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AQA A LEVEL BIOLOGY PAPER 1, 2 & 3 QUESTION PAPERS & MARK SCHEMES JUNE 2021 COMPLETE BUNDLE [ 7402/1, 7402/2 & 7402/3 ] OFFICIAL PAPERS ⭐⭐⭐⭐⭐
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AQA A Level Biology Paper 1 7402/1 Question Paper June 2021 Version 1.0 Final
AQA A-Level Biology Paper 1 7402-1 Question Paper June 2021 / AQA A-Level Biology Paper 1 7402/1 Question Paper June 2021 / AQA A-Level Biology Paper 1 Qu... [Show More] estion Paper June 2021 / A Level Biology Paper 1 7402/1 Question Paper June 2021 / A Level Biology Paper 1 Question Paper June 2021 IB/H/Jun21/E14 7402/1 For Examiner’s Use Question Mark 1 2 3 4 5 6 7 8 9 10 TOTAL 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. •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 91. 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 1 AQA 2 *02* IB/H/Jun21/7402/1 Do not write outside the box Answer all questions in the spaces provided. 0 1 . 1 Describe the induced-fit model of enzyme action and how an enzyme acts as a catalyst. [3 marks] 0 1 . 2 Scientists investigated the action of the enzyme ATP synthase. They made reaction mixtures each containing: •ATP synthase •buffer (to control pH) •substrates. One of the substrates required in these reaction mixtures is inorganic phosphate (Pi). Tick () one box to show which other substrate the scientists must add to the reaction mixtures to produce ATP. [1 mark] Adenine Adenosine diphosphate Glucose Ribose 3 *03* Turn over ► IB/H/Jun21/7402/1 Do not write outside the box 0 1 . 3 The scientists investigated the effect of concentration of inorganic phosphate (Pi) on ATP synthase activity. After 2 minutes, they stopped each reaction and then measured the concentration of ATP. Figure 1 shows the scientists’ results. Figure 1 Suggest and explain a procedure the scientists could have used to stop each reaction. [2 marks] 0 1 . 4 Explain the change in ATP concentration with increasing inorganic phosphate concentration. [2 marks] 8 4 *04* IB/H/Jun21/7402/1 Do not write outside the box 0 2 . 1 Explain the advantage for larger animals of having a specialised system that facilitates oxygen uptake. [2 marks] 5 *05* Turn over ► IB/H/Jun21/7402/1 Do not write outside the box Figure 2 shows two models of oxygen uptake found in animals. Figure 2 0 2 . 2 Suggest how the environmental conditions have resulted in adaptations of systems using Model A rather than Model B. [2 marks] Question 2 continues on the next page 6 *06* IB/H/Jun21/7402/1 Do not write outside the box 0 2 . 3 Figure 3 shows changes in concentration of oxygen in two gas exchange systems. Figure 3 A student studied Figure 3 and concluded that the fish gas exchange system is more efficient than the human gas exchange system. Use Figure 3 to justify this conclusion. [2 marks] 0 2 . 4 Explain how the counter-current principle allows efficient oxygen uptake in the fish gas exchange system. [2 marks] 7 *07* Turn over ► IB/H/Jun21/7402/1 Do not write outside the box 0 2 . 5 Table 1 shows features of two mammals. Bats are flying mammals; shrews are ground-living mammals. Table 1 Mammal Mean body mass / kg Mean lung volume / cm3 Bat 0.096 12.48 Shrew 0.024 0.72 Calculate how many times the lung volume per unit of body mass of the bat is greater than that of the shrew. Give your answer to an appropriate number of significant figures. Give one suggestion to explain this difference. [3 marks] Answer Explanation 11 8 *08* IB/H/Jun21/7402/1 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 9 *09* Turn over ► IB/H/Jun21/7402/1 Do not write outside the box 0 3 . 1 Describe how one amino acid is added to a polypeptide that is being formed at a ribosome during translation. [3 marks] Question 3 continues on the next page 10 *10* IB/H/Jun21/7402/1 Do not write outside the box Table 2 shows: • mRNA codons and the amino acid coded for by each codon • the type of bond formed by the R group of some of the amino acids. Table 2 0 3 . 2 Crystallin is a structural protein found in the human eye. An inherited disease that leads to blindness is caused by changes in properties of crystallin. The replacement of the amino acid Arg with the amino acid Gly causes these changes. Use information in Table 2 to suggest why this amino acid replacement changes the properties of crystallin. [2 marks] 11 *11* Turn over ► IB/H/Jun21/7402/1 Do not write outside the box 0 3 . 3 The amino acid replacement of Arg with Gly is caused by a single base substitution mutation in the DNA. The non-mutant DNA triplet is TCC. Complete Table 3. Give: • the mRNA codon complementary to the non-mutant DNA triplet • the mutated mRNA codon that could cause the change from Arg to Gly in the crystallin protein • the DNA triplet complementary to this mutated mRNA codon. [2 marks] Table 3 mRNA codon for the non-mutant triplet Mutated mRNA codon Mutated DNA triplet Turn over for the next question 7 12 *12* IB/H/Jun21/7402/1 Do not write outside the box 0 4 A student dissected an organ from a mammal to observe blood vessels. He dissected a slice of the organ and identified two blood vessels. Figure 4 shows a photograph of his dissection. Figure 4 Figure 5 shows a drawing of the blood vessels from his dissection. Figure 5 13 *13* Turn over ► IB/H/Jun21/7402/1 Do not write outside the box 0 4 . 1 Suggest two ways the student could improve the quality of his scientific drawing of the blood vessels in this dissection. [2 marks] 1 2 0 4 . 2 Identify the type of blood vessel labelled as X and the type of blood vessel labelled as Y in Figure 4. Describe one feature that allowed you to identify the blood vessels. [2 marks] Blood vessel X Blood vessel Y Feature 0 4 . 3 Describe two precautions the student should take when clearing away after the dissection. [2 marks] 1 2 6 14 *14* IB/H/Jun21/7402/1 Do not write outside the box 0 5 . 1 Describe how a sample of chloroplasts could be isolated from leaves. [4 marks] 15 *15* Turn over ► IB/H/Jun21/7402/1 Do not write outside the box 0 5 . 2 Scientists grew two groups of plants: • control plants with all the inorganic ions needed • iron-deficient plants with all the inorganic ions needed but without iron ions. After 1 week, the scientists measured the mass of protein and the mass of chlorophyll in the chloroplasts isolated from samples of leaves of these two groups of plants. Table 4 shows the scientists’ results. Table 4 Mass of protein / percentage of control Mass of chlorophyll / percentage of control 40 10 Some proteins found inside the chloroplast are synthesised inside the chloroplast. Give one feature of the chloroplast that allows protein to be synthesised inside the chloroplast and describe one difference between this feature in the chloroplast and similar features in the rest of the cell. [2 marks] Feature Structural difference 0 5 . 3 The ratio of protein to chlorophyll in control plants is 9:1 Use the information in Table 4 to calculate the ratio of protein to chlorophyll in iron-deficient plants. [1 mark] Ratio Question 5 continues on the next page 16 *16* IB/H/Jun21/7402/1 Do not write outside the box 0 5 . 4 The scientists also observed the chloroplasts from the samples of leaves using an electron microscope. Figure 6 shows a chloroplast from a control plant (image A) and a chloroplast from an iron-deficient plant (image B). Figure 6 Use Figure 6 to suggest why iron-deficient plants have a reduced growth rate. [3 marks] 10 17 *17* Turn over ► IB/H/Jun21/7402/1 Do not write outside the box Turn over for the next question DO NOT WRITE ON THIS PAGE ANSWER IN THE SPACES PROVIDED 18 *18* IB/H/Jun21/7402/1 Do not write outside the box 0 6 . 1 Figure 7 shows the mean distance between centromeres and the poles (ends) of the spindle during mitosis. Figure 7 Calculate the rate of movement of the centromeres during phase E. Give your answer in μm minute–1 and to 3 decimal places. [2 marks] μm minute–1 19 *19* Turn over ► IB/H/Jun21/7402/1 Do not write outside the box 0 6 . 2 Name the three phases of mitosis shown by C, D and E on Figure 7. Describe the role of the spindle fibres and the behaviour of the chromosomes during each of these phases. [5 marks] C D E Turn over for the next question 7 20 *20* IB/H/Jun21/7402/1 Do not write outside the box 0 7 . 1 ADCs are molecules made of a monoclonal antibody linked to a cancer drug. Figure 8 shows how an ADC enters and kills a tumour cell. The process of entering the cell and the breakdown of the antibody to release the drug is very similar to phagocytosis. Figure 8 Use your knowledge of phagocytosis to describe how an ADC enters and kills the tumour cell. [3 marks] 21 *21* Turn over ► IB/H/Jun21/7402/1 Do not write outside the box 0 7 . 2 Some of the antigens found on the surface of tumour cells are also found on the surface of healthy human cells. Use this information to explain why treatment with an ADC often causes side effects. [2 marks] Question 7 continues on the next page 22 *22* IB/H/Jun21/7402/1 Do not write outside the box Scientists investigated whether one type of ADC could be used to treat human breast cancer. This ADC is a monoclonal antibody combined with a drug to inhibit mitosis. The monoclonal antibody binds to a protein found on human breast cancer cells. The scientists placed small pieces of human breast cancer tissue under the skin of mice. The scientists then randomly divided the mice into three groups. They treated the groups as follows on day 0. Group G – control Group H – injected with monoclonal antibody only Group J – injected with ADC (monoclonal antibody + drug). Every few days, the scientists measured the volume of the tumours formed from the human breast cancer tissue. Figure 9 shows the scientists’ results. Figure 9 23 *23* Turn over ► IB/H/Jun21/7402/1 Do not write outside the box 0 7 . 3 Mice in Group H were injected with 2 mg kg–1 of monoclonal antibody. The monoclonal antibody was in a solution of concentration 500 mg dm–3 Calculate the volume of antibody solution that the scientists would have injected into a 23 g mouse. Give your answer in dm3 and in standard form. [2 marks] dm3 0 7 . 4 Suggest one reason why there are no data for Group G and Group H after day 8 [1 mark] 0 7 . 5 Suggest and explain two further investigations that should be done before this ADC is tested on human breast cancer patients. [2 marks] 1 2 Turn over for the next question 10 24 *24* IB/H/Jun21/7402/1 Do not write outside the box 0 8 . 1 Describe how a triglyceride molecule is formed. [3 marks] 25 *25* Turn over ► IB/H/Jun21/7402/1 Do not write outside the box 0 8 . 2 Table 5 shows some properties of four fatty acids. Table 5 Fatty acid Number of carbon atoms in the R group Number of double bonds in the R group Caprylic acid 8 0 Palmitoleic acid 16 1 Stearic acid 18 0 Linoleic acid 18 2 Figure 10 shows diagrams of these fatty acids. Figure 10 Put a tick () in one box that contains correct information about one of these fatty acids. [1 mark] Caprylic acid is an unsaturated fatty acid represented by diagram L. Linoleic acid is a saturated fatty acid represented by diagram N. Palmitoleic acid is an unsaturated fatty acid represented by diagram K. Stearic acid is a saturated fatty acid represented by diagram M. 26 *26* IB/H/Jun21/7402/1 Do not write outside the box The percentage of saturated fatty acids compared with unsaturated fatty acids found in lipid stores in seeds differs in different populations. Scientists investigated two populations of the plant, Helianthus annuus. The scientists grew young plants from seeds collected from each population. They placed the seeds on wet tissue paper so that the root growth was visible. They grew seeds from each population at two temperatures: • warm temperature of 24 °C • cool temperature of 10 °C After 10 days, the scientists measured the length of each root. Table 6 shows some of the properties of the two populations and the scientists’ results. Table 6 Population Temperature in natural environment In the seed – Mean percentage of fatty acids that are saturated Mean length of root after 10 days at 24 °C / mm (± 2 x standard deviation) Mean length of root after 10 days at 10 °C / mm (± 2 x standard deviation) 1 Warm 10.9 8.2 (±1.0) 3.1 (±0.3) 2 Cool 6.1 5.5 (±0.9) 4.3 (±0.2) The mean ±2 × standard deviation includes 95% of the data. 0 8 . 3 The scientists used a data logger to measure the length of the root rather than a ruler. Suggest one reason why they used a data logger and explain why this was important in this investigation. [1 mark] 27 *27* Turn over ► IB/H/Jun21/7402/1 Do not write outside the box 0 8 . 4 It is known that: • during respiration saturated fatty acids yield more energy than unsaturated fatty acids • saturated fatty acids have higher melting points than unsaturated fatty acids • lipases in seeds act more rapidly on liquid substrates. Use this information and Table 6 to show how each population is better adapted for its natural environment when compared with the other population. [4 marks] 0 8 . 5 Although these two populations are completely separate and show genetic variation, they are both called Helianthus annuus. Explain why they are both given this name. [1 mark] 10 28 *28* IB/H/Jun21/7402/1 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 29 *29* Turn over ► IB/H/Jun21/7402/1 Do not write outside the box 0 9 . 1 Complete Table 7 with ticks () to show which elements are found in the following biological molecules. [2 marks] Table 7 Biological molecules Element Carbon Nitrogen Oxygen Phosphorus Galactose Phospholipid RNA Sucrose Question 9 continues on the next page 30 *30* IB/H/Jun21/7402/1 Do not write outside the box After Watson and Crick proposed the model of DNA structure, scientists investigated the possible mechanisms for DNA replication. Two scientists grew a bacterial population, providing them with a nitrogen source containing only the heavy isotope of nitrogen, 15 N. As soon as all the DNA in this population contained 15 N, the scientists changed the nitrogen source to one containing only the lighter isotope of nitrogen, 14 N. They changed the nitrogen source at 0 hours. During the investigation, the scientists measured the size of the population of bacterial cells. Figure 11 shows the scientists’ results. Figure 11 31 *31* Turn over ► IB/H/Jun21/7402/1 Do not write outside the box 0 9 . 2 The generation time for a population of bacteria is the time taken for all the bacteria to divide once by binary fission. Use Figure 11 and the following equation to calculate the generation time for this population of bacteria. Give your answer in hours. [2 marks] Number of generations= log10size of population at time +4 hourssize of population at time – 4 hourslog10 2 Generation time hours Question 9 continues on the next page 32 *32* IB/H/Jun21/7402/1 Do not write outside the box At intervals during this investigation, the scientists removed samples of the bacterial population, isolated the DNA and measured the density of the DNA. DNA made using 15 N has a higher density than DNA made using 14 N. Figure 12 shows the scientists’ results. Figure 12 33 *33* Turn over ► IB/H/Jun21/7402/1 Do not write outside the box There are three possible models of DNA replication. These models are shown in Figure 13. Figure 13 0 9 . 3 Which of these models, P, Q or R, is supported by the results shown in Figure 12? Give the letter and name of the model supported and explain why the results do not support the other models. [3 marks] Model Name Explanation for first unsupported model Explanation for second unsupported model 7 34 *34* IB/H/Jun21/7402/1 Do not write outside the box 1 0 . 1 Describe the structure of DNA. [5 marks] 35 *35* Turn over ► IB/H/Jun21/7402/1 Do not write outside the box 1 0 . 2 Name and describe five ways substances can move across the cell-surface membrane into a cell. [5 marks] Question 10 continues on the next page 36 *36* IB/H/Jun21/7402/1 Do not write outside the box Figure 14 shows transmission electron micrographs of two cells, one animal cell and one prokaryotic cell. Figure 14 37 *37* IB/H/Jun21/7402/1 Do not write outside the box 1 0 . 3 Contrast the structure of the two cells visible in the electron micrographs shown in Figure 14. [5 marks] END OF QUESTIONS 15 38 *38* IB/H/Jun21/7402/1 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 39 *39* IB/H/Jun21/7402/1 Do not write outside the box Question number Additional page, if required. Write the question numbers in the left-hand margin. 40 *40* IB/H/Jun21/7402/1 Do not write outside the box Question number Additional page, if required. Write the question numbers in the left-hand margin. [Show Less]
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AQA A Level Biology 7402/1 Paper 1 Mark Scheme June 2021 Version 1 Final
AQA A-Level Biology 7402/1 Paper 1 Mark Scheme June 2021 Version 1.0 Final / AQA A-Level Biology 7402-1 Paper 1 Mark Scheme June 2021 Version 1.0 Final / A... [Show More] QA A-Level Biology Paper 1 Mark Scheme June 2021 Version 1.0 Final A-level BIOLOGY 7402/1 Paper 1 Mark scheme June 2021 Version: 1.0 Final AQA MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – 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 – 7402/1 – 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 whatis acceptable or not worthy of credit or, in discursive answers, to give an overview of the areain which 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 – 7402/1 – 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 – 7402/1 – JUNE 2021 5 Question Marking Guidance Mark Comments 01.1 1.Substrate binds to the active site/enzyme OR Enzyme-substrate complex forms; 2.Active site changes shape (slightly) so it iscomplementary to substrate OR Active site changes shape (slightly) so distorting/breaking/forming bonds in the substrate; 3.Reduces activation energy; 3 1.Accept for ‘binds’,fits Question Marking Guidance Mark Comments 01.2 Adenosine diphosphate; 1 Question Marking Guidance Mark Comments 01.3 Mark in pairs, 1 and 2 OR 3 and 4 OR 5 and 6 1.Boil OR Add (strong) acid/alkali; 2.Denatures the enzyme/ATP synthase; OR 3.Put in ice/fridge/freezer; 4.Lower kinetic energy so no enzyme-substratecomplexes form; OR 5.Add high concentration of inhibitor; 6.Enzyme-substrate complexes do not form; 2 1.Accept heat at >50oC OR at veryhigh temperatures 2.Accept for'denatures', adescription ofdenaturation 4.Accept ES forenzyme substratecomplex MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2021 6 Question Marking Guidance Mark Comments 01.4 1. (With) increasing Pi concentration, more enzyme-substrate complexes are formed; 2. At or above 40 (mmol dm-3) all active sites occupied OR At or above 40 (mmol dm-3) enzyme concentration is a limiting factor; 2 MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2021 7 Question Marking Guidance Mark Comments 02.1 1. Large(r) organisms have a small(er) surface area:volume (ratio); OR Small(er) organisms have a large(r) surface area:volume (ratio); 2. Overcomes long diffusion pathway OR Faster diffusion; 2 2. Accept short diffusion pathway 2. Accept for ‘faster’, more Question Marking Guidance Mark Comments 02.2 Mark in pairs, 1, and 2 OR 3. and 4. 1. Water has low(er) oxygen partial pressure/concentration (than air); 2. So (system on outside) gives large surface area (in contact with water) OR So (system on outside) reduces diffusion distance (between water and blood); 3. Water is dense(r) (than air); 4. (So) water supports the systems/gills; 2 MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2021 8 Question Marking Guidance Mark Comments 02.3 1. In fish, blood leaving (V) has more oxygen than water leaving (E); 2. (But) in humans, blood leaving (V) has less oxygen than air leaving (E); 3. Difference in oxygen (concentration) between artery and vein is greater in fish than in humans; 4. (So) fish remove a greater proportion from the oxygen they take in; 2 max Question Marking Guidance Mark Comments 02.4 1. Blood and water flow in opposite directions; 2. Diffusion/concentration gradient (maintained) along (length of) lamella/filament; 2 Accept for 2 marks, suitably labelled diagram Question Marking Guidance Mark Comments 02.5 1. and 2. Correct answer for 2 marks, 4.3 (times greater);; Accept for 1 mark, 4.333333333 (correct answer not given to 2 significant figures) OR Evidence of 130 (cm3 kg-1) and 30 (cm3 kg-1) Correct explanation for 1 mark, 3. Provides more oxygen for respiration; 3 MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2021 9 Question Marking Guidance Mark Comments 03.1 1. tRNA brings specific amino acid (to ribosome); 2. Anticodon (on tRNA) binds to codon (on mRNA); 3. Amino acids join by condensation reaction (using ATP) OR Amino acids join to form a peptide bond (using ATP); 3 Question Marking Guidance Mark Comments 03.2 1. Hydrogen bonds form instead of ionic bonds; 2. Changes the tertiary structure (of the crystallin); 2 2. Ignore reference to active site Question Marking Guidance Mark Comments 03.3 3 correct = 2 marks;; 2 correct = 1 mark; 0 or 1 correct = 0 marks mRNA codon for the non-mutant triplet AGG Mutated mRNA codon GGG Mutated DNA triplet CCC 2 MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2021 10 Question Marking Guidance Mark Comments 04.1 1. Only use single lines/do not use sketching (lines)/ensure lines are continuous/connected; 2. Add labels/annotations/title; 3. Add magnification/scale (bar); 4. Draw all parts to same scale/relative size; 5. Do not use shading/hatching; 2 max Question Marking Guidance Mark Comments 04.2 1. Blood vessel X – artery/arteriole and Blood vessel Y – vein/venule; 2. (Difference in) lumen size OR (Difference in) wall thickness; 2 Ignore name of blood vessel, eg. (pulmonary) artery Question Marking Guidance Mark Comments 04.3 1. Carry/wash sharp instruments by holding handle OR Carry/wash sharp instruments by pointing away (from body)/down; 2. Disinfect instruments/surfaces; 3. Disinfect hands OR Wash hands with soap (and water); 4. Put organ/gloves/paper towels in a (separate) bag/bin/tray to dispose; 2 max 1. and 2. Accept for ‘instruments’, a suitable named example, eg. scalpel 2. and 3. Accept for ‘disinfect’, sanitise OR use antiseptic MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2021 11 Question Marking Guidance Mark Comments 05.1 1. Break open cells/tissue and filter OR Grind/blend cells/tissue/leaves and filter; 2. In cold, same water potential/concentration, pH controlled solution; 3. Centrifuge/spin and remove nuclei/cell debris; 4. (Centrifuge/spin) at high(er) speed, chloroplasts settle out; 4 1. Accept homogenise and filter 2. Accept for ‘same water potential/ concentration’, isotonic 2. Accept for ‘pH controlled’, buffered Question Marking Guidance Mark Comments 05.2 Mark in pairs, 1 and 2 OR 3 and 4 1. DNA; 2. Is not associated with protein/histones but nuclear DNA is OR Is circular but nuclear DNA is linear OR Is shorter than nuclear DNA; 3. Ribosomes; 4. Are smaller than cytoplasmic ribosomes; 2 4. Accept: 70S ribosomes in chloroplast, but 80S ribosomes in cytoplasm MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2021 12 Question Marking Guidance Mark Comments 05.3 Correct answer for 1 mark, 36:1; 1 Question Marking Guidance Mark Comments 05.4 1. Less (thylakoid) membrane OR Fewer/smaller grana; 2. Smaller surface area (of membrane in chloroplast)/less chlorophyll; 3. (Less chlorophyll so) reduced light absorption; 4. (So) slower rate of photosynthesis; 3 max 4. Accept reduced rate of any named biochemical process in photosynthesis; eg. reduced production of ATP/reduced NADP MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2021 13 Question Marking Guidance Mark Comments 06.1 Correct answer for 2 marks, 1.286;; Accept for 1 mark, 1.28571429 (correct answer not to 3 decimal places) OR 1.285 (incorrect rounding to 3 decimal places) OR Evidence of 0.02142857 OR Evidence of 19 and 4 and 700 OR Evidence of 15 and 1800 and 2500 OR Evidence of 15 and 700 2 Question Marking Guidance Mark Comments 06.2 1. C = prophase and D = metaphase and E = anaphase; 2. (In) prophase, chromosomes condense; 3. (In) prophase OR metaphase, centromeres attach to spindle fibres; 4. (In) metaphase, chromosomes/pairs of chromatids at equator/centre of spindle/cell; 5. (In) anaphase, centromeres divide; 6. (In) anaphase, chromatids (from each pair) pulled to (opposite) poles/ends (of cell); 7. (In) prophase/metaphase/anaphase, spindle fibres shorten; 5 max If mark point 1 is not credited = 4 max Do not carry forward error from 1. Accept letters for stages as indicated in 1. 2. Accept chromatin for ‘chromosomes’ and for ‘condense’, shorten and thicken 6. Accept for ‘chromatids’, chromosomes but reject homologous chromosomes 7. Accept for ‘shorten’, contract MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2021 14 Question Marking Guidance Mark Comments 07.1 1. Cell ingests/engulfs the antibody/ADC OR Cell membrane surrounds the antibody/ADC (to take it inside the cell); 2. Lysosomes fuse with vesicle/phagosome (containing ADC); 3. Lysozymes breakdown/digest the antibody/ADC to release the drug; 3 1. Accept endocytosis for ingest/engulf 3. Accept hydrolytic enzyme for lysozyme Question Marking Guidance Mark Comments 07.2 1. ADC will bind to non-tumour/healthy cells; 2. Cause death/damage of non-tumour/healthy cells OR Cause damage to other organs/systems; 2 1. Reject reference to active site Question Marking Guidance Mark Comments 07.3 Correct answer for 2 marks, 9.2 x 10‒5;; Accept for 1 mark, 0.046 (correct mass injected into 23g mouse) 0.000092 (correct answer but not in standard form) 2 Question Marking Guidance Mark Comments 07.4 Mice died OR Not ethical to continue; 1 MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2021 15 Question Marking Guidance Mark Comments 07.5 1. Tested on other mammals to check for safety/side effects; 2. Tested on (healthy) humans to check for safety/side effects; 3. See if repeat doses stop the tumours regrowing (in Group J); 4. Investigate different concentrations of ADC to find suitable/safe dosage; 2 max 1. Accept named mammal, eg rat 2. Accept: Tested on (healthy) human tissue/cells to check for no side-effects MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2021 16 Question Marking Guidance Mark Comments 08.1 1.One glycerol and three fatty acids; 2.Condensation (reactions) and removal ofthree molecules of water; 3.Ester bond(s) (formed); 3 Accept all marks in suitably labelled diagram OR in a balanced equation Question Marking Guidance Mark Comments 08.2 Palmitoleic acid is an unsaturated fatty acid represented by diagram K; 1 Question Marking Guidance Mark Comments 08.3 1.To increase accuracy/resolution becausedifferences/lengths are small; 2.To increase accuracy because reduces risk ofhuman error; 3.To increase accuracy because roots are less(likely to be) damaged; 4.To reduce error/uncertainty becausedifferences/lengths are small; 1 max Ignore 'precision' MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2021 17 Question Marking Guidance Mark Comments 08.4 1. Population 1 grew longer roots in warm temperatures and population 2 grew longer roots in cool temperatures; 2. Standard deviations do not overlap so difference (in mean) unlikely to be/not due to chance; 3. Population 1 (is better adapted to warm conditions because it) has more saturated fatty acids so more energy available (and more growth); 4. Population 2 (is better adapted to cool conditions because it) has more unsaturated/liquid fatty acids so more lipase activity (and more growth); 4 2. Accept: 'Standard deviations do not overlap showing difference (in mean likely to be) significant' 3. and 4. Accept for ‘fatty acids’, fat Question Marking Guidance Mark Comments 08.5 Same species OR (If mated) can produce fertile offspring OR (It is) genus and species name; 1 MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2021 18 Question Marking Guidance Mark Comments 09.1 4 rows correct = 2 marks;; 2 or 3 rows correct = 1 mark; 0 or 1 row correct = 0 marks Biological molecules Element Carbon Nitrogen Oxygen Phosphorous Galactose ✓ ✓ Phospholipid ✓ ✓ ✓ RNA ✓ ✓ ✓ ✓ Sucrose ✓ ✓ 2 Question Marking Guidance Mark Comments 09.2 Correct answer for 2 marks, 0.8376308/0.84/0.8 (hours);; Accept for 1 mark, Evidence of 4 x 106 and 3 x 109 (written in any format, for correct readings from graph) OR Evidence of 9.550746785 (correct number of generations) OR Evidence of 1.1938443348 (correct generations/ hour) OR Evidence of 50.26 (correct generation time in minutes) Incorrect reading of graph, 3 x 106 and 2 x 109 Accept for 1 mark, calculation carried out correctly Evidence of 9.380821784 (correct calculation of number of generations) 2 Accept correct rounding to any number of decimal places MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2021 19 OR Evidence of 1.172602723 (correct calculation of generations/ hour) OR Evidence of 51.16822503 (correct calculation of generation time in minutes) OR Evidence of 0.8528037505 (correct calculation of generation time in hours) Incorrect reading of graph, 106.4 and 109.3 OR 106.3 and 109.2 Accept for 1 mark, calculation carried out correctly Evidence of 9.633591475 (correct calculation of number of generations) OR Evidence of 1.204198934 (correct calculation of generations/ hour) OR Evidence of 49.82565445 (correct calculation of generation time in minutes) OR Evidence of 0.8304275742 (correct calculation of generation time in hours) OR Evidence of 0.83, with no other working MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2021 20 Question Marking Guidance Mark Comments 09.3 1. (Model) Q and (Name) Semi-conservative (replication); Explanation 2. (Model) P (is unsupported because) There should be two peaks in generation 1 OR (Only) one peak is shown in generation 1 OR There should be 3:1 (ratio) of peaks in generation 2 OR There should not be an intermediate/15N 14N peak in generation 1/2/3 OR The original/generation 0/15N peak should be in generation 1/2/3; 3. (Model) R (is unsupported because) There should be >2 peaks in generation 2/3 OR There should be one wide/overlapping peak in generation 3; 3 Accept answers 2. and 3. in either order Accept for ‘peak’, density OR distribution 3. Accept for ‘>2’, many OR several MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2021 21 Question Marking Guidance Mark Comments 10.1 1. Polymer of nucleotides; 2. Each nucleotide formed from deoxyribose, a phosphate (group) and an organic/nitrogenous base; 3. Phosphodiester bonds (between nucleotides); 4. Double helix/2 strands held by hydrogen bonds; 5. (Hydrogen bonds/pairing) between adenine, thymine and cytosine, guanine; 5 1. Accept ‘Polynucleotide’ 1. Accept for ‘phosphate’. phosphoric acid Question Marking Guidance Mark Comments 10.2 1. (Simple) diffusion of small/non-polar molecules down a concentration gradient; 2. Facilitated diffusion down a concentration gradient via protein carrier/channel; 3. Osmosis of water down a water potential gradient; 4. Active transport against a concentration gradient via protein carrier using ATP; 5. Co-transport of 2 different substances using a carrier protein; 5 For any answer accept a correct example If no reference to 'small/ non-polar' for 1. accept this idea from 'large/charged' given in description of 2. 2. Reject if active rather than passive 5. For ‘carrier protein’ accept symport OR co-transport protein MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2021 22 Question Marking Guidance Mark Comments 10.3 1.Magnification (figures) show A is bigger than B; 2.A has a nucleus whereas B has free DNA; 3.A has mitochondria whereas B does not; 4.A has Golgi body/endoplasmic reticulum whereas Bdoes not; 5.A has no cell wall whereas B has a murein/glycoproteincell wall; 6.A has no capsule whereas B has a capsule; 7.A has DNA is bound to histones/proteins whereas Bhas DNA not associated with histones/proteins OR A has linear DNA whereas B has circular DNA; 8.A has larger ribosomes; 5 max Accept in all marking points, animal/eukaryote for A and prokaryote/ bacterium for B 5.Acceptpeptidoglycan [Show Less]
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AQA A Level Biology 7402/2 Paper 2 Question Paper June 2021 Version 1.0 Final
AQA A-Level Biology 7402/2 Paper 2 Question Paper June 2021 Version 1.0 Final / AQA A-Level Biology 7402-2 Paper 2 Question Paper June 2021 Version 1.0 Fin... [Show More] al / AQA A-Level Biology Paper 2 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. •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 91. 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 2 AQA 2 *02* IB/H/Jun21/7402/2 Do not write outside the box Answer all questions in the spaces provided. 0 1 . 1 Describe the process of glycolysis. [4 marks] 0 1 . 2 Malonate inhibits a reaction in the Krebs cycle. Explain why malonate would decrease the uptake of oxygen in a respiring cell. [2 marks] 6 3 *03* Turn over ► IB/H/Jun21/7402/2 Do not write outside the box 0 2 . 1 Explain how a resting potential is maintained across the axon membrane in a neurone. [3 marks] 0 2 . 2 Explain why the speed of transmission of impulses is faster along a myelinated axon than along a non-myelinated axon. [3 marks] Question 2 continues on the next page 4 *04* IB/H/Jun21/7402/2 Do not write outside the box 0 2 . 3 A scientist investigated the effect of inhibitors on neurones. She added a respiratory inhibitor to a neurone. The resting potential of the neurone changed from –70 mV to 0 mV. Explain why. [3 marks] 9 5 *05* Turn over ► IB/H/Jun21/7402/2 Do not write outside the box 0 3 Figure 1 shows an investigation into growth factors in plants. Figure 1 0 3 . 1 Use your knowledge of indoleacetic acid (IAA) to explain the growth curvature shown in Figure 1. [3 marks] Question 3 continues on the next page 6 *06* IB/H/Jun21/7402/2 Do not write outside the box A bioassay is a method to determine the concentration of a substance by its effect on living tissues. Figure 2 shows the practical procedure used in a growth curvature bioassay to determine the concentration of IAA in shoot tips. Figure 2 Figure 3 shows the calibration curve for this growth curvature bioassay. Figure 3 7 *07* Turn over ► IB/H/Jun21/7402/2 Do not write outside the box 0 3 . 2 Using the procedure in Figure 2 and the calibration curve in Figure 3, describe how you could compare the IAA concentration in shoot tips from two different plant species. In your answer you should refer to all the variables that should be controlled to produce a valid comparison. [5 marks] Question 3 continues on the next page 8 *08* IB/H/Jun21/7402/2 Do not write outside the box A scientist investigated the effect of a directional light stimulus on the distribution of IAA in shoot tips. The scientist set up three experiments as shown in Figure 4. All variables were controlled apart from exposure to light. Figure 4 She then used the growth curvature bioassay to compare the IAA concentrations in the agar blocks from: • experiment 1 • experiment 2 • experiment 3 section A • experiment 3 section B. Table 1 shows the scientist’s results. Table 1 Experiment Degree of curvature in Bioassay / degrees 1 17.69 2 17.61 3A 11.22 3B 6.50 9 *09* Turn over ► IB/H/Jun21/7402/2 Do not write outside the box 0 3 . 3 State two conclusions about IAA that you can make from the results shown in Table 1. [2 marks] 1 2 Turn over for the next question 10 10 *10* IB/H/Jun21/7402/2 Do not write outside the box 0 4 In fruit flies, males have the sex chromosomes XY and the females have XX. In fruit flies, a gene for eye colour is carried on the X chromosome. The allele for red eyes, R, is dominant to the allele for white eyes, r. 0 4 . 1 Male fruit flies are more likely than female fruit flies to have white eyes. Explain why. [2 marks] 0 4 . 2 A female fruit fly with white eyes was crossed with a male fruit fly with red eyes to produce a large number of offspring. Tick () one box next to the statement which correctly describes the phenotypes produced from this cross. [1 mark] All offspring red-eyed All females red-eyed, all males white-eyed All males red-eyed, all females white-eyed All males white-eyed, females red-eyed and females white-eyed 11 *11* Turn over ► IB/H/Jun21/7402/2 Do not write outside the box In fruit flies, the genes for body colour and for wing development are not on the sex chromosomes. The allele for grey body colour, G, is dominant to the allele for black body colour, g. The allele for long wings, L, is dominant to the allele for short wings, l. A geneticist carried out a cross between fruit flies with grey bodies and long wings (heterozygous for both genes) and fruit flies with black bodies and short wings. Table 2 shows the results of this cross. Table 2 Phenotype of offspring Number of offspring Grey body and long wings 223 Black body and short wings 218 0 4 . 3 Explain the results in Table 2. [3 marks] Question 4 continues on the next page 12 *12* IB/H/Jun21/7402/2 Do not write outside the box 0 4 . 4 The first generation of a population of fruit flies had 50 females. Calculate how many female fruit flies would be produced from this population in the fifth generation. You can assume: •each female produces 400 offspring each generation •half the offspring produced each generation are female •there is no immigration or emigration •no flies die before reproducing. Show your working. Give your answer in standard form. [3 marks] Answer 9 13 *13* Turn over ► IB/H/Jun21/7402/2 Do not write outside the box 0 5 . 1 Neonatal diabetes is a disease that affects newly born children. The disease is caused by a change in the amino acid sequence of insulin. This change prevents insulin binding to its receptor. Explain why this change prevents insulin binding to its receptor. [2 marks] Question 5 continues on the next page 14 *14* IB/H/Jun21/7402/2 Do not write outside the box Phosphoinositide 3-kinase (PI3K) is an enzyme in several metabolic processes. Figure 5 shows the role of PI3K in the control of blood glucose concentration. Figure 5 0 5 . 2 A decrease in the activity of PI3K can cause type II diabetes. Use Figure 5 to explain why. [3 marks] 15 *15* Turn over ► IB/H/Jun21/7402/2 Do not write outside the box 0 5 . 3 Using your knowledge of the kidney, explain why glucose is found in the urine of a person with untreated diabetes. [3 marks] Turn over for the next question 8 16 *16* IB/H/Jun21/7402/2 Do not write outside the box 0 6 Myelodysplastic syndromes (MDS) are a group of malignant cancers. In MDS, the bone marrow does not produce healthy blood cells. Haematopoietic stem cell transplantation (HSCT) is one treatment for MDS. In HSCT, the patient receives stem cells from the bone marrow of a person who does not have MDS. Before the treatment starts, the patient’s faulty bone marrow is destroyed. 0 6 . 1 For some patients, HSCT is an effective treatment for MDS. Explain how. [3 marks] 0 6 . 2 MDS can develop from epigenetic changes to tumour suppressor genes. In some patients, the drug AZA has reduced the effects of MDS. AZA is an inhibitor of DNA methyltransferases. These enzymes add methyl groups to cytosine bases. Suggest and explain how AZA can reduce the effects of MDS in some patients. [3 marks] 17 *17* Turn over ► IB/H/Jun21/7402/2 Do not write outside the box Scientists investigated the effectiveness of AZA in patients with MDS. A total of 360 patients were randomised in the ratio of 1:1 to receive AZA or conventional drugs (control). Figure 6 shows the scientists’ results. Figure 6 0 6 . 3 The control patients were treated with conventional drugs. Give two reasons why. [2 marks] 1 2 0 6 . 4 Use Figure 6 and the information provided to calculate the difference in the number of patients surviving at 10 months after treatment with AZA compared with conventional drugs. [2 marks] Answer 10 18 *18* IB/H/Jun21/7402/2 Do not write outside the box 0 7 Hepatitis B is a life-threatening liver infection caused by the hepatitis B virus (HBV). Figure 7 shows the structure of HBV. Figure 7 0 7 . 1 HBV infects a liver cell. The liver cell is 25 μm in diameter. Use Figure 7 to calculate how many times larger in diameter this cell is than HBV. You should use the lipid layer to measure the diameter of HBV. [2 marks] Answer times larger 19 *19* Turn over ► IB/H/Jun21/7402/2 Do not write outside the box Scientists investigated the effectiveness of two types of RNA interference (RNAi) molecules on reducing HBV replication. These molecules were: • short hairpin RNA (shRNA) • long hairpin RNA (IhRNA). The scientists infected mouse liver cells with HBV and transferred either shRNA or lhRNA into these cells. Then they determined the concentration of the attachment proteins, HBsAg, in these cells. The concentration of HBsAg is a measure of HBV replication. Figure 8 shows the scientists’ results. The error bars represent ±2 standard deviations from the mean, which includes over 95% of the data. Figure 8 0 7 . 2 One method of transferring RNAi molecules into cells involves combining these molecules with a lipid. Suggest why this increases uptake of RNAi molecules into cells. [1 mark] Question 7 continues on the next page 20 *20* IB/H/Jun21/7402/2 Do not write outside the box 0 7 . 3 Using all the information provided, evaluate the use of the two types of RNAi in treating hepatitis B in humans. Do not refer in your answer to how RNAi reduces HBV replication. [5 marks] 8 21 *21* Turn over ► IB/H/Jun21/7402/2 Do not write outside the box 0 8 . 1 Describe and explain how the polymerase chain reaction (PCR) is used to amplify a DNA fragment. [4 marks] Figure 9 shows the number of DNA molecules produced using a PCR. Figure 9 0 8 . 2 Explain the shape of the curve in Figure 9. [2 marks] 6 22 *22* IB/H/Jun21/7402/2 Do not write outside the box 0 9 A coral reef is an underwater ecosystem formed as a ridge of mainly calcium carbonate deposits. Algae are photosynthesising organisms. Some algae grow on coral reefs. Succession results in a wide variety of fish living on coral reefs. 0 9 . 1 Describe a method that could be used to determine the mean percentage cover of algae on a coral reef. Do not include information on the difficulties of using your method underwater. [3 marks] 0 9 . 2 Explain how succession results in a wide variety of fish living on coral reefs. Do not describe the process of succession in your answer. [2 marks] 23 *23* Turn over ► IB/H/Jun21/7402/2 Do not write outside the box Ecologists investigated the effect of two fish species, the redband parrotfish and the ocean surgeonfish, on algal growth on an artificial reef. They made this artificial reef by submerging many large concrete blocks at a depth of 16–18 metres off the coast of Florida. They attached four large wire cages, A, B, C and D, to each block and populated the cages as shown. A – No fish B – Two redband parrotfish C – Two ocean surgeonfish D – One redband parrotfish and one ocean surgeonfish After 34 weeks, the ecologists measured the mean percentage cover of all algae within each set of wire cages. The ecologists used a statistical test to find out whether the mean for each set of cages was significantly lower than the mean for set A. Table 3 shows the probability (P) values that the ecologists obtained using this statistical test. Table 3 Set of cages P value B =0.841 C <0.001 D =0.634 0 9 . 3 Using all the information, evaluate the effect of the two fish species on algal growth on coral reefs. [5 marks] 10 24 *24* IB/H/Jun21/7402/2 Do not write outside the box 1 0 Read the following passage. Lake Malawi in East Africa has more species of fish than any other lake in the world. Many of these species have evolved from a common ancestor. Lake Malawi is one of the largest lakes in the world and was formed several million years ago. Since then, the water level has fluctuated greatly. As a result, what is now a large lake was at one time many smaller, separate lakes. The country of Malawi has a total area of 118 000 km2. The actual land area is only 94 080 km2, because approximately one-fifth of the country is Lake Malawi. In December 1990, forests covered 41.4% of the actual land area of Malawi. In December 2016, forests covered 26.4% of the actual land area of Malawi. Deforestation and farming along the shores of Lake Malawi have caused increased soil erosion and loss of nutrients into the lake. This has resulted in a decrease in some fish populations. The mark-release-recapture method can be used to estimate the size of a fish population. However, this method can produce unreliable results in very large lakes. Use the information in the passage and your own knowledge to answer the following questions. 5 10 15 1 0 . 1 Lake Malawi in East Africa has more species of fish than any other lake in the world (line 1). Suggest and explain how this speciation may have occurred. [4 marks] 25 *25* Turn over ► IB/H/Jun21/7402/2 Do not write outside the box 1 0 . 2 The percentage of forest cover in Malawi decreased between December 1990 and December 2016 (lines 9–10). Calculate the mean loss of forest cover in km2 per week during this time period. [2 marks] Answer km2 per week 1 0 . 3 Loss of nutrients into Lake Malawi has resulted in a decrease in some fish populations (lines 12–13). Explain why. [4 marks] 26 *26* IB/H/Jun21/7402/2 Do not write outside the box 1 0 . 4 The mark-release-recapture method can be used to estimate the size of a fish population (lines 13–14). Explain how. [4 marks] 1 0 . 5 Suggest why the mark-release-recapture method can produce unreliable results in very large lakes (lines 14–15). [1 mark] END OF QUESTIONS 15 27 *27* IB/H/Jun21/7402/2 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 28 *28* IB/H/Jun21/7402/2 Do not write outside the box Question number Additional page, if required. Write the question numbers in the left-hand margin. 29 *29* IB/H/Jun21/7402/2 Do not write outside the box Question number Additional page, if required. Write the question numbers in the left-hand margin. 30 *30* IB/H/Jun21/7402/2 Do not write outside the box Question number Additional page, if required. Write the question numbers in the left-hand margin. 31 *31* IB/H/Jun21/7402/2 Do not write outside the box Question number Additional page, if required. Write the question numbers in the left-hand margin. 32 *32* IB/H/Jun21/7402/2 Do not write outside the There ar box e no questions printed on this page DO NOT WRITE ON THIS PAGE ANSWER IN THE SPACES PROVIDED [Show Less]
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AQA A Level Biology 7402/2 Paper 2 Mark Scheme June 2021 Version 1.0 Final
AQA A-Level Biology 7402/2 Paper 2 Mark Scheme June 2021 Version 1.0 Final / AQA A-Level Biology 7402-2 Paper 2 Mark Scheme June 2021 Version 1.0 Final / A... [Show More] QA A-Level Biology Paper 2 Mark Scheme June 2021 Version 1.0 Final A-level BIOLOGY 7402/2 Paper 2 Mark scheme June 2021 Version: 1.0 Final AQA MARK SCHEME – A-LEVEL BIOLOGY – 7402/2 – 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 – 7402/2 – JUNE 2021 3 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 – 7402/2 – JUNE 2021 4 Question Marking Guidance Mark Comments 01.1 1.Phosphorylation of glucose using ATP; 2.Oxidation of triose phosphate to pyruvate; 3.Net gain of ATP; 4.NAD reduced; 4 max Accept all mark points in diagrams. 2.Accept removal ofhydrogen from triosephosphate foroxidation. 3.Accept anydescription thatindicates a net gaine.g., 4 produced, 2used. 4.AcceptNADH/NADH2/NADH+H+ produced. Question Marking Guidance Mark Comments 01.2 1.Less/no reduced NAD/coenzymes OR Fewer/no hydrogens/electrons removed (andpassed to electron transfer chain); 2.Oxygen is the final/terminal (electron) acceptor; 2 1.Accept less/no FADreduced. MARK SCHEME – A-LEVEL BIOLOGY – 7402/2 – JUNE 2021 5 Question Marking Guidance Mark Comments 02.1 1. Higher concentration of potassium ions inside and higher concentration of sodium ions outside (the neurone) OR potassium ions diffuse out OR sodium ions diffuse in; 2. (Membrane) more permeable to potassium ions (leaving than sodium ions entering) OR (Membrane) less permeable to sodium ions (entering than potassium ions leaving); 3. Sodium ions (actively) transported out and potassium ions in; 3 1. Accept ‘more’ for ‘higher concentration’. 1. Accept ‘sodium ions can’t diffuse in (due to alternative explanation). 2. Accept for ‘less permeable to sodium ions’ is ‘impermeable to sodium ions’ or ‘sodium gates/channels are closed’ (alternative explanation). 1, 2 and 3 reference to ions or Na+ and K+ is required. If mentioned once allow for all mark points. 1, 2 and 3. If an answer provides two or three of these mark points without any reference to ions – award one maximum mark. 3. Accept 3 Na+ out and 2 K+ in but reject if numbers used are incorrect. Question Marking Guidance Mark Comments 02.2 1. Myelination provides (electrical) insulation; 2. (In myelinated) saltatory (conduction) OR (In myelinated) depolarisation at nodes (of Ranvier); 3. In non-myelinated depolarisation occurs along whole/length (of axon); 3 1. Reject thermal insulation. 1. Accept description of (electrical) insulation. 3. Accept action potentials for depolarisation. 2 and 3. ‘Messages’ or ‘signals’ disqualifies first of these marks credited. MARK SCHEME – A-LEVEL BIOLOGY – 7402/2 – JUNE 2021 6 Question Marking Guidance Mark Comments 02.3 1. No/less ATP produced; 2. No/less active transport OR Sodium/potassium pump inhibited; 3. Electrochemical gradient not maintained OR (Facilitated) diffusion of ions causes change to 0 mV OR (Results in) same concentration of (sodium and potassium) ions (either side of membrane) OR No net movement of (sodium and potassium) ions; 3 2. Accept Na+ not/fewer moved out and K+ not/fewer moved in. 3. Accept reaches electrical equilibrium/balance. 3. Accept concentration gradient of sodium and potassium ions not maintained. MARK SCHEME – A-LEVEL BIOLOGY – 7402/2 – JUNE 2021 7 Question Marking Guidance Mark Comments 03.1 1. Tip produces IAA; 2. IAA diffuses (into shoot); 3. (More) elongation of cells on one side (than other); 3 1. Accept source/release for produces but ignore contains/stores IAA. 1 and 2. Accept auxin for IAA. 2. Accept IAA diffuses down. 3. Accept (more) elongation of cells on left side. 3. Reject any reference to shaded/dark side or away from light. Question Marking Guidance Mark Comments 03.2 1. Size of shoot/tip; 2. Number of shoot tips; 3. Size/type of agar (block); 4. (Shoots) at same stage of growth/development; 5. Time (period) tips kept on agar OR Time (period) agar/block kept on (cut shoot) OR Time (period shoots) kept in dark; 6. Temperature; 7. (Repeat several times and) calculate a mean; 8. Compare/read degree of curvature (on calibration curve) to determine (IAA) concentration OR Higher the degree of curvature the higher the IAA concentration; 5 max Mark points 1 to 6 = max 3. 1 to 6. Ignore pH, species, carbon dioxide, humidity, nutrients, water and light. 3. Accept ‘amount of agar’. 4. Accept (Shoots/plants) are same age. MARK SCHEME – A-LEVEL BIOLOGY – 7402/2 – JUNE 2021 8 Question Marking Guidance Mark Comments 03.3 1. (IAA) is not broken down by light OR (IAA) is produced in the dark OR Light/dark does not affect (IAA) production; 2. (IAA) moves away from light OR (IAA) moves to shaded side; 2 2. IAA accumulates on shaded side is not enough on its own, idea of movement is required. Question Marking Guidance Mark Comments 04.1 1. Males have one allele; 2. Females need two recessive alleles OR Females must be homozygous recessive OR Females could have dominant and recessive alleles OR Females could be heterozygous/carriers; 2 1 Accept males only need one allele. 1 and 2. Ignore references to X and Y chromosomes. 1 and 2. Accept r as recessive allele and R as dominant allele. If no reference to allele, accept for one mark male needs one recessive gene whereas females need two recessive genes. Question Marking Guidance Mark Comments 04.2 1. Box 2. All females red-eyed, all males white-eyed. 1 Reject if more than one box with tick. Ignore crossed-out ticks. Question Marking Guidance Mark Comments 04.3 1. The (two) genes are linked OR Autosomal linkage; 3 1. Accept that the genes are on the same chromosome. MARK SCHEME – A-LEVEL BIOLOGY – 7402/2 – JUNE 2021 9 2. No crossing over (occurs) OR (Linked) genes are close together; 3. No Gl and no gL (gametes produced) OR No Ggll and no ggLl (offspring produced) OR Only GL and gl (gametes produced); 1. Accept ‘Alleles are linked’ (accept symbols for alleles) but reject if context suggests alleles of the ‘same gene’. 2. Accept crossing over less likely to occur. 1 2, and 3. Ignore reference to independent assortment. Question Marking Guidance Mark Comments 04.4 1. Correct answer of 8 × 1010 = 3 marks;;; 2. Correct answer not in standard form = 2 marks OR 1.6 × 1013 = 2 mark OR 1.6 × 1011 = 2 mark OR 6.4 × 1011 = 2 mark OR Shows 8 × 1010 in the working = 2 marks;; 3. 1.28 × 1012 = 1mark OR 3.2 × 1011 = 1 mark OR 8 × 1011 = 1 mark OR 8 × 109 = 1 mark OR Shows 1.6 × 1011 in the working = 1 mark OR Shows 2004 in the working = 1 mark; 3 If no other mark is credited accept for one mark working which shows multiplication by 200 for 4 generations. This could be shown in a variety of ways e.g. multiplied by 400 divided by 2 for 4 generations. MARK SCHEME – A-LEVEL BIOLOGY – 7402/2 – JUNE 2021 10 Question Marking Guidance Mark Comments 05.1 1. Changes tertiary structure; 2. No longer complementary (to receptor); 2 1. Reject change in tertiary structure of receptor. 2. Reject ‘active site’ or reference to enzyme or substrate. Question Marking Guidance Mark Comments 05.2 1. Less/no AKT activated; 2. Fewer/no vesicles move to membrane OR Fewer/no (channel) proteins in membrane; 3. Less/no glucose diffuses into cell (so high blood glucose); 3 2. Accept ‘fuse with membrane’. Question Marking Guidance Mark Comments 05.3 1. High concentration of glucose in blood/filtrate; 2. Not all the glucose is (re)absorbed at the proximal convoluted tubule; 3. Carrier/co-transport proteins are working at maximum rate OR Carrier/co-transport proteins/ are saturated; 3 1. Accept tubule for filtrate. 2. Reject no glucose is (re)absorbed. 3. Accept all carrier/co-transport proteins are ‘in use’ but reject all carriers are ‘used up’. 3. Accept symport for carrier protein. 3. Accept not enough carrier proteins to absorb all the glucose. MARK SCHEME – A-LEVEL BIOLOGY – 7402/2 – JUNE 2021 11 Question Marking Guidance Mark Comments 06.1 1. Produce healthy (blood) cells; 2. No MDS/faulty/cancerous (blood) cells; 3. Stem cells divide/replicate; 3 1 and 2. Produce only healthy/normal (blood) cells = two marks. 1. Accept produce ‘normal’ /non-MDS cells. 2. Accept no (cancerous) tumour. 1 and 3. Ignore reference to totipotent/pluripotent/ multipotent/unipotent 3. Accept ‘clone’ for divide. Question Marking Guidance Mark Comments 06.2 1. (AZA) reduces methylation (of DNA/cytosine/gene); 2. (Tumour suppressor) gene is transcribed/expressed; 3. Prevents rapid/uncontrollable cell division OR Cell division can be controlled/stopped/slowed; 3 1. Reject any reference to mutation. 2. Accept mRNA produced for transcription/transcribed. 2. Ignore gene is ‘switched on’ or activated but allow protein is formed. 3. Ignore growth. Question Marking Guidance Mark Comments 06.3 1. Effect of AZA can be compared; 2. Unethical not to treat (control group); 2 1. Comparison on its own is not enough for a mark. Question Marking Guidance Mark Comments 06.4 1. Correct answer of 29/28.8 = 2 marks;; 2. Working shows 0.74 and 0.58 = 1 mark OR 58/57.6 = 1 mark OR 28 = 1 mark; 2 MARK SCHEME – A-LEVEL BIOLOGY – 7402/2 – JUNE 2021 12 Question Marking Guidance Mark Comments 07.1 1. Correct answer of 625 = 2 marks;; 2. Shows 625 but decimal point incorrect = 1 mark OR Working shows 40 = 1 mark OR 1600/1.6 = 1 mark OR 667/666.6 = 1 mark; 2 Question Marking Guidance Mark Comments 07.2 (Cell/membrane has a) phospholipid bilayer OR No channel/carrier protein (for uptake) OR No need for channel/carrier protein (for uptake); 1 Question Marking Guidance Mark Comments 07.3 1. Both are more effective than the control; 2. Differences in the means not (likely to be) due to chance OR Significant difference (in effectiveness between both types); 3. (As) SDs do not overlap; 4. HBsAg (reduced), not zero OR Replication (reduced), not zero; 5. Not (investigated in) humans OR (Investigated in) mice; 6. shRNA (more effective as) 7.5% (of control) compared with 50% for lhRNA; 7. No indication of sample size/number; 5 max Mark points 4 to 10 = 4 max. 1. Accept both (results) are below the control. 2. Reject ‘results are significant’. 2. Accept significantly higher or significantly lower in correct context. 3. Accept error bars do not overlap. 6. Accept 42.5% difference. 6. Accept (mean) concentration for %. MARK SCHEME – A-LEVEL BIOLOGY – 7402/2 – JUNE 2021 13 8. Long term effects not known OR Side effects not known; 9. No statistical test to determine significance; 10. (Investigated) in vitro OR Not (investigated) in vivo; 8. Accept ‘could be toxic’ for side effects not known. 10. Accept not done inside an organism or not done in liver (organ) but ‘only tested in liver cells’ is insufficient unless qualified. Ignore only ‘one study’ or ‘no repeats’. Question Marking Guidance Mark Comments 08.1 1. (Requires DNA fragment) DNA polymerase, (DNA) nucleotides and primers; 2. Heat to 95 °C to break hydrogen bonds (and separate strands); 3. Reduce temperature so primers bind to DNA/strands; 4. Increase temperature, DNA polymerase joins nucleotides (and repeat method); 4 1 and 4. Accept Taq polymerase for DNA polymerase. 2. Accept temperature in range 90 to 95 °C. 3. Accept temperature in range 40 to 65 °C. 4. Accept temperature in range 70 to 75 °C. Question Marking Guidance Mark Comments 08.2 1. (Initially) number (of molecules) doubling is low OR Doubles each cycle to produce exponential increase; 2. Plateaus as no more nucleotides/primers; 2 1. First alternative relates to idea of low numbers i.e., 2, 4, 8, 16, 32 etc. 2. Accept ‘levels out’ or ‘flattens’ for plateaus. 2. Accept enzyme/polymerase (eventually) denatures. MARK SCHEME – A-LEVEL BIOLOGY – 7402/2 – JUNE 2021 14 Question Marking Guidance Mark Comments 09.1 1. Method of randomly determining position (of quadrats) e.g. random numbers table/generator; 2. Large number/sample of quadrats; 3. Divide total percentage by number of quadrats/samples/readings; 3 1. Ignore line/belt transect. 2. Accept many/multiple/lots but ignore several. 2. Ignore point quadrat. 2. Accept squares/frames (of a grid) for quadrats. 2. If a specified number is given, it must be 10 or more. Question Marking Guidance Mark Comments 09.2 1. Increase in variety/diversity of species/plants/animals; OR Increase in number of species/populations; 2. Provides more/different habitats/niches OR Provides greater variety/types of food; 2 1. Accept increase in biodiversity or species richness. 2. Ignore shelter/homes/environments. 2. Ignore ‘more food’ but accept ‘more food sources’. 2. Accept ‘less hostile’ (environment). Question Marking Guidance Mark Comments 09.3 1. Significant (difference/decrease) with C (compared with A); 2. No significant (difference/decrease) with B and D (compared with A); 3. Reference to less than 5%/0.05 probability that difference is (less likely) due to chance OR Reference to more than 95%/0.95 probability that difference is not due to chance; 4. Species of algae not known OR Species of algae may differ (on other reefs); 5. Only done off (coast of) Florida OR Not done on other reefs; 5 max Mark points 4 to 9 = 4 max. 1 and 2. Accept names of fish species present as alternatives to sets B, C and D. 1 and 2. Award both marks if answer states only C is significantly (different/lower). 1 and 2. ‘Results are significant/not significant’ disqualifies first of these marks credited. 3. Accept equal to specified probabilities. MARK SCHEME – A-LEVEL BIOLOGY – 7402/2 – JUNE 2021 15 6.Only done at 16 to 18 metres OR Not done (on reefs) at other depths; 7.Only 34 weeks; 8.Concrete/artificial reef could affectresults/growth OR Natural reef results/growth may differ; 9.Cage may allow other fish/animals to enter; 8.Accept any reference tocomposition of reef beingdifferent (from natural). Question Marking Guidance Mark Comments 10.1 1.Variation/differences due to mutation/s; 2.(Reference to) allopatric (speciation); 3.Smaller/different lakes have differentenvironmental conditions OR Smaller/different lakes have differentselection pressures; 4.Reproductive separation/isolation OR No gene flow OR Gene pools remain separate; 5.Different alleles passed on/selected OR Change in frequency of allele/s; 6.Eventually different species/populationscannot breed to produce fertile offspring; 4 max 2.Ignore sympatricspeciation. 3.Accept differentpopulations for differentlakes. Question Marking Guidance Mark Comments 10.2 1.Correct answer of 10/10.4 = 2 marks;; 2.Working shows 14,112 = 1 mark OR 13.09/13.1 = 1 mark; 2 1.Ignore any numbersafter 10.4 MARK SCHEME – A-LEVEL BIOLOGY – 7402/2 – JUNE 2021 16 Question Marking Guidance Mark Comments 10.3 1.(Growth/increase of) algae/surfaceplants/algal bloom blocks light; 2.Reduced/no photosynthesis so(submerged) plants die; 3.Saprobiotic (microorganisms) aerobicallyrespire OR Saprobiotic (microorganisms) use oxygen inrespiration; 4.Less oxygen for fish to respire; 4 3.Accept:Saprobiont/saprophyte/saprotroph 3.Neutral: decomposer Question Marking Guidance Mark Comments 10.4 1.Capture/collect sample, mark and release; 2.Ensure marking is not harmful (to fish) OR Ensure marking does not affect survival (offish); 3.Allow (time for) fish to (randomly) distributebefore collecting a second sample; 4.(Population =) number in first sample ×number in second sample divided bynumber of marked fish in secondsample/number recaptured; 4 2.Accept examples e.g.,marking should not betoxic. Question Marking Guidance Mark Comments 10.5 1.Less chance of recapturing fish OR Unlikely fish distribute randomly/evenly; 1 Accept ‘harder to capture marked fish’ (recaptured fish) but ignore ‘harder to capture fish’. Accept that fish may remain in one area. Accept fish may congregate. [Show Less]
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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 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]
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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 / 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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