows the scientists’ results. Figure 8 0 5 . 1 Calculate the total uptake of carbon dioxide between 11.40 and 13.15 hours in trees exposed to full sun in a forest that is 12 000 m2 in area. Give your answer in standard form. Show your working. [3 marks] Answer µmol 17 *17* Turn over ► IB/M/Jun22/7402/3 Do not write outside the box 0 5 . 2 Figure 8 shows there is a small difference in the mean uptake of carbon dioxide between 08.30 and 09.40 hours by trees in full sun and by trees in the shade. When the scientists performed a statistical test on these data, they calculated P > 0.5 State what this P value tells you about this difference. Explain your answer using the words probability and chance. [2 marks] 0 5 . 3 In this species of tree, very high light intensities can inhibit the release of electrons from chlorophyll. Suggest how this could explain the results shown in Figure 8 for 11.40 to 13.15 hours. [4 marks] Question 5 continues on the next page 18 *18* IB/M/Jun22/7402/3 Do not write outside the box Tomato plants grow best in high light intensities. To increase the yield of tomato plants, a farmer uses LED lightbulbs to provide additional light. The increase in dry mass (D) produced when using additional light can be calculated using this equation. L D = 0.4F Where L = light used in photosynthesis F = GPP to NPP conversion factor for tomato plants Table 5 shows some of these values for LED lightbulbs. Table 5 L / MJ m–2 h–1 F / MJ kg–1 2.87 × 10–2 20 0 5 . 4 Use the equation and Table 5 to calculate the increase in dry mass produced when using LED lightbulbs. Give your answer in standard form and give the units. [2 marks] Answer Units 19 *19* Turn over ► IB/M/Jun22/7402/3 Do not write outside the box 0 5 . 5 Mature leaves from slow-growing, shade-tolerant plants produce poisonous chemicals that are a defence against being eaten by herbivores. Suggest how this benefits slow-growing, shade-tolerant plant [Show More]

Last updated: 2 years ago

Preview 1 out of 36 pages