9700_s10_qp_41
A paper of Biology, 9700
Questions:
10
Year:
2010
Paper:
4
Variant:
1
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1
The flatback turtle, Natator depressus, is an endangered species that nests on northern Australian beaches. shows a flatback turtle. Each female lays approximately 50 eggs per nest, which is a smaller number than all other species of marine turtle. The eggs are buried in the sand and when the hatchlings emerge each has a mass of approximately 43 g. Unlike most marine turtles, flatback turtles spend most of their time in coastal waters. This is where they feed and mate. shows the numbers of female flatback turtles nesting on a beach in northern Australia between 1993 and 2002. 1993 1994 1996 1997 1999 2000 number of female turtles nesting year For Examiner’s Use Calculate the mean rate of decrease in the numbers of females nesting between 1993 and 2002. Show all the steps in your calculation. Answer Suggest ways in which the flatback turtle could be protected.
18. Classification, biodiversity and conservation  →  18.3. Conservation
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2
For Examiner’s Use The glycoproteins CD28 and CD40 are found on the surface of T-lymphocytes (T-cells). They are binding sites for cell-signalling molecules and are essential for triggering the cloning of T-cells in an immune response. A monoclonal antibody (mAb), which could block the CD40 signalling pathway, was produced from hamsters using the hybridoma method. Outline the procedure, starting with a hamster, for producing mAbs suitable for use in another mammal, such as a mouse. For Examiner’s Use The ability of the mAb produced in to prevent rejection of transplanted hearts in mice was compared with that of a protein, P, which blocks the CD28 signalling pathway. Four groups of mice were treated as follows: • group A – no treatment • group B – treated with protein P only • group C – treated with mAb only • group D – treated with both mAb and protein P. shows the percentage survival of the transplanted hearts in the four groups of mice over a period of 80 days. time / days percentage survival of transplanted hearts D B C A With reference to describe the effectiveness of the four different treatments For Examiner’s Use suggest an explanation for the differences in survival of the transplanted hearts in groups B and D. Examination of the transplanted hearts showed that the hearts in group A had significant damage to their coronary arteries, whereas in group D these blood vessels appeared normal. Explain the importance of the coronary arteries of the heart. State two uses of mAbs in humans, other than preventing rejection of transplanted tissue. 1. 2.
11. Immunity  →  11.2. Antibodies and vaccination
14. Homeostasis  →  14.1. Homeostasis in mammals
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4
For Examiner’s Use is a photomicrograph of a transverse section through the leaf of a C4 plant. J K L Identify structures J to L. J K L Outline how this leaf anatomy adapts the plant for high rates of carbon fixation at high temperatures. For Examiner’s Use Sorghum is a C4 plant and Sorghum bicolor is a major food crop in dry tropical regions. The leaves of S. bicolor are covered with a layer of wax made up of a mixture of esters and free fatty acids, with a melting point of 77– 85 °C. Waxes from the leaves of non- tropical plants tend to have melting points lower than this. For example, wax from the bayberry, Myrica sp., has a melting point of 45 °C. Suggest how the wax on sorghum leaves helps the plant to survive in dry, tropical regions. An investigation was carried out into the response of sorghum to being kept at a low temperature for a short period of time. Soybean plants, which are better adapted than sorghum for growth in subtropical and temperate climates, were used for comparison. Plants of sorghum and soybean were kept at 25 °C for several weeks and then at 10 °C for three days. The temperature was then increased to 25 °C again for seven days. Day length, light intensity and carbon dioxide concentration were kept constant throughout. The uptake of carbon dioxide, as mg CO2 absorbed per gram of leaf dry mass, was measured • at 25 °C before cooling • on each of the three days at 10 °C • for seven days at 25 °C. The results are shown in Table 4.1. Table 4.1 plant carbon dioxide uptake / mg CO2 g–1 at 25 °C, before cooling at 10 °C at 25 °C (mean over days 4 to 10) day 1 day 2 day 3 sorghum 48.2 5.5 2.9 1.2 1.5 soybean 23.2 5.2 3.1 1.6 6.4 For Examiner’s Use Compare the changes in carbon dioxide uptake in sorghum and soybean during the three days at 10 °C. During the cooling period, the ultrastructure of the sorghum chloroplasts changed. The membranes of the thylakoids moved closer together, eliminating the spaces between them. The size and number of grana became reduced. Explain how these changes could be responsible for the low rate of carbon dioxide uptake by sorghum even when returned to a temperature of 25 °C.
13. Photosynthesis  →  13.2. Investigation of limiting factors
2. Biological molecules  →  2.2. Carbohydrates and lipids
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5
For Examiner’s Use Anole lizards are found throughout the Caribbean and the surrounding mainland. There are many species. Each species is found only on one island or a small group of islands, apart from Anolis carolinensis which is found in mainland Florida. shows the distribution of four species of anole lizards. Florida U.S.A. Cuba Andros Acklins Anolis smaragdinus Anolis brunneus Anolis carolinensis Anolis porcatus 50 100 km An investigation was carried out into the relationships between these four lizard species, using DNA analysis. The base sequences of a region of mitochondrial DNA from the four species were compared. The results are shown in Table 5.1. The smaller the number, the smaller the differences between the base sequences of the two species. For Examiner’s Use Table 5.1 A. brunneus A. brunneus A. smaragdinus A. smaragdinus 12.1 A. carolinensis A. carolinensis 16.7 15.0 A. porcatus A. porcatus 11.3 8.9 13.2 With reference to Table 5.1, state the species to which A. brunneus appears to be most closely related. The researchers put forward the hypothesis that the three species, A. brunneus, A. smaragdinus and A. carolinensis, have originated from three separate events in which a few individuals of A. porcatus spread directly from Cuba to three different places. Explain how the results in Table 5.1 support the researchers’ hypothesis. Explain how a population of A. porcatus that became isolated on an island could evolve into a new species.
17. Selection and evolution  →  17.3. Evolution
18. Classification, biodiversity and conservation  →  18.1. Classification
18. Classification, biodiversity and conservation  →  18.2. Biodiversity
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6
For Examiner’s Use Explain what is meant by the term gene mutation. Rickets is a childhood disorder involving the softening and weakening of bones. It is usually caused by a lack of vitamin D, calcium ions or phosphate ions. A rare form of rickets that cannot be successfully treated with vitamin D therapy is caused by a mutant allele on the X chromosome. shows a pedigree chart for a family that has a history of this condition. key: male without rickets female without rickets male with rickets female with rickets Using the symbols XR for the mutant allele on the X chromosome Xr for the non-mutant allele on the X chromosome state the genotypes of the following individuals. For Examiner’s Use The gene in which this mutation occurs codes for a protein found in the cells of the proximal convoluted tubule of the kidney. This protein is involved in phosphate ion transport across membranes. Suggest why individuals with this mutant allele show symptoms of rickets.
16. Inheritance  →  16.2. The roles of genes in determining the phenotype
19. Genetic technology  →  19.2. Genetic technology applied to medicine
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7
For Examiner’s Use The initial stages of respiration convert one molecule of glucose into two molecules of a 3C compound. State the name given to these initial stages where these stages occur in cells the total number of ATP molecules formed during these stages. Most of the ATP formed in respiration is produced within the mitochondria by oxidative phosphorylation. State the location, in the mitochondrion, of oxidative phosphorylation. Outline the process of oxidative phosphorylation. For Examiner’s Use In an investigation, mammalian liver cells were homogenised (broken up) and the resulting homogenate centrifuged. Samples of the complete homogenate and samples containing only nuclei, only ribosomes, only mitochondria or only the remaining cytosol were incubated with: glucose pyruvate glucose and cyanide pyruvate and cyanide Cyanide inhibits oxidative phosphorylation. After incubation the presence or absence of carbon dioxide and lactate in each sample was determined. The results are summarised in Table 7.1. Table 7.1 samples of homogenate complete only nuclei only ribosomes only mitochondria only cytosol carbon dioxide lactate carbon dioxide lactate carbon dioxide lactate carbon dioxide lactate carbon dioxide lactate 1 glucose ✓ ✓ ✘ ✘ ✘ ✘ ✘ ✘ ✘ ✓ 2 pyruvate ✓ ✓ ✘ ✘ ✘ ✘ ✓ ✘ ✘ ✓ 3 glucose and cyanide ✘ ✓ ✘ ✘ ✘ ✘ ✘ ✘ ✘ ✓ 4 pyruvate and cyanide ✘ ✓ ✘ ✘ ✘ ✘ ✘ ✘ ✘ ✓ ✘ = absent ✓ = present With reference to Table 7.1, name the two organelles not involved in respiration. 1. 2. For Examiner’s Use Explain why carbon dioxide is produced when mitochondria are incubated with pyruvate but not when they are incubated with glucose. Explain why, in the presence of cyanide, lactate is produced but carbon dioxide is not.
12. Energy and respiration  →  12.2. Respiration
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8
For Examiner’s Use Cystic fibrosis (CF) is an inherited disease. Explain briefly how two parents who do not have CF may have a child with CF. Describe two ways in which CF affects the lungs. One potential way of treating CF is by using gene therapy. Outline, with reference to CF, what is meant by gene therapy. Describe one possible advantage and one possible disadvantage of using gene therapy to treat CF. advantage disadvantage
19. Genetic technology  →  19.2. Genetic technology applied to medicine
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