OCR A-Level Biology: Synoptic & Practical Skills

Excess amino acids cannot be stored, so they are broken down in the liver. The amino group is first removed and converted to ammonia, which is highly toxic and very soluble. A series of enzyme-controlled reactions in liver cells then combines ammonia with carbon dioxide to form urea, which is far less toxic and is excreted by the kidneys.

Explain how the properties of enzymes allow the liver to convert toxic ammonia into urea, and why this conversion is important for the mammal. In your answer you should link the structure and mode of action of enzymes to the process of excretion.

(9 marks)

When a nerve impulse arrives at a cholinergic synapse, the neurotransmitter acetylcholine is released and the postsynaptic neurone may reach threshold and produce an action potential. Both the resting potential and the action potential depend on proteins in the cell surface membrane.

Explain how different membrane proteins are involved in transmission across a cholinergic synapse and in generating an action potential in the postsynaptic neurone. In your answer you should link the structure and transport roles of membrane proteins to the events of synaptic transmission and the action potential.

(9 marks)

Some lichens are very sensitive to the gas sulfur dioxide, which is released in vehicle exhaust. A student investigated whether the number of lichen species growing on tree trunks was related to distance from a busy road. At eight sites at different distances from the road, they recorded the number of lichen species on the trunks. The data are shown below.

Site	Distance from road / m	Number of lichen species
A	10	3
B	25	4
C	40	6
D	60	5
E	90	9
F	130	11
G	180	10
H	250	14

The student decided to test for a correlation using Spearman's rank correlation coefficient:

$r_s = 1 - \frac{6 \sum d^2}{n(n^2 - 1)}$rs​=1−n(n2−1)6∑d2​

The critical value of $r_s$rs​ at the 0.05 probability level for $n = 8$n=8 is 0.738.

Calculate Spearman's rank correlation coefficient for these data and interpret your result. State the null hypothesis you are testing.

(6 marks)

Leaves growing in bright sunlight often have more stomata per unit area than leaves of the same species growing in shade. A student counted the stomatal density (stomata per mm²) of five leaves from a plant grown in full sun and five leaves from a plant of the same species grown in shade.

Leaf	Stomatal density (sun) / mm⁻²	Stomatal density (shade) / mm⁻²
1	62	44
2	58	48
3	65	41
4	60	46
5	70	50

The student calculated the mean stomatal density of each group as 63.0 mm⁻² (sun) and 45.8 mm⁻² (shade), with sample standard deviations of 4.69 mm⁻² (sun) and 3.49 mm⁻² (shade). They used an unpaired Student's t-test to compare the two means:

$t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\dfrac{s_1^{\,2}}{n_1} + \dfrac{s_2^{\,2}}{n_2}}}$t=n1​s12​​+n2​s22​​​xˉ1​−xˉ2​​

The critical value of $t$t at the 0.05 probability level for 8 degrees of freedom is 2.31.

Calculate the value of $t$t and use it to reach a conclusion about whether the difference in mean stomatal density is significant. State the null hypothesis you are testing.

(6 marks)

A student investigated the effect of glucose concentration on the growth of a yeast population. They set up five flasks of nutrient broth, each with a different glucose concentration, and inoculated each with yeast. After 48 hours they estimated the number of yeast cells in each flask using a haemocytometer (a counting chamber viewed under a microscope). For each flask they took a single sample, placed it under the haemocytometer without diluting it, and counted the cells in one of the chamber's grid squares. They concluded that the flask with the highest glucose concentration produced the largest yeast population.

Evaluate this method. In your answer you should identify the independent, dependent and control variables, comment on the reliability, accuracy and validity of the cell counts, identify sources of error, and suggest improvements.

(6 marks)

Some antibiotics kill bacteria by binding to an enzyme the bacterium needs. A random mutation in the gene for that enzyme can change a single amino acid in the enzyme, so the antibiotic no longer binds, and the bacterium survives treatment. When the antibiotic is used widely, resistant strains become increasingly common.

Explain how a gene mutation can make a bacterium resistant to an antibiotic, and how natural selection causes resistant bacteria to become more common in a population. In your answer, link gene mutation and protein structure to natural selection.

(5 marks)