Biology 2024 · Oct/Nov

Turn on the blender for 15 seconds and then filter the extract to remove all the small pieces of leaf.

Place the leaf extract in a centrifuge and spin at low speed.

Pour of f the supernatant that contains the chloroplasts. Keep this stock chloroplast suspension ice cold and in the dark.

Prepare 5 dif ferent concentrations of the chloroplast suspension using 10% sucrose solution. The percentage concentrations are 10%, 20%, 30%, 40% and 50% of the stock chloroplast suspension. 7 Put 10 cm and add 1 cm in colour. * 0000800000002 *

Place 1 of the tubes beneath a light source as shown in Fig. 1.2. min s chloroplast suspension

Start a timer . Remove the black plastic film from the tube. Record the time taken for the DCPIP to decolourise so that the chloroplast suspension is green. rate = 1000 t = time taken in seconds for the chloroplast suspension to reach a green colour when all the DCPIP is decolourised.

Repeat step 8 to step 10 for the other tubes.

Suggest two suitable control experiments that the student should carry out as part of the preliminary experiment. 1 2 [2] * 0000800000003 *

The results of the preliminary experiment are shown in Table 1.1. Table 1.1 percentage concentration of chloroplast suspension time taken for DCPIP to decolourise / s of photosynthesis / s 10 351 20 59 30 21 40 10 50 5 for each concentration of chloroplast suspension. Give your answers to one decimal place. [2] * 0000800000004 *

Plot a line graph of the data in Table 1.1 on the grid in Fig. 1.3 to show the ef fect of changing of photosynthesis. [3] * 0000800000005 * 10% sucrose to make the stock chloroplast suspension in step 1. The DCPIP did not decolourise. Suggest why the procedure worked when 10% sucrose solution was used but did not work when distilled water was used in step 1.

The preliminary experiment was carried out at 17 °C. photosynthesis, the student added some details to the instructions in steps 7, 8 and 9 to make sure that accurate results were recorded. Describe an investigation that the student could follow to determine the ef fect of temperature the investigation. Y our method should be set out in a logical order and be detailed enough to allow another person to follow it. [Total: 15] * 0000800000007 * BLANK PAGE * 0000800000008 * northern Europe and other parts of the world. The wings of these beetles are covered by two tough structures known as elytra, as shown in Fig. 2.1. protective elytra A. bipunctata is an important predator of insect pests such as aphids. These insect pests feed on plants, including many crop species. Population numbers of A. bipunctata fluctuate in response to changes in the population numbers of their prey. The elytra of A. bipunctata show phenotypic variation. The three most common morphs (forms) found in northern Europe are known as typica (T), quadrimaculata (Q) and sexpustulata (S).red black typica (T) non-melanic melanic quadrimaculata (Q) sexpustulata (S)red black In northern Europe, the populations of A. bipunctata hibernate (are inactive) during the winter. As the temperature increases in early spring, the populations become active. The populations remain active and produce three generations before the next winter begins. * 0000800000009 *

It was observed that in some areas in populations of A. bipunctata, the proportion of the year in the autumn. Researchers carried out a study in one area of Germany to compare the proportion of generations had been produced. The data were collected over a period of 12 years. The results are shown in Fig. 2.3. sampling times percentage frequency in sampled populations of non-melanic and melanic phenotypes of A. bipunctata 1929 spring autumn 1930 spring autumn 1931 spring autumn 1932 spring autumn 1933 spring autumn 1934 spring autumn 1935 spring autumn 1936 spring autumn 1937 spring autumn 1938 spring autumn 1939 spring autumn 1940 spring autumnKey non-melanic melanic no dataND NDND Fig. 2.3 * 0000800000010 *

Outline a method that could be followed to collect the data shown in Fig. 2.3.

The researchers concluded that there was a change in the proportion of the two showing a higher percentage frequency in the spring. Suggest two pieces of extra information about the investigation that researchers should provide to improve confidence in their conclusion.

Environmental temperature is one of many dif ferent selection pressures that could be acting on populations of A. bipunctata. The melanic phenotypes, morphs Q and S, are considered to be better adapted to the climate in northern Europe as they absorb heat energy more readily Scientists investigated the ef fect of temperature on the body temperature of the two dif ferent phenotypes. (inactive) and were kept outdoors for six weeks before the investigation. same mass. switched off at 5 minutes. The results are shown in Fig. 2.4. morphs. Explain how the information shown in Fig. 2.4 supports this conclusion.

State two further conclusions that can be made from the results shown in Fig. 2.4. 0 1 2 3 4 5 6 7 8 9 temperature / °C time / min 0 1 2 3 4 5 6 7 8 9 temperature / °C time / min Q and T males mass = 10.1 g and 10.0 g Q and T females mass = 13.8 g and 13.9 gKey melanic phenotype non-melanic phenotype air temperature heat lamp on heat lamp off Fig. 2.4 * 0000800000013 *

The scientists noticed that there appeared to be a relationship between body mass and the increase in body temperature. The scientists therefore continued to take results from another 17 individuals. They recorded the maximum dif ference between body temperature and the air temperature for each individual in their investigation. All the results of the investigation are shown in Table 2.1 and Fig. 2.5. Table 2.1 phenotype morph sex mass / g maximum temperature difference / °C male 8.4 1.1 male 9.2 2.7 male 10.0 2.9 male 10.8 4.0 female 11.5 5.4 female 12.0 5.9 female 13.1 5.3 female 13.9 7.3 melanic Q male 8.4 3.5 female 8.6 3.7 male 10.1 5.3 female 10.4 4.8 female 11.0 8.5 male 11.7 7.5 female 13.1 7.3 female 13.8 10.2 S male 8.6 4.1 male 9.1 3.7 male 10.8 5.9 female 13.6 10.0 * 0000800000014 * 7 8 9 10 mass / g maximum temperature difference / °C 11 12 13 14 State a statistical test that could be used to analyse the relationship shown in Table 2.1 and Fig. 2.5. Justify your choice. statistical test justification [3] * 0000800000015 *

The researchers assumed that the appearance of the elytra was controlled by a single gene with two alleles. Breeding experiments showed that the melanic phenotype is dominant to the Researchers made the hypothesis: Some melanics are homozygous dominant, and some melanics are heterozygous. Outline a breeding experiment that could be carried out to test this hypothesis, and state the results you would expect if the hypothesis is supported. [Total: 15] The boundaries and names shown, the designations used and the presentation of material on any maps contained in this question paper/insert do not imply official endorsement or acceptance by Cambridge Assessment International Education concerning the legal status of any country , territory, or area or any of its authorities, or of the delimitation of its frontiers or boundaries. reasonable ef fort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Local Examinations Syndicate (UCLES), which is a department of the University of Cambridge. * 0000800000016 *