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AQA A-Level Biology 7402/3 Paper 3 Question Paper June 2021 Version 1.0 Final / AQA A-Level Biology 7402-3 Paper 3 Question Paper June 2021 Version 1.0 Fin... [Show More] al / AQA A-Level Biology Paper 3 Question Paper June 2021 Version 1.0 Final 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 ofthis 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. 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 AQA 2 *02* IB/H/Jun21/7402/3 Do not write outside the box Section A Answer all questions in this section. You are advised to spend no more than 1 hour and 15 minutes on this section. 0 1 In one species of squirrel, Sciurus carolinensis, fur colour is controlled by one gene, with two codominant alleles. CG represents the allele for grey fur colour, and CB represents the allele for black fur colour. Table 1 shows the three possible phenotypes. Table 1 Genotype Phenotype CGCG Grey fur CGCB Brown-black fur CBCB Black fur 0 1 . 1 In a population of 34 S. carolinensis, 2 had black fur. Use the Hardy–Weinberg equation to estimate how many squirrels in this population had brown-black fur. Show your working. [2 marks] Answer 3 *03* Turn over ► IB/H/Jun21/7402/3 Do not write outside the box 0 1 . 2 The actual number of squirrels in this population that had brown-black fur was 16. Use all of the information to calculate the actual frequency of the CG allele. Do not use the Hardy–Weinberg equation in your calculation. Give your answer to 2 decimal places. [1 mark] Answer 0 1 . 3 S.carolinensis were first introduced to the UK from North America in the 1870s. They are now widely distributed across the UK. S.carolinensis from both North America and the UK show exactly the same genotypicand phenotypic variation. An identical mutation causing black fur has also been found in several other species closely related to S. carolinensis. Use this information to deduce which one of the following conclusions is most likely true. Tick (✓) one box. [1 mark] A The mutation that caused black fur happened after S.carolinensis was introduced to the UK from North America. B The mutation that caused black fur happened in a common ancestor of S. carolinensis and other closely related species. C The mutation that caused black fur happened independently in S. carolinensis and all other closely related species. D The phenotypic variation shown in S. carolinensis and other closely related species is caused by genetic drift. Question 1 continues on the next page 4 *04* IB/H/Jun21/7402/3 Do not write outside the box The mutation that caused the CB allele was due to a 24 base-pair deletion from the CG allele. 0 1 . 4 The protein coded for by the CB allele is 306 amino acids long. Calculate the percentage reduction in size of the protein coded for by the CB allelecompared with the protein coded for by the CG allele. Give your answer to 3 significant figures and show your working. [2 marks] Answer In S. carolinensis, fur colour depends on the distribution and relative amounts of light pigments and dark pigments in the hairs of the fur. Figure 1 shows how the protein produced from the CG allele can result in the production of a light pigment or a darkpigment. Figure 1 5 *05* Turn over ► IB/H/Jun21/7402/3 Do not write outside the box The deletion mutation in the CB allele results in the production of a receptor protein that does not have glutamic acid. The lack of glutamic acid in the receptor protein has the same effect as αMSH leaving the receptor protein. 0 1 . 5 Use Figure 1 and this information to suggest why S. carolinensis with the genotype CBCB have black fur rather than grey fur. [3 marks] Turn over for the next question 9 6 *06* IB/H/Jun21/7402/3 Do not write outside the box 0 2 . 1 Describe how the human immunodeficiency virus (HIV) is replicated once inside helper T cells (TH cells). [4 marks] 7 *07* Turn over ► IB/H/Jun21/7402/3 Do not write outside the box HIV-1 is the most common type of HIV. HIV-1 binds to a receptor on TH cells called CCR5. Current treatment for HIV-1 involves the use of daily antiretroviral therapy (ART) to stop the virus being replicated. Only 59% of HIV-positive individuals have access to ART. Scientists have found that two HIV-1-positive patients (P and Q) have gone into remission (have no detectable HIV-1). This happened after a blood stem cell transplant (BSCT). • Patient P was given two BSCTs, and patient Q was given one BSCT. • All BSCTs came from a donor with TH cells without the CCR5 receptor. • In addition, patient P had radiotherapy, and patient Q had chemotherapy. Both of these treatments are toxic. • Both patients (P and Q) stopped receiving ART 16 months after BSCT. 18 months after stopping ART, both patients had no HIV-1 RNA in their plasma, no HIV-1 DNA in their TH cells and no CCR5 on their TH cells. 0 2 . 2 Use the information given to evaluate the use of BSCT to treat HIV infections. [5 marks] 9 8 *08* IB/H/Jun21/7402/3 Do not write outside the box 0 3 Scientists investigated movement in adult pine beetles. Adult beetles emerge from cracks in tree bark. The scientists released a newly emerged adult beetle, G, from the centre of a sample area that had a single light source coming from one direction. They made a drawing of the beetle’s path of walking. They repeated this with three more beetles, J, P and R. Figure 2 shows the scientists’ results. Figure 2 0 3 . 1 Name the type of behaviour shown by beetles G, J, P and R, and suggest one advantage to adult beetles of the type of behaviour shown. [2 marks] Behaviour Advantage 9 *09* Turn over ► IB/H/Jun21/7402/3 Do not write outside the box Question 3 continues on the next page DO NOT WRITE ON THIS PAGE ANSWER IN THE SPACES PROVIDED 10 *10* IB/H/Jun21/7402/3 Do not write outside the box At higher temperatures and higher light intensities, adult pine beetles normally • move more • fly rather than walk. When preparing to fly, these adult beetles walk slowly. The scientists investigated the movement of adult beetles at different temperatures, and in the light and the dark. They created a box that was half in the light and half in the dark. They released an adult beetle at the midpoint of the central dividing line between light and dark areas. They recorded the path of the beetle’s movement and its location after 5 minutes. From this, they calculated the mean speed of movement. They repeated the experiment with many beetles and at several temperatures. Figure 3 shows the scientists’ results. Figure 3 11 *11* Turn over ► IB/H/Jun21/7402/3 Do not write outside the box 0 3 . 2 After studying these experiments, a student concluded: •there is a significant change in movement between 35 °C and 37.5 °C •between 35 °C and 37.5 °C, more beetles move away from the light •between 35 °C and 37.5 °C, more beetles have a slower walking speed. Suggest reasons why these conclusions might not be valid. [3 marks] Turn over for the next question 5 12 *12* IB/H/Jun21/7402/3 Do not write outside the box 0 4 Freshwater marshes have one of the highest rates of gross primary production (GPP) and net primary production (NPP) of all ecosystems. Carbon use efficiency (CUE) is the ratio of NPP:GPP. Freshwater marshes have a high CUE. 0 4 . 1 Use your knowledge of NPP to explain why freshwater marshes have a high CUE and the advantage of this. Do not refer to abiotic factors in your answer. [2 marks] Explanation Advantage 0 4 . 2 Freshwater marsh soils are normally waterlogged. This creates anaerobic conditions. Use your knowledge of the nitrogen cycle to suggest why these soils contain relatively high concentrations of ammonium compounds and low concentrations of nitrite ions and nitrate ions. [2 marks] 13 *13* Turn over ► IB/H/Jun21/7402/3 Do not write outside the box A student investigated the growth rate of a freshwater marsh plant. The growth rate (R) of a plant can be determined using this equation. R = (InW2 − In W1)t Where ln = natural logarithm t = duration of the investigation in days W1 = plant biomass at the start of the investigation W2 = plant biomass at the end of the investigation The student used the equation above; however, she substituted height for biomass. This was because she did not want to destroy the plants to measure their biomass. 0 4 . 3 State the assumption the student has made and suggest why this assumption might not be valid. [2 marks] 0 4 . 4 At the end of the investigation, the student noted the freshwater marsh plant had grown 268 mm in height, and now measured 387 mm. She calculated the rate of growth (R) to be 0.097 mm m–1 day–1 Use this information and, substituting height for biomass, use the equation to calculate the duration of the student’s investigation. Give your answer to the nearest full day. Show your working. [2 marks] days 8 14 *14* IB/H/Jun21/7402/3 Do not write outside the box 0 5 . 1 The action of endopeptidases and exopeptidases can increase the rate of protein digestion. Describe how. [2 marks] 0 5 . 2 As humans age, there is a decrease in body protein. Give the name of one body protein that could have resulted in: [2 marks] reduced muscle power reduced immunity Scientists investigated the effect of two types of dietary protein on the ability of old men to produce body proteins. Table 2 shows information about the two types of dietary protein investigated. Table 2 Physiological factor Name of dietary protein Casein Whey Rate of absorption of dietary protein / mmol dm–3 amino acids in blood plasma h–1 3.05 4.33 Stimulation of protein synthesis Higher rate Lower rate Breakdown of body proteins No effect Inhibitory effect 15 *15* Turn over ► IB/H/Jun21/7402/3 Do not write outside the box Figure 4 shows the percentage of protein absorbed that becomes body protein in old men following a meal of casein or whey. Figure 4 A statistical test confirmed that the difference between the results shown in Figure 4 was significant. 0 5 . 3 Suggest which type of dietary protein would be better for old men to eat to cause a net gain of body proteins. Use the information provided to explain your answer. [3 marks] 7 16 *16* IB/H/Jun21/7402/3 Do not write outside the box 0 6 Plants transport sucrose from leaves to other tissues for growth and storage. SUT1 is a sucrose co-transporter protein. Scientists investigated whether the cells of tobacco plant leaves used SUT1 to transport sucrose to other tissues. 0 6 . 1 The scientists used a radioactively labelled DNA probe to show that the cells of tobacco plant leaves contained the SUT1 gene. Describe how they would do this. Do not include PCR in your answer. [4 marks] 17 *17* Turn over ► IB/H/Jun21/7402/3 Do not write outside the box 0 6 . 2 To study the role of SUT1 in tobacco plants, scientists reduced the expression of the SUT1 gene. When the SUT1 gene is transcribed, the SUT1 mRNA produced is called ‘sense’ SUT1 mRNA. The scientists genetically modified plants by inserting an extra gene so that this also allowed the production of ‘antisense’ SUT1 mRNA. The scientists had two types of tobacco plants: • type A – plants that were genetically modified • type B – plants that were not genetically modified. Suggest how the production of ‘antisense’ SUT1 mRNA in type A plants would reduce the expression of the SUT1 gene. [4 marks] Question 6 continues on the next page 18 *18* IB/H/Jun21/7402/3 Do not write outside the box 0 6 . 3 The scientists hypothesised that lower rates of sucrose transport from leaves would cause reduced growth. To test this hypothesis, the scientists provided leaves of type A and type B plants with labelled carbon dioxide (14CO2). To estimate sucrose transport out of leaves, they measured the percentage of 14C remaining in the leaves for 16 hours. Figure 5 shows their results. Figure 5 Calculate the ratio of percentage of 14C remaining in leaves of type B to type A plants16 hours after providing 14CO2 [1 mark] Answer 19 *19* Turn over ► IB/H/Jun21/7402/3 Do not write outside the box 0 6 . 4 In type B plants, the percentage of 14C remaining in the leaves does not reach zero per cent, as shown in Figure 5. Suggest two reasons why. [2 marks] 1 2 Question 6 continues on the next page 20 *20* IB/H/Jun21/7402/3 Do not write outside the box The scientists measured physiological differences between type A plants and type B plants. Table 3 shows the scientists’ results as they presented them. Table 3 Physiological factor Type of tobacco plant Type A Type B Rate of sucrose transport from leaf cells / μmol m–2 s–1 0.1 3.7 Leaf sucrose concentration / mmol m–2 22 4 Ratio of shoot:root dry mass 6:1 2:1 Rate of photosynthesis / μmol glucose m–2 s–1 4 14 21 *21* Turn over ► IB/H/Jun21/7402/3 Do not write outside the box Sucrose is able to inhibit the production and activity of rubisco in leaves of a plant. Type A plants have decreased dry mass compared with type B plants. 0 6 . 5 Use all the information to suggest and explain how the physiological factors in Table 3 would contribute to the decreased dry mass observed in type A plants. [4 marks] Turn over for Section B 15 22 *22* IB/H/Jun21/7402/3 Do not write outside the box Section B Answer one question. You are advised to spend no more than 45 minutes on this section. 0 7 Write an essay on one of the topics below. Either 0 7 . 1 The importance of complementary shapes of molecules in organisms [25 marks] Or 0 7 . 2 The importance of ions in metabolic processes [25 marks] 23 *23* Turn over ► IB/H/Jun21/7402/3 Do not write outside the box 24 *24* IB/H/Jun21/7402/3 Do not write outside the box 25 *25* Turn over ► IB/H/Jun21/7402/3 Do not write outside the box 26 *26* IB/H/Jun21/7402/3 Do not write outside the box 27 *27* Turn over ► IB/H/Jun21/7402/3 Do not write outside the box 28 *28* IB/H/Jun21/7402/3 Do not write outside the box 29 *29* Turn over ► IB/H/Jun21/7402/3 Do not write outside the box 30 *30* IB/H/Jun21/7402/3 Do not write outside the box 31 *31* Turn over ► IB/H/Jun21/7402/3 Do not write outside the box 32 *32* IB/H/Jun21/7402/3 Do not write outside the box END OF QUESTIONS 25 33 *33* IB/H/Jun21/7402/3 Do not write outside the box There are no questions printed on this page DO NOT WRITE ON THIS PAGE ANSWER IN THE SPACES PROVIDED 34 *34* IB/H/Jun21/7402/3 Do not write outside the box Question number Additional page, if required. Write the question numbers in the left-hand margin. 35 *35* IB/H/Jun21/7402/3 Do not write outside the box Question number Additional page, if required. Write the question numbers in the left-hand margin. 36 *36* IB/H/Jun21/7402/3 Do not write outside the box Question number Additional page, if required. Write the question numbers in the left-hand margin. [Show Less]
AQA A-Level Biology 7402/3 Paper 3 Mark Scheme June 2021 Version 1.0 Final / AQA A-Level Biology 7402-3 Paper 3 Mark Scheme June 2021 Version 1.0 Final / A... [Show More] QA A-Level Biology Paper 3 Mark Scheme June 2021 Version 1.0 Final A-level BIOLOGY 7402/3 Paper 3 Mark scheme June 2021 Version: 1.0 Final AQA MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2021 2 Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all associates participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the students’ responses to questions and that every associate understands and applies it in the same correct way. As preparation for standardisation each associate analyses a number of students’ scripts. Alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Examiner. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of students’ reactions to a particular paper. Assumptions about future mark schemes on the basis of one year’s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper. Further copies of this mark scheme are available from aqa MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2021 3 Mark scheme instructions to examiners 1.General The mark scheme for each question shows: •the marks available for each part of the question •the total marks available for the question •the typical answer or answers which are expected •extra information to help the examiner make his or her judgement and help to delineate what isacceptable or not worthy of credit or, in discursive answers, to give an overview of the area inwhich a mark or marks may be awarded. The extra information in the ‘Comments’ column is aligned to the appropriate answer in the left-hand part of the mark scheme and should only be applied to that item in the mark scheme. At the beginning of a part of a question a reminder may be given, for example: where consequential marking needs to be considered in a calculation; or the answer may be on the diagram or at a different place on the script. In general the right-hand side of the mark scheme is there to provide those extra details which confuse the main part of the mark scheme yet may be helpful in ensuring that marking is straightforward and consistent. 2.Emboldening 2.1 In a list of acceptable answers where more than one mark is available ‘any two from’ is used, with the number of marks emboldened. Each of the following bullet points is a potential mark. 2.2 A bold and is used to indicate that both parts of the answer are required to award the mark. 2.3 Alternative answers acceptable for the same mark are indicated by the use of OR. Different terms in the mark scheme are shown by a / ; eg allow smooth / free movement. 3.Marking points 3.1 Marking of lists This applies to questions requiring a set number of responses, but for which students have provided extra responses. The general principle to be followed in such a situation is that ‘right + wrong = wrong’. Each error / contradiction negates each correct response. So, if the number of errors / contradictions equals or exceeds the number of marks available for the question, no marks can be awarded. However, responses considered to be neutral (often prefaced by ‘Ignore’ in the ‘Comments’ column of the mark scheme) are not penalised. MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2021 4 3.2 Marking procedure for calculations Full marks can be given for a correct numerical answer, without any working shown. However, if the answer is incorrect, mark(s) can usually be gained by correct substitution / working and this is shown in the ‘Comments’ column or by each stage of a longer calculation. 3.3 Interpretation of ‘it’ Answers using the word ‘it’ should be given credit only if it is clear that the ‘it’ refers to the correct subject. 3.4 Errors carried forward, consequential marking and arithmetic errors Allowances for errors carried forward are most likely to be restricted to calculation questions and should be shown by the abbreviation ECF or consequential in the mark scheme. An arithmetic error should be penalised for one mark only unless otherwise amplified in the mark scheme. Arithmetic errors may arise from a slip in a calculation or from an incorrect transfer of a numerical value from data given in a question. 3.5 Phonetic spelling The phonetic spelling of correct scientific terminology should be credited unless there is a possible confusion with another technical term. 3.6 Brackets (…..) are used to indicate information which is not essential for the mark to be awarded but is included to help the examiner identify the sense of the answer required. 3.7 Ignore/Insufficient/Do not allow Ignore or insufficient is used when the information given is irrelevant to the question or not enough to gain the marking point. Any further correct amplification could gain the marking point. Do not allow means that this is a wrong answer which, even if the correct answer is given, will still mean that the mark is not awarded. MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2021 5 Question Marking Guidance Mark Comments 01.1 1. Answer of 12/13 = 2 marks;; 2. 0.36(48)/0.365/0.37 = 1 mark OR 36(.48)/36.5/37% = 1 mark OR q2= 0.06/0.059/0.0588 = 1 mark OR or q = 0.2/0.24/0.243 = 1 mark; 2 For 1 mark accept q2 = 6%/5.9%/5.88% Question Marking Guidance Mark Comments 01.2 0.71 1 Question Marking Guidance Mark Comments 01.3 Second box ticked/answer key: B: The mutation that caused black fur happened in a common ancestor of S. carolinensis and other closely related species. 1 Question Marking Guidance Mark Comments 01.4 1. 2.55% = 2 marks;; 2. 2.61% = 1 mark (question misread ie 8/306x100) OR Evidence of dividing by 314 or 942 = 1 mark OR Answers not given to three significant figures = 1 mark; 2 MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2021 6 Question Marking Guidance Mark Comments 01.5 1. Mutation/lack of glutamic acid leads to (permanent) activation of the receptor/protein; 2. (Because) the receptor/protein does not require the binding/leaving of αMSH (to become activated); 3. ASIP (might) not (be) able to bind to the receptor/protein; 4. (Only) the dark pigment is produced 3 max 2. Answer must convey the idea that binding/leaving is not required MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2021 7 Question Marking Guidance Mark Comments 02.1 1. RNA converted into DNA using reverse transcriptase; 2. DNA incorporated/inserted into (helper T cell) DNA/chromosome/genome/nucleus; 3. DNA transcribed into (HIV m)RNA; 4. (HIV mRNA) translated into (new) HIV/viral proteins (for assembly into viral particles); 4 1. Reject ‘messenger’ or ‘m’ before RNA 3. Accept descriptions of transcription 4. Accept descriptions of translation 4. Accept named viral protein, eg capsid 4. Reject viral cells MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2021 8 Question Marking Guidance Mark Comments 02.2 For 1. (There appears to be) no virus/ HIV(-1)/RNA/DNA, so could be a cure/effective; 2. No CCR5/receptor, so not get HIV(-1) in the future OR No CCR5/receptor, so nothing for HIV(-1) to bind to; 3. Only one transplant/BSCT needed (shown by patient Q) 4. Would not need (daily) ART (16 months after BSCT); Against 5. Don’t know if chemotherapy/radiotherapy is needed OR Do not know if BSCT alone would be effective; OR Do not know which treatment is having the effect OR Could be due to chemotherapy/radiotherapy; 6. Only for HIV-1; 7. Don’t know if it would work in all people OR Only worked/tried in 2 cases; 8. Might not be long term OR Only 18 months; 9. HIV-1 may mutate and be able to bind to a different receptor (on TH cells); 10. Might be a lack of (suitable stem cell/BSCT) donors; 5 max Max 4 for reasons for or against 1. Ignore virus is killed 2. Reject less CCR5/less HIV(-1) bind 5. Accept: chemotherapy/radiotherapy is toxic/harmful/has side-effects 6. Accept: Might not work in other types of HIV 10. Accept stem cells/BSCT (might be) rejected MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2021 9 Question Marking Guidance Mark Comments 03.1 Behaviour 1. (Positive photo) taxis; Advantage 2. Accept any suitable suggestion, eg to avoid competition, to find a mate, increase dispersal, to avoid predators; 2 1. Reject negative (photo) taxis 2. Neutral – to move into the open or to move out of the tree bark Question Marking Guidance Mark Comments 03.2 1. No stats test, so do not know if change (in movement away from light) is significant; 2. Between 35 °C and 36.5 °C more than half of beetles are still found on the light side; 3. (At higher temperatures/above 35 °C) beetles might be flying (not walking) OR (Y-axis) states speed of movement, might not just be walking speed; 4. Slowing of movement happens before 35 °C; 5. Slowing of movement could be due to beetles preparing to fly (and not temperature); 6. Speed (of movement) not recorded above 35 °C/ between 35 and 37.5 °C/between 35 and 40 °C; OR Speed (of movement) not recorded at 37.5 °C 7. (Mean speed could mean) some might walk very quickly and others stay still/not move; 3 max MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2021 10 Question Marking Guidance Mark Comments 04.1 1. Low respiration; 2. More growth/biomass/colonisation; 2 1. Accept less energy lost in respiration 2. Allow examples of more carbon-containing molecules eg glucose Question Marking Guidance Mark Comments 04.2 1. Less nitrification OR Fewer/less active nitrifying bacteria; OR Nitrification/nitrifying bacteria require oxygen/aerobic conditions; 2. (Less) oxidation/conversion of ammonium (ions) to nitrite (ions) and to nitrate (ions); 3. More denitrification OR More/more active denitrifying bacteria OR Denitrification/denitrifying bacteria do not require oxygen OR Denitrification/denitrifying bacteria require anaerobic conditions; 4. (So more) nitrate (ions) reduced/converted to nitrogen (gas); 2 max 2. Order must be nitrite then nitrate 2. Accept ammonia for ammonium ions 2. Accept correct chemical formulae for ions, eg there will be little oxidation/conversion of NH4+ → NO2- → NO3- 2. Ignore ‘breakdown’ for oxidation/conversion 4. Accept correct chemical formulae eg So more NO3-reduced/converted to N2; MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2021 11 Question Marking Guidance Mark Comments 04.3 1. Assumed that height is (directly) proportional to biomass; 2. (Plants may put biomass into) other named aspect of growth (other than height) OR Height does not include the roots OR Some increase in height results from water gain; 2 1. Accept descriptions of ‘is proportional to’, eg correlates to, is equivalent to 2. Examples of other named aspects of growth could include root growth, flower/seed/fruit formation, lateral growth, wider leaves Question Marking Guidance Mark Comments 04.4 1. Answer of 12 days = 2 marks;; 2. 12.16 (12.15774433) = 1 mark OR 4 days (used 387 and 268, ie not calculated starting length) = 1 mark; 2 MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2021 12 Question Marking Guidance Mark Comments 05.1 1. Exopeptidases hydrolyse peptide bonds at the ends of a polypeptide/protein AND endopeptidases hydrolyse internal peptide bonds within a polypeptide/protein; 2. More ‘ends’ OR More surface area; 2 1. Reference to 'hydrolyse' required at least once 2. Accept even if via action of incorrect enzyme Question Marking Guidance Mark Comments 05.2 1. Actin/myosin/tropomyosin; 2. Antibodies; 2 1. Accept troponin 1. Accept ATP synthase/hydrolase 2. Accept immunoglobulins 2. Accept lysozyme Question Marking Guidance Mark Comments 05.3 Whey (no mark) as it: 1. Is absorbed quicker OR It has a faster/higher/greater/the highest/the greatest/the fastest rate of absorption; 2. Still stimulates/increases protein synthesis (even if lower than casein); 3. Prevents/inhibits/limits breakdown of body proteins; 4. Significantly more becomes body protein; 3 max If student selects casein allow 1 mark only for ‘as it stimulates a higher rate of protein synthesis’ 1. and 4. Accept use of data to show differences MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2021 13 Question Marking Guidance Mark Comments 06.1 1. Extract DNA and add restriction endonucleases/restriction enzymes; 2. Separate fragments using electrophoresis; 3. (Treat DNA to) form single strands OR (Treat DNA to) expose bases; 4. The probe will bind to/hybridise/base pair with the SUT1/gene; 5. Use autoradiography (to show the bound probe); 4 max 3. Ignore method used to separate strands 5. Accept use photographic or X ray film (to show the bound probe) 5. X rays alone is not sufficient Question Marking Guidance Mark Comments 06.2 1. Antisense mRNA is complementary to 'sense' mRNA; 2. Antisense mRNA would bind/base pair to (sense) mRNA; OR Double stranded (m)RNA forms; 3. Ribosomes would not be able to bind; 4. Preventing/less translation (of mRNA) OR Preventing/less production of SUT1 (protein); 4 4. Accept descriptions of translation Question Marking Guidance Mark Comments 06.3 :1; 1 Accept any suitable rounding MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2021 14 Question Marking Guidance Mark Comments 06.4 1. Some (14CO2) used to make cellulose/cell walls; 2. Some (14CO2) converted into starch (which remains in the leaf); 3. Not all (14CO2) fixed/used in photosynthesis; OR Not enough RuBP (to combine with all of the 14CO2); 4. Some used to reform RuBP OR Some (is still) in glycerate 3-phosphate/GP/triose phosphate/in the Calvin cycle; 2 max 1. Accept some becomes lipids/ proteins/DNA/RNA/ nucleotides 2. Accept some (14CO2) converted into glucose 3. Accept descriptions of this MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2021 15 Question Marking Guidance Mark Comments 06.5 1. Reduced SUT1 expression/less SUT 1 (protein) means less sucrose exported (so concentration increases in leaves); 4 max 2. Accept less rubisco or less active rubisco for ‘inhibits rubisco’ 2. (Increased sucrose in leaves) inhibits rubisco, so less 14CO2 fixed into GP; OR (Increased sucrose in leaves) inhibits rubisco, so less 14CO2 combines with RuBP; OR (Increased sucrose in leaves) inhibits rubisco, so less Calvin cycle/light independent reaction/s; 3. Less sucrose transported to roots, so roots do not develop/grow (as shown by larger shoot to root dry mass ratio); 4. Roots less developed so fewer minerals available for growth 4. Accept: roots less developed so less water available for photosynthesis 5. Less growth means less dry mass; 5. Accept: less photosynthesis/light independent reaction/s means less dry mass; MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2021 16 Question 7 Level of response marking guidance Level of response marking instructions Level of response mark schemes are broken down into levels, each of which has a descriptor. The descriptor for the level shows the average performance for the level. There are marks in each level. Before you apply the mark scheme to a student’s answer read through the answer and annotate it (as instructed) to show the qualities that are being looked for. You can then apply the mark scheme. Step 1 Determine a level Start at the lowest level of the mark scheme and use it as a ladder to see whether the answer meets the descriptor for that level. The descriptor for the level indicates the different qualities that might be seen in the student’s answer for that level. If it meets the lowest level then go to the next one and decide if it meets this level, and so on, until you have a match between the level descriptor and the answer. With practice and familiarity you will find that for better answers you will be able to quickly skip through the lower levels of the mark scheme. When assigning a level you should look at the overall quality of the answer and not look to pick holes in small and specific parts of the answer where the student has not performed quite as well as the rest. If the answer covers different aspects of different levels of the mark scheme you should use a best fit approach for defining the level and then use the variability of the response to help decide the mark within the level, ie if the response is predominantly level 3 with a small amount of level 4 material it would be placed in level 3 but be awarded a mark near the top of the level because of the level 4 content. Step 2 Determine a mark Once you have assigned a level you need to decide on the mark. The descriptors on how to allocate marks can help with this. The exemplar materials used during standardisation will help. There will be an answer in the standardising materials which will correspond with each level of the mark scheme. This answer will have been awarded a mark by the Lead Examiner. You can compare the student’s answer with the example to determine if it is the same standard, better or worse than the example. You can then use this to allocate a mark for the answer based on the Lead Examiner’s mark on the example. You may well need to read back through the answer as you apply the mark scheme to clarify points and assure yourself that the level and the mark are appropriate. Indicative content in the mark scheme is provided as a guide for examiners. It is not intended to be exhaustive and you must credit other valid points. Students do not have to cover all of the points mentioned in the Indicative content to reach the highest level of the mark scheme. An answer which contains nothing of relevance to the question must be awarded no marks. MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2021 17 21–25 Extended Abstract Generalised beyond specific context Response shows holistic approach to the question with a fully integrated answer which makes clear links between several different topics and the theme of the question. Biology is detailed and comprehensive A-level content, uses appropriate terminology, and is very well written and always clearly explained. No significant errors or irrelevant material. For top marks in the band, the answer shows evidence of reading beyond specification requirements. 16–20 Relational Integrated into a whole Response links several topics to the main theme of the question, to form a series of interrelated points which are clearly explained. Biology is fundamentally correct A-level content and contains some points which are detailed, though there may be some which are less well developed, with appropriate use of terminology. Perhaps one significant error and/or, one irrelevant topic which detracts from the overall quality of the answer. 11–15 Multistructural Several aspects covered but they are unrelated Response mostly deals with suitable topics but they are not interrelated and links are not made to the theme of the question. Biology is usually correct A-level content, though it lacks detail. It is usually clearly explained and generally uses appropriate terminology. Some significant errors and/or, more than one irrelevant topic. 6–10 Unistructural Only one or few aspects covered Response predominantly deals with only one or two topics that relate to the question. Biology presented shows some superficial A-level content that may be poorly explained, lacking in detail, or show limited use of appropriate terminology. May contain a number of significant errors and/or, irrelevant topics. 1–5 Unfocused Response only indirectly addresses the theme of the question and merely presents a series of biological facts which are usually descriptive in nature or poorly explained and at times may be factually incorrect. Content and terminology is generally below A-level. May contain a large number of errors and/or, irrelevant topics. 0 Nothing of relevance or no response. MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2021 18 Commentary on terms and statements in the levels mark scheme The levels mark scheme for the essay contains a number of words and statements that are open to different interpretations. This commentary defines the meanings of these words and statements in the context of marking the essay. Many words and statements are used in the descriptions of more than one level of response. The definitions of these remain the same throughout. Levels mark scheme word/statement Definition Holistic Synoptic, drawing from different topics (usually sections of the specification) A fully integrated answer which makes clear links between several different topics and the theme of the question. All topics relate to the title and theme of the essay; for example, explaining the biological importance of a process. When considering, for example, the importance of a process, the explanation must be at A-level standard. ‘Several’ here is defined as at least four topic areas from the specification covered. This means some sentences, not just a word or two. It does not mean using many examples from one topic area. Biology is detailed and comprehensive A-level content, uses appropriate terminology, and is very well written and always clearly explained. Detailed and comprehensive A-level content is the specification content. Terminology is that used in the specification. Well written and clearly explained refers mainly to biological content and use of terminology. Prose, handwriting and spelling are secondary considerations. Phonetic spelling is accepted, unless examiners are instructed not to do so for particular words; for example, glucagon, glucose and glycogen. No significant errors or irrelevant material. A significant error is one which significantly detracts from the biological accuracy or correctness of a described example. This will usually involve more than one word. Irrelevant material is several lines (or more) that clearly fails to address the title, or the theme of the title. For top marks in the band, the answer shows evidence of reading beyond specification requirements. An example that is relevant to the title and is not required in the specification content. The example must be used at A-level standard. Response mostly deals with suitable topics but they are not interrelated and links are not made to the theme of the question. Not addressing the biological theme of the essay (eg importance) at A-level standard. MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2021 19 Question Marking Guidance Mark 07.1 The importance of complementary shapes of molecules in organisms • 3.1.4.2 Many proteins are enzymes • 3.1.5.1 Structure of DNA and RNA • 3.1.5.2 DNA replication • 3.1.6 ATP • 3.2.2 All cells arise from other cells • 3.2.3 Transport across cell membranes • 3.2.4 Cell recognition and the immune system • 3.3.3 Digestion and absorption • 3.4.1 DNA, genes and chromosomes • 3.4.2 DNA and protein synthesis • 3.4.3 Genetic diversity can arise as a result of mutation or during meiosis • 3.5.1 Photosynthesis • 3.5.2 Respiration • 3.6.1.2 Receptors • 3.6.2.1 Nerve impulses • 3.6.2.2 Synaptic transmission • 3.6.3 Skeletal muscles are stimulated to contract by nerves and act as effectors • 3.6.4.2 Control of blood glucose concentration • 3.6.4.3 Control of blood water potential • 3.8.1 Alteration of the sequence of bases in DNA can alter the structure of proteins • 3.8.2.2 Regulation of transcription and translation • 3.8.2.3 Gene expression and cancer [25 marks] In order to fully address the question and reach the highest mark bands students must also include at least four topics in their answer, to demonstrate a synoptic approach to the essay. Students may be able to show the relevance of other topics from the specification. Note; other topics from beyond the specification can be used, providing they relate to the title and contain factually correct material of at least an A-level standard. Credit should not be given for topics beyond the specification which are below A-level standard. MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2021 20 Question Marking Guidance Mark 07.2 The importance of ions in metabolic processes •3.1.4.2 Many proteins are enzymes (H and denaturation) •3.1.5.2 DNA replication •3.1.6 ATP •3.1.8 Inorganic ions •3.2.3 Transport across cell membranes •3.3.3 Digestion and absorption •3.3.4.1 Mass transport in animals •3.3.4.2 Mass transport in plants •3.4.2 DNA and protein synthesis •3.5.1 Photosynthesis •3.5.2 Respiration •3.5.4 Nutrient cycles •3.6.1.1 Survival and response •3.6.1.2 Receptors •3.6.2.1 Nerve impulses •3.6.2.2 Synaptic transmission •3.6.3 Skeletal muscles are stimulated to contract by nervesand act as effectors •3.6.4.3 Control of blood water potential •3.8.4.3 Genetic fingerprinting [25 marks] In order to fully address the question and reach the highest mark bands students must also include at least four topics in their answer, to demonstrate a synoptic approach to the essay. Students may be able to show the relevance of other topics from the specification. Note, other topics from beyond the specification can be used, providing they relate to the title and contain factually correct material of at least an A-level standard. Credit should not be given for topics beyond the specification which are below A-level standard. [Show Less]
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