Biology Paper 3 past paper questions

A student who saw these results concluded that in any future trials of this kinesin inhibitor with people, a concentration of 100 nmol dm−3 would be most appropriate to use. Do these data support the student's conclusion? Give reasons for your answer. (4 marks)

1. (No, because) at 100 there are still some (7%) cancer cells dividing/undergoing mitosis; 2. So, cancer not destroyed/may continue to form tumours 3. Best concentration may be between 100 and 1000/need trials between 100 and 1000; 4. This research in culture, don't know effect of KI on people; OR 5. (Yes, because) above 100 produces little increase in % of cells not dividing/undergoing mitosis 6. Above 100 may be harmful (to body) 7. Higher concentrations more expensive; 8. (Above 100) will have more effect on (rapidly dividing) cancer cells;

The scientists concluded that an increase in phosphate in the embryo was linked to growth of the embryo. Suggest two reasons why an increase in phosphate can be linked to growth of the embryo.

1. (Phosphate required) to make RNA; 2. (Phosphate required) to make DNA; 1 and 2. 3. (Phosphate required) to make ATP/ADP; 4. (Phosphate required) to make membranes 5. (Phosphates required) for phosphorylation.

At the start of their investigation, the scientists made a solution of kinesin inhibitor (KI) with a concentration of 10 000 nmol dm. They used this to make the other concentrations by a series of dilutions with water. Describe how they made 100 cm3 of 1000 nmol dm−3 solution of kinesin inhibitor. (2 marks)

1. 10 cm3 of 10 000 nmol dm−3/ (original) solution; 2. 90 cm3 of water.

The scientists carried out a statistical test to see if the difference in the distribution of each species between the canopy and understorey was due to chance. The P values obtained are shown in the table. Explain what the results of these statistical tests show. (3 marks)

1. For Zaretis itys, difference in distribution is probably due to chance / probability of being due to chance is more than 5%; 2. For all species other than Zaretis itys, difference in distribution is (highly) unlikely to be due to chance; 3. Because P < 0.001 which is highly significant / is much lower than 5%.

By how many times is the species diversity in the canopy greater than in the understorey? Show your working. Use the formula to calculate species diversity, where N is the total number of organisms of all species and n is the total number of organisms of each species. (3 marks)

1. Index for canopy is 3.73; 2. Index for understorey is 3.30; 3. Index in canopy is 1.13 times bigger.

Explain what the standard deviations suggest about the difference in mean total number of dung beetles between the different types of farm. (2 marks)

1. No overlap in standard deviations 2. Difference in mean total significant /is not due to chance

Suggest why the development of a monopolar mitotic spindle would prevent successful mitosis. (2 marks)

1. No separation of chromatids/chromosomes/centromeres; 2. Chromatids/ chromosomes all go to one pole/ end/ sides of cell/ not pulled to opposite poles; 3. Doubles chromosome number in cell/one daughter cell gets no chromosomes or chromatids.

The behaviour of the herbivores in having their heads up has a benefit but it also has costs. The benefit is being able to see, and escape from, predators. Suggest and explain one cost to the herbivores of this behaviour. (2 marks)

1. Raising head makes them more visible to predators; 2. So more likely to be attacked/eaten/killed.

On the intensive farms, the farmers had removed hedges to increase land for grazing. This resulted in a decrease in the diversity of birds on these farms. Explain why the removal of hedges caused a decrease in the diversity of birds. (3 marks)

1. Removes species/types of plant/insect 2. Fewer food sources; 3. Fewer habitats/niches;

The chilling requirement of seeds of certain plant species is considered to be an adaptation for survival in countries with seasonal changes in environmental conditions. Suggest how this adaptation may enable these plant species to survive and respond to seasonal changes. (3 marks)

1. Seeds/embryo remain dormant/inactive in winter/cold 2. Seeds/plants develop in spring/summer 3. Plant photosynthesise (in spring/when warm); 4. Produce (more) seeds/offspring in spring/growing season;

Suggest how this student would obtain data to give a more precise value for the index of diversity of this habitat. (2 marks)

1. Take more samples and find mean; 2. Method for randomised samples described.

What is the mean species richness for dung beetles on the rough grazing farms? (1 mark)

14

Calculate the index of diversity of this sample. Show your working. Use the following formula to calculate the index of diversity where N is the total number of organisms of all species and n is the total number of organisms of each species. (2 marks)

2.68(6)

What is the species richness of this sample? (1 mark)

4 species richness = how many species

Calculate the ratio of the mean mass of phosphate found at 5 °C to the mean mass of phosphate found at 25 °C after 9 weeks of chilling.

4:1 (anywhere from 3.7:1 - 4.1:1)

The scientists observed both groups of animals for 75 hours. Use data from the graph to calculate the difference in the mean number of hours spent by each species looking around in the area where there were many predators. Show your working. (2 marks)

75 x 0.11 = 8.25 75 x 0.23 = 17.25 17.25 - 8.25 = 9 hours

The letters A, B, C, D and E represent stages in mitosis. 3 A - anaphase B - interphase C - metaphase D - prophase E - telophase Write one of the letters, A to E, in the box to complete the following statement - Chromosomes line up on the equator of the mitotic spindle in: (1 mark)

C - metaphase

In addition to the information provided in Figures 1 and 2, what other measurement is required to calculate an index of diversity for dung beetles? (1 mark)

Number of individuals in each species (of dung beetle)

Species richness and an index of diversity can be used to measure biodiversity within a community. What is the difference between these two measures of biodiversity? (1 mark)

Species richness measures only number of (different) species / does not measure number of individuals.

The traps in the canopy were set at 16-27 m above ground level. Suggest why there was such great variation in the height of the traps. (1 mark)

Trees vary in height.

The scientists concluded that these herbivores spend more time looking for predators in areas where there are many predators. Do these data support this conclusion? Give reasons for your answer. (4 marks)

Yes because 1. each species mean time looking around is greater when predators are present 2. differences appear to be significant because SDs do not overlap; OR No because 3. Wildebeest spend same (mean) time looking around where many predators as impalas where few predators; 4. Don't know what they are looking for (when heads up); 5. Habitats might be different in different areas (which could affect the behaviour);

Phospholipids are one of the storage molecules found in cherry seeds. Name the type of reaction used to break down phospholipids to release phosphate. (1 mark)

hydrolysis

The scientists placed traps to collect the dung beetles at sites chosen at random. Explain the importance of the sites being chosen at random. (1 mark)

no bias