AQA A Level Biology Paper 1 June 2019 question paper 7402/1

Describe how a non-competitive inhibitor can reduce the rate of an enzyme-controlled reaction.

Pectin is a substance found in some fruit and vegetables.

A scientist investigated the effect of pectin on the hydrolysis of lipids by a lipase enzyme.

His results are shown in Figure 1.

The scientist concluded that pectin is a non-competitive inhibitor of the lipase enzyme.

Use Figure 1 to explain why the scientist concluded that pectin is a non-competitive inhibitor.

The scientist also found that pectin stops the action of bile salts. He prepared two suspensions:

• suspension A – lipid and bile salts

• suspension B – lipid, bile salts and pectin.

He did not add lipase to either suspension.

He observed samples from the suspensions using an optical microscope. Figure 2 shows what he saw in a typical sample from each suspension.

Calculate the maximum length of the large lipid droplet marked X in Figure 2.

Using a ruler with millimetre intervals always includes an uncertainty in the measurement. Use the uncertainty in your measurement to determine the uncertainty of your calculated maximum length.

You can assume there is no uncertainty in the magnification.

No large lipid droplets are visible with the optical microscope in the samples from suspension A.

Explain why.

Table 1 shows cell wall components in plants, algae, fungi and prokaryotes. Complete Table 1 by putting a tick (✓) where a cell wall component is present.

Cell walls make up much of the fibre that people eat.

Scientists investigated the relationship between the mass of fibre people ate each day and their risk of cardiovascular disease (CVD).

They gathered data from a large sample of people and used this to calculate a relative risk.

A relative risk of 1 means there is no difference in risk between the sample and the whole population.

A relative risk of < 1 means CVD is less likely to occur in the sample than in the whole population.

A relative risk of > 1 means CVD is more likely to occur in the sample than in the whole population.

Their results are shown in Figure 3. A value of ± 2 standard deviations from the mean includes over 95% of the data.

A student concluded from Figure 3 that eating an extra 10 g of fibre per day would significantly lower his risk of cardiovascular disease.

Evaluate his conclusion.

The scientists estimated the mean mass of fibre eaten per day using a food frequency questionnaire (FFQ).

The FFQ asks each person how often they have eaten many types of food over the past year.

An alternative method to calculate fibre eaten is for a nurse to ask each person detailed questions about what they have eaten in the last 24 hours.

Suggest one advantage of using the FFQ method and one disadvantage of using the FFQ method compared with the alternative method.

A group of students investigated biodiversity of different areas of farmland. They collected data in each of these habitats:

• the centre of a field

• the edge of a field

• a hedge between fields.

Their results are shown in Figure 4.

What data would the students need to collect to calculate their index of diversity in each habitat?

Do not include apparatus used for species sampling in your answer.

Give two ways the students would have ensured their index of diversity was representative of each habitat.

Modern farming techniques have led to larger fields and the removal of hedges between fields.

Use Figure 4 to suggest why biodiversity decreases when farmers use larger fields.

Farmers are now being encouraged to replant hedges on their land.

Suggest and explain one advantage and one disadvantage to a farmer of replanting hedges on her farmland.

Scientists collected data on 800 000 human births. The data showed the mass of each baby at birth and whether the baby needed to be transferred to a special care unit for very ill babies.

Their results are shown in Figure 5.

Use Figure 5 to explain how human mass at birth is affected by stabilising selection.

The scientists studied 1500 more births. They recorded the mass at birth of each baby and the nature of the KIR gene in the mother’s genome.

Some of their results are shown in Table 2.

The scientists used a statistical test to test the following null hypothesis:

‘The presence of KIR2DS1 in the mother’s genome does not affect the frequency of births above 4500 g’

Tick (✓) one box that gives the name of the statistical test that the scientists should use with the data in Table 2 to test this null hypothesis.

The scientists calculated a P value of 0.03 when testing their null hypothesis.

What can you conclude from this result? Explain your answer.

Describe the structure of the human immunodeficiency virus (HIV).

Some people infected with HIV do not develop AIDS. These people are called HIV controllers.

Scientists measured the number of HIV particles (the viral load) and the number of one type of T helper cell (CD4 cells) in the blood of a group of HIV controllers and also in a group of HIV positive patients who had symptoms of AIDS.

The median values and the range of their results are shown in Table 3.

A test sample of 500mm³ of blood is taken from an HIV controller to determine the viral load.

Tick (✓) one box that shows the number of virus particles that would be present in a test sample of blood taken from an HIV controller with the median viral load.

Use the data in Table 3 and your knowledge of the immune response to suggest why HIV controllers do not develop symptoms of AIDS.

Scientists investigated the cell cycle in heart cells taken from mice 6 days before their birth and then at 4, 14 and 21 days after their birth.

Their results are shown in Table 4. Age 0 days = day of birth.

Describe and explain the data in Table 4.

The scientists determined the percentage of heart cells undergoing DNA replication by using a chemical called BrdU. Cells use BrdU instead of nucleotides containing thymine during DNA replication.

Describe how BrdU would be incorporated into new DNA during semi-conservative replication.

Cells with BrdU in their DNA are detected using an anti-BrdU antibody with an enzyme attached.

Use your knowledge of the ELISA test to suggest and explain how the scientists identified the cells that have BrdU in their DNA.

Ulva lactuca is an alga that lives on rocks on the seashore. It is regularly covered by seawater.

Figure 6 shows a diagram of one Ulva lactuca alga.

Unlike plants, Ulva lactuca does not have xylem tissue.

Suggest how Ulva lactuca is able to survive without xylem tissue.

Ulva lactuca has a haploid and a diploid form.

Figure 7 shows the life cycle of Ulva lactuca.

On Figure 7 complete each box with an appropriate letter to show the type of cell division happening between each stage in the life cycle. Use ‘ T ’ to represent mitosis and ‘ E ’ to represent meiosis.

Ulva prolifera also produces haploid, mobile single cells that can fuse to form a zygote.

Suggest and explain one reason why successful reproduction between Ulva prolifera and Ulva lactuca does not happen.

The water potential of leaf cells is affected by the water content of the soil.

Scientists grew sunflower plants. They supplied different plants with different volumes of water.

After two days, they determined the water potential in the leaf cells by using an instrument that gave a voltage reading.

The scientists generated a calibration curve to convert the voltage readings to water potential.

Figure 8 shows their calibration curve.

The scientists needed solutions of known water potential to generate their calibration curve.

Table 5 shows how to make a sodium chloride solution with a water potential of

Complete Table 5 by giving all headings, units and volumes required to make 20 cm³ of this sodium chloride solution.

Table 6 shows some of the concentrations of sodium chloride solution the scientists used and the water potential of each solution.

There is a linear relationship between the water potential and the concentration of sodium chloride solution.

Use the data in Table 6 to calculate the concentration of sodium chloride solution with a water potential of 3.41MPa

In addition to determining the water potential in the leaf cells, the scientists measured the growth of the leaves.

They recorded leaf growth as a percentage increase of the original leaf area.

Their results are shown in Figure 9.

One leaf with an original area of 60cm² gave a voltage reading of 7 µV

Use Figure 8 (on page 28) and Figure 9 to calculate by how much this leaf increased in area. Give your answer in cm²

Sunflowers are not xerophytic plants. The scientists repeated the experiment with xerophytic plants.

Suggest and explain one way the leaf growth of xerophytic plants would be different from the leaf growth of sunflowers in Figure 9.

Use your knowledge of gas exchange in leaves to explain why plants grown in soil with very little water grow only slowly.

A scientist investigated the affinity for oxygen of horse haemoglobin and mouse haemoglobin.

Some of their results are shown in Table 7.

Plot the haemoglobin saturation data from Table 7 and use these points to sketch the full oxyhaemoglobin dissociation curves for a horse and a mouse.

The following equation can be used to estimate the metabolic rate of an animal.

BM = body mass in grams

Use this equation to calculate how many times faster the metabolic rate of a mouse is than the metabolic rate of a horse.

The data in Table 7 show differences between the oxyhaemoglobin dissociation curve for a mouse and the oxyhaemoglobin dissociation curve for a horse.

Suggest how these differences allow the mouse to have a higher metabolic rate than the horse.

Mammals such as a mouse and a horse are able to maintain a constant body temperature.

Use your knowledge of surface area to volume ratio to explain the higher metabolic rate of a mouse compared to a horse.

Explain five properties that make water important for organisms.

Describe the biochemical tests you would use to confirm the presence of lipid, non-reducing sugar and amylase in a sample.

Describe the chemical reactions involved in the conversion of polymers to monomers and monomers to polymers.

Give two named examples of polymers and their associated monomers to illustrate your answer.