Cambridge A Level Biology 9700 — 2021 May/June Paper 5 · Variant 3

1 When bananas ripen, they turn brown in colour. This is due to the formation of a brown product called melanin. The enzyme catechol oxidase acts on its substrate, catechol, leading to the formation of a brown product, melanin, as shown in Fig. 1.1. catechol oxidase catechol melanin (pale pink substrate) (brown product) Fig. 1.1 As melanin is produced, the colour of the reaction mixture changes to brown. The intensity of the brown colour produced is proportional to the concentration of melanin. The more active the enzyme, the more intense the brown colour. A colorimeter is used to measure the absorbance of the reaction mixture. Absorbance is a measure of the light absorbed by a coloured solution. With the reaction shown in Fig. 1.1, the more intense the brown colour, the higher the absorbance. Catechol oxidase can be extracted from bananas. (a) Some students were asked to extract catechol oxidase by grinding a slice of banana with sand and water, using a mortar and pestle. The resulting mixture was then filtered through a piece of cloth. Suggest a reason for: grinding with sand .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... filtering through cloth. .......................................................................................................................................................... .......................................................................................................................................................... .................................................................................................................................................... [2] (b) The students investigated the effect of the substrate concentration on the enzyme-catalysed reaction shown in Fig. 1.1. The students were provided with: • the extracted catechol oxidase enzyme solution prepared in step (a), which was kept cold until needed • a stock solution of 1.0% catechol solution made up in a buffered solution of pH 7.0 • a buffer solution of pH 7.0. The procedure used by the students is outlined in steps 1 to 3. 1. The extracted catechol oxidase enzyme solution was mixed with the 1.0% catechol solution. 2. After 2 minutes a colorimeter was used to measure the absorbance of this mixture. 3. Step 1 and step 2 were repeated using different concentrations of catechol solution. (i) Suggest why the students used a colorimeter to measure the absorbance rather than judging the intensity of the colour by eye. ........................................................................................................................................... ..................................................................................................................................... [1] (ii) Identify the independent variable and the dependent variable in this investigation. independent variable ......................................................................................................... dependent variable ...................................................................................................... [2] The students used the absorbance values at 2 minutes as the initial rates of reaction. Fig. 1.2 shows the results. initial rate of reaction percentage concentration of catechol Fig. 1.2 (c) The students decided to investigate the effect of inhibitors on the reaction shown in Fig. 1.1. The students planned to add an inhibitor, inhibitor Y, to reaction mixtures containing different concentrations of catechol solution. (i) Describe a method the students could use to collect the data needed to test the effect of inhibitor Y in reaction mixtures containing different concentrations of catechol solution. The description of your method should be set out in a logical way and be detailed enough for another person to follow. You should not repeat the details from (a) describing how the extract of catechol oxidase is prepared. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [8] Fig. 1.3 shows the results when no inhibitor Y was added. no inhibitor Y initial rate of reaction percentage concentration of catechol Fig. 1.3 The students suggested that inhibitor Y was acting as a non-competitive inhibitor. (ii) On Fig. 1.3, sketch the curve expected if inhibitor Y was acting as a non-competitive inhibitor. [2] Vmax is the maximum initial rate of reaction of the enzyme. The Michaelis-Menten constant, Km, is the substrate concentration at which the initial rate of reaction is half its maximum value. (iii) Draw on Fig. 1.3 the positions of Vmax and Km of the enzyme when no inhibitor Y is present. [2] (iv) Use your graph to describe the effect of the addition of inhibitor Y on the Km of this enzyme. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [1] (d) Hypertension is a condition where people have long-term high blood pressure. Two main types of drug are used to lower blood pressure in people with hypertension: • ACE inhibitors • diuretics. Scientists investigated the effectiveness of the drugs when used separately and in combination. 505 people with hypertension were split into four groups as shown in Table 1.1. Table 1.1 group treatment 1 placebo (no active drug given) 2 ACE inhibitor only 3 diuretic only 4 ACE inhibitor and diuretic The results are shown in Fig. 1.4. 2.0 1.5 mean reduction 1.0 in blood pressure / kPa 0.5 0.0 group 1 group 2 group 3 group 4 Fig. 1.4 The scientists analysed the data and concluded that a combination of an ACE inhibitor and a diuretic should be used to lower blood pressure in the treatment of hypertension. With reference to the data in Table 1.1 and Fig. 1.4, discuss the conclusion that a combination of an ACE inhibitor and a diuretic should be used to lower blood pressure in the treatment of hypertension. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] [Total: 21]

Mark scheme: 1(a) idea of: grinding with sand to break open the cells / release enzyme ; filtering through cloth to remove, sand / debris / AW ; 2 1(b)(i) (idea that result / it, is) quantitative / AW ; 1 1(b)(ii) independent variable: concentration of catechol / substrate ; dependent variable : absorbance (on colorimeter) ; 2 Question Answer Marks 1(c)(i) any eight from: 1 ref. to preparing / using, at least five dilutions concentrations of catechol ; 2 stated range of at least five concentrations from 1% downwards with % units ; 3 method of keeping catechol oxidase / enzyme solution / banana extract cold ; 4 idea of bringing enzyme and substrate to temperature / equilibrating, before mixing ; 5 method of keeping reactants / reaction mixture, at, constant / controlled / stated / standardised, temperature ; 6 same / constant / stated / standardised / known, volume / concentration / mass of, Y / inhibitor ; 7 same / constant / stated / standardised / known, volume of, catechol / substrate ; 8 same / constant / stated / standardised / known, volume / concentration, of catechol oxidase / enzyme / banana extract ; 9 ref. to measuring absorbance / intensity of colour (on colorimeter) after one set time ; 10 additional detail of use of colorimeter ; 11 repeat at least twice / three replicates, and calculate mean ; 12 idea of repeating, whole experiment / range of concentrations, without, Y / inhibitor ; 13 named hazard and risk and precaution ; 8 1(c)(ii) curve should be to the right of the original curve ; curve should plateau below the original curve ; 2 1(c)(iii) correct placement of Vmax and Km ; 2 1(c)(iv) (reaction with non-competitive inhibitor) has same Km value than with no inhibitor ; 1 Question Answer Marks 1(d) any three from: 1 combination of drugs gives a larger reduction in blood pressure ; 2 no error bars ; 3 no, statistical test / t-test, done to compare groups ; 4 no data on long term results ; 5 AVP ; ; any two from: • no information on blood pressure prior to trial • no information on type of blood pressure • no information on number of patients in each group • no data on side effects • ref. to reduction in blood pressure brought about by the placebo alone • no data on, age / gender 3

More questions on Factors that affect enzyme action

2 Inheritance of wing shape and eye colour in the fruit fly, Drosophila melanogaster, is controlled by two genes. • Gene N/n controls wing shape. Allele N for wrinkled wings is dominant to allele n for normal wings. • Gene E/e controls eye colour. Allele E for rosy eyes is dominant to allele e for red eyes. A biologist predicted that, if the genes are on different chromosomes, the ratio of the phenotypes of the F2 generation would be 9:3:3:1. The biologist carried out a breeding experiment. • Homozygous dominant fruit flies with wrinkled wings and rosy eyes were crossed with homozygous recessive fruit flies with normal wings and red eyes. • All the F1 fruit flies had wrinkled wings and rosy eyes. • The F1 fruit flies were crossed with each other. Table 2.1 shows the results for the F2 generation. Table 2.1 F2 phenotype frequency wrinkled wings 44 rosy eyes wrinkled wings 2 red eyes normal wings 2 rosy eyes normal wings 16 red eyes total 64 (a) The chi-squared test (χ 2 test) was used to analyse the data in Table 2.1. (i) State one reason why the chi-squared test (χ 2 test) was used. ........................................................................................................................................... ..................................................................................................................................... [1] (ii) State the null hypothesis that the biologist would use for this test. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [1] (iii) Complete Table 2.2 and calculate the value of χ 2 for the results of the F2 generation. The equation for the calculation of χ 2 is: O = observed result (O – E)2 χ 2 = ! E E = expected result ! = sum of Table 2.2 (O – E )2 offspring phenotype O E E wrinkled wings 44 rosy eyes wrinkled wings 2 red eyes normal wings 2 rosy eyes normal wings 16 red eyes χ 2 = [3] Table 2.3 shows some critical values of χ 2 at different probability levels. Table 2.3 degrees of probability (p) freedom 0.10 0.05 0.01 0.001 1 2.71 3.84 6.64 10.83 2 4.61 5.99 9.21 13.82 3 6.25 7.82 11.34 16.27 4 7.78 9.49 13.28 18.46 (iv) State why the student used the critical value of 7.82 in this investigation. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [1] (v) Use your calculated value of χ 2 to: • explain whether the null hypothesis should be accepted or rejected .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... • suggest a conclusion the biologist could make about the inheritance of the genes controlling wing shape and eye colour in fruit flies. .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .............................................................................................................................. [3] [Total: 9]

Mark scheme: 2(a)(i) any one from: idea of compare / AW, observed and expected (data / results / ratios / phenotypes) ; or data is, categoric / discrete / discontinuous / nominal ; 1 2(a)(ii) idea of no (significant) difference between the observed and the expected (results) ; or idea of difference between observed and expected (results) is due to chance ; 1 2(a)(iii) offspring phenotype O E ( ) 2 O E E − wrinkled wings rosy eyes 44 36 (36.000) 1.78 (1.77778) wrinkled wings red eyes 2 12 (12.000) 8.33 (8.33333) normal wings rosy eyes 2 12 (12.000) 8.33 (8.33333) normal wings red eyes 16 4 ; (4.000) 36.00 ; (36.00000) χ 2 = 54.44 ; (54.44444) apply ecf as needed 3 2(a)(iv) n–1 and probability of 0.05 / one less degree of freedom than number of categories and probability of 0.05 ; 1 Question Answer Marks 2(a)(v) apply ecf as needed 1 reject (null hypothesis) and state that, the calculated value / χ 2, is higher than, 7.82 / the critical value ; 2 If null hypothesis rejected: (between O and E): there is a significant difference / the difference is not due to chance idea that there is a less than 5 % probability that the difference is due to chance / there is a more than 95% probability that the difference is not due to chance ; 3 From mp1 If null hypothesis rejected or mp2 If there is a significant difference / AW, stated idea that: genes, are on the same chromosome / are not on different chromosomes / are linked / show autosomal linkage / do not assort independently ; A idea of, ‘may be / can be’, throughout mp3 3

More questions on Passage of information from parents to offspring