All Questions: AS & A Level Biology - 9700 Paper 4 2023 Winter Zone 3
Theory
MCQ
01.
Theory 15 Marks
CH13 - PHOTOSYNTHESIS, CH18 - BIODIVERSITY, CLASSIFICATION AND CONSERVATION

Fig. 1.1 shows some structural features of a chloroplast and some processes that occur within it.

(a) (i) Identify the structures labelled A in Fig. 1.1.
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(ii) Explain how the structure and appearance of the granum, and the components labelled A, relate to their function.
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(b) (i) Identify the metabolic pathway labelled cycle C in Fig. 1.1.
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(ii) Explain why pathway C is described as a cycle.
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(c) (i) Identify the products of photosynthesis labelled B and D in Fig. 1.1.
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(ii) Suggest and explain the importance of glucose and the product labelled B in Fig. 1.1 to ecosystems.
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02.
Theory 17 Marks
CH18 - BIODIVERSITY, CLASSIFICATION AND CONSERVATION, CH17 - SELECTION AND EVOLUTION

Biodiversity can be assessed at three different levels. One of these is the genetic variation within each species.
(a) Outline two other levels at which biodiversity can be measured.
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To calculate the genetic variation that exists within a species, scientists:
• obtain DNA sequences from many individuals of one species
• count the number of nucleotides that differ when the sequences of two individuals are compared
• repeat this with different pairs of individuals.
This allows scientists to calculate the mean number of differences at every nucleotide position along the sequence (mean number of nucleotide differences per site).
(b) Explain why scientists use databases and computers to calculate the mean number of nucleotide differences per site.
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(c) Table 2.1 shows the mean number of nucleotide differences per site of some species.

Table 2.1

speciesmean number of nucleotide differences per site
Drosophila melanogaster, fruit fly0.0087
Anopheles gambiae, mosquito vector of malaria0.0301
Plasmodium falciparum, malarial pathogen0.0015
Zea mays, wild maize0.0139

(i) State the genus name of the species that shows the most genetic variation.
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(ii) State how many kingdoms of organisms are represented in Table 2.1.
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(d) Genetic variation is considered important in the conservation of species. Low genetic variation is assumed to decrease the chance of the long-term survival of a species.
(i) Give reasons why low genetic variation may decrease the long-term survival of a species.
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Fig. 2.1 shows how the International Union for the Conservation of Nature (IUCN) categorises species according to their conservation status.
Common species with the lowest conservation status (least risk of extinction) are categorised as Least Concern (LC).

![Fig. 2.1](https://via.placeholder.com/600x200)

(ii) Question 2(d) states that ‘low genetic variation is assumed to decrease the chance of the long-term survival of a species’.
Predict the relationship between genetic variation and conservation status if this assumption is true.
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Fig. 2.2 shows the mean number of nucleotide differences per site of some species and sub-species of mammals and their conservation status.

![Fig. 2.2](https://via.placeholder.com/600x400)

(iii) Assess whether the data in Fig. 2.2 provide support for the prediction you made in 2(d)(ii).
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03.
Theory 16 Marks
CH16 - INHERITED CHANGE, CH14 - HOMEOSTASIS

(a) Fig. 3.1 shows Ipomoea purpurea, the common morning glory.



The gene that determines flower colour in I. purpurea has two alleles:
• a dominant allele that results in purple flowers
• a recessive allele that results in red flowers.

A student recorded the flower colour of all the I. purpurea plants in a field.

The student recorded 660 plants with purple flowers and 440 plants with red flowers.

Assuming the Hardy-Weinberg principle applies to this population, calculate the number of plants in the field that are heterozygous.

Use the equations:

\( p + q = 1 \)
\( p^2 + 2pq + q^2 = 1 \)

Show your working and give your answer to the nearest whole number.

(b) The Japanese morning glory, I. nil, has over 20 different flower colour phenotypes, including shades of blue, purple, red and pink.

The flower colour of I. nil is controlled by at least four genes. The flower colour can change gradually after the flowers open each morning and can change with fluctuations in the carbon dioxide concentration of the surrounding air.

A student concluded that the flower colour phenotype in I. nil shows continuous variation.

Suggest two reasons why the student made this conclusion.

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(c) Scientists investigated the response of stomata to changing carbon dioxide (CO₂) concentrations in the beach morning glory, I. pes-caprae.

The scientists placed I. pes-caprae plants in chambers. They measured the width of open stomata (stomatal apertures) after the plants had been exposed to different CO₂ concentrations for 40 minutes. Light intensity and temperature were kept constant.

The relationship between CO₂ concentration and the mean width of stomatal apertures is shown in Fig. 3.2.



(i) In 2016, a study measured the atmospheric CO₂ concentration as 400 µmol mol⁻¹.

In the future, climate change may reduce water availability and increase atmospheric CO₂ concentrations in some habitats.

Suggest how the stomatal response shown in Fig. 3.2 would allow I. pes-caprae to survive the effects of climate change.

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(c) (ii) Under certain conditions, the closure of stomata is controlled by abscisic acid.

Describe how abscisic acid causes the closure of stomata.

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(d) (i) Scientists are researching whether abscisic acid can be used in crop treatment to increase yield. Evidence suggests that abscisic acid modifies the effect of auxin on elongation growth in plants.

Scientists investigated the effect of different concentrations of abscisic acid on root elongation in seedlings of thale cress, Arabidopsis thaliana.

The seedlings were divided into four groups:
• a control group (0.0 µmol abscisic acid)
• three experimental groups, each treated with a different concentration of abscisic acid: 0.1 µmol, 1.0 µmol, or 10.0 µmol.

For each group of seedlings, root length was measured for six days during treatment. The rate of root elongation was calculated each day.

The results are shown in Fig. 3.3.

Key
⚫ control (no abscisic acid)
◐ 0.1 µmol of abscisic acid
✖ 1.0 µmol of abscisic acid
△ 10.0 µmol of abscisic acid



With reference to Fig. 3.3, describe the effect of treatment with abscisic acid on the rate of root elongation.

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(d) (ii) The passage outlines the role of auxin in elongation growth in plants.

Complete the passage by using the most appropriate scientific terms.

The binding of auxin to receptors causes ........................... to be pumped into cell walls.
This activates proteins called expansins, which disrupt the links between ........................... microfibrils. The cell walls are then able to expand.


04.
Theory 7 Marks
CH19 - GENETIC TECHNOLOGY

(a) (i) Suggest what is meant by recombinant DNA technology.
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(ii) Suggest why the scientists created two different types of GM potato plant.
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(iii) State why the scientists also performed the test on non-GM potato plants.
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(b) Discuss how the results in Table 4.1 provide information that could help to solve the global demand for food.
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05.
Theory 7 Marks
CH12 - ENERGY AND RESPIRATION

(a) The respiratory quotient (RQ) values for different respiratory substrates can be calculated.

Table 5.1 shows:
• the formulae of four respiratory substrates
• the number of oxygen molecules (O$_2$) needed to completely respire each substrate to carbon dioxide and water.

$\text{Table 5.1}$

[Table_1]

Using Table 5.1, name the respiratory substrate with the highest RQ and the respiratory substrate with the lowest RQ.

respiratory substrate with the $\textbf{highest}$ RQ ............................................................

respiratory substrate with the $\textbf{lowest}$ RQ ............................................................

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(b) In anaerobic conditions, the production of ATP in mammals and yeast involves glycolysis and fermentation.

Describe the similarities $\textbf{and}$ differences between fermentation in mammals and in yeast.

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06.
Theory 11 Marks
CH16 - INHERITED CHANGE

(a) (i) The fish breeder concluded that the new transparent phenotype had occurred because of a mutation.
Explain how the results in Table 6.1 support this conclusion.
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(ii) State the approximate whole-number ratio shown by the results in Table 6.1.
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(iii) Explain what the results in Table 6.1 show about the genes and alleles that determine the wild-type, albino and the two different transparent phenotypes in tiger barbs.
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(b) The F1 tiger barbs all looked the same but the F2 offspring showed variation. The F2 offspring showed four different phenotypes.
Describe the processes that occurred during meiosis in the F1 fish that allowed this variation to occur.
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07.
Theory 9 Marks
CH15 - CONTROL AND CO-ORDINATION

When an impulse arrives at a neuromuscular junction, it stimulates a muscle fibre of striated muscle to contract.
(a) (i) Outline the similarities in structure between a neuromuscular junction and a cholinergic synapse.
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(ii) The hydrolysis of ATP during muscle contraction releases inorganic phosphate ($P_i$).
Calcium ions ($Ca^{2+}$) can combine with $P_i$ in the sarcoplasmic reticulum to form insoluble calcium phosphate. This may result in fewer power strokes occurring in sarcomeres.
Suggest why calcium phosphate formation in the sarcoplasmic reticulum may result in fewer power strokes occurring in sarcomeres.
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(b) Adrenaline is a hormone that can affect muscle contraction. Adrenaline binds to G-protein-coupled receptors on T-tubule membranes.
The cell signalling pathway that occurs in response to the binding of adrenaline is similar to the pathway that occurs in liver cells in response to the binding of glucagon.
Fig. 7.1 is an outline of the cell signalling pathway of adrenaline.

Identify the molecules represented by A and B in Fig. 7.1.
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08.
Theory 6 Marks
CH13 - PHOTOSYNTHESIS, CH19 - GENETIC TECHNOLOGY

Environmental conditions such as light intensity affect plant physiology.
(a) Use the letter X to identify a point on the sketch graph in Fig. 8.1 where light intensity is acting as a limiting factor on the rate of photosynthesis. [1]


(b) An experiment investigated how light intensity affected gene expression in kale, _Brassica oleracea sabellica_.

- Two groups of kale plants were grown, with one group in high light intensity and one group in low light intensity. All other conditions were standardised.
- After the same period of time, messenger RNA (mRNA) was extracted from each group of kale plants.
- The mRNA was used to produce cDNA.
- The cDNA was hybridised with probes for 89621 kale genes on a microarray.
- The microarrays showed which genes were switched on (expressed) in each set of conditions.

The results showed that expression of 18% of the genes was affected by light intensity.

- 14% of the genes were switched on only in high light intensity.
- 4% were switched on only in low light intensity.

(i) State the name of the enzyme that produces cDNA from an mRNA template.
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(ii) State the name of the type of proteins that control gene expression in plants.
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(iii) Suggest why _more_ genes were switched on only in plants growing in _high_ light intensity compared to fewer genes that were switched on only in plants growing in low light intensity.
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09.
Theory 10 Marks
CH17 - SELECTION AND EVOLUTION

Green lacewings are a family of insects with more than 1300 species.

The common green lacewing, Chrysoperla carnea, is shown in Fig. 9.1.



(a) Green lacewings have sense organs, known as tympanal organs, that detect sound. The tympanal organ of green lacewings has evolved to detect the high frequency sounds that bats make when they are hunting. Bats eat green lacewings. When a green lacewing senses the presence of a bat, it moves away or closes its wings in flight to escape.

(i) Outline how the tympanal organ of green lacewings could have evolved by natural selection.
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(ii) When high frequency sound is detected, the receptor cells in the tympanal organ stimulate the transmission of impulses in sensory neurones. Describe the sequence of events that results in an action potential in a sensory neurone. The first event in the sequence has been given for you.

Calcium ions enter the cytoplasm of the receptor cell ....................................................
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(b) Two species of green lacewing, C. carnea and C. downesi, evolved from a common ancestor. The two species have populations with overlapping distributions in parts of North America.

Table 9.1 shows a comparison of the characteristics of overlapping populations of the two species.

[Table_1]

Suggest how speciation occurred to produce the two different species of green lacewing.
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10.
Theory 7 Marks
CH19 - GENETIC TECHNOLOGY

Medicine is defined as the diagnosis, treatment and prevention of disease.
Outline the different ways in which genetic technology can be applied to medicine, with reference to named diseases.