Cambridge A Level Biology 9700 — 2021 May/June Paper 5 · Variant 2

1 The enzyme catalase catalyses the reaction shown in Fig. 1.1. catalase hydrogen peroxide water + oxygen Fig. 1.1 A student had read that copper(II) sulfate can act as an inhibitor, reducing the activity of catalase. The student wanted to investigate the effect of copper(II) sulfate as an inhibitor on the activity of catalase to test the hypothesis: The higher the concentration of the inhibitor copper(II) sulfate, the lower the volume of oxygen produced in the presence of catalase. (a) State the independent variable and dependent variable in this investigation. independent variable ................................................................................................................ ................................................................................................................................................... dependent variable ................................................................................................................... ................................................................................................................................................... [2] (b) (i) The student was provided with a 0.04 mol dm–3 solution of copper(II) sulfate. Describe how the student could use serial dilution to make a suitable range of concentrations to test their hypothesis. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) The student used the apparatus shown in Fig. 1.2 to collect the data needed to test the effect of different concentrations of copper(II) sulfate on the production of oxygen in the presence of catalase. graduated tube 5 cm3 of 10% to collect oxygen hydrogen peroxide mixture of catalase and copper(II) sulfate water Fig. 1.2 Describe a method the student could use to collect the data needed to test the effect of different concentrations of copper(II) sulfate on the production of oxygen in the presence of catalase. 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 (b)(i) describing how to dilute the 0.04 mol dm–3 copper(II) sulfate solution or how to set up the apparatus shown in Fig. 1.2. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... 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[6] (iii) The hypothesis the student tested was: The higher the concentration of the inhibitor copper(II) sulfate, the lower the volume of oxygen produced in the presence of catalase. Complete Fig. 1.3 by: • adding axes labels and units • sketching a graph of the results you would expect if the hypothesis is correct. Fig. 1.3 [3] (c) The student carried out a further investigation to find out if copper(II) sulfate was a competitive or non-competitive inhibitor. • The student measured the oxygen produced in the presence and absence of copper(II) sulfate at different concentrations of hydrogen peroxide. • The student processed the data to find the rate of oxygen production. The student then plotted these rates on the graph shown in Fig. 1.4. withoutwithout copper(II)copper(II) sulfatesulfate rate of oxygen withwith copper(II)copper(II) sulfatesulfate production percentage concentration of hydrogen peroxide Fig. 1.4 (i) Draw on Fig. 1.4 to show how you could derive the Km value of catalase without copper(II) sulfate. [3] (ii) The student concluded that copper(II) sulfate was a non-competitive inhibitor of catalase. State the evidence in Fig. 1.4 that supports this conclusion. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 18]

Mark scheme: 1(a) independent variable: concentration of, copper (II) sulfate ; dependent variable: volume of, oxygen ; 2 1(b)(i) 1 minimum of five stated dilutions and correct units ; 2 method of serial dilution for a minimum of 2 intermediate (concentrations) ; 2 Question Answer Marks 1(b)(ii) any six from: 1 maintaining a constant temperature and a suitable method ; 2 maintaining a constant pH and use a buffer ; mp 3 and 4 and 5: if ‘amount’ rather than ‘volume’, penalise only once and ecf the following ones 3 same / stated, volume of catalase / enzyme or same / stated, concentration of catalase / enzyme ; 4 same / stated, volume of each copper sulfate / inhibitor solution ; 5 measure / record, volume of oxygen / gas collected, in set time / set intervals or time taken to reach a set volume ; 6 state a suitable control ; 7 use new / fresh (named) solutions for each replicate / experiment ; 8 repeat at least twice / three replicates, and finding mean for at least one concentration ; 9 named hazard and risk and precaution ; e.g. hazard risk precaution catalase / enzyme irritant / allergy wear gloves / goggles / mask PPA moderate / medium, hazard hydrogen peroxide irritant / allergy moderate / medium, hazard oxidising agent copper(II) sulfate irritant / allergy moderate / medium, hazard corrosive toxic do not pour down sink 10 AVP ; 6 Question Answer Marks 1(b)(iii) 1 axes correctly orientated with labels ; 2 axes labels have correct units ; 3 line shows downward trend as concentration of copper(II) sulfate increases ; e.g. 3 1(c)(i) All three lines correctly drawn / indicated ; Vmax or 1 2 Vmax correctly labelled for lines drawn ; Km correctly labelled (on correct line) ; 3 1(c)(ii) plateaus below Vmax / does not reach Vmax / Vmax is lower (than without inhibitor) ; same / almost same / similar, Km ; 2 volume of oxygen / mm3 or cm3 concentration of copper sulfate / mol dm–3

More questions on Factors that affect enzyme action

2 (a) Inheritance of petal colour and pollen shape in sweet pea plants are controlled by two genes. • Gene P/p controls petal colour. Allele P for purple petals is dominant to allele p for red petals. • Gene G/g controls the shape of pollen grains. Allele G for long pollen grains is dominant to allele g for round pollen grains. Fig. 2.1 shows the structures inside a flower from a sweet pea plant. petals stigma – female anther – male ovary containing female gametes Fig. 2.1 Sweet pea plants are normally self-fertilising. Pollen grains containing the male gametes are produced in the anthers and released into the flower. The pollen grains fall on the stigma in the flower, leading to fertilisation of the female gametes. A plant breeder cross-fertilised two sweet pea plants, A and B. A homozygous dominant plant with purple petals and long pollen grains (plant A) was crossed with a homozygous recessive plant with red petals and round pollen grains (plant B). The breeder transferred pollen grains from the flowers on plant A to the flowers on plant B. (i) Suggest one way in which the breeder could prevent the self-fertilisation of plant B. ........................................................................................................................................... ..................................................................................................................................... [1] (ii) Suggest one way in which the breeder could transfer pollen grains from an anther to a stigma. ........................................................................................................................................... ..................................................................................................................................... [1] (b) The plant breeder then: • collected the seeds from plant B • grew the seeds in a glasshouse. The environmental conditions were standardised. State three variables that should have been standardised in the glasshouse. 1 ................................................................................................................................................ ................................................................................................................................................... 2 ................................................................................................................................................ ................................................................................................................................................... 3 ................................................................................................................................................ ................................................................................................................................................... [3] (c) All the F1 plants grown from the seeds collected from plant B had purple petals and long pollen grains. The plant breeder then carried out a test cross by crossing the F1 plants with homozygous recessive plants with red petals and round pollen grains. Table 2.1 shows the results of the test cross. Table 2.1 offspring phenotypes frequency purple petals 102 long pollen grains red petals 112 round pollen grains purple petals 14 round pollen grains red petals 20 long pollen grains total 248 The plant breeder expected the test cross to result in a 1:1:1:1 phenotypic ratio.

Mark scheme: 2(a)(i) remove / cover, anthers (from plant B) ; 1 2(a)(ii) method of transferring pollen (from anther) to stigma ; or shake anthers, on / over, stigmas ; 1 Question Answer Marks 2(b) any three from: light, intensity / duration / amount / strength / wavelength / colour ; temperature ; frequency of watering / amount of water ; humidity ; CO2 / O2 concentration ; growing medium / soil composition ; planting density ; 3 2(c)(i) offspring phenotype O E ( ) 2 − O E E purple petals long pollen grains 102 62 25.81 red petals round pollen grains 112 62 40.32 purple petals round pollen grains 14 62 37.16 red petals long pollen grains 20 62 28.45 ; χ 2 = 131.74 ; 2 Question Answer Marks 2(c)(ii) 1 reject and state that their calculated value / χ2, is higher than 7.82 / critical value ; any two from: 2 (if the null hypothesis is rejected) There is a significant difference between the expected and observed values or there is a less than 5% / 0.05 probability that the difference has occurred due to chance ; 3 genes, are on the same chromosome / are not on different chromosomes / are linked / do not assort independently or autosomal linkage or alleles (likely to be) inherited together ; 4 the two genes are close together (on the same chromosome) or small number of recombinants ; 3 2(d) 1 yes in copper sensitive plants and no in copper tolerant plants / only in copper sensitive plants ; 2 decrease, in percentage / %, viability, in sensitive plants / ora or slight increase, in percentage / %, viability, in tolerant plants ; 3 in sensitive plants, there is no overlap of (error) bars, which means the difference may be significant or in tolerant plants, there is an overlap of (error) bars, which means difference is not significant ; 4 no statistical test was carried out (in copper-sensitive plants) ; 5 growth of pollen tube does not indicate pollen grain is viable ; 6 no reference to concentration of copper ions used in the investigation ; 2