AQA GCSE Biology Paper 1 Higher November 2020 question paper 8461/1H

This question is about photosynthesis.

Complete the word equation for photosynthesis.

Describe how energy for the photosynthesis reaction is gained by plants.

Students investigated the effect of temperature on the rate of photosynthesis. The students shone light from a lamp onto pondweed and measured the volume of oxygen produced per hour. Table 1 shows the results.

Calculate mean value X.

The students identified one anomalous result in Table 1.

Draw a ring around the anomalous result in Table 1.

Suggest one possible cause of the anomalous result.

How did the students deal with the anomalous result?

Give one factor the students should have kept constant in this investigation.

Table 1 is repeated below.

Why did the rate of photosynthesis decrease from 35 °C to 45 °C?

Complete Figure 1 using data from Table 1. You should:

• label the y-axis

• use a suitable scale for the y-axis

• plot the mean data from Table 1 for temperatures from 25 °C to 45 °C

• draw a line of best fit.

Diffusion is an important process in animals and plants.

What is meant by the term diffusion?

Figure 2 shows part of a leaf.

Molecules of carbon dioxide diffuse from the air into the mesophyll cells. Which two changes will increase the rate at which carbon dioxide diffuses into the mesophyll cells? Tick (✓) two boxes.

Diffusion also happens in the human lungs. Figure 3 shows the human breathing system.

Explain how the human lungs are adapted for efficient exchange of gases by diffusion.

Figure 4 shows a root hair cell.

Name the process by which water molecules enter the root hair cell.

Nitrate ions need a different method of transport into the root hair cell. Explain how the nitrate ions in Figure 4 are transported into the root hair cell. Use information from Figure 4 in your answer.

Figure 5 shows three types of cell.

Give two similarities between the prokaryotic cell and the eukaryotic cells in Figure 5.

Give three differences between the prokaryotic cell and the eukaryotic cells in Figure 5.

Calculate the ratio of the size of the bacterial cell to the size of the mesophyll cell.

Name the type of cell division that produces genetically identical body cells for growth and repair.

Figure 6 shows a cell cycle.

What percentage of the time for one cell cycle is represented by stage 2 and stage 3 together? Tick (✓) one box.

Describe what happens during each stage of the cell cycle.

Lipases break down lipids. Which two products are formed when lipids are broken down? Tick (✓) two boxes.

One model used to explain enzyme action is the ‘lock and key theory’.

Figure 7 shows a model of the theory.

Explain the ‘lock and key theory’ of enzyme action. Use information from Figure 7 in your answer.

There are many different types of lipase in the human body. Why does each different type of lipase act on only one specific type of lipid molecule?

Students investigated the presence of starch and glucose in the leaves of

geranium plants. This is the method used.

1. Place two identical geranium plants on a bench near a sunny window for two days.

2. After two days:

leave one plant near the window for two more days.

place one plant in a cupboard with no light for two more days.

3. Remove one leaf from each plant.

4. Crush each leaf to extract the liquid from the cells.

5. Test the liquid from each leaf for glucose and for starch.

Describe how the students would find out if the liquid from the leaf contained glucose.

Describe how the students would find out if the liquid from the leaf contained starch.

Table 2 shows the students’ results.

Explain why the leaf in the light for four days contained both glucose and starch.

Explain why the leaf left in a cupboard with no light for two days did contain glucose but did not contain starch.

Suggest one way the students could develop the investigation to find out more about glucose and starch production in plants.

Many plants have evolved defence mechanisms. Figure 8 shows part of a gorse plant and part of a deadly nightshade plant.

The gorse plant has evolved to have sharp thorns. What type of defence response are thorns?

How do thorns defend the gorse plant?

The deadly nightshade plant has poisonous berries. What type of defence response are poisonous berries?

A scientist noticed that in one area the gorse plants had yellow leaves and had stunted growth. One reason for yellow leaves and stunted growth is a deficiency of nitrate ions in the soil. Explain two other possible reasons for the yellow leaves and stunted growth. Do not refer to nitrate ions in your answer.

The gorse plant has nodules on its roots.

The nodules are part of the living root tissue. Bacteria which convert nitrogen gas into soluble nitrate ions live in the nodule tissue. Figure 9 shows the nodules on the roots.

Suggest how the nodules benefit the bacteria.

Explain how the nodules benefit the gorse plant.

For many years drugs have been extracted from plants. Which plant material was chewed as a painkiller? Tick (✓) one box.

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Data from ‘The Million Women’ survey in the UK was collected for over 15 years. Scientists analysed the data to study the effect of consuming alcohol on liver disease. The scientists:

• included 400 000 women who regularly consumed alcohol

• included 400 000 women who did not consume alcohol

• excluded women who already had a liver disease.

Age and gender were two factors controlled in this analysis. Many other factors were also controlled. Suggest two other factors which the scientists would have controlled.

The data was analysed for:

• women who drank alcohol with meals

• women who drank alcohol not with meals

• women who did not drink alcohol.

During the survey approximately 1500 women developed a liver disease called cirrhosis of the liver. Scientists calculated the relative risk of developing cirrhosis of the liver for each group who consumed alcohol. A relative risk of 1.0 means there was no statistical difference between the groups who did consume alcohol and the group who did not consume alcohol. Figure 10 shows a summary of the results.

A woman drinks 150 g of alcohol per week not with meals. The woman decides to change to drinking 150 g of alcohol per week with meals. Calculate the percentage decrease in relative risk of developing cirrhosis of the liver for this woman.

One glass of wine contains 12 g of alcohol. A different woman drinks two glasses of wine each day with her meals. Calculate the relative risk of developing cirrhosis of the liver for this woman.

Figure 10 is repeated below.

Consuming alcohol with meals instead of not with meals decreases the relative risk of developing cirrhosis of the liver. Give two other conclusions about the relative risk of developing cirrhosis of the liver related to alcohol consumption. Use data from Figure 10 in your answer.

Suggest two reasons why the data is considered to be valid.

Suggest one aspect of the survey which might reduce validity.

Cirrhosis of the liver leads to liver failure. Describe the effects of liver failure on the human body.

Monoclonal antibodies (mAbs) are usually made using mouse lymphocytes. Candida albicans infection produces serious symptoms in patients with a poor immune system. Recently scientists have produced mAbs to Candida albicans using human lymphocytes produced naturally after an infection.

Candida albicans lives in the throat of infected patients. A sample is taken from the throat of a patient with a suspected Candida albicans infection. The sample is transferred onto a microscope slide. Describe how the mAbs and a fluorescent dye could be used to see any Candida albicans pathogens on the slide.

In a laboratory the human lymphocyte mAbs were injected into animals infected

with Candida albicans. The mAbs caused increased phagocytosis of the Candida albicans pathogens. Doctors intend to start a trial to give the mAbs to patients severely ill with Candida albicans.

Explain how increased phagocytosis of the Candida albicans pathogen will help the patient.

It has been shown that this mAbs treatment is effective in the laboratory using both:

• infected tissue culture cells

• infected live animals.

The mAbs treatment for Candida albicans is now ready for clinical trials on people. Describe how the clinical trials should be carried out.

Scientists have also used human lymphocytes to make mAbs to other pathogens and to some types of cancer cells. Suggest one reason why these new mAbs have been more successful in treating diseases in humans than mAbs made using mice.