Download AQA A LEVEL BIOLOGY PAPER 3 QP 2023 and more Exams Biology in PDF only on Docsity! 7402/3IB/M/Jun23/ E12 AQA A LEVEL BIOLOGY PAPER 3 QP 2023 7402/3IB/M/Jun23/ E12 A-level BIOLOG Y Paper 3 Wednesday 21 June 2023 Morning Time allowed: 2 hours Materials For this paper you must have: a ruler with millimetre measurements a scientific calculator. Instructions Use black ink or black ball-point pen. Fill in the boxes at the top of this page. Answer all questions in Section A. Answer one question from Section B. You must answer the questions in the spaces provided. Do not write outside the box around each page or on blank pages. If you need extra space for your answer(s), use the lined pages at the end of this book. Write the question number against your answer(s). Show all your working. Do all rough work in this book. Cross through any work you do not want to be marked. Information The marks for the questions are shown in brackets. The maximum mark for this paper is 78. For Examiner’s Use Questio n Mark 1 2 3 4 5 6 7 TOTAL Please write clearly in block capitals. Centre number Candidate number Surname Forename(s) Candidate signatu re 3 Do not write outside the *03 * IB/M/ Jun23/7402/3 2 3 6 . Use Figure 1 to describe what is meant by the all-or-nothing principle. [2 marks] box . On Figure 1, from 0.6 ms to 4.0 ms, no new generator potential could be produced. What is this time period called? [1 mark] Turn over for the next question Turn over ► 10 10 4 Do not write outside the *04 * IB/M/ Jun23/7402/3 A student investigated the effect of different sugars on the rate of respiration in yeast. Yeast normally respires glucose. Figure 2 shows the method she used for her first experiment. Figure 2 box 0 2 . 1 Other than those stated, suggest two variables the student needed to keep constant in her investigation. [1 mark] 1 2 0 2 5 Do not write outside the *05 * IB/M/ Jun23/7402/3 Figure 3 shows the result she obtained for yeast in glucose solution. Figure 3 Turn over ► box 0 2 . 2 Use Figure 2 and Figure 3 to calculate the rate of carbon dioxide production in mm s–1 for yeast in glucose solution. Give your answer in standard form and to 2 significant figures. Show your working. [2 marks] Answer mm s–1 8 Do not write outside the IB/M/ Jun23/7402/3 Turn over ► 1 9 Do not write outside the IB/M/ Jun23/7402/3 . Below are four statements about the structure of prokaryotic cells. 1. No prokaryotic cell has DNA that is associated with proteins. 2. No prokaryotic cell has membrane-bound organelles. 3. All prokaryotic cells have one or more flagella. 4. All prokaryotic cells have smaller ribosomes than eukaryotic cells. Which statements about the structure of prokaryotic cells are correct? Tick (✓) one box. [1 mark] box A statements 1, 2 and 3 B statements 1, 2 and 4 C statements 2, 3 and 4 D statements 1, 2, 3 and 4 A student investigated the effect of two antibiotics on the growth of the bacterium Micrococcus luteus. During the investigation, the student: transferred 9 cm3 of a liquid culture of M. luteus into each of three bottles added the antibiotic chloramphenicol to the first bottle added the antibiotic novobiocin to the second bottle added no antibiotic to the third bottle. After 24 hours, he diluted the contents of each bottle by 1 in 100 000 (10– 5). He then transferred 0.25 cm3 samples from the first bottle onto each of 3 separate agar plates. He repeated this with 0.25 cm3 samples from the second bottle and the third bottle, resulting in 9 agar plates in total. He incubated the plates for 48 hours. Table 1 shows the number of colonies of bacteria he counted on each plate after 48 hours’ incubation. 30 1 0 Do not write outside the IB/M/ Jun23/7402/3 *0 * 13 Do not write outside the *13* IB/M/ Jun23/7402/3 Turn over for the next question DO NOT WRITE ON THIS PAGE ANSWER IN THE SPACES PROVIDED box Turn over ► 14 Do not write outside the *14* IB/M/ Jun23/7402/3 1 Following a body injury, bone marrow stem cells move to the site of damage and undergo cell differentiation. Figure 5 shows how this differentiation occurs. Figure 5 box . Suggest how SCFR is destroyed by a lysosome. [2 marks] After a heart attack, cardiomyocytes (cardiac muscle cells) die, and become infarcted tissue. Infarcted tissue cannot contract. Stem cells in bone marrow cannot move to the infarcted tissue and differentiate into cardiomyocytes. Scientists used laboratory rats to investigate if bone marrow stem cell transplants could be used to repair infarcted tissue resulting from a heart attack. They split the rats into three groups. Control group did not get a transplant of bone marrow stem cells. c-KIT+ group got a transplant of bone marrow stem cells with a functioning c-KIT gene. c-KIT– group got a transplant of bone marrow stem cells with no functioning c-KIT gene. After 9 days, the scientists measured the mean ventricular blood pressure of each of the three groups. 40 0 4 A gene called c-KIT codes for stem cell growth factor receptor protein (SCFR) on the surface membrane of the stem cell A stem cell factor (SCF) binds to SCFR and activates tyrosine kinase (TK) TK phosphorylat es cell signalling molecules to begin cell differentiatio n After the cell has differentiated , SCFR is enclosed within a vesicle and destroyed by a lysosome 15 Do not write outside the *15* IB/M/ Jun23/7402/3 Figure 6 shows their results. The differences between the groups were all statistically significant. Figure 6 Turn over ► 0 4 . 2 Using all of the information, suggest explanations for the results for the Control group and the c-KIT– group shown in Figure 6. [4 marks] Control c-KIT– 18 Do not write outside the *18* IB/M/ Jun23/7402/3 Turn over for the next question DO NOT WRITE ON THIS PAGE ANSWER IN THE SPACES PROVIDED box Turn over ► 19 Do not write outside the *19* IB/M/ Jun23/7402/3 Scientists investigated the effect that the release of heated water into a river from a power station had on the biodiversity of a local fish community over 29 years. They measured the species richness and the number of fish of each species at the same site in October every year. The scientists used this information to calculate an index of diversity (d) of fish for each year. Figure 7 and Figure 8 show their results. Figure 7 Figure 8 box0 5 20 Do not write outside the IB/M/ Jun23/7402/3 1 2 . The scientists used the following formula to calculate the index of diversity (d) of fish. N (N–1) d = Σn (n–1) box where N = total number of fish of all species and n = total number of fish of each species In some years, the values were N = 624 and Σn (n–1) = 64 792 Which years had these values? Use Figure 8 and the formula above to work out your answer. [1 mark] Answer . In 1997, the scientists recorded the highest species richness, but the lowest value of d over the 29 years. Describe and explain how these results for 1997 were possible. [2 marks] Question 5 continues on the next page 50 50 23 Do not write outside the IB/M/ Jun23/7402/3 *1 * 24 Do not write outside the *24* IB/M/ Jun23/7402/3 Turn over for the next question DO NOT WRITE ON THIS PAGE ANSWER IN THE SPACES PROVIDED box Turn over ► 25 Do not write outside the *25* IB/M/ Jun23/7402/3 2 box An autoimmune disease causes the immune system to attack healthy body tissues. Scientists investigated the immune responses of healthy mice and mice with autoimmune disease. The chemical OXA causes an immune response in mice and can make their skin swell. Mice had olive oil applied to their left ear and OXA in olive oil applied to their right ear. The immune response was recorded in two ways: the cellular response by measuring the mean increase in ear thickness 24 hours after exposure to OXA the humoral response by measuring the mean concentration of anti- OXA antibody in blood 14 days after exposure to OXA. Table 2 shows the results of this investigation. The values in the brackets show ± 2 standard deviations. A value of ± 2 standard deviations from the mean includes over 95% of the data. Table 2 Type of mice Sex of mice Mean increase in ear thickness / cm × 10–3 Mean concentration of anti-OXA antibody / arbitrary units Healthy Male 17.9 (±4.1) 16 (±3) Female 18.5 (±2.9) 14 (±4) Autoimmu ne disease Male 25.9 (±4.5) 14 (±2) Female 16.7 (±3.0) 26 (±7) . Suggest and explain one reason why olive oil was applied to the left ear of the mice. [1 mark] 60 0 6 . 1 Give two types of cell that can stimulate an immune response. [2 marks] 1 2 28 Do not write outside the *28* IB/M/ Jun23/7402/3 4. What can you conclude about the effects of autoimmune disease on the cellular response and the humoral response in male and female mice? box Use the data to justify your conclusions. [3 marks] 60 29 Do not write outside the *29* IB/M/ Jun23/7402/3 5. Some studies have shown that in humans, oestrogen has the opposite effect on two different autoimmune diseases. Oestrogen: accelerates the progression of systemic lupus erythematosus (SLE) prevents the progression of rheumatoid arthritis (RA). The scientists investigated the effect of oestrogen on the immune response in healthy mice and mice with autoimmune disease. Table 3 shows the scientists’ results. Table 3 Type of mice Effect of oestrogen on humoral response Effect of oestrogen on cellular response Healthy No effect No effect Autoimmu ne disease Increase in response Decrease in response A student studying these data made the following conclusions. 1.In humans, SLE is caused by an overproduction of antibodies. 2.In humans, RA is caused by an overproduction of cytotoxic T cells (TC cells). box Evaluate the student’s conclusions. [4 marks] Question 6 continues on the next page Turn over ► 60 30 Do not write outside the *30* IB/M/ Jun23/7402/3 6. In mice, one type of autoimmune disease is inherited as a dominant allele. Would the Hardy–Weinberg principle hold true for a population of mice, some of which had this autoimmune disease? box Explain your answer. [2 marks] 60 15 33 Do not write outside the box IB/M/ Jun23/7402/3 *27* Turn over ► 34 Do not write outside the box IB/M/ Jun23/7402/3 *2 * 35 Do not write outside the box *35* IB/M/ Jun23/7402/3 Turn over ► 38 Do not write outside the box *38* IB/M/ Jun23/7402/3 39 Do not write outside the box *39* IB/M/ Jun23/7402/3 Turn over ► 40 Do not write outside the box *40* IB/M/ Jun23/7402/3 43 Do not write outside the IB/M/ Jun23/7402/3 *37* boxQuestio n number Additional page, if required. Write the question numbers in the left-hand margin. 44 Do not write outside the IB/M/ Jun23/7402/3 Questio n number Additional page, if required. Write the question numbers in the left-hand margin. *3 * box 45 Do not write outside the IB/M/ Jun23/7402/3 *39* boxQuestio n number Additional page, if required. Write the question numbers in the left-hand margin.