9700_w09_qp_41
A paper of Biology, 9700
Questions:
10
Year:
2009
Paper:
4
Variant:
1
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2
Use Asellus aquaticus is a small freshwater crustacean. 200 A. aquaticus were released into a pond where there had previously been none. The pond was favourable for their growth and reproduction. Describe and explain the expected changes in the population size of A. aquaticus over the following few months. In order for natural selection to occur a population must show phenotypic variation. Explain why variation is important in natural selection.
17. Selection and evolution  →  17.2. Natural and artificial selection
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Use Proteases that work in alkaline conditions are made in large quantities for use in the detergent industry. The microorganism that is generally used for this is the bacterium Bacillus subtilis. An investigation was carried out to compare three potential production methods: ● using free cells of B. subtilis ● using B. subtilis cells immobilised in cubes of agar ● using B. subtilis cells immobilised in beads of sodium alginate. To immobilise the cells in agar, the agar was dissolved and cooled. A suspension of B. subtilis was then added. The agar-bacterium mixture was poured into sterile dishes and allowed to solidify. It was then cut into cubes with sides of 2 mm. Explain why the agar was cooled before the suspension of B. subtilis was added. Describe how cells of B. subtilis could be immobilised in beads of alginate. A liquid medium containing glucose, a nitrogen source and various mineral ions was made up, and 50 cm3 placed into each of three flasks. Samples of a culture of free cells of B. subtilis, agar cubes containing immobilised B. subtilis and alginate beads containing B. subtilis were placed in the three flasks. Each flask contained the same number of bacteria. All the flasks were incubated at 37 °C for 48 hours. Samples of the liquid medium in each flask were taken at six hourly intervals and the concentration of protease measured. The results are shown in . Examiner’s Use free cells cells in agar cubes cells in alginate beads Key: time / hours concentration of protease / arbitrary units With reference to , compare the results for the free cells of B. subtilis and cells immobilised in alginate beads. Examiner’s Use Suggest why lower concentrations of protease were produced by B. subtilis immobilised in agar cubes than B. subtilis immobilised in alginate beads. Two new cultures of immobilised B. subtilis were set up as described in . However, this time a repeat batch fermentation method was used, in which the liquid medium was replaced every 24 hours. This was continued until the cubes or beads had begun to disintegrate. The results are shown in Table 3.1. Table 3.1 number of batches before cubes or beads disintegrated total fermentation time / hours total protease produced / arbitrary units mean productivity of protease / arbitrary units per hour agar cubes 12.44 alginate beads 15.11 With reference to Table 3.1 calculate the percentage increase in the total protease produced when the bacteria were immobilised in alginate rather than agar. Show your working. explain why using bacteria immobilised in alginate rather than agar would be a more cost-effective production of protease.
3. Enzymes  →  3.2. Factors that affect enzyme action
3. Enzymes  →  3.1. Mode of action of enzymes
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Use Modern varieties of wheat have developed from numerous hybridisation events between different species of wild grasses. shows some of the possible steps that are believed to have been involved in the development of bread wheat, Triticum aestivum. The letters A, B and C represent three different sets of seven chromosomes. einkorn Triticum urartu 14 chromosomes AA × goat grass 1 Aegilops speltoides 14 chromosomes BB hybridisation and doubling of chromosome number emmer wheat Triticum turgidum 28 chromosomes AABB × goat grass 2 Aegilops tauschii 14 chromosomes CC hybridisation and doubling of chromosome number bread wheat Triticum aestivum 42 chromosomes Complete by writing letters to represent the sets of chromosomes in bread wheat. Write your answer on . Examiner’s Use Explain why hybridisation between emmer wheat and goat grass 2 would have produced a sterile hybrid, if doubling of chromosome number had not occurred. With reference to , suggest why Triticum urartu and Triticum turgidum are classified as different species. Triticum turgidum emerged as a new species without being geographically isolated from Triticum urartu. Outline how geographical isolation may result in speciation.
16. Inheritance  →  16.1. Passage of information from parents to offspring
17. Selection and evolution  →  17.2. Natural and artificial selection
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Use Hormones are secreted by endocrine glands. Explain what is meant by the term endocrine gland. shows the changes in concentration in the blood of follicle stimulating hormone (FSH) and luteinising hormone (LH) during the first half of the menstrual cycle. days from start of menstrual cycle FSH LH concentration of hormone in blood /arbitrary units With reference to , describe, the changes that take place in the ovary during this time, as a result of the action of FSH the role of LH. Examiner’s Use In preparation for in-vitro fertilisation (IVF), women are injected with FSH. Explain why treatment with FSH is a necessary preparation for IVF. The standard treatment with FSH and clomiphene causes significant side-effects. Clomiphene occupies oestrogen receptors, blocking a negative feedback mechanism. Explain briefly what is meant by negative feedback. Outline the feedback mechanism that is blocked by clomiphene. Recently a so-called ‘mild’ treatment has been introduced in the hope of avoiding the side-effects of the standard treatment. This treatment does not use clomiphene. Instead, an antagonist to LH secretion is used. The days in the first half of the menstrual cycle on which injections of FSH and clomiphene are given in the two treatments are shown by asterisks (*) in . days from start of menstrual cycle * * * * * * * * FSH antagonist to LH secretion * * * * * * * * * * * * * * * * FSH mild treatment standard treatment clomiphene With reference to the concentrations of LH shown in , show, using an asterisk on when the antagonist to LH secretion should first be given. Put your asterisk into the grey area on . Examiner’s Use Suggest why an antagonist to LH secretion forms part of the mild treatment. The average dose of FSH given in the mild treatment is 1300 international units (IU), compared with an average dose of 1800 IU in the standard treatment. This could lead to the mild treatment being less effective. The outcomes of an investigation into the two treatments are shown in Table 5.1. Table 5.1 mild treatment standard treatment mean number of oocytes harvested per treatment cycle 6.7 8.5 mean number of embryos produced per treatment cycle 2.8 3.8 percentage of pregnancies resulting in live birth 43.4 44.7 With reference to Table 5.1, compare the effectiveness of the two treatments. FSH consists of two polypeptide chains which are encoded by genes on different chromosomes. The two genes, together with their promoters, have been inserted into bacteria to produce the hormone used in fertility treatments. Explain briefly why promoters need to be transferred into the recipient bacteria together with the two genes for the FSH polypeptides.
14. Homeostasis  →  14.1. Homeostasis in mammals
15. Control and coordination  →  15.1. Control and coordination in mammals
11. Immunity  →  11.2. Antibodies and vaccination
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Use A husband and wife who already have a child with cystic fibrosis (CF) elected to have their second child tested for the condition while still a fetus in very early pregnancy. The results of the test, a DNA banding pattern, were discussed with a genetic counsellor. The relevant DNA banding pattern produced by electrophoresis is shown in . father first child fetus mother direction of movement of DNA fragments electrophoresis gel band of DNA With reference to , explain why, the fetus will develop CF, the positions of the bands of DNA of the first child and of the fetus indicate that the mutant allele for CF has a deletion in comparison with the normal allele. Explain briefly the need to discuss the result of the test with a genetic counsellor.
19. Genetic technology  →  19.1. Principles of genetic technology
19. Genetic technology  →  19.2. Genetic technology applied to medicine
16. Inheritance  →  16.2. The roles of genes in determining the phenotype
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Use The fruit fly, Drosophila melanogaster, feeds on sugars found in damaged fruits. A fly with normal features is called a wild type. It has a striped body and its wings are longer than its abdomen. There are mutant variations such as an ebony coloured body or vestigial wings. These three types of fly are shown in . wild type ebony body vestigial wing Wild type features are coded for by dominant alleles, A for wild type body and B for wild type wings. Explain what is meant by the terms allele and dominant. allele dominant Examiner’s Use Two wild type fruit flies were crossed. Each had alleles A and B and carried alleles for ebony body and vestigial wings. Draw a genetic diagram to show the possible offspring of this cross. Examiner’s Use When the two heterozygous fruit flies in were crossed, 384 eggs hatched and developed into adult flies. A chi-squared (χ2) test was carried out to test the significance of the differences between observed and expected results. χ2 =  (O – E)2 E where  = sum of O = observed value E = expected value Complete the missing values in Table 7.1. Table 7.1 phenotypes of Drosophila melanogaster grey body long wing grey body vestigial wing ebony body long wing ebony body vestigial wing observed number (O) expected ratio expected number (E) O – E -9 - 4 (O – E)2 (O – E)2 E 0.38 0.22 1.50 Calculate the value for χ2. χ2 = Examiner’s Use Table 7.2 relates χ2 values to probability values. As four classes of data were counted the number of degrees of freedom was 4 – 1 = 3. Table 7.2 gives values of χ2 where there are three degrees of freedom. Table 7.2 probability greater than 0.50 0.20 0.10 0.05 0.01 0.001 values for χ2 2.37 4.64 6.25 7.82 11.34 16.27 Using your value for χ2, and Table 7.2, explain whether or not the observed results were significantly different from the expected results.
16. Inheritance  →  16.2. The roles of genes in determining the phenotype
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Use shows the results from two experiments carried out to investigate the effect of light intensity and carbon dioxide concentration on the rate of photosynthesis. light intensity / arbitrary units rate of photosynthesis /arbitrary units experiment 1 (25 °C 0.04% CO2) experiment 2 (25 °C 0.4% CO2) Describe and explain the results shown in for experiment 1. Describe and explain the difference between the results for experiment 1 and experiment 2. Examiner’s Use The optimum temperature for many plants living in temperate regions is approximately 25 °C. Explain why the rate of photosynthesis in these plants decreases at temperatures above 25 °C.
13. Photosynthesis  →  13.2. Investigation of limiting factors
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