BREAKING
Mitochondria finally takes a day off, cell in chaos★Student draws plant cell, forgets the cell wall, again★Required practical goes wrong, cress wins★
  1. Home
  2. A Level
  3. Biology
  4. AQA
  5. Classification

AQA A Level Biology

Topic questions Evolution and biodiversity Classification

Classification

9 questions · 79 marks · about 2 hours

Classification questions

AQA 7402 Classification 9 questions

Paper 3 June 2025, Question 3 8 marks

Tansy, Tanacetum vulgare, is a plant that grows in dense patches. The tansy beetle, Chrysolina graminis, feeds on, and completes its life cycle on T. vulgare.

C. graminis is 8–10 mm long and green in colour.

Two other green beetles can also be found in the same habitat.

• The dock beetle, Gastrophysa viridula, which is 4–7 mm long.

• The mint leaf beetle, Chrysolina herbacea, which is 6–10 mm long.

Figure 2 shows the life cycle of C. graminis.

Figure for question 3
3.1

How many different genera of beetles are there in the information provided?

[1]
Reveal mark scheme Mark scheme for question 3.1
3.2

Ecologists investigated the population size of C. graminis using the

mark-release-recapture method.

Table 2 shows the ecologists’ results.

Figure for question 3

The ecologists captured and marked the first sample of beetles during the second week of August. They identified the species of beetle before marking them.

Use the information provided to suggest:

two reasons why the first sample of beetles was captured during the second week of August

[3]
Reveal mark scheme Mark scheme for question 3.2
3.3

one reason why the beetles had to be identified before being marked.

Use the data provided in Table 2 to estimate the population size of C. graminis.

Show your working.

[2]
Reveal mark scheme Mark scheme for question 3.3
3.4

The ecologists repeated the investigation over a five–year period to monitor the population size of C. graminis.

State two precautions they would have taken each year to obtain comparable data.

[2]
Reveal mark scheme Mark scheme for question 3.4

Paper 1 June 2024, Question 9 6 marks

Courtship behaviour in the frog species, Xenopus laevis, involves male frogs calling to:

• attract sexually active females – these are advertisement calls

• start and continue mating – these are mating calls

• signal when a male is not sexually active – these are rasping calls.

Scientists investigated frog courtship behaviour by feeding a population of sexually active male frogs a diet containing the hormone EE2. The scientists also fed a separate control population of sexually active male frogs a diet without EE2.

They determined the percentage of males making advertisement calls or rasping calls in each population.

Table 6 shows their results.

Figure for question 9
9.1

There were 800 males in the control population.

Each male made one type of call.

Use this information and Table 6 to calculate the number of males making mating calls in the control population.

[1]
Reveal mark scheme Mark scheme for question 9.1
9.2

Suggest one change the scientists could make to both frog populations to increase the number of mating calls.

[1]
Reveal mark scheme Mark scheme for question 9.2
9.3

The scientists also investigated the effect on female frog courtship behaviour of feeding EE2 to male frogs.

Table 7 shows their results.

Figure for question 9

EE2 is contained in human contraceptive pills. Some EE2 is released in human urine and collects in sewage. Untreated sewage pollutes the water in frog habitats.

Suggest and explain the effect EE2 pollution in frog habitats will have on frog breeding.

Use information from Table 6 and Table 7 in your answer.

Effect on frog breeding

[4]
Reveal mark scheme Mark scheme for question 9.3

Paper 1 June 2023, Question 3 9 marks

3.1

The human disease, malaria, is caused by infection with a single-celled eukaryotic organism.

Figure 2 shows a diagram of Plasmodium vivax, one of the species that can cause malaria.

Figure for question 3

Other than the Golgi apparatus, name one structure in Figure 2 which shows that P. vivax is a eukaryote.

[1]
Reveal mark scheme Mark scheme for question 3.1
3.2

Describe two functions of the Golgi apparatus in a eukaryotic cell.

[2]
Reveal mark scheme Mark scheme for question 3.2
3.3

P. vivax evolved from a common ancestor in Africa. As humans migrated around the world, new strains of P. vivax evolved.

P. vivax is now extremely rare in Africa but there are several different strains of P. vivax in other parts of the world.

Figure 3 shows a phylogenetic diagram of the evolution of these different strains.

Figure for question 3

What does Figure 3 suggest is the order of human migration out of Africa?

Tick (✓) one box.

Figure for question 3
[1]
Reveal mark scheme Mark scheme for question 3.3
3.4

There are an estimated 229 million cases of human malaria worldwide per year. 94% of these cases are found in Africa, but are not caused by P. vivax. P. vivax does cause 61% of the cases of human malaria outside Africa.

Use this information to calculate the number of cases worldwide caused by P. vivax each year.

Question 3, shown as an image
[1]
Reveal mark scheme Mark scheme for question 3.4
3.5

In Africa today, most of the human population are resistant to malaria caused by P. vivax.

Use your knowledge of natural selection to explain why this resistance is so common in Africa.

[4]
Reveal mark scheme Mark scheme for question 3.5

Paper 1 June 2023, Question 6 8 marks

6.1

Define genome and proteome.

[2]
Reveal mark scheme Mark scheme for question 6.1
6.2

The classification system used in the early 20th century grouped different species of bacteria according to the position and shape of flagella on bacterial cells and by the number of flagella per cell. These were observed using an optical microscope. Each species of bacterium has a characteristic cell shape and arrangement of flagella. These characteristics may be shared with other species within a genus. Flagella are fragile, difficult to stain and may extend from the cell at any angle.

Consider the accuracy and limitations of the early classification of bacteria using the arrangement of flagella.

[3]
Reveal mark scheme Mark scheme for question 6.2
6.3

Suggest why several bacterial species have been renamed in recent years.

[1]
Reveal mark scheme Mark scheme for question 6.3
6.4

Figure 6 shows an image from an optical microscope of a single bacterial cell.

Figure for question 6

This bacterial cell is 2.3 μm long (excluding the flagellum).

Calculate the magnification of this image.

Show your working.

[2]
Reveal mark scheme Mark scheme for question 6.4

Paper 3 June 2022, Question 5 15 marks

Also involves Plant disease and defence

Scientists investigated the effect of full sun and shade on the rate of photosynthesis in a species of shade-tolerant tree.

To estimate the rate of photosynthesis, the scientists measured uptake of carbon dioxide by trees in a forest. They measured uptake of carbon dioxide during two parts of the day:

• 08.30 – 09.40 hours

• 11.40 – 13.15 hours.

Figure 8 shows the scientists’ results.

Figure 8

Figure 8 not reproduced here due to third-party copyright restrictions

5.1

Calculate the total uptake of carbon dioxide between 11.40 and 13.15 hours in trees exposed to full sun in a forest that is 12 000 m2 in area.

Give your answer in standard form. Show your working.

[3]
Reveal mark scheme Mark scheme for question 5.1
5.2

Figure 8 shows there is a small difference in the mean uptake of carbon dioxide between 08.30 and 09.40 hours by trees in full sun and by trees in the shade. When the scientists performed a statistical test on these data, they calculated P > 0.5

State what this P value tells you about this difference.

Explain your answer using the words probability and chance.

[2]
Reveal mark scheme Mark scheme for question 5.2
5.3

In this species of tree, very high light intensities can inhibit the release of electrons from chlorophyll.

Suggest how this could explain the results shown in Figure 8 for 11.40 to 13.15 hours.

[4]
Reveal mark scheme Mark scheme for question 5.3
5.4

Tomato plants grow best in high light intensities. To increase the yield of tomato plants, a farmer uses LED lightbulbs to provide additional light.

The increase in dry mass (D) produced when using additional light can be calculated using this equation.

Question 5, shown as an image

Where L = light used in photosynthesis F = GPP to NPP conversion factor for tomato plants

Table 5 shows some of these values for LED lightbulbs.

Figure for question 5

Use the equation and Table 5 to calculate the increase in dry mass produced when using LED lightbulbs.

Give your answer in standard form and give the units.

[2]
Reveal mark scheme Mark scheme for question 5.4
5.5

Mature leaves from slow-growing, shade-tolerant plants produce poisonous chemicals that are a defence against being eaten by herbivores.

Suggest how this benefits slow-growing, shade-tolerant plants.

[2]
Reveal mark scheme Mark scheme for question 5.5
5.6

Scientists measured the concentration of poisonous chemicals produced by

shade-tolerant plant species in six taxa. They compared this with the mean concentration of poisonous chemicals produced by all plants and the phylogenetic relationships between the six taxa.

Figure 9 shows the scientists’ results.

Figure for question 5

A journalist published the following summary of these results.

‘The more recently a shade-tolerant plant species evolved, the greater the concentration of poisonous chemicals it produces.’

Do the data in Figure 9 support this summary? Justify your answer.

[2]
Reveal mark scheme Mark scheme for question 5.6

Paper 1 November 2021, Question 8 10 marks

Also involves Adaptations

8.1

Describe how a triglyceride molecule is formed.

[3]
Reveal mark scheme Mark scheme for question 8.1
8.2

Table 5 shows some properties of four fatty acids.

Figure for question 8

Figure 10 shows diagrams of these fatty acids.

Figure for question 8

Put a tick () in one box that contains correct information about one of these fatty acids.

Figure for question 8
[1]
Reveal mark scheme Mark scheme for question 8.2
8.3

The percentage of saturated fatty acids compared with unsaturated fatty acids found in lipid stores in seeds differs in different populations.

Scientists investigated two populations of the plant, Helianthus annuus.

The scientists grew young plants from seeds collected from each population. They placed the seeds on wet tissue paper so that the root growth was visible.

They grew seeds from each population at two temperatures:

• warm temperature of 24 °C

• cool temperature of 10 °C

After 10 days, the scientists measured the length of each root.

Table 6 shows some of the properties of the two populations and the scientists’ results.

Figure for question 8

The mean ±2 × standard deviation includes 95% of the data.

The scientists used a data logger to measure the length of the root rather than a ruler.

Suggest one reason why they used a data logger and explain why this was important in this investigation.

[1]
Reveal mark scheme Mark scheme for question 8.3
8.4

It is known that:

• during respiration saturated fatty acids yield more energy than unsaturated fatty acids

• saturated fatty acids have higher melting points than unsaturated fatty acids

• lipases in seeds act more rapidly on liquid substrates.

Use this information and Table 6 to show how each population is better adapted for its natural environment when compared with the other population.

[4]
Reveal mark scheme Mark scheme for question 8.4
8.5

Although these two populations are completely separate and show genetic variation, they are both called Helianthus annuus.

Explain why they are both given this name.

[1]
Reveal mark scheme Mark scheme for question 8.5

Paper 1 June 2019, Question 7 5 marks

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.

Figure for question 7
7.1

Unlike plants, Ulva lactuca does not have xylem tissue.

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

[1]
Reveal mark scheme Mark scheme for question 7.1
7.2

Ulva lactuca has a haploid and a diploid form.

Figure 7 shows the life cycle of Ulva lactuca.

Figure for question 7

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.

[2]
Reveal mark scheme Mark scheme for question 7.2
7.3

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.

[2]
Reveal mark scheme Mark scheme for question 7.3

Paper 1 June 2018, Question 3 11 marks

Also involves Sampling and fieldwork

Bees are flying insects that feed on nectar made in flowers. There are many different species of bee.

Scientists investigated how biodiversity of bees varied in three different habitats during a year. They collected bees from eight sites of each habitat four times per year for three years.

The scientists’ results are shown in Figure 4 in the form they presented them.

Figure for question 3
3.1

What is meant by ‘species richness’?

[1]
Reveal mark scheme Mark scheme for question 3.1
3.2

From the data in Figure 4, a student made the following conclusions.

1. The natural habitat is most favourable for bees.

2. The town is the least favourable for bees.

Do the data in Figure 4 support these conclusions? Explain your answer.

1. The natural habitat is most favourable for bees.

[4]
Reveal mark scheme Mark scheme for question 3.2
3.3

2. The town is the least favourable for bees.

The scientists collected bees using a method that was ethical and allowed them to identify accurately the species to which each belonged.

In each case, suggest one consideration the scientists had taken into account to make sure their method

[2]
Reveal mark scheme Mark scheme for question 3.3
3.4

2. allowed them to identify accurately the species to which each belonged.

Suggest and explain two ways in which the scientists could have improved the method used for data collection in this investigation.

[2]
Reveal mark scheme Mark scheme for question 3.4
3.5

Three of the bee species collected in the farmland areas were Peponapis pruinosa, Andrena chlorogaster and Andrena piperi.

What do these names suggest about the evolutionary relationships between these bee species? Explain your answer.

[2]
Reveal mark scheme Mark scheme for question 3.5

Paper 1 June 2017, Question 8 7 marks

Figure 3 shows two different ways of classifying the same three species of snake.

maths symbol Classification X is based on the frequency of observable characteristics

maths symbol Classification Y is based on other comparisons of genetic characteristics.

All three species of snake belong to the Python family.

Figure for question 8
8.1

What do these classifications suggest about the evolutionary relationships between these species of snake?

[2]
Reveal mark scheme Mark scheme for question 8.1
8.2

Complete Table 4 below to show the missing names of the taxa when classifying these snakes.

Figure for question 8
[1]
Reveal mark scheme Mark scheme for question 8.2
8.3

There is a debate about the name of one of these species of snake. Some scientists name it Liasis papuana and other scientists name it Apodora papuana.

Give the name of the taxon about which the scientists disagree.

[1]
Reveal mark scheme Mark scheme for question 8.3
8.4

State three comparisons of genetic diversity that the scientists used in order to generate Classification Y.

[3]
Reveal mark scheme Mark scheme for question 8.4

Report a problem with a question on this page
What is wrong?

Goes to whoever maintains the question split. No account needed and nothing is published on this page.

Memes

Certified Biology memes. Peer-reviewed by nobody.

Laugh now. Cry in the exam hall later. Then come back and do a past paper.

Learn the whole of osmosis
Write ‘water moves’ and leave
Me on the 6-marker about potatoes
Knowing the mitochondria is the powerhouse
Knowing what ATP stands for
Drawing the heart the right way round
Explaining active transport without crying
Biology enlightenment, ranked

Mum 💖

how's biology going?? 🥰

learning about the heart rn

that's my baby 💖

📱 Screen time this week · Snapchat5h 52m

The heart, in fact, was broken

Biology starter pack

Microscope you can't focusPotato chips in salt waterCress that died in week 1This website
Collect all four