AQA A Level Biology Paper 2 2017, Exams of Advanced Education

Download AQA A Level Biology Paper 2 2017 and more Exams Advanced Education in PDF only on Docsity! AQA A Level Biology Paper 2 2017 1.1) Exercise causes an increase in heart rate. Describe the role of receptors and of the nervous system in this process. - 1. Chemoreceptors detect rise in CO2/H+ /acidity/carbonic acid/fall in pH OR Baro/pressure receptors detect rise in blood pressure. 2. Send impulses to cardiac centre/medulla. 3. More impulses to SAN. 4. By sympathetic (nervous system for chemoreceptors/CO2) OR By parasympathetic (nervous system for baro/pressure receptors/blood pressure); 1.2) Using Figure 1, explain the benefit of activation of AMPK during exercise. - 1. Less/no malonyl-CoA. 2. (More) fatty acids transported/moved into mitochondria. 3. Respiration/oxidation of fatty acids provides ATP. 2.1) Explain how using the SIT could reduce transmission of dengue. - 1. Compete (with fertile males) to mate/for food/resources. 2. Do not reproduce. 2.2) Describe how the mark-release-recapture method could be used to determine the population of A. aegypti at the start of the investigation. - 1. Sample, mark and release. 2. Leave time for mosquitoes/Aedes to disperse before second sampling. 3. Population = number in first sample × number in second sample divided by number of marked in second sample/number recaptured. 2.3) The release of radiation-sterilised A. aegypti has not been very successful in controlling the transmission of dengue. Suggest one reason why - Radiation affects their life span. 2.4) Suggest why the scientists released more transgenic males every week. - To maintain competition as they die. 2.5) The release of transgenic males proved successful in reducing the number of A. aegypti. Describe how the results in Figure 2 support this conclusion. - Number of mosquitoes in treated area is lower after 12/13/14/15/16 weeks. 3.1) Succinic acid dehydrogenase is an enzyme used in the Krebs cycle. Suggest one reason for the difference in the staining between the muscle fibres of the control mice and the trained mice. - Increase in aerobic respiration. 3.2) The scientists then compared the length of time that the control mice and the trained mice could carry out prolonged exercise. The trained mice were able to exercise for a longer time period than control mice. Explain why. - 1. More aerobic respiration produces more ATP. 2. Anaerobic respiration delayed. 3. Less or no lactate. 3.3) The scientists determined the mean diameter of muscle fibres in trained mice using an optical microscope to examine sections of muscle tissue. The circular area (πr 2 ) of one field of view was 1.25 mm2 . The diameter of this area was equal to the diameter of 15 muscle fibres. Using this information, calculate the mean diameter in µm (micrometres) of muscle fibres in this section of tissue. - 84.2 3.4) Describe two differences between these samples of muscle fibres. - - Young fibres range from 14-48 um and adult fibres 18-88 um. - Young fibres most frequent at lower diameter. 4.1) The solution that the student used to produce the chloroplast suspension had the same water potential as the chloroplasts. Explain why it was important that these water potentials were the same - - Osmosis does not occur. - Chloroplast does not burst. 4.2) Explain why the student set up Tube 1. - 1. To show light does not affect DCPIP. 2. To show chloroplasts are required. 4.3) Explain the results in Tube 3. - Reduction of DCPIP by electrons from light dependent reaction. 8.4) A quantitative comparison can be made of the amount of RNA in samples A and B. This involves determining the number of cycles required to reach 50% maximum concentration of DNA (C). The amount of RNA in a sample can be measured as: 1 C Use this information to calculate the ratio for RNA content in sample A : RNA content in sample B. - 1.4:1 8.5) Suggest one reason why DNA replication stops in the polymerase chain reaction. - Limited number of primers. 8.6) Scientists have used the RT-PCR method to detect the presence of different RNA viruses in patients suffering from respiratory diseases. The scientists produced a variety of primers for this procedure. Explain why. - 1. Base sequences differ. 2. Different complementary primers required. 9.1) What is a gene pool? - All the alleles in a population. 9.2) Lord Howe Island in the Tasman Sea possesses two species of palm tree which have arisen via sympatric speciation. The two species diverged from each other after the island was formed 6.5 million years ago. The flowering times of the two species are different. Using this information, suggest how these two species of palm tree arose by sympatric speciation. - 1. Occurs in the same environment. 2. Mutations cause different flowering times. 3. Reproductive isolation. 4. Different alleles passed on. 5. Eventually different species cannot interbreed to produce fertile offspring. 10.1) Assuming no one with AD died in 2014, calculate the annual percentage increase in AD cases in America for 2014 (lines 2-4). - 19.4% 10.2) Explain how donepezil could improve communication between nerve cells (lines 7-9). - 1. Less acetylcholine broken down. 2. Acetylcholine attaches to receptors. 3. More Na+ enter to reach threshold. 10.3) Suggest and explain two reasons why there is a high frequency of the E280A mutation in Yaramul (lines 13-15). - 1. Isolated so inbreeding. 2. Allele inherited from common ancestor. 10.4) Explain why natural selection has not reduced the frequency of the E280A mutation in the population (lines 16-17). - 1. AD develops late. 2. Already reproduced. 10.5) The age at which the E280A mutation is expressed to cause AD can vary (lines 11-12). Suggest and explain one reason for this. - 1. Environment. 2. Methylation of genes or acetylation. 10.6) One scientific study which analysed chromosome 14 involved 102 individuals. The scientists recorded a sample size of 204. In this sample they detected 75 E280A mutations but only 74 potential AD cases (lines 19-21). Suggest explanations for the figures the scientists recorded. - One person was homozygous dominant. 10.7) Suggest why a DNA probe for the mutated triplet was not considered a suitable method for detection of the E280A mutation (lines 22-23). - 1. Triplet is found in other places. 2. Would not know if it was the mutation/allele/gene.