Paper 1 Higher

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AQA GCSE Biology

Paper 1 Higher

June 2025
100 marks · 1 hour 45 minutes · 8 questions

A real paper, in its real order. Each description and answer is ours, so you can see how each mark is earned. Not made or approved by AQA.

The real exam gives you an equations sheet. See the equations

PrepWise guide: AQA GCSE Biology Paper 1 HigherQuestion 1

Measles is a disease that spreads between people. A survey recorded how serious the illness was in children who had been vaccinated and in children who had not (Figure 1).

Figure 1
Bar chart on a grid with two groups of bars, Mild and Severe, each with a vaccinated bar and an unvaccinated bar, and the percentage of children on the vertical axis.
01.1[1 mark]

Name the type of pathogen that causes measles.

See how you would score on Pathogens and Disease Transmission
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A virus. Measles is caused by a virus, not by a bacterium, a fungus or a protist.

1 mark

How the marks are earned

  1. A virus.
Revise Pathogens and Disease Transmission
01.2[1 mark]

State one way that measles passes between people.

See how you would score on Pathogens and Disease Transmission
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Droplets from a cough or a sneeze. An infected person breathes out droplets that carry the virus, and another person breathes them in.

1 mark

How the marks are earned

  1. Any one of: droplets, saliva or mucus; sneezes; coughs.
  2. Other ways of passing on a pathogen, given with no more detail (touch or direct contact are examples), earn nothing, but they do not cancel a correct route given with them.
  3. Transfer through sexual intercourse is not accepted.
Revise Pathogens and Disease Transmission
01.3[1 mark]

Aspirin treats pain in adults. Name the plant that aspirin first came from.

See how you would score on Antibiotics and Drug Resistance
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Willow. The pain-relieving chemical was first found in willow bark, and aspirin was developed from it.

1 mark

How the marks are earned

  1. Willow.
  2. Naming a part of the willow, such as the bark or the leaves, neither adds to the answer nor takes away from it.
Revise Antibiotics and Drug Resistance
01.4[1 mark]

Name one symptom of measles that is not pain.

See how you would score on Pathogens and Disease Transmission
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A fever, meaning a high body temperature. A red skin rash is another symptom.

1 mark

How the marks are earned

  1. Fever (a high temperature) or a rash.
  2. Also credited: sweating, spots, lumps or boils (red or not), red or itchy skin, a cough or sneeze, a runny or blocked nose, a sore throat, red, watery or sore eyes, a headache.
  3. Flu-like symptoms with no detail, and pain with no detail, earn nothing.
  4. Nausea and diarrhoea are not accepted.
Revise Pathogens and Disease Transmission
01.5[2 marks]

Explain why vaccinating many children reduces how far the measles pathogen spreads through the population.

See how you would score on Vaccination and Herd Immunity
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Vaccinated children make antibodies quickly if the virus enters, so they are much less likely to become ill. Far fewer people in the population carry the virus.

So an unvaccinated person is much less likely to meet someone who is infectious, and the virus is less likely to be passed on to them.

2 marks

How the marks are earned

  1. One mark: after vaccination a smaller number of people in the population carry the pathogen, the measles virus, or have the disease. Saying vaccination gives herd immunity is also credited here, and a named type of pathogen, such as a virus, counts as the pathogen.
  2. One mark: so the pathogen, measles or the disease is less likely to be passed on, spread or caught. Saying a person is less likely to meet someone who is infected also earns this.
  3. Both marks go to a single answer saying fewer people have measles and so spread is less likely. Being less likely to meet an infected person is also accepted for the second mark.
  4. Saying only that the spread of the disease is reduced repeats the question and earns nothing.

Why it scores

The first mark is for the smaller number of infected people. This is the idea behind herd immunity.

The second mark is for the effect on spread. It must say that passing on or catching the pathogen is less likely, and not only that spread is reduced.

Revise Vaccination and Herd Immunity
01.6[3 marks]

The survey covered 240 children. From Figure 1, find the number of children without the vaccine whose measles was severe.

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Read the grey bar for severe infection: 45% of the children.

number of children = 45 ÷ 100 × 240

number of children = 108

3 marks

How the marks are earned

  1. One mark: read the percentage for unvaccinated children with severe measles from Figure 1, which is 45.
  2. One mark: that percentage of 240, set out as 45 ÷ 100 × 240, or as 0.45 × 240.
  3. One mark: the final number of children, 108.
  4. Full marks for a correct answer of 108 on its own, with no working shown.
  5. If the wrong bar was read, one mark is allowed for 7 or 7.2 where the working shows 3 per cent was used, or for 88, 88.8 or 89 where the working shows 37 per cent was used.

Why it scores

Reading the right bar from Figure 1 earns the first mark. The unvaccinated children are the grey bars, and the severe group is the right-hand pair.

Taking this percentage of the whole survey earns the second mark. 0.45 × 240 is also accepted.

The final value earns the third mark. A correct 108 with no working earns all three.

Revise Vaccination and Herd Immunity
01.7[2 marks]

Back the advice to vaccinate every child against measles, using numbers from Figure 1.

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Severe measles was rarer in the vaccinated children. Of all the children surveyed, 3% were vaccinated and had severe measles, but 45% were unvaccinated and had severe measles.

The unvaccinated share is 42 percentage points higher, or 15 times as large. So vaccination makes measles less severe.

2 marks

How the marks are earned

  1. One mark: a statement that vaccinated children were less often severely ill (a smaller percentage), or that unvaccinated children were more often severely ill, or that the vaccine lessens how serious measles is.
  2. The converse is accepted, and so is an answer given as a number of children or as the chance of getting the disease.
  3. One mark: figures from Figure 1 used in a comparison, for example 42 per cent higher in the unvaccinated or 15 times higher.
  4. Correct raw values used in a comparison also earn this mark.

Why it scores

The first mark is for saying that vaccination goes with less severe measles. The four bars in Figure 1 together make up all the children surveyed, so each bar is a share of everyone. The statement alone is not enough for the data mark.

The second mark is for data used as a comparison. A difference or a multiple works, and so do the two raw values placed side by side.

Revise Vaccination and Herd Immunity
AQA owns the copyright in its exam questions, so we cannot copy their wording. These descriptions and answers are ours. Not made or approved by AQA.[11 marks]
PrepWise guide: AQA GCSE Biology Paper 1 HigherQuestion 2

This question links diet, obesity and enzymes. Table 1 lists sugar intake and obesity rate in seven countries, and Figure 3 relates sugar intake to tooth decay. A student mixed starch with amylase at four temperatures. Five minutes later, iodine solution, yellow-brown to begin with, was added to each mixture.

Table 1
Table listing seven countries, lettered A to G, each with a sugar figure and an obesity figure in two number columns.
Figure 2
Blank squared grid with a numbered scale along the bottom for sugar eaten, nothing written beside the upright axis, and a pair of crosses named A and B.
Figure 3
A graph with an arrow on each axis and no numbers. Many small separate dots fill the space between the axes.
Table 2
Two column table pairing four temperatures with the colour seen in each test.
Table 3
Table with columns for protease enzyme, optimum pH and organ where the enzyme is active, and rows for pepsin and trypsin, with the organ column left empty.
02.1[1 mark]

Obesity raises the risk of certain diseases. Give one of them.

See how you would score on The Heart and Circulation
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Type 2 diabetes. Being obese raises the chance of it, and of heart disease and some cancers.

1 mark

How the marks are earned

  1. Diabetes, specifically its type 2 form.
  2. Also credited: high blood pressure, heart disease or CHD (the word coronary is optional), cardiovascular disease (CVD), stroke, heart attack, arthritis (including osteoarthritis), cancer, sleep apnoea, liver disease, or any other disease that obesity is truly linked to.
  3. Diabetes with no type named earns nothing, and type 1 diabetes is not accepted.
  4. High cholesterol and heart failure earn nothing.
Revise The Heart and Circulation
02.2[4 marks]

Finish Figure 2 from Table 1: title and number the vertical axis, plot countries C to G, add a best-fit line.

See how you would score on Cell Biology Practical Investigations
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Mean obesity rate, arbitrary units

scale 0, 5, 10, 15, 20, 25, 30, 35: one large square for 5, one small square for 1

Plot a cross at each of (172, 25), (150, 23), (110, 15), (144, 19) and (192, 32).

Line of best fit: one ruled straight line through the middle of all seven crosses, about 9 at 80 g and about 30 at 192 g

4 marks

How the marks are earned

  1. One mark: the vertical axis labelled mean obesity rate and the unit, arbitrary units, together with a correct even scale that suits the data. The crosses already drawn for A and B must sit at their Table 1 values.
  2. Two marks for placing the five new crosses (C to G) accurately, each within half a small square of its true spot. Three or four placed correctly earn one mark.
  3. Letters beside the points are ignored.
  4. One mark: a line of best fit. A line carried on past the first and last point is ignored.
  5. A line joining the points one to the next with straight sections earns nothing.

Why it scores

The label for the vertical axis names the measure and its unit. With a sensible even scale, it earns the first mark.

The scale has to match crosses A and B, which are already drawn, and has to reach the highest value.

In each pair the first number is sugar intake across and the second is obesity rate up. Correct plotting of all five new points earns two marks. Each cross must be within half a small square, and three or four right points earn one mark.

The line of best fit earns the last mark. It is a single straight line through the trend, not a line joined from point to point.

Revise Cell Biology Practical Investigations
02.3[1 mark]

What kind of graph is Figure 3?

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A scatter graph. Each dot is one measurement of the two variables, and the dots are not joined.

1 mark

How the marks are earned

  1. The word scatter is the essential part. Graph or diagram after it is optional, and scattergram is accepted.
Revise Cell Biology Practical Investigations
02.4[1 mark]

From Figure 3, say how the amount of sugar eaten is related to the number of decayed teeth. Tick one box.

See how you would score on Cell Biology Practical Investigations
Inversely proportional
No correlation
Positive correlation
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Positive correlation. As the mean sugar intake rises, the mean number of decayed teeth rises too.

1 mark

How the marks are earned

  1. Tick positive correlation: both variables rise together.
Revise Cell Biology Practical Investigations
02.5[4 marks]

Table 2 gives the iodine colour for each temperature. Give the reason for the colour seen with the 30 °C and 90 °C mixtures.

See how you would score on Enzymes in Digestion
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30 °C: the iodine stayed yellow-brown, so no starch was left. The amylase was working well at this temperature and broke all the starch down.

90 °C: the iodine turned blue-black, so starch was still there. The high temperature denatured the amylase, because its active site changed shape, so it could not break the starch down.

4 marks

How the marks are earned

  1. 30 °C, first mark: no starch remained, so the iodine stayed yellow-brown. Also credited: starch was broken down, with or without the word all.
  2. 30 °C, second mark: because amylase or the enzyme broke the starch down. Also credited: the amylase or enzyme was active or working, or this was the optimum or best temperature, or close to it, for the enzyme.
  3. 90 °C, third mark: blue-black iodine, because starch was still present. Also credited: the starch was not broken down, or the amylase or enzyme had no effect on the starch.
  4. 90 °C, fourth mark: because the amylase or enzyme had denatured. A description of denaturing is credited, and so is saying it was destroyed or damaged.
  5. Saying the enzyme was killed is not accepted.

Why it scores

Two marks: the colour shows the starch was gone, and the cause is the enzyme breaking it down near its best temperature.

Two marks: the colour shows the starch was not broken down, and the cause is a denatured enzyme. Blue-black means starch is present.

Revise Enzymes in Digestion
02.6[1 mark]

Which change to the investigation would find a more precise best temperature for amylase? Tick one box.

See how you would score on Enzymes in Digestion
Take extra readings from 10 °C up to 30 °C
Take extra readings from 10 °C up to 60 °C
Take extra readings from 30 °C up to 60 °C
Take extra readings from 30 °C up to 90 °C
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Add readings across the range from 10 °C up to 60 °C. Only the 30 °C mixture kept the iodine yellow-brown, so the best temperature lies in the gap around it, and closer readings there narrow it down.

1 mark

How the marks are earned

  1. Tick the option whose extra readings span 10 °C up to 60 °C.
Revise Enzymes in Digestion
02.7[2 marks]

Fill in Table 3: name the part of the digestive system in which pepsin works and the part in which trypsin works.

See how you would score on The Human Digestive System
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Pepsin: the stomach, where the pH is about 2.

Trypsin: the small intestine, where the pH is about 8.

2 marks

How the marks are earned

  1. One mark: pepsin works in the stomach.
  2. One mark: trypsin works in the small intestine. The ileum or the duodenum is also accepted.

Why it scores

Pepsin has an acid optimum, which matches the acid conditions in the stomach.

Trypsin has an alkaline optimum, which matches the small intestine, where the conditions are alkaline.

Revise The Human Digestive System
AQA owns the copyright in its exam questions, so we cannot copy their wording. These descriptions and answers are ours. Not made or approved by AQA.[14 marks]
PrepWise guide: AQA GCSE Biology Paper 1 HigherQuestion 3

A student studied osmosis using potato. A cube was cut, weighed and left in salt solution of concentration 0.4 mol/dm3. When taken out later it was weighed again and the change in mass worked out. Two other salt concentrations were then tested in the same way.

03.1[6 marks]

Judge how good this potato method is, and say what the student should change to make the results more valid.

See how you would score on Cell Transport
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The method is a reasonable start, but the results would not be fully valid because too many variables are not controlled. Its strengths are that every piece is a cube, the mass is recorded before and after, the change in mass is worked out and salt solutions of different concentrations are used.

To improve validity, cut every cube from the same potato and peel off the skin, because different potatoes and skin change how fast water moves. Cut the cubes to the same size and surface area, because a larger surface area lets water move faster.

Leave each cube in the solution for the same time, keep the temperature steady in a water bath and use the same volume of solution that covers the cube, because time, temperature and volume of solution all change how much osmosis happens.

Blot each cube dry before weighing, because water on the surface adds false mass. Work out the percentage change in mass, because the cubes start with different masses.

Repeat each concentration to spot anomalies, and find a mean, because no mean was worked out. Add a boiled cube as a control, because none was included, and test more than three concentrations, because only three were used.

Levels of response, 6 marks

How the marks are earned

  1. This is level marked. For 5 to 6 marks, give a judgement that is strongly linked to a good range of correct reasons, and include a strength and improvements with reasons. For 3 to 4 marks, give some reasons that link up, with or without a simple judgement. For 1 to 2 marks, make relevant points that do not link up. An answer with nothing relevant in it earns no marks.
  2. Strengths worth stating: the same potato used throughout; every piece cut as a cube; weighing at the start and the end; working out the change in mass; testing several salt concentrations.
  3. Improvements and the reason for each: use one potato for all the cubes, because different potatoes may have been used; peel off the skin, because skin may still be on; blot the cube dry before weighing, because water on the surface adds mass; use a set size or surface area, because the size of the cubes was not controlled.
  4. More improvements and reasons: repeat the whole investigation, or each concentration, since the test was done once only, and to spot anomalies; find a mean from the repeats, because the student never worked one out; give every cube the same soaking time, since none was set; hold the temperature steady using a water bath, since nothing controlled it; give every cube an equal amount of solution, or make sure every cube is completely covered, because the amount of solution was never fixed.
  5. More improvements and reasons: calculate the percentage change in mass, because each cube started with a different mass; add a control with a boiled cube, because no control was included; use more than three concentrations, because only three were tested.

Why it scores

Level 3 needs a judgement and a strength. This paragraph gives the judgement first and then the strengths that are really in the method.

Each improvement comes with its reason. Using one potato, removing skin and fixing the size are all controls on the potato itself.

These are controls on the conditions around the potato. The reason is that each one affects the amount of osmosis.

These two points make the measurement fair. The percentage change lets cubes with different starting masses be compared.

Repeats show up anomalies, and a mean reduces the effect of random error. More concentrations give a better picture. Each point is given with its reason.

Revise Cell Transport
03.2[1 mark]

Why does a potato cube put in distilled water gain mass? Tick one box.

See how you would score on Cell Transport
The potato cell contents had a lower concentration than pure water
Pure water was more dilute than the contents of the potato cells
In total the cells held a bigger volume of water than the beaker of distilled water did
In total the cells held a smaller volume of water than the beaker of distilled water did
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The contents of the potato cells are more concentrated than the pure water outside, so water moves into the cells by osmosis and the cube gains mass.

1 mark

How the marks are earned

  1. Tick the option saying pure water was more dilute than the contents of the potato cells.
Revise Cell Transport
03.3[1 mark]

State one difference between active transport and osmosis.

See how you would score on Cell Transport
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Active transport needs energy from respiration and can move substances against a concentration gradient, from low to high concentration. Osmosis needs no energy.

1 mark

How the marks are earned

  1. Any one: active transport needs energy; it moves substances against or up a concentration gradient, from a low to a high concentration; it moves minerals or ions, or another named substance such as glucose (water need not be mentioned).
  2. Saying it does not move water is also credited, and so is saying it is not passive.
  3. Saying only that it is active earns nothing, and neither does saying it goes along or across the concentration gradient.
  4. Moving down a concentration gradient is not accepted.
Revise Cell Transport
AQA owns the copyright in its exam questions, so we cannot copy their wording. These descriptions and answers are ours. Not made or approved by AQA.[8 marks]
PrepWise guide: AQA GCSE Biology Paper 1 HigherQuestion 4

Pondweed gives off bubbles when it photosynthesises. A student tested how temperature changes its rate. Figure 4 shows a single tube. Five of these tubes were left for two minutes in water baths at five different temperatures, then each was timed until 10 bubbles had come off. Table 4 lists the times.

Figure 4
Test tube of water holding a sprig of pondweed, with a few bubbles at the top, and labels for bubble, pondweed and water.
Table 4
Two column table pairing five temperatures with the number of seconds each piece of weed took to give off its bubbles.
04.1[1 mark]

Name the independent variable in the pondweed investigation.

See how you would score on Photosynthesis
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Temperature. It is the variable the student chooses and changes, and the time for 10 bubbles is what is measured.

1 mark

How the marks are earned

  1. Temperature, meaning the temperature of the water bath.
Revise Photosynthesis
04.2[4 marks]

From Table 4, work out how fast the pondweed photosynthesised at 20 °C. Round to 3 significant figures and state the unit.

See how you would score on Photosynthesis
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time at 20 °C = 43 s

rate = 10 ÷ 43

rate = 0.23255...

rate = 0.233 bubbles/s (3 s.f.)

4 marks

How the marks are earned

  1. One mark: rate as bubbles ÷ time, with the time for 20 °C read from Table 4, set out as 10 ÷ time, or as 1 ÷ time × 10.
  2. One mark: the unrounded value, 0.23255 and so on.
  3. One mark: the value to 3 significant figures, 0.233. A wrongly worked value of the student's own, correctly rounded to 3 significant figures, is allowed, and so is a calculation using a different time from Table 4.
  4. One mark: the unit, bubbles per second. A correct calculation in another unit is allowed, such as bubbles per minute (14.0) or bubbles per hour (837). The unit must match the student's own value. The same unit written bubbles/s is accepted.
  5. A correct answer on its own, 0.233 bubbles per second, or the same rate correctly worked in another unit, earns all 4 marks with no working shown. A rate worked correctly in another unit can earn every mark, not only the unit mark. A correct final answer reached from wrong working does not earn full marks.

Why it scores

This is not a mark by itself, but it is the value the first mark needs.

The first mark is for setting out the bubbles divided by the time, using the right time for 20 °C.

The second mark is for the unrounded value from the calculator.

The third mark is for rounding to 3 significant figures, and the fourth is for the unit. A rate is the amount per unit of time, so the unit is bubbles per second.

Revise Photosynthesis
04.3[3 marks]

Using Table 4, give reasons why photosynthesis runs faster or slower at 40 °C than at 10 °C.

See how you would score on Photosynthesis
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The rate is higher at 40 °C. Pondweed took only 25 s to release 10 bubbles at 40 °C, but 84 s at 10 °C.

At 40 °C the reactant molecules and the enzymes have more kinetic energy, so they move faster.

So they collide more often, and more collisions are successful. Photosynthesis therefore goes faster.

3 marks

How the marks are earned

  1. One mark: the rate is higher at 40 °C. Saying the rate rises as the temperature rises is also credited, and so is the opposite for 10 °C if it is clearly stated.
  2. One mark: at the higher temperature the particles, reactants, molecules or enzymes have more energy (the word kinetic is not needed). Saying they move faster is also credited.
  3. One mark: so they collide more frequently. Also credited: more collisions in the same period of time, or more successful collisions.

Why it scores

The first mark is for saying the rate is higher at 40 °C. Quoting the times from Table 4 supports the answer.

The second mark is for the cause, which is more kinetic energy and faster movement.

The third mark is for the effect on collisions. It needs frequent collisions in the same time, not just a fast reaction.

Revise Photosynthesis
04.4[2 marks]

Apart from temperature, name two things that can speed up or slow down photosynthesis.

See how you would score on Photosynthesis
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1. Light intensity. 2. Carbon dioxide concentration.

2 marks

How the marks are earned

  1. Any two, one mark each: light intensity or level; the colour or wavelength of the light; carbon dioxide (the word concentration is optional); chlorophyll (the amount or mass is optional). The number of chloroplasts is also credited.
  2. An extra wrong answer cancels a correct one, so a list that includes an error can lose marks.
  3. If neither light intensity nor the colour or wavelength of light is credited, light with no more detail earns one mark, and it can pair with another listed factor.
  4. Water and humidity are ignored, and wind is not accepted.

Why it scores

Each correct factor earns one mark, so two factors score both. Light intensity and carbon dioxide are two of the factors that limit photosynthesis; the amount of chlorophyll would also earn a mark.

Revise Photosynthesis
04.5[1 mark]

Describe one change to the method that would give more accurate photosynthesis rates.

See how you would score on Photosynthesis
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Collect the oxygen in a gas syringe and measure its volume for a set time.

1 mark

How the marks are earned

  1. Trap the gas (oxygen) given off, in a gas syringe or in an upturned measuring cylinder.
  2. Saying bubbles in place of gas or oxygen is allowed, but naming the wrong gas is not accepted.
  3. Measuring the volume of gas in a syringe or in an inverted cylinder is allowed. Saying only use a gas syringe, collect the gas in a measuring cylinder, or use an inverted cylinder earns nothing.
  4. Any other correct method is allowed.

Why it scores

Bubbles vary in size, so counting them is not accurate. A measured volume of gas is a more accurate measure of the rate.

Revise Photosynthesis
04.6[2 marks]

Some of a gardener's plants are growing poorly. Give two methods the gardener could use to find out which disease it is.

See how you would score on Plant Diseases and Defenses
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1. Compare the symptoms with the pictures in a gardening manual or a gardening website. 2. Take a diseased plant to a laboratory or an expert to be tested.

2 marks

How the marks are earned

  1. Any two, one mark each: check the symptoms in a gardening manual, a website or an app; send diseased plants to a lab, or to a scientist or expert; use a kit built on monoclonal antibodies to test for the pathogen.
  2. Saying only use or search the internet earns nothing.
  3. An extra wrong answer cancels a correct one, so a list that includes an error can lose marks.

Why it scores

Each way of finding out which disease the plant has earns one mark: comparing the symptoms with a reference, and having a sample tested by an expert or in a laboratory.

Revise Plant Diseases and Defenses
04.7[1 mark]

A plant's leaves showed purple spots, then dropped from the plant. Name the disease that fits best.

See how you would score on Plant Diseases and Defenses
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Black spot of roses, a fungal disease. It causes purple or black spots on the leaves, which can turn yellow and fall early. Less photosynthesis happens, so growth is poorer.

1 mark

How the marks are earned

  1. Black spot (of roses), a fungal disease. The word rose is optional.
  2. Another correct disease, such as rust, is allowed.
  3. A nitrate or magnesium deficiency is not accepted, and neither is tobacco mosaic virus (TMV).
Revise Plant Diseases and Defenses
04.8[2 marks]

Leaves of another plant had turned yellow and showed no spotting. Explain what could cause leaves to turn yellow.

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The plant may lack magnesium ions in the soil. Without them it cannot make enough chlorophyll, so the leaves lose their green colour and turn yellow. This yellowing is called chlorosis.

2 marks

How the marks are earned

  1. One mark: a shortage of magnesium ions. A shortage of nitrate ions or iron ions is also credited.
  2. One mark: this leads to chlorosis, or to chlorophyll being made in smaller amounts, since magnesium is a part of chlorophyll production. Likewise credited: nitrate or iron being needed for chlorophyll production.
  3. Or, one mark for a lack of light, and one mark for it causing chlorosis or the breakdown of chlorophyll.
  4. Chlorosis on its own earns no mark, and neither does fewer chloroplasts.

Why it scores

The first mark is for the mineral shortage and the second is for the effect on chlorophyll. Chlorophyll is the green pigment, so less of it leaves the leaves yellow. Lack of light, which breaks down chlorophyll, would also earn both marks.

Revise Plant Diseases and Defenses
AQA owns the copyright in its exam questions, so we cannot copy their wording. These descriptions and answers are ours. Not made or approved by AQA.[16 marks]
PrepWise guide: AQA GCSE Biology Paper 1 HigherQuestion 5

This question covers blood, the heart and its vessels. Figure 5 is an enlarged red blood cell. Figure 6 shows a heart with four vessels lettered, and Table 5 lists their pressures. An athlete ran on a treadmill until reaching maximum effort, then stopped; Figure 7 follows oxygen use and blood lactic acid.

Figure 5
A red blood cell drawn as a rounded, flattened disc, with a double-headed arrow beneath it marking its diameter and a printed label.
Figure 6
Front view of a heart in grey, with four blood vessels at its top marked with the letters A, B, C and D.
Table 5
Table with two columns, pressure in arbitrary units and blood vessel letter, and four rows, one of them already filled in.
Figure 7
Graph with time in minutes along the bottom and no scale numbers, a solid line for use of oxygen and a dashed line for blood lactic acid concentration, with arrows marking when exercise started and stopped.
05.1[1 mark]

Give one job of blood plasma, other than carrying blood cells or oxygen.

See how you would score on Blood Components and Vessels
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Plasma carries carbon dioxide from the body cells to the lungs. It also carries other dissolved substances such as glucose, urea and hormones.

1 mark

How the marks are earned

  1. Plasma moves a named substance round the body, for example glucose, urea, carbon dioxide, hormones, enzymes, antibodies, proteins or ions. Any other correctly named substance is accepted, and so is moving platelets. Naming only nutrients, waste or oxygen is ignored.
Revise Blood Components and Vessels
05.2[5 marks]

Figure 5 is a red blood cell image magnified 7200 times. Work out the real diameter of the cell in micrometres.

See how you would score on Microscopy
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magnification = (image size) ÷ (real size)

real size = (image size) ÷ magnification

real diameter = (6.3 cm) ÷ 7200

real diameter in cm = 0.000875

0.000875 cm × 10 000 = 8.75 µm

5 marks

How the marks are earned

  1. One mark: write the magnification equation, magnification = image size over real size. Using diameter in place of size is accepted. An equation triangle on its own is ignored.
  2. One mark: rearrange so real size = image size over magnification. The diameter form is accepted.
  3. One mark: substitute 6.3 over 7200 into the rearranged equation (the 6.3 cm is printed on Figure 5). Substituting 63 000 over 7200 after converting first is also accepted, and a wrongly converted value still earns this mark.
  4. One mark: carry out the division to get 0.000875 cm. An answer that follows from a wrongly converted value still earns this mark, and so does dividing 63 000 by 7200 if the conversion came first.
  5. One mark: convert to micrometres (µm) to reach 8.75. The conversion can come at any stage. 8.8 is also accepted, and any further rounding is ignored.
  6. A correct answer on its own, 8.75 micrometres (8.8 also accepted), earns all 5 marks with no working shown. A correct final answer reached from wrong working does not earn full marks. A step left out may still earn its mark if a later step is correct.

Why it scores

First mark: the equation that links the three quantities. Writing diameter instead of size is fine.

Second mark: rearrange so the quantity you want is on its own.

Third mark: put in the image diameter read from Figure 5, in cm, and the magnification 7200. A wrongly converted value here still earns the mark.

Fourth mark: divide 6.3 by 7200 to get 0.000875 cm. An answer that follows from a wrongly converted value still earns this mark.

Fifth mark: 1 cm is 10 000 micrometres, so multiply by 10 000. Changing the units at an earlier step also works, and 8.8 micrometres is accepted.

Revise Microscopy
05.3[2 marks]

Complete Table 5 by writing the correct Figure 6 letter beside each pressure. One row is already done.

See how you would score on The Heart and Circulation
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Pressure 5 goes with vessel A.

Pressure 18 goes with vessel B.

Pressure 120 goes with vessel C.

2 marks

How the marks are earned

  1. Two marks for all three letters right: pressure 5 goes with A, pressure 18 with B, pressure 120 with C.
  2. One mark for one or two of those letters right. Names of the vessels written in the cells are ignored.

Why it scores

A is the vena cava, a vein carrying blood back to the heart, so its pressure is very low. The other vein, D, is also low.

B is the pulmonary artery. The blood only has to reach the nearby lungs, so it needs much less pressure than blood sent round the body.

C is the aorta. The left ventricle pumps blood through it to the whole body, so it has the highest pressure of the four.

Revise The Heart and Circulation
05.4[1 mark]

Give the name of the vessel labelled D on the heart in Figure 6.

See how you would score on The Heart and Circulation
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D is the pulmonary vein, which brings oxygenated blood from the lungs to the left side of the heart.

1 mark

How the marks are earned

  1. The pulmonary vein.
Revise The Heart and Circulation
05.5[1 mark]

Describe the structure of an artery's walls.

See how you would score on Blood Components and Vessels
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An artery wall has a thick layer of muscle and elastic fibres, which lets it cope with blood at high pressure.

1 mark

How the marks are earned

  1. The wall is thick elastic tissue or fibres, or thick muscle (the word layer is optional, but thick is needed). A thick wall with no detail is ignored, and so is saying there are no valves. Cell walls are not accepted.
Revise Blood Components and Vessels
05.6[2 marks]

During the run the athlete used both types of respiration. Explain why anaerobic respiration is the less efficient one.

See how you would score on Respiration
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Anaerobic respiration releases less energy from each glucose molecule, so more glucose is needed to release the same amount of energy.

This is because the glucose is only partly broken down, into lactic acid. Aerobic respiration, with oxygen, breaks it right down to carbon dioxide and water.

2 marks

How the marks are earned

  1. One mark: less energy is released or transferred; adding that this is per glucose molecule is optional. Also accepted: to get an equal amount of energy, more glucose must be used up. Not accepted: energy produced, made or created; no energy being released at all; less energy released for respiration.
  2. One mark: the glucose is incompletely broken down, or incompletely oxidised, so only part of its energy is released. Also accepted: glucose is broken down only as far as lactic acid.

Why it scores

First mark. Say energy is released or transferred, never made or created, and compare per glucose molecule.

Second mark: the cause is the incomplete breakdown of glucose, which keeps most of its energy locked in the lactic acid.

Revise Respiration
05.7[2 marks]

Figure 7 shows oxygen use and blood lactic acid during and after exercise. Explain why oxygen use stays high once exercise has stopped.

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Hard exercise used oxygen faster than the blood could deliver it, so the muscles respired anaerobically and lactic acid built up. The athlete is left in oxygen debt.

After exercise, extra oxygen is needed to break down (oxidise) the lactic acid, so oxygen use stays high until the lactic acid has gone.

2 marks

How the marks are earned

  1. One mark: an oxygen debt has built up. Also accepted: the lactic acid concentration is still high.
  2. One mark: extra oxygen is needed to oxidise the lactic acid. Also accepted: extra oxygen is needed to remove it or to break it down. Oxygen turning lactic acid back into glucose in the liver is ignored, and so is naming the liver as the place where lactic acid is oxidised.

Why it scores

First mark: the oxygen debt, or a lactic acid concentration that is still high.

Second mark: link the extra oxygen to getting rid of the lactic acid.

Revise Respiration
05.8[2 marks]

Some athletes train in conditions that raise their red blood cell count. Explain how more red blood cells help an athlete.

See how you would score on Blood Components and Vessels
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Red blood cells contain haemoglobin, which binds oxygen. More red blood cells give more haemoglobin, so more oxygen is carried to the muscles.

The muscles can then respire aerobically for longer, releasing more energy for contraction and delaying the muscle fatigue that lactic acid causes.

2 marks

How the marks are earned

  1. One mark: more red blood cells mean more haemoglobin, so more oxygen is carried or bound. More oxyhaemoglobin forming is also accepted.
  2. One mark, by either route: the athlete respires aerobically more, so more energy is released; or aerobic respiration goes up so the muscles tire or cramp less. Also accepted: anaerobic respiration falls, and it would have released less energy; or anaerobic respiration falls, and it would otherwise have brought on tiredness or cramp. Not accepted: the athlete doing more anaerobic respiration; energy produced, made or created; energy released for respiration.

Why it scores

First mark: more haemoglobin means more oxygen carried (or more oxyhaemoglobin formed).

Second mark: link the extra oxygen to more aerobic respiration, with either more energy released or less fatigue and cramp.

Revise Blood Components and Vessels
AQA owns the copyright in its exam questions, so we cannot copy their wording. These descriptions and answers are ours. Not made or approved by AQA.[16 marks]
PrepWise guide: AQA GCSE Biology Paper 1 HigherQuestion 6

This question is about gas exchange in fish. Figure 8 shows a fish with one enlarged part of a gill. Some fish spend time resting on the sea floor.

Figure 8
A fish with its gills marked and an enlarged oval showing part of a gill: filaments branching from a blood vessel, with an arrow for blood flow and an arrow for the direction of water flow.
06.1[1 mark]

State what is meant by diffusion.

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Diffusion is the spreading out of particles from a region of higher concentration to a region of lower concentration, down a concentration gradient.

1 mark

How the marks are earned

  1. Particles spread out, moving down a concentration gradient from where there are more of them to where there are fewer. Substances, chemicals, gases or liquids are accepted in place of particles. Mentioning a membrane is ignored. Movement against the concentration gradient is not accepted.
Revise Cell Transport
06.2[4 marks]

Explain how the gill in Figure 8 is built to move as much oxygen as possible from water into blood.

See how you would score on Gas Exchange in Humans
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The gill has many thin filaments, which give a large surface area for oxygen to diffuse into the blood.

Each filament has a network of capillaries, a good blood supply. The blood carries the oxygen away, which keeps a steep concentration gradient.

The walls of the filaments and capillaries are very thin, only one cell thick, so the diffusion distance is short.

The blood supply is right next to the water at the gill surface, so oxygen has only a short way to diffuse.

4 marks

How the marks are earned

  1. One mark: the gills, or the gill filaments, give a large surface area for diffusion (or absorption).
  2. One mark: a good blood supply, or a network of capillaries, keeps up the concentration gradient, for example by carrying oxygen away. Blood vessels are accepted for capillaries, but arteries or veins are not.
  3. One mark: the walls of the filaments or capillaries are thin, or one cell thick, giving a short diffusion distance or time. Blood vessels are accepted for capillaries, but arteries or veins are not. Saying that membranes, filaments or capillaries themselves (not their walls) are one cell thick is ignored. Cell walls are not accepted.
  4. One mark: the water passes close to the blood supply, giving a short diffusion distance or time. Blood supply close to the gill surface is also accepted. Blood flowing the opposite way to the water, which keeps the concentration gradient up, is also accepted for this mark.

Why it scores

A larger surface area lets more oxygen cross at once.

Keeping the oxygen concentration low in the blood means diffusion carries on.

A short distance means faster diffusion.

Closeness of blood and water is a separate mark from thin walls.

Revise Gas Exchange in Humans
06.3[4 marks]

Explain why metabolic rate rises when a fish swims instead of resting.

See how you would score on Respiration
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A swimming fish moves its body much more than a resting fish.

Its muscles contract more, and contracting muscles need more energy.

That energy is released by respiration, so more respiration takes place in the muscle cells.

Metabolic rate is the rate of all the chemical reactions in the body, including respiration, so it is higher when the fish swims.

4 marks

How the marks are earned

  1. One mark: movement (more movement is fine). Swimming named on its own is ignored. For every mark in this part, the opposite argument is accepted if it is clearly about the resting fish.
  2. One mark: movement needs energy (saying more energy is fine), or energy is provided, for example by respiration. Not accepted: energy produced, made or created; energy for respiration.
  3. One mark: more respiration takes place (the word so is not needed).
  4. One mark: muscle contraction, with or without the word increased (muscles contracting is accepted). It does not have to be linked to the energy.

Why it scores

Increased movement is the first mark. Saying only that it is swimming earns nothing.

Muscle contraction and the need for more energy are two further marks.

Say energy is released by respiration, never made or created.

Ties the extra respiration back to the question about metabolic rate.

Revise Respiration
AQA owns the copyright in its exam questions, so we cannot copy their wording. These descriptions and answers are ours. Not made or approved by AQA.[9 marks]
PrepWise guide: AQA GCSE Biology Paper 1 HigherQuestion 7

A patient with a badly failed liver might be given embryonic stem cells by injection. A drug that helps stop a donor liver being rejected is made of monoclonal antibodies aimed at an antigen on liver cells. Figure 9 shows some steps in making them. The liver also makes bile.

Figure 9
Flow diagram: a mouse, an arrow down to a group of cells marked cell type A, a second group marked cell type B joining by an arrow labelled fusion, and a third group marked cell type C.
07.1[1 mark]

Name the property of embryonic stem cells that makes them useful for treating liver disease.

See how you would score on Stem Cells and Cell Differentiation
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Embryonic stem cells can differentiate, so any kind of cell, liver cells included, can form from them.

1 mark

How the marks are earned

  1. The property is that they differentiate: they can form any type of cell. Differentiating into liver cells is also accepted. Saying they specialise is ignored.
Revise Stem Cells and Cell Differentiation
07.2[1 mark]

Suggest one ethical objection a patient might have to receiving embryonic stem cells.

See how you would score on Stem Cells and Cell Differentiation
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Embryonic stem cells come from embryos, and taking them destroys the embryo, which is a potential human life.

1 mark

How the marks are earned

  1. Any one of: taking the cells destroys an embryo, a life or a potential life; the embryo cannot give consent (the embryo has no choice is accepted); the treatment goes against religious, cultural or personal beliefs (against someone's faith is accepted). Against God's will is ignored, and religion with no detail is ignored.
Revise Stem Cells and Cell Differentiation
07.3[3 marks]

Figure 9 shows three cell types, A, B and C, used in making monoclonal antibodies. Name each one.

See how you would score on Monoclonal Antibodies
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A is a lymphocyte, the cell that makes the antibody for the antigen.

B is a tumour cell.

C is a hybridoma cell.

3 marks

How the marks are earned

  1. One mark: A is a lymphocyte, the white blood cell that makes antibody to the antigen. White blood cell on its own is ignored.
  2. One mark: B is a tumour cell. Myeloma cell, or another named type of tumour cell, is also accepted. Cancer cell on its own is ignored.
  3. One mark: C is a hybridoma cell, formed by fusing A and B.

Why it scores

Use lymphocyte, as white blood cell alone is too general.

Tumour cells divide again and again, which the fused cell needs.

It is made by fusing the lymphocyte with the tumour cell.

Revise Monoclonal Antibodies
07.4[1 mark]

What is done to the mouse so that it makes cell type A?

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The mouse is injected with the liver antigen, or with liver cells, so its lymphocytes make antibodies against it.

1 mark

How the marks are earned

  1. The mouse receives an injection of antigens (liver antigens also fine). Giving the mouse liver cells by injection is also accepted, and there the word liver is needed. Injecting a pathogen is not accepted.
Revise Monoclonal Antibodies
07.5[3 marks]

The patient takes this antibody drug after the transplant. Explain how it protects the donor liver cells from the patient's own white blood cells.

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The drug's antibodies attach to antigens sitting on the donor liver cells.

They can do this because their shape is complementary to the shape of the liver antigen, like a lock and key.

So the antibodies made by the patient's lymphocytes cannot bind to the liver cells, and phagocytes cannot detect the cells to engulf them.

3 marks

How the marks are earned

  1. One mark: the monoclonal antibodies bind to antigens or cells (the word liver is optional). The drug sticking to them is also accepted.
  2. One mark: the antibody and the antigen have shapes that fit together, so they are complementary. Describing the fit is also accepted. Saying only that the antibodies are specific is ignored. Saying the antibody has the same shape as the antigen is not accepted.
  3. One mark: the patient's own antibodies from white blood cells are then unable to attach to antigens or cells (the word liver is optional). Antibodies from lymphocytes is also accepted, and so is phagocytes being unable to identify or detect the cells they would engulf. Antibodies from phagocytes is not accepted.

Why it scores

First mark: the drug attaches to the liver antigens.

Second mark: complementary shape, not the same shape.

Third mark: the antigens are covered, so the patient's own immune response cannot attach to the liver cells.

Revise Monoclonal Antibodies
07.6[1 mark]

State a use of monoclonal antibodies other than in pregnancy tests.

See how you would score on Monoclonal Antibodies
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Monoclonal antibodies can carry a cancer drug to tumour cells, so the drug acts on the cancer cells.

1 mark

How the marks are earned

  1. Any one of: delivering a drug or chemical to cells, toxic or not (chemotherapy or immunotherapy is accepted); locating or detecting cancer, or cancer cells; locating or identifying particular molecules inside a cell or tissue; tests for hormones, chemicals or drugs; tests for pathogens (a COVID test, a lateral flow test or an LFT is accepted). Treating a named disease is accepted. Saying only that it diagnoses or treats disease is ignored.
Revise Monoclonal Antibodies
07.7[1 mark]

Which organ stores bile? Tick one box.

See how you would score on The Human Digestive System
Gall bladder
Kidney
Pancreas
Small intestine
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Bile is made in the liver and stored in the gall bladder.

1 mark

How the marks are earned

  1. Tick gall bladder.
Revise The Human Digestive System
07.8[6 marks]

Explain why severe liver disease makes it hard to digest fats.

See how you would score on Enzymes in Digestion
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The liver makes bile. In severe liver disease it makes little or no bile.

Bile emulsifies fat, breaking large fat globules into tiny droplets. With little bile the fat stays in large globules, so its surface area is small.

Lipase can only work on the surface of the fat, so with a small surface area it digests the fat into fatty acids and glycerol more slowly.

Bile is also alkaline. It neutralises the acid that comes from the stomach, so without bile the small intestine stays too acidic.

An acidic small intestine is not the optimum pH for lipase, so lipase works poorly or is denatured. Less fat is digested.

Levels of response, 6 marks

How the marks are earned

  1. This is level marked. For 5 to 6 marks, give reasons in enough detail, linked into a clear account, and cover lipase activity, emulsification and the effect of pH. For 3 to 4 marks, give reasons with some attempt at linking, but the account is not fully clear. For 1 to 2 marks, give simple statements that are not linked.
  2. Emulsification: the damaged liver makes little or no bile, so fat is not broken into small droplets, the surface area of the fat does not increase, and lipase digests the fat more slowly, or digests less of it.
  3. pH: bile is alkaline and normally neutralises the acid from the stomach, so without it the small intestine is not at the best pH for lipase, lipase denatures or works poorly, and less fat is digested.

Why it scores

Start with the cause: the liver is where bile is made.

Emulsification is the first route. Link the missing bile to a smaller surface area.

Linking the small surface area to slower lipase action is one part of the top band (5 to 6 marks). The top band also needs the effect of pH.

pH is the second route and has to be covered for the top level.

Finish by linking the wrong pH to the lipase and the fat. A top answer covers both routes.

Revise Enzymes in Digestion
AQA owns the copyright in its exam questions, so we cannot copy their wording. These descriptions and answers are ours. Not made or approved by AQA.[17 marks]
PrepWise guide: AQA GCSE Biology Paper 1 HigherQuestion 8

Figure 10 is a drawing of a bacterial cell beside a yeast cell. A scientist grew one type of bacterium in three cultures, each with a different antibiotic and each starting with equal numbers of bacteria. Figure 11 shows how the number of bacteria per cm3 changed with time.

Figure 10
Two unlabelled cells side by side, a rod-shaped bacterial cell on the left and an oval yeast cell on the right, with their inner parts drawn and a note that they are not to scale.
Figure 11
Line graph on a grid of bacteria per cm^3 against time in hours, with three lines for antibiotics A, B and C, a key, and an arrow showing when the bacteria were added. The vertical scale has no numbers.
08.1[1 mark]

Choose the correct statement about bacterial and yeast cells. Tick one box.

See how you would score on Cell Structure
Both are eukaryotes.
Both are prokaryotes.
Bacteria are eukaryotes and yeast are prokaryotes.
Bacteria are prokaryotes and yeast are eukaryotes.
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Bacteria are prokaryotes, while yeast cells are eukaryotes.

1 mark

How the marks are earned

  1. Tick the box that calls the bacterium prokaryotic and the yeast eukaryotic.
Revise Cell Structure
08.2[2 marks]

Using Figure 10, name two structures that a yeast cell has and a bacterial cell lacks.

See how you would score on Cell Organelles
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A yeast cell has a nucleus and mitochondria, and it also has a large vacuole. A bacterial cell has none of these.

2 marks

How the marks are earned

  1. Any two, one mark each: a nucleus; a vacuole; mitochondria. Chromosomes, or larger ribosomes, are also accepted. If no other mark is earned, one mark is given for membrane-bound structures inside the cell (sub-cellular structures).

Why it scores

Any two of the three earn the two marks. Compare the two drawings carefully, since both cells have cytoplasm, a cell wall and a membrane.

Revise Cell Organelles
08.3[4 marks]

Compare the effectiveness of antibiotics A, B and C, using Figure 11.

See how you would score on Antibiotics and Drug Resistance
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For about the first 4 hours all three antibiotics have much the same effect, and the bacteria increase in every culture.

Antibiotic C is the most effective. Its bacteria stop rising first, at about 4 to 5 hours, and peak at the lowest number. They start to fall soonest and are the only ones to reach zero, at about 15 hours.

Antibiotic B is in between. Its bacteria peak at about 6 hours, so they start to fall before those with A, and they fall further and faster than with A. They start to fall later than with C, and a few are still left at 16 hours.

Antibiotic A is the least effective. Its bacteria peak last, at about 8 hours, and at the highest number. They fall most slowly, and at 16 hours there are far more than with B.

Levels of response, 4 marks

How the marks are earned

  1. This is level marked. An answer worth 3 to 4 marks compares all three antibiotics, says clearly how they are alike or differ, and where useful says by how much. An answer worth 1 to 2 marks picks out relevant features and notes differences.
  2. Early on: for roughly the first 4 or 5 hours all three antibiotics have a similar effect, or none.
  3. Antibiotic A is the least effective: its numbers start to fall last, fall most slowly, peak the highest, never fall as far as with B or C, and are still large at 16 hours.
  4. Antibiotic B is in the middle: it works better than A but worse than C, starting to reduce numbers after C but before A, and reducing them more and faster than A but more slowly than C.
  5. Antibiotic C is the most effective: its numbers start to fall first (falling soonest, or fastest, is also credited), peak the lowest, and it is the only one to take the bacteria to zero.

Why it scores

Starting with what is alike makes the comparison clear.

Use words such as most, fastest and lowest, and refer to features of the graph.

Every antibiotic has to be compared with the others to reach the top level.

Giving the size of the difference, such as the highest peak and the numbers left, shows the magnitude.

Revise Antibiotics and Drug Resistance
08.4[2 marks]

A crowded culture cannot be counted directly. Ten samples of 0.02 cm3 are taken. Describe how to estimate the bacteria per cm3.

See how you would score on Cell Biology Practical Investigations
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Dilute each sample so the bacteria are spread out enough to count, then count the bacteria in each sample under a microscope.

Work out the mean count for a single sample of 0.02 cm3.

1 ÷ 0.02 = 50

bacteria per cm3 = mean number in a sample × 50

2 marks

How the marks are earned

  1. One mark: dilute each sample, then count the bacteria in it, for example under a microscope.
  2. One mark: work out the mean count for one sample, then scale up to 1 cm3. Also accepted: the mean × 50, or the total of all ten counts × 5.

Why it scores

First mark: dilute, then count. Counting works only when the cells are not crowded together.

Second mark starts here: the mean of the ten counts reduces the effect of any odd sample.

One cm3 holds 50 samples' worth.

Scaling up gives the estimate.

Revise Cell Biology Practical Investigations
AQA owns the copyright in its exam questions, so we cannot copy their wording. These descriptions and answers are ours. Not made or approved by AQA.[9 marks]
END OF QUESTIONS100 marks total
See how you would score on Enzymes in Digestion

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