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In the following passage, the numbered spaces can be filled with biological terms. Glycolysis is the first stage of respiration. Itoccurs inthe (1) ~—_—soof cells andi an anaerobic process. Glycolysis involves the (2) of glucose to glucose phosphate, using ATP. The oxidation of triose phosphate to (3) results in the net gain of ATP and reduced (4) ; Write the correct biological term beside each number below that matches the space ir the passage. [2 marks The equation for the aerobic respiration of glucose is: CeH1206 + 6 02 —————___» 6C0. +6 H20 Six carbon dioxide molecules are produced from each glucose molecule. Use your knowledge of the link reaction and Krebs cycle to explain why. [2 marks An inhibitor of the electron transfer chain was added to aerobically respiring yeast cells. As aerobic respiration was inhiblted, the following changes occurred: e oxygen uptake decreased e ATP production decreased e ethanol was produced, Explain the changes in the yeast cells following the addition of the inhibitor. [3 marks] Oxygen uptake decreased se cereeeseeneettencemeemnting ATP production decreased Ethanol was produced Turn over for the next question Blood pH above 7.45 Blood pH 7.35 to 7.45 Blood pH below 7.35 Figure 1 Decrease in Decrease in stimulation of » | rate of chemoreceptors ventilation Increase in Increase in Stimulation of | |——> |rate of chemoreceptors ventilation Figure 1 shows one of the physiological systems involved in controlling blood pH. Blood pH 7.35 to 7.45 Use the information in Figure 1 to outline the principles of negative feedback. [2 marks] Use Figure 1 to describe and explain these changes. Vigorous exercise causes a change in blood pH and in the rate of ventilation. i uipment used was an [3].[4] A student determined the dry mass of lettuce leaves. The equip oven and a weighing balance. Describe how the student could use this equipment to determine the dry mass of lettuce leaves. [2 marks] The student then used a calorimeter to determine the chemical energy stored in a sample of lettuce leaves with a dry mass of 1.50 g The 1.50 g sample was fully combusted in a calorimeter, The volume of water in the calorimeter was 150 cm? The increase in temperature of the water was 18,2 °C 4.18 J of heat energy are re quired to increase the temperature of 4 cms by 1°C of water Use this information to calculate the heat energy released, in kJ g-', of dry mass of lettuce leaves, Show your working. [2 marks] Answer a oo (ole. [1] People with the inherited blood disorder haemophilia A (HA) have a reduced ability to clot blood. One variant of HA is due to an inversion mutation in the F8 gene. This mutation causes the production of a non-functional form of the protein, factor VIII. Factor Vill is not an enzyme. Explain how an inversion mutation can cause the production of non-functional factor Mille [4 marks -[2] Currently, the most effective treatment for HA is weekly or monthly injections of factor VIIl. Scientists investigated the effectiveness of a gene therapy product, Roctavian, as a treatment for HA. Roctavian contains a viral vector that has the DNA coding for factor Vill. The scientists: * selected 112 male volunteers with severe HA, who were at least 18 years of age * treated each volunteer with a single injection of Roctavian into their blood * recorded the number of bleeding episodes (internal bleeding) i Ing) in th over a two-year period (set A tenuis) ( g) in these volunteers * compared these data with the number of bleedi i i vith Ng episodes per year volunteers before receiving Roctavian (set B rec ee Table 1 shows the scientists’ results. A value of +2 x SD from the mean includes over 95% of the data. Table 1 Mean number of bleeding episodes Volunteer set per yoar (+2 x $0) A 4 (Received Roctavian) mee em) i 4.8 + (0.6) (Received factor VIII) Use all the information provided to evaluate the use of Roctavian as a treatment for HA. [5 marks Question 4 continues on the next page 3 Explain what is meant by the term phenotype. [2 marks] In genetic crosses, the observed phenotypic ratios obtained in the offspring are often not the same as the expected ratios. Suggest two reasons why. Do not refer to sex-linkage or autosomal linkage in your answer. [2 marks] 10 10 | 5 [3] In sweet pea plants, the F allele for purple flowers is dominant to the f allele for red flowers. The L allele for long pollen is dominant to the | allele for round pollen. A scientist carried out a genetic cross between a plant with purple flowers and long pollen (heterozygous for both genes) and a plant with red flowers and round pollen. Table 2 shows the results of this cross. Table 2 Phenotype of offspring Percentage of total offspring Purple flowers and long pollen 48.6 Red flowers and round pollen 47.5 Purple flowers and round pollen 2.1 Red flowers and long pollen 1.8 Use your knowledge of autosomal linkage to explain the results in Table 2. Your answer should include: * an explanation of what is meant by autosomal linkage * a comment about the relative proportions of the different genotypes of gametes produced in this cross. [3 marks] 11 3] The parasympathetic nervous system (PNS) uses acetylcholine as its neurotransmitter, Atropine Is an anticholinergic drug that attaches to postsynaptic receptors, Atropine can be used to increase tho heart rate In patients with bradycardia (a slower than normal heart rate). Use the information provided to explain how atropine causes an increase in the heart rate. (3 marks) Question 6 continues on the next page 13 , [4] One study has suggested that the atropine dosage for severely obese male patients should be 0.01 mg kg-' of lean body weight (LBW). LBW is determined by using this equation. LBW = 9.27 x 10° x TBW (6.68 x 10°) + (216 x BMI) where TBW = total body weight BMI = body mass index Calculate the suggested dosage of atropine, in mg, for a severely obese male patient whose: © TBW = 115.2 kg * BMI=41 Give your answer to 3 significant figures. Show your working. [ - 2 marks Answer mg 14 Figure 4 A C 0% | 50% light intensity light intensity B Ve) 100% light intensity light intensity The student stated that ‘the proce dure used in this investigation helps to ensure the validity of any conclusions made’, Suggest and explain three features of this investigation that justify this statement. [3 marks] 4 .[2] Table 3 shows the student's results. Table 3 Segment of choice Percentage light Number of G. pulex counted chamber intensity after 10 minutes A 0 15 B 25 8 Cc 50 4 D 100 3 The student observed directional movement by G. pulex. Name the type of behavi of this behaviour. Use the information in Table 3 in your answer. Type of behaviour four shown by G. pulex and suggest an advantage to G. pulex [2 marks] Advantage Be ee 17 Using the calculated Y’ value of 11.87 and the information in Table 4, what can you conclude from this investigation? You should include the terms probability and chance in your answer, [4 marks 19 (oje}.[4] A study of cells from a human embryo found that 2.66 x 10-7 % of the genome of one cell had changed due to mutation. You can assume that: * the genome contains 3 x 10° nucleotides * the mean rate of mutation is 1.14 mutations per cell cycle * the cell cycle lasts 24 hours. Use this information to calculate the age of the embryo in days. Show your working. [2 marks] Answer days The genome of the bacterium Escherichia coli is approximately 4.6 million base pairs in length and contains around 4000 genes. The analysis of the genome of E. coli allows the proteome to be determined more easily than that of eukaryotes. Explain why. [2 marks] 20 
