Biology 9700/52 May/June 2021: question paper and mark scheme

Question

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. 5 cm3 of 10% graduated tube hydrogen to collect oxygen 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. …… [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. without copper(II) sulfate rate of oxygen with copper(II) sulfate 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]

Question

(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. (i) Complete Table 2.2 and calculate the value of chi-squared (χ2) for the results of the test cross. The equation for the calculation of χ2 is: (O - E) 2 O = observed result χ2 = / E E = expected result / = sum of Table 2.2 offspring phenotype O E (O - E) 2 E purple petals 102 62 long pollen grains red petals 112 62 round pollen grains purple petals 14 62 round pollen grains red petals 20 62 long pollen grains χ2 = [2] The null hypothesis of this chi-squared test (χ2) is: There is no difference between the observed and expected results. The critical value at p = 0.05 is 7.82 at 3 degrees of freedom. (ii) Use your calculated value of χ2 to: • explain whether the null hypothesis should be accepted or rejected • suggest a conclusion the plant breeder could make about the inheritance of the genes controlling petal colour and pollen grain shape in sweet pea plants. …… [3] (d) Some individuals of the plant species Mimulus guttatus show copper tolerance as they are able to grow on land heavily polluted with copper. Copper tolerance is controlled by an allele of a single gene. Plants that do not have this allele are sensitive to high concentrations of copper(II) ions in the soil and show poor growth. Scientists wanted to investigate if pollen grains are sensitive to copper(II) ions. Pollen viability is the ability of pollen to produce male gametes that can fertilise a female gamete. The scientists investigated the viability of pollen grains by placing them into a culture medium. Normally, after several days pollen grains start to grow pollen tubes. Pollen tubes grow down into the ovary to carry the male gametes to the female gametes. • Pollen grains were collected from copper-sensitive and copper-tolerant plants of M. guttatus. • Pollen grains were placed in culture media as shown in Table 2.3. Table 2.3 number of pollen grains added to culture medium type of plant culture medium containing culture medium containing no copper(II) ions copper(II) ions copper-tolerant 200 200 copper-sensitive 200 200 All other variables were standardised. The results are shown in Fig. 2.2. 100 Key no copper(II) ions 80 with copper(II) ions error bars show standard error 60 percentage of pollen grains that grew pollen tubes 40 20 0 copper- copper- sensitive tolerant Fig. 2.2 The scientists concluded that the presence of copper(II) ions has a significant effect in reducing the viability of pollen grains. Explain whether or not the scientists are justified in this conclusion. …… [2] [Total: 12]