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OCR A Level Biology

Topic questions Ecology Adaptations

Adaptations

6 questions · 79 marks · about 1.5 hours

Adaptations questions

OCR H420 Adaptations 6 questions

Unified Biology June 2024, Question 3 16 marks

Xylem vessels maintain the transpiration stream by transporting water up plant stems.

(a)

The sentences describe the role of meristems in producing xylem vessels.

Complete the sentences using the most appropriate words or phrases.

Meristem cells are located between xylem and........ tissues in

plants. Cells of either of these vascular tissues can be produced when meristem cells divide and

[2]
Reveal mark scheme Mark scheme for question 3(a)
(b)

A pair of students dissect the vascular tissue of the primrose plant, Primula vulgaris.

(b)(i)

State and explain one safety precaution that the students should take when dissecting the vascular tissue.

[2]
Reveal mark scheme Mark scheme for question 3(b)(i)
(b)(ii)

This is a light micrograph of a transverse section through the stem of P. vulgaris.

Figure for question 3

The diameter of one of the vascular bundles is shown by the line labelled V.

Calculate the actual diameter of the vascular bundle labelled V.

Give your answer in mm and to 2 significant figures.

[2]
Reveal mark scheme Mark scheme for question 3(b)(ii)
(c)

A scientist plans to investigate the effect of temperature on the rate of transpiration in P. vulgaris.

The scientist uses a potometer in their investigation, as shown in the diagram below.

Figure for question 3
Question 3, shown as an image
Figure for question 3
(c)(i)

The scientist plans to calculate the rate of transpiration by measuring the distance moved by the air bubble.

The scientist carries out a preliminary trial by measuring the distance moved by the air bubble in 30 minutes at 20 maths symbolC.

The bubble moves 18 mm in 30 minutes. The diameter of the capillary tubing in the potometer is 1 mm.

Calculate the rate of transpiration, in mm3 h–1, during this preliminary trial.

Use the formula: Volume of cylinder = maths symbolr2l

Rate of transpiration =........ mm3 h–1

[2]
Reveal mark scheme Mark scheme for question 3(c)(i)
(c)(ii)

* Describe how the scientist could use the potometer to collect the data needed to assess the effect of temperature on the rate of transpiration in P. vulgaris.

Details of how to set up and use a potometer are not needed.

In your answer you should name an appropriate statistical test that could be used to analyse the data.

Extra answer space if required.

[6]
Reveal mark scheme Mark scheme for question 3(c)(ii)
(d)

Many plants, such as Erica cinerea (bell heather), have evolved adaptations to limit water loss by transpiration.

This is a drawing of a leaf of E. cinerea.

Figure for question 3

State and explain how the leaf of E. cinerea is adapted to limit water loss by transpiration.

[2]
Reveal mark scheme Mark scheme for question 3(d)

Unified Biology June 2023, Question 4 15 marks

The Convention on International Trade in Endangered Species (CITES) came into effect in 1975.

(a)
Figure for question 4 Figure for question 4
[6]
Reveal mark scheme Mark scheme for question 4(a)
(b)

The kakapo, shown in Fig. 4.3, is an endangered species of flightless bird that lives in New Zealand. The population size of kakapos has experienced a large decrease over the past few hundred years. There are now fewer than 250 kakapos living in the wild.

Figure for question 4
(b)(i)

State the term for a large decrease in population size that reduces the gene pool.

[1]
Reveal mark scheme Mark scheme for question 4(b)(i)
(b)(ii)

Adaptations can be categorised into three different types:

• anatomical

• behavioural

• physiological.

The table lists four traits that kakapos have evolved.

Complete the table by naming the type of adaptation represented by each of the four

Figure for question 4
[2]
Reveal mark scheme Mark scheme for question 4(b)(ii)
(b)(iii)

The kakapo and two other species of bird, the kaka and the kea, evolved from a common ancestor approximately 70 million years ago.

The evolutionary timeline of the three species is shown in Fig. 4.4.

Fig. 4.4

Emergence of the Southern Alps mountain range

Figure for question 4
Question 4, shown as an image

• The kakapo cannot fly. It forages for leaves and roots on the ground in forests and grasslands.

• The kaka can fly. It eats seeds, fruit, and occasionally the eggs of other birds in forest habitats.

• The kea can fly. It eats plants, larvae and other small animals. It lives in mountainous forest habitats.

• All three species live on the South Island of New Zealand and had overlapping ranges until the population size of kakapos started to decrease. Populations of kakas also live on the North Island of New Zealand.

A student studied the information and suggested that all three species evolved by sympatric speciation.

Evaluate the student’s conclusion.

[4]
Reveal mark scheme Mark scheme for question 4(b)(iii)
(b)(iv)

New Zealand has a high species biodiversity compared to many countries.

Species biodiversity includes the concepts of species richness and species evenness.

Explain the difference between species richness and species evenness.

[2]
Reveal mark scheme Mark scheme for question 4(b)(iv)

Biological Processes November 2021, Question 19 15 marks

Also involves Osmosis · Cell fractionation and ultracentrifugation

(a)

The Hill reaction is a model system used to study the light-dependent stage of photosynthesis. It uses a blue dye, DCPIP, which is colourless when reduced.

A student was provided with the following method:

• Cut three small spinach leaves into small pieces and place in a mortar containing 20 cm3 of 0.5 mol dm–3 sucrose solution.

• Homogenise by grinding vigorously.

• Filter the mixture and pour the filtrate into a centrifuge tube.

• Centrifuge at high speed for 5 minutes.

• Gently pour the supernatant (liquid part) into a clean tube labelled A.

• Resuspend the pellet (sediment) in 20 cm3 of pH 7.0 buffer.

• Add 5 cm3 of resuspended pellet to each of tubes labelled B – D.

• Boil tube B for 5 minutes and then cool.

• Add 10 cm3 DCPIP solution to each of tubes A – E.

• Transfer tube C to a dark cupboard.

• Observe colour of the tubes after 5 minutes.

The table shows the results of the experiment.

Tube ATube B
Tube C
Tube D
Tube E
Contentssupernatantresuspended resuspended resuspended distilled pellet pellet pellet water
Boiled for 5 minutesnoyes no no no
DCPIPyesyes yes yes yes
Illuminationlightlight dark light light
Colour after 5 minutesblueblue blue colourless blue
(a)(i)

State the name of the final electron acceptor in the light-dependent stage of photosynthesis.

[1]
Reveal mark scheme Mark scheme for question 19(a)(i)
(a)(ii)

DCPIP is reduced in the Hill reaction.

Suggest and explain the function of DCPIP in the Hill reaction.

[2]
Reveal mark scheme Mark scheme for question 19(a)(ii)
(a)(iii)

Using the results shown in the table, explain what can be concluded from each tube, or pair of tubes, about the light-dependent stage of photosynthesis.

Tubes C & D........

[4]
Reveal mark scheme Mark scheme for question 19(a)(iii)
(a)(iv)

The student knew that it was important to use sucrose solution when homogenising the leaves.

Explain why it was important that the pellet was suspended in buffer solution and why it did not contain sucrose.

[2]
Reveal mark scheme Mark scheme for question 19(a)(iv)
(a)(v)

Suggest and explain two improvements that would increase the validity of the method.

[4]
Reveal mark scheme Mark scheme for question 19(a)(v)
(b)

Terrariums are popular for growing houseplants.

A terrarium is a glass container containing soil and small plants.

Once established, a terrarium can be sealed and the plants will be able to grow for months or even years despite not being in contact with the outside atmosphere.

The terrarium maintains moist conditions for the plants.

(b)(i)

Suggest one other reason why the plants in a sealed terrarium continue to grow.

[1]
Reveal mark scheme Mark scheme for question 19(b)(i)
(b)(ii)

Cacti are popular house plants.

Suggest why cacti do not grow well in a terrarium.

[1]
Reveal mark scheme Mark scheme for question 19(b)(ii)

Unified Biology June 2019, Question 1 11 marks

Figure for question 1
(a)(i)

Identify the cellular components shown at A and B.

[2]
Reveal mark scheme Mark scheme for question 1(a)(i)
(a)(ii)

The real size of the line between C and D on Fig. 1.1 is 1.4 × 10−6 m.

Calculate the magnification that was used to produce the image in Fig. 1.1.

Give your answer to 2 significant figures.

[2]
Reveal mark scheme Mark scheme for question 1(a)(ii)
(a)(iii)

Fig. 1.2 shows a student’s drawing of another sago pondweed cell, which was observed under a light microscope. The student used a sharp pencil but did not label the drawing.

Figure for question 1

Describe two other ways in which the drawing could be improved.

[2]
Reveal mark scheme Mark scheme for question 1(a)(iii)
(a)(iv)

The student stained a sago pondweed sample to improve the contrast between cellular components when viewed under a microscope.

The student used the following procedure to stain the sample:

• Use forceps to place the sample on a glass slide.

• Use a pipette to place two drops of the stain in the centre of the sample.

• Carefully lower a cover slip onto the sample, ensuring that the cover slip is parallel with the slide as it is lowered.

Describe two improvements the student should make to their staining procedure.

[2]
Reveal mark scheme Mark scheme for question 1(a)(iv)
(b)

Sago pondweed has evolved many adaptations to its aquatic environment. Three such adaptations are described below.

Explain the advantage of each adaptation.

Adaptation 1: No waxy cuticle

Adaptation 2: Stem tissue that contains air spaces

Adaptation 3: A thin, flexible stem

[3]
Reveal mark scheme Mark scheme for question 1(b)

Unified Biology June 2019, Question 3 12 marks

Icefish live in very cold water.

Icefish contain biological molecules that allow them to tolerate cold temperatures.

(a)

Adaptations can be grouped into three general categories.

Which category of adaptation is represented by cold-tolerant molecules?

[1]
Reveal mark scheme Mark scheme for question 3(a)
(b)
Figure for question 3
[2]
Reveal mark scheme Mark scheme for question 3(b)
(c)
Figure for question 3
(c)(i)

Suggest and explain two improvements that would increase the validity of the students’ investigation.

[4]
Reveal mark scheme Mark scheme for question 3(c)(i)
(c)(ii)

Suggest appropriate units to use to represent the rate of enzyme activity in this investigation.

[1]
Reveal mark scheme Mark scheme for question 3(c)(ii)
(c)(iii)

The students recorded the temperature that produced the fastest reaction rate in each of the five replicates. These results are shown in Table 3.

Figure for question 3

One of the students made the following statement:

I think the mean is a more accurate measure than the median or mode for these results.

Evaluate the student’s statement.

[2]
Reveal mark scheme Mark scheme for question 3(c)(iii)
(c)(iv)

The students wanted to know whether there was a difference between the reaction rates of the two forms of trypsin at 30 °C.

They performed a statistical test on the mean of the five replicates for human trypsin and the five replicates for icefish trypsin.

Suggest the most appropriate statistical test for the students to use and explain why this test is appropriate.

[2]
Reveal mark scheme Mark scheme for question 3(c)(iv)

Unified Biology June 2017, Question 3 10 marks

Tigers, Panthera tigris, are predatory mammals. They have evolved striped patterns on their fur (as shown in Fig. 3.1a on the insert), which provide camouflage in their habitats.

(a)(i)

Adaptations can be divided into three types.

State the type of adaptation represented by the tiger’s stripes.

[1]
Reveal mark scheme Mark scheme for question 3(a)(i)
(a)(ii)

* Describe and explain how a tiger with striped fur may have evolved from a non-striped ancestor.

In your answer you should discuss the different types of genes that might be involved in the creation of the striped pattern in the tiger’s fur.

[6]
Reveal mark scheme Mark scheme for question 3(a)(ii)
(b)

One subspecies of tiger is the Bengal tiger. One in 10 000 Bengal tiger births results in a white Bengal tiger.

White Bengal tigers (shown in Fig. 3.1b on the insert) have black stripes but lack orange fur.

The allele that causes white fur is recessive and is a result of a mutation to a gene called SLC45A2.

According to the Hardy-Weinberg principle, the following equations can be used to estimate allele frequency within a population:

p2 + 2pq + q2 = 1

p + q = 1

Use the Hardy-Weinberg equations to calculate the percentage of Bengal tigers that are heterozygous for the SLC45A2 gene.

Give your answer to one significant figure.

Show your working.

[3]
Reveal mark scheme Mark scheme for question 3(b)

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