2024_AQA: AS BIOLOGY Paper 2 (Merged Question paper and marking scheme) Thursday 23 May 2024

2024_AQA: AS BIOLOGY Paper 2 (Merged Question paper and marking scheme) Thursday 23 May 2024 Please write clearly in block capitals. Centre number Surname Forename(s) Candidate signat ure Candidate number I declare this is my own work. AS BIOLOGY Paper 2 Thursday 23 May 2024 Materials For this paper you must have:  a ruler with millimetre measurements  a scientific calculator. Instructions Morning  Use black ink or black ball-point pen.  Fill in the boxes at the top of this page.  Answer all questions. Time allowed: 1 hour 30 minutes For Examiner’s Use Question Mark 1 2 3 4  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 75. 5 6 7 8 9 TOTAL AS Biology Paper 2: Key Areas to Revise This paper covers more specialized topics in biology, focusing on practical skills, application of knowledge, and understanding of biological principles in greater depth. The key areas to revise include: 1. Transport in Plants and Animals: Understand how plants and animals transport materials. Revise the structure and function of xylem and phloem in plants, including transpiration and translocation. For animals, focus on the circulatory system, blood vessels, and the heart’s role in pumping blood. Understand the factors affecting blood pressure and the transport of oxygen and carbon dioxide. 2. The Immune System: Review the structure and function of the immune system. Know the different types of white blood cells (e.g., phagocytes, lymphocytes), how they respond to infection, and the role of antibodies and antigens. Be familiar with the concept of vaccines, immunity, and how the body defends itself against pathogens. 3. Muscle Contraction: Study the mechanism of muscle contraction, focusing on the sliding filament theory, and the role of actin and myosin filaments in muscle fibers. Understand the processes of excitation-contraction coupling and how ATP is involved. 4. Photosynthesis and Respiration: Be able to explain the processes of photosynthesis and respiration, including the light-dependent and light-independent reactions (Calvin cycle) of photosynthesis, as well as aerobic and anaerobic respiration. Understand how energy is transferred and stored in living organisms. 5. Genetic Diversity and Adaptation: Revise the concepts of genetic variation, mutations, and natural selection. Understand how these contribute to adaptation in populations and the development of new species. Be familiar with the Hardy-Weinberg principle and how it applies to genetic equilibrium. 6. Practical Skills and Data Analysis: Be prepared for questions that assess your practical understanding of biology. This may involve interpreting experimental data, identifying variables, and evaluating methods. Understand common laboratory techniques and equipment used in biology experiments. Focusing on these areas will help you strengthen your knowledge and analytical skills for AS Biology Paper 2. IB/M/Jun24/G4001/E7 7401/2 2 Do not write outside the box Answer all questions in the spaces provided. IB/M/Jun24/7401/2 0 1 . 1 Water has a high heat capacity and a large latent heat of vaporisation. Describe the importance of each of these properties to living organisms. [2 marks] High heat capacity Large latent heat of vaporisation 0 1 . 2 Figure 1 shows that water loss from a porous pot can cause the upward movement of water. Figure 1 3 Do not write outside the box Biologists have concluded that the experiment in Figure 1 supports the IB/M/Jun24/7401/2 0 1 . 3 An air bubble was introduced into the glass tubing in Figure 1. The air bubble moved a distance (d) of 1.5 cm in 120 minutes. The radius of the lumen (hole) of the glass tubing was 0.6 cm Use this information and the formula πr2d to calculate the rate of water movement in the glass tubing in cm3 hour–1. Use π = 3.14 in your calculation. [1 mark] Answer cm3 hour–1 6 cohesion–tension theory of water transport in the xylem. Explain how this conclusion is supported by the experiment in Figure 1. [3 marks] Turn over ► 4 Do not write outside the box IB/M/Jun24/7401/2 0 2 . 1 Figure 2 shows an image of a red blood cell at a magnification of × 5500 Figure 2 Calculate the actual diameter in μm of the red blood cell between points P and Q. Show your working. [2 marks] Answer μm 5 Do not write outside the box IB/M/Jun24/7401/2 0 2 . 2 A haemocytometer is a special microscope slide that can be used to determine the mean number of red blood cells in 0.004 mm3 of blood.  A researcher prepared a 10–3 dilution of a sample of blood from an adult.  Using a haemocytometer, the researcher determined that the mean number of red blood cells in 0.004 mm3 of the diluted blood sample was 21 The volume of blood in the body of the adult was 4.8 dm3 Calculate the total number of red blood cells in the body of this adult. Show your working. [2 marks] Answer 0 2 . 3 The solution used to dilute the blood had to have the same water potential as the blood. Explain why. [2 marks] Question 2 continues on the next page Turn over ► 6 Do not write outside the IB/M/Jun24/7401/2 4 0 2 . 5 Using Table 1, give the blood groups of people who could accept a donation of blood group O without causing agglutination of the donated red blood cells. [1 mark] Blood groups 9 . There are four main blood groups in the human ABO blood group system. Table 1 shows the basis on which each of these blood groups is classified. Table 1 ABO blood group Antigens present on red blood cells Antibodies present in blood plasma A A Anti-B B B Anti-A AB A and B No anti-A and no anti-B O No A and no B Anti-A and anti-B A transfusion of blood from a blood group A donor to a blood group B recipient would cause agglutination of the donated red blood cells. box Use information in Table 1 to explain why. [2 marks] 2 0 7 Do not write outside the box IB/M/Jun24/7401/2 0 3 . 1 Describe how you would use cell fractionation and ultracentrifugation to obtain a sample of mitochondria from muscle tissue. [4 marks] Question 3 continues on the next page Turn over ► 8 Do not write outside the box IB/M/Jun24/7401/2 0 3 . 2 Mitochondrial respiratory capacity is a measure of maximum ATP production in a mitochondrion. Scientists investigated the effect of a resistance exercise training (RET) programme on the respiratory capacity of mitochondria in skeletal muscle tissue. RET develops muscle strength. The scientists:  took samples of muscle tissue from 11 young males before and after a 12-week RET programme  recorded the respiratory capacity of the mitochondria in the samples of muscle tissue. Figure 3 shows some of the scientists’ results. The error bars represent ± 2 standard deviations from the mean, which includes over 95% of the data. Figure 3 9 Do not write outside the box Using all the information, evaluate whether 12 weeks of RET would improve athletic IB/M/Jun24/7401/2 performance in the general population. [5 marks] Turn over for the next question Turn over ► 9 10 Do not write outside the box IB/M/Jun24/7401/2 [1 mark] 0 4 . 2 The genome of an adenovirus is a single, linear molecule of double-stranded DNA. Adenoviruses use eukaryotic host cells to transcribe their genes in protein synthesis. The process of transcription of adenovirus genes is similar to the process of transcription of genes in eukaryotes. Scientists investigated the process of transcription of this viral DNA. Figure 4 shows one of the experiments carried out by these scientists. Figure 4 0 4 . 1 Where does transcription occur in a eukaryotic cell? 11 Do not write outside the box Looking at the results of the experiment in Figure 4, the scientists concluded that IB/M/Jun24/7401/2 splicing had taken place. Use Figure 4 to describe and explain why the scientists’ conclusion was justified. [3 marks] 0 4 . 3 Describe and explain how the results of the experiment in Figure 4 would differ if the scientists had used prokaryotic DNA. [2 marks] 0 4 . 4 Errors in the precise location of splicing in the DNA molecule can lead to mutations. Explain why. [1 mark] Turn over ► 7 12 Do not write outside the box IB/M/Jun24/7401/2 0 5 . 1 Orchids form a large family of flowering plants. Scientists analysed genes coding for ribosomal RNA in orchids. Explain how this analysis allowed the scientists to determine the phylogenetic relationships between species of orchids. [2 marks] 0 5 . 2 Suggest one reason why analysing genes coding for ribosomal RNA allows phylogenetic relationships to be studied between all cellular organisms. [1 mark] 0 5 . 3 Most orchid species are found in tropical rainforests. Two common features found in many of these species are:  they grow up trees to reach the upper branches  they only open their stomata at night. Suggest how each of these two features benefits these orchids. [2 marks] They can reach the upper branches of trees They only open their stomata at night 13 Do not write outside the box IB/M/Jun24/7401/2 0 5 . 4 A student used an optical microscope to observe a piece of tissue from the lower surface of an orchid leaf. The piece of leaf tissue observed was very thin. Explain why this was important. [2 marks] 0 5 . 5 The student produced a biological drawing of the leaf tissue they viewed through an optical microscope. Give three ways the student could ensure they produce a correct biological drawing of the leaf tissue. Assume the student uses a sharp pencil. [3 marks] 1 2 3 Turn over ► 10 14 Do not write outside the IB/M/Jun24/7401/2 1 . Disinfectants are used to kill microorganisms on non-living surfaces. A student investigated the effect of different concentrations of disinfectant X on the growth of Bacillus subtilis. The student:  added 5 cm3 of a different concentration of disinfectant X to 5 different test tubes  added 5 cm3 of distilled water to another test tube  added 2 cm3 of a culture of B. subtilis to all 6 test tubes  incubated the test tubes at 25 °C for 24 hours  used a colorimeter to record the percentage of light absorbed by the contents of each tube. Table 2 shows the student’s results. Table 2 Percentage concentration of disinfectant X 100 Percentage light absorbance 10 The student prepared the different concentrations of disinfectant X. Describe how the student made 5 cm3 of the 60% concentration using distilled water and undiluted disinfectant. box [1 mark] 6 0 15 Do not write outside the box IB/M/Jun24/7401/2 0 6 . 2 The student used a sterile pipette with 0.1 cm3 graduations to transfer 2 cm3 of B. subtilis into each test tube. What is the uncertainty in measuring 2 cm3 with this pipette? Calculate the percentage uncertainty of this 2 cm3 measurement. [2 marks] Uncertainty cm3 Percentage uncertainty 0 6 . 3 Use Table 2 to evaluate the effectiveness of disinfectant X at killing microorganisms on non-living surfaces. [4 marks] Turn over ► 7 16 Do not write outside the box IB/M/Jun24/7401/2 0 7 . 1 Edwards’ syndrome is a condition resulting from an extra chromosome 18. A chromosome mutation in the second meiotic division is the most frequent cause of Edwards’ syndrome. Explain how a chromosome mutation in the second meiotic division could result in an extra chromosome 18. In your answer, name the type of chromosome mutation which would result in the extra chromosome. [2 marks] 0 7 . 2 Complete trisomy 18 is the most common type of Edwards’ syndrome. This occurs when all the cells of the body have an extra chromosome 18. Explain why all the cells of the body have an extra chromosome 18. [2 marks] 17 Do not write outside the box IB/M/Jun24/7401/2 0 7 . 3 Mosaic trisomy 18 is another type of Edwards’ syndrome. This occurs due to a chromosome mutation after fertilisation. In mosaic trisomy, the body has cells with an extra chromosome 18 and cells with the correct number of chromosomes. Explain how cells with different numbers of chromosomes are produced in mosaic trisomy. [1 mark] 0 7 . 4 The age of the female parent is a factor linked to the risk of a child having Edwards’ syndrome. Which statistical test should be used to test whether this link is statistically significant? Tick () one box. [1 mark] Correlation coefficient Chi-squared Student’s t-test Question 7 continues on the next page Turn over ► 18 Do not write outside the box IB/M/Jun24/7401/2 8 0 7 . 5 A ventricular septal defect (VSD) is a common feature of Edwards’ syndrome. A VSD is a hole in the wall between the two ventricles of the heart. A VSD can cause higher blood pressure in the lungs. Explain how a VSD can cause higher blood pressure in the lungs. [2 marks] 19 Do not write outside the box IB/M/Jun24/7401/2 0 8 Lyme disease is most frequently caused by the bacterium Borrelia burgdorferi. Lyme disease can be difficult to diagnose. Figure 5 shows an ELISA test that is used to find out if a person has antibodies to B. burgdorferi. Figure 5 A false positive in this test is a result which incorrectly indicates that antibodies to B. burgdorferi are present. 0 8 . 1 Failure to thoroughly wash the well in Step 4 can result in a false positive. Explain why. [2 marks] Turn over ► 20 Do not write outside the box IB/M/Jun24/7401/2 0 8 . 2 A false positive can be produced if a person has been infected by another bacterium that causes a disease called syphilis. Suggest why. [1 mark] 0 8 . 3 A false negative in this test is often produced if a person is tested within 2 weeks of being infected with B. burgdorferi. Explain why. [2 marks] 0 8 . 4 Sometimes, symptoms of Lyme disease can persist for 6 months following antibiotic treatment. This condition is known as Post-Treatment Lyme Disease Syndrome (PTLDS). Scientists investigated the symptoms experienced by a large number of PTLDS patients and a control group. During a 2-week period, they asked all the participants:  if they had experienced symptoms of PTLDS  to record the intensity of these symptoms. The scientists used a statistical test to determine if there was a difference in the intensity of symptoms of PTLDS between these two groups. 21 Do not write outside the box Table 3 shows some of the scientists’ results, including the probability (P) values IB/M/Jun24/7401/2 9 obtained using the statistical test. Table 3 Symptom Percentage of PTLDS group experiencing symptom Percentage of control group experiencing symptom P value for difference in the intensity of symptoms Fatigue 100 57 <0.001 Joint pain 96 32 <0.001 Depression 40 4 <0.005 Fever 35 3 <0.005 Muscle pain 86 62 <0.001 The scientists concluded that more PTLDS patients than the control group experienced:  symptoms  greater intensity of symptoms. Evaluate the scientists’ conclusions. [4 marks] Turn over ► 22 Do not write outside the box IB/M/Jun24/7401/2 0 9 . 1 Describe the role of haemoglobin (Hb) in the loading, transport and unloading of oxygen. [5 marks] 23 Do not write outside the box IB/M/Jun24/7401/2 10 [5 marks] END OF QUESTIONS 0 9 . 2 Describe how a polypeptide is produced by translation of mRNA. 24 Do not write outside the box There are no questions printed on this page IB/M/Jun24/7401/2 DO NOT WRITE ON THIS PAGE ANSWER IN THE SPACES PROVIDED 25 Question number Additional page, if required. Write the question numbers in the left-hand margin. IB/M/Jun24/7401/2 Do not write outside the box 26 Question number Additional page, if required. Write the question numbers in the left-hand margin. IB/M/Jun24/7401/2 Do not write outside the box 27 Question number Additional page, if required. Write the question numbers in the left-hand margin. IB/M/Jun24/7401/2 Do not write outside the box 28 Do not write outside the box There are no questions printed on this page IB/M/Jun24/7401/2 DO NOT WRITE ON THIS PAGE ANSWER IN THE SPACES PROVIDED Copyright information For confidentiality purposes, all acknowledgements of third-party copyright material are published in a separate booklet. This booklet is published after each live examination series and is available for free download from Permission to reproduce all copyright material has been applied for. In some cases, efforts to contact copyright-holders may have been unsuccessful and AQA will be happy to rectify any omissions of acknowledgements. If you have any queries please contact the Copyright Team. Copyright © 2024 AQA and its licensors. All rights reserved. AS BIOLOGY 7401/2 Paper 2 Mark scheme June 2024 Version: 1.0 Final MARK SCHEME – AS BIOLOGY – 7401/2 – JUNE 2024 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 Copyright information AQA retains the copyright on all its publications. However, registered schools/colleges for AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to schools/colleges to photocopy any material that is acknowledged to a third party even for internal use within the centre. Copyright © 2024 AQA and its licensors. All rights reserved. 2 MARK SCHEME – AS BIOLOGY – 7401/2 – JUNE 2024 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 is acceptable or not worthy of credit or, in discursive answers, to give an overview of the area in 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 2.2 2.3 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. A bold and is used to indicate that both parts of the answer are required to award the mark. 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. 3 MARK SCHEME – AS BIOLOGY – 7401/2 – JUNE 2024 3.2 3.3 3.4 3.5 3.6 3.7 4 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. 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. 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. Phonetic spelling The phonetic spelling of correct scientific terminology should be credited unless there is a possible confusion with another technical term. 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. 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 – AS BIOLOGY – 7401/2 – JUNE 2024 Question Marking Guidance Mark Comments 01.1 1. Buffers changes in temperature; 2. (Provides a) cooling effect (via evaporation); 2 (2 x AO1) 1. Accept descriptions of buffering changes in temperature, eg resists temperature changes Question Marking Guidance Mark Comments 01.2 1. Evaporation/transpiration (from the porous pot); 2. Tension created moves water (upwards); 3. Cohesion maintains the column of water OR Cohesion is due to hydrogen bonds between water (molecules); 3 (1 x AO2, 2 x AO3) 2. Accept ‘negative pressure’ for tension or ‘water pulled’ or ‘suction’. Ignore adhesion but reject mark point 3 if adhesion used to describe attraction between water molecules. Question Marking Guidance