Aqa a level 2020 biology paper 1 question paper 2020, Exams of Biology

Download Aqa a level 2020 biology paper 1 question paper 2020 and more Exams Biology in PDF only on Docsity! AQA~ A-LEVEL 2020 BIOLOGY PAPER 1 QUESTION PAPER(2020) AQA Please write clearly in block capitals. Centre number Candidate number Surname Forename(s) Candidate signature | declare this is my own work. A-level BIOLOGY Paper 1 Thursday 4 June 2020 Morning Time allowed: 2 hours Materials For Examiner's Use For this paper you must have: - * aruler with millimetre measurements Question [i Matk e ascientific calculator. 1 2 Instructions 3 ¢ Use black ink or black ball-point pen. ¢ Fill in the boxes at the top of this page. 4 ¢ Answer all questions. 5 ¢ You must answer the questions in the spaces provided. Do not write 6 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 7 this book. Write the question number against your answer(s). 8 ¢ Show all your working. 9 © Do all rough work in this book. Cross through any work you do not want 10 to be marked. TOTAL Information ¢ The marks for the questions are shown in brackets. ¢ The maximum mark for this paper is 91. *JUN207402101* IB/M/Jun20/E 16 7402/1 Do not write outside the Figure 2 is a diagram of one SGLT1 carrier protein. box Figure 2 Hydrophilic region / [ Hydrophobic region / 0000000000 000000000 0000000000 0000000000 QQ00000000 0000000000 0000000000 0000000000 QQQN00000) 0000000000 0000000000 “Hydrophilic region COOH NH i] Draw phospholipids on Figure 2 to show how the carrier protein, SGLT1, would fit into the cell-surface membrane. Do not draw more than eight phospholipids. [2 marks] *O4* IB/M/Jun20/7402/1 Do not write outside the [o]1}.[5] Figure 2 shows the SGLT1 polypeptide with NH2at one end and COOH at the box other end. Describe how amino acids join to form a polypeptide so there is always NHz2 at one end and COOH at the other end. You may use a diagram in your answer. [2 marks] Space for diagram: 10 Turn over for the next question Furn-over- > *O5S* IB/M/Jun20/7402/1 Do not write outside the box To study lipid digestion, a scientist placed a tube into the gut of a healthy 20-year-old man. The end of the tube passed through the stomach but did not reach as far as the ileum. The scientist fed the man a meal containing triglycerides through the tube. The scientist also used the tube to remove samples from the man’s gut at intervals after the meal. The scientist measured the type of lipid found in the samples. Some of her results are shown in Table 1. Table 1 Sample | Time of collection Concentration of Concentration of after meal / min | fatty acids / mg cm’ | triglycerides / mg cm™® A 45 2.7 0.6 B 75 3.3 0.0 [o[2].[4] Use your knowledge of lipid digestion to explain the differences in the results for samples A and B shown in Table 1. You should assume that no absorption had occurred. [3 marks] *O6G* IB/M/Jun20/7402/1 Do not write outside the lols |[2] At Q on Figure 3 there is a small increase in pressure and in rate of blood flow in the box aorta. Explain how this happens and its importance. [2 marks] 0 3 By A student correctly plotted the right ventricle pressure on the same grid as the left ventricle pressure in Figure 3. Describe one way in which the student's curve would be similar to and one way it would be different from the curve shown in Figure 3. [2 marks] Similarity Difference Lols].[4] Use information from Figure 3 to calculate the heart rate of this dog. [1 mark] Heart rate. beats minute” 7 Furn-over > *OOQ* IB/M/Jun20/7402/1 (oJ4].[4] | 2) *10* 10 Anthocyanins are coloured pigments found in the cell vacuole of some plant cells. Anthocyanins cannot move across undamaged cell membranes. A student investigated how to extract anthocyanins from blueberries. She mixed 10 g of crushed, fresh blueberries with 100 cm’ of extraction solvent for 1 hour. She investigated three different extraction solvents: ¢ E- Ethanol, water and acid ¢ F- Ethanol and water ¢ G—Water When making up extraction solvent E, the student used a volume ratio of 70:30:1 ethanol:water:acid. Tick (“) one box that shows the most appropriate volumes she would use to make up 100 cm* of extraction solvent E. [1 mark] 63.6cm* ethanol, 27.3cm° water, 9.1 cm® acid 69.3cm* ethanol, 29.7cm° water, 1.0cm® acid 70.0cm* ethanol, 30.0 cm° water, 1.0cm® acid 70.7cm* ethanol, 30.3 cm° water, 1.0cm® acid The student kept constant: ¢ the mass of fresh blueberries ¢ the volume of extraction solvent ¢ the time for the mixture to stand. Name two other variables the student should have kept constant during this investigation. [2 marks] IB/M/Jun20/7402/1 Do not write outside the box 11 [o[4].[3] After 1 hour, the student filtered the samples. She placed the filtrate in a colorimeter and measured the light absorbance. Her results are shown in Figure 4. 10.0 8.0 6.0 Absorbance / arbitrary units " 4.0 2.0 0.0 Figure 4 E F G Extraction solvent Use your knowledge of membrane structure to explain the results in Figure 4. [4 marks] Turn over > *~1T1* IB/M/Jun20/7402/1 Do not write outside the box Do not write outside the box 14 [2 marks] [o[5].[1] Describe the role of DNA polymerase in the semi-conservative replication of DNA. Figure 5 shows the percentage of rat cells undergoing DNA replication. Some cells contained a protein called cyclin D and some cells did not contain cyclin D. All cells were in early interphase at time 0 Figure 5 60 50 ! : ‘Ny 40 i . ' + | Hi S Es Percentage of i | cells undergoing 30 t NS DNA replication ‘i oS I i 1 20 i i i i } i 1 ! t 10 cs tid 0 0 10 20 Time /h Key - With cyclin D — Without cyclin D IB/M/Jun20/7402/1 *~14* 15 Do not write outside the [o[5].[2] It took less time for 25% of cells with cyclin D to be undergoing DNA replication than box for 25% of cells without cyclin D. Use Figure 5 to calculate this time difference as a percentage decrease. Show your working. [2 marks] Answer % [o[5|.[3] Cyclin D stimulates the phosphorylation of DNA polymerase, which activates the DNA polymerase. Describe how an enzyme can be phosphorylated. [2 marks] [o[5].[4] Some tumour cells contain higher than normal concentrations of cyclin D. Use Figure 5 to suggest why higher than normal concentrations of cyclin D could result in a tumour. [2 marks] 8 Turn over > *15* IB/M/Jun20/7402/1 16 Do not write outside the box [o[6].[1] Particulate matter is solid particles and liquid particles suspended in air. Polluted air contains more particulate matter than clean air. A high concentration of particulate matter results in the death of some alveolar epithelium cells. If alveolar epithelium cells die inside the human body they are replaced by non-specialised, thickened tissue. Explain why death of alveolar epithelium cells reduces gas exchange in human lungs. [3 marks] *~16* IB/M/Jun20/7402/1 Lo | [3] A tick is a small animal that bites humans and feeds on their blood. This results in 19 proteins from the tick saliva entering the human body. Scientists have suggested one hypothesis for the allergic reaction to alpha-gal in red meat. They think that an earlier immune response to a tick bite can cause a person to have an allergic reaction to alpha-gal in red meat. Suggest how one antibody can be specific to tick protein and to alpha-gal. [2 marks] Question 7 continues on the next page *19% Turnover & IB/M/Jun20/7402/1 Do not write outside the ‘box 20 Do not write rl outside the box -olz [a] Scientists took blood samples from one man over several weeks and measured the concentration of antibody in the man’s blood. During this time, the man had two tick bites and had an allergic reaction to alpha-gal in red meat. The scientists’ results are shown in Figure 7. Figure 7 600 x 500 t Key Hosts : —— Monoclonal antibody 400 | specific ta alpha-gal Antibody | / it Peene te alpne-s concentration é ROMER ono Total antibody inthe blood; 300 i <q . . arbitrary units i # Mt —-» Exposure to tick protein 200 F * Allergic reaction in i response to eating 100 7 alpha-gal in red meat / 0 0 1 2 3 4 5 6 7 Time / weeks The scientists’ hypothesis was that an earlier immune response to tick protein causes the allergic reaction. Consider whether Figure 7 supports this hypothesis. [3 marks] 10 IB/M/Jun20/7402/1 *DO* 21 0|8 G) Complete Table 2 to show three differences between DNA in the nucleus of a plant cell and DNA in a prokaryotic cell. [3 marks] Table 2 DNA in the nucleus of a plant cell DNA in a prokaryotic cell [o/8|.[2] Scientists investigated the genetic diversity between several species of sweet potato. They studied non-coding multiple repeats of base sequences. Define ‘non-coding base sequences’ and describe where the non-coding multiple repeats are positioned in the genome. [2 marks] Question 8 continues on the next page Turn over > *DIT* IB/M/Jun20/7402/1 Do not write outside the box 24 Do not write outside the ‘box There are no questions printed on this page DO NOT WRITE THIS PAGE ANSWER IN THE SPACES PROVIDED *DA* IB/M/Jun20/7402/1 25 0|9 Scientists investigated stomatal density on leaves of one species of tree. Figure 9 shows three examples of the square fields of view the scientists used to calculate a mean stomatal density. Figure 9 _ | * Stomata * Zz | White lines show the counting field for stomata | (each edge of white square = 250 ~m) [o[9].[4] Calculate the mean stomatal density in the three fields of view in Figure 9. Give your answer as number of stomata per mm? Show your working. [2 marks] Stomatal density per mm? Question 9 continues on the next page Turn over > *DS* IB/M/Jun20/7402/1 Do not write outside the box 26 Do not write . . . ee . i, itside the The scientists used leaves from individual trees that had grown in different areas of oe nox the world in different years. Each tree had grown in an area and year with known carbon dioxide concentration. Their results are shown in Figure 10. Figure 10 230 220 210 200 Stomatal density / 190 * PY number of stomata [Sl per mm? 180 170 160 150 40 + 290 300 310 320 330 340 350 360 370 Carbon dioxide concentration in the atmosphere / parts per million Key Each plotted point represents mean stomatal density from 10 leaves from one tree Line shows line of best fit, which shows a statistically significant change [2] Give a null hypothesis for this investigation and name a statistical test that would be appropriate to test your null hypothesis. [2 marks] Null hypothesis Statistical test *DE6* IB/M/Jun20/7402/1 29 [1[0].[2] Describe how a polypeptide is formed by translation of mRNA. [6 marks] Question 10 continues on the next page Turn over > *DOx* IB/M/Jun20/7402/1 Do not write outside the box 30 Do not write outside the [1[0].[3] Define ‘gene mutation’ and explain how a gene mutation can have: box * no effect on an individual * apositive effect on an individual. [4 marks] 15 END OF QUESTIONS *Z30* IB/M/Jun20/7402/1 31 Do not write outside the ‘box There are no questions printed on this page DO NOT WRITE THIS PAGE ANSWER IN THE SPACES PROVIDED *231* IB/M/Jun20/7402/1 34 Question number Additional page, if required. Write the question numbers in the left-hand margin. *34* IB/M/Jun20/7402/1 Do not write outside the ‘box 35 Question Additional page, if required. number Write the question numbers in the left-hand margin. IB/M/Jun20/7402/1 Do not write outside the box 36 Do not write outside the ‘box There are no questions printed on this page DO NOT WRITEAN 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 www.aqa.org.uk. 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 © 2020 AQA andits licensors. All rights reserved. * *206A7402/1* IB/M/Jun20/7402/1