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Question 1

Topic: 15.1

(a) Fig. 1.1 is a drawing of a longitudinal section (LS) of a human kidney.

Use the letters A, B, C and D in Fig. 1.1 to complete Table 1.1.
Each letter may be used once, more than once or not at all.
For each description, list all the letters that are correct.

(b) The volume and water potential of the urine produced by the kidney vary according to the water potential of the blood. This is a result of osmoregulation.
Describe the role of aquaporins in osmoregulation.

(c) Describe the role of the brain in osmoregulation when the water potential of the blood increases above the set point.

▶️ Answer/Explanation

Solution

(a)
Answer to Table 1.1

Explanation: The letters correspond to kidney structures: A (renal cortex), B (renal medulla), C (renal pelvis), and D (ureter). The table is completed by matching each description to the correct structure(s). For example, “contains glomeruli” applies to the cortex (A), while “collects urine” refers to the renal pelvis (C).

(b)

Explanation: Aquaporins are water channel proteins in the collecting duct and distal convoluted tubule. They increase membrane permeability to water, allowing reabsorption into the blood when ADH is present. This regulates urine concentration based on hydration needs.

(c)

Explanation: When blood water potential rises (e.g., after drinking), hypothalamic osmoreceptors detect the change and reduce ADH secretion from the posterior pituitary. Less ADH decreases aquaporin insertion in the kidney, producing dilute urine to restore balance.

Question 2

Topic: 19.1

Interferon-alpha (IFN- \(\alpha\)) can be produced as a recombinant human protein to treat some types of cancer. The gene IFNA2 codes for IFN- \(\alpha\).

One method of producing recombinant IFN- \(\alpha\) uses genetically engineered Escherichia coli bacteria that contain recombinant plasmids. Each recombinant plasmid contains:

the gene IFNA2
three regulatory sequences of the lac operon (promoter, operator and \(lacI\))
a gene for antibiotic resistance, \(AMP^R\).

Each of the sequences for the lacI gene and \(AMP^R\) gene contains its own promoter. As a result, these genes are always expressed in E. coli bacteria that contain this recombinant plasmid.

Fig. 2.1 is a diagram of the recombinant plasmid. The promoter regions of the lacI gene and \(AMP^R\) gene are not shown.

(a) The start of transcription of the gene IFNA2 by E. coli with the recombinant plasmid shown in Fig. 2.1 needs to be controlled to obtain an optimum yield of IFN- \(\alpha\).

Scientists investigated the effect of two inducers of transcription on the production of recombinant IFN- \(\alpha\):

lactose, which is converted to allolactose in E. coli
IPTG, which is a synthetic molecule with a very similar structure to allolactose. IPTG cannot be broken down by E. coli.

The scientists grew three cultures of \(E\). coli containing the recombinant plasmid in the same growth medium. The growth medium contained glucose, amino acids, essential vitamins and minerals. The growth medium did not contain lactose.

After four hours, either lactose or IPTG at the same concentration was added to two of the cultures of E. coli. As a control, the third culture of E. coli was grown without adding lactose or IPTG.

The concentration of recombinant IFN- \(\alpha\) in the cultures was measured at different times over a period of 28 hours. The results are shown in Fig. 2.2.

(i) The regulatory sequences of the lac operon contained in the recombinant plasmid are involved in the control of transcription of the gene IFNA2.

Explain the role of the gene lacI in the control of transcription of the IFNA2 gene between 0 hours and 4 hours.

(ii) With reference to Fig. 2.2, describe the changes in the concentration of recombinant IFN- \(\alpha\) in the culture containing IPTG from when IPTG was added at \(\mathbf{4}\) hours to the end of the experiment at \(\mathbf{28}\) hours.

(iii) Suggest one reason for the difference between the concentration of recombinant IFN- \(\alpha\) in the culture at \(\mathbf{8}\) hours in the presence of lactose and the concentration of recombinant IFN- \(\alpha\) in the culture at \(\mathbf{8}\) hours in the presence of IPTG.

(iv) Suggest one reason for the change in the concentration of recombinant IFN- \(\alpha\) in the culture containing IPTG from 12 hours to 16 hours.

(b) The gene \(AMP^R\) in the plasmid shown in Fig. 2.1 codes for a protein that provides resistance to the antibiotic ampicillin.

Suggest how \(AMP^R\) allows genetically engineered \(E\). coli containing the recombinant plasmid to be identified.

(c) Bacteria can evolve antibiotic resistance through natural processes.
Outline how bacteria can evolve to become resistant to antibiotics.

▶️ Answer/Explanation

Solution

(a)(i)

Explanation: The lacI gene codes for the lac repressor protein, which binds to the operator region of the lac operon in the absence of an inducer (allolactose or IPTG). Between 0–4 hours, the repressor prevents RNA polymerase from binding to the promoter, inhibiting transcription of IFNA2.

(a)(ii)

Explanation: After IPTG addition at 4 hours: (1) IFN-α concentration rises steeply, peaking at 8 hours. (2) The concentration then declines gradually but remains above baseline by 28 hours. (3) The peak concentration (~0.8 μg mL⁻¹) is higher than the initial level (~0.1 μg mL⁻¹).

(a)(iii)

Explanation: IPTG yields higher IFN-α than lactose because: (1) IPTG is not metabolized by E. coli, maintaining a constant concentration, whereas lactose is broken down into glucose and galactose. (2) IPTG binds more stably to the repressor.

(a)(iv)

Explanation: The decline after 12 hours may occur due to IFN-α degradation or depletion of cellular resources (e.g., amino acids) for protein synthesis.

(b)

Explanation: \(AMP^R\) allows selection of transformed E. coli by growing them on ampicillin-containing medium. Only bacteria with the plasmid (and \(AMP^R\)) survive.

(c)

Explanation: Antibiotic resistance evolves via: (1) Random mutations in bacterial DNA (e.g., in drug target genes). (2) Selection pressure from antibiotics favors resistant mutants. (3) Resistant bacteria survive and reproduce. (4) Resistance genes spread via horizontal gene transfer (conjugation/transduction). (5) Overuse of antibiotics accelerates this process.

Question 3

Topic: 17.1

Salmon can be genetically modified (GM) to produce increased quantities of growth hormone, which is a protein. GM salmon modified in this way have a faster growth rate and reach their maximum body mass at a younger age than non-GM salmon.

(a) Within any population of salmon there is variation in body mass. This is an example of continuous variation.
Explain what is meant by continuous variation and how it can be caused.

(b) Scientists investigated whether injection of very young non-GM salmon with recombinant growth hormone could cause an increase in the growth rate of the salmon.

The scientists used two groups of non-GM salmon:
• a control group of salmon that were not injected with recombinant growth hormone
• an experimental group of salmon that were injected with 1.0µg of recombinant growth hormone at the start of the experiment and once a week for the next six weeks.

The mean body mass of the salmon in the two groups at the start of the experiment was the same (5.3g).
After six weeks, the body mass of every salmon was measured again. The results are summarised in Table 3.1.

A student decided that a t-test should be performed on the results shown in Table 3.1.

(i) Calculate the value of \(t\) for the results shown in Table 3.1 using the formula for the \(t\)-test:
\[ t=\frac{\left|\bar{x}_1-\bar{x}_2\right|}{\sqrt{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)}} \]
Give your answer to two decimal places.
Show your working.

(iii) Suggest one advantage, other than cost, of farming GM salmon that produce increased quantities of growth hormone instead of farming non-GM salmon that are injected with recombinant growth hormone each week.

▶️ Answer/Explanation

Solution

(a) Continuous variation shows a range of intermediate phenotypes (e.g., body mass) that follow a normal distribution. It’s caused by polygenic inheritance (many genes with additive effects) and environmental influences.

Explanation: Unlike discrete traits (e.g., blood type), continuous traits like body mass have many possible values because multiple genes and environmental factors (e.g., food availability) contribute.

(b)(i) Calculation of t-value:
\[ t = \frac{|7.7 – 9.4|}{\sqrt{\frac{0.4^2}{28} + \frac{1.1^2}{27}}} = \frac{1.7}{\sqrt{0.0057 + 0.0448}} = \frac{1.7}{0.2249} = 7.56 \]

Explanation: The large t-value (7.56) indicates a significant difference between the groups. This is calculated by comparing the means (7.7g vs 9.4g) while accounting for sample sizes and variances.

(b)(iii) GM salmon don’t require weekly injections, reducing stress and infection risks.

Explanation: Genetic modification provides continuous hormone production internally, eliminating the need for repeated handling and injections that can stress fish and introduce pathogens.

Question 4

Topic: 19.2

Array comparative genome hybridisation (aCGH) is a technique involving the use of a microarray to analyse a genome or sections of a genome.

(a) Outline the steps required to prepare the genome of an individual so that the genome is ready for analysis using a microarray chip.

(b) DiGeorge syndrome is a dominant inherited disease in humans. DiGeorge syndrome is caused by deletion of a large number of nucleotides from chromosome 22. The number of nucleotides deleted varies between individuals in a range from 800000 to 3100000. The largest deletions can cause the removal of up to 46 protein-coding genes from the chromosome.

Fig. 4.1 shows the results of aCGH using a microarray specific for the section of chromosome 22 within which the DiGeorge syndrome deletion occurs. The microarray analysed DNA from two individuals:

• one with DiGeorge syndrome
• one who did not have DiGeorge syndrome (control DNA for comparison). In the aCGH results shown in Fig. 4.1:
• Each small circle represents the results from a single probe on the microarray.
• The x-axis shows the position of each probe on chromosome 22. The position is shown as distance along the chromosome in millions of nucleotides.
• A result close to 100% fluorescence on the y-axis means that the DNA from the individual with DiGeorge syndrome fluoresces at the same intensity as the control DNA for that probe.
• A result close to 50% fluorescence on the y-axis means that the DNA from the individual with DiGeorge syndrome fluoresces half as much as the control DNA for that probe.

(i) With reference to Fig. 4.1, estimate the number of nucleotides deleted from the affected chromosome 22 in the individual with DiGeorge syndrome.
Give your answer to the nearest 100000 nucleotides.

(ii) Explain how the microarray technique works to give the results shown in Fig. 4.1.

(iii) Suggest why the phenotypes of two individuals with DiGeorge syndrome can be different.

▶️ Answer/Explanation

Solution

(a)

Extract DNA from the individual’s cells/tissue.
Fragment the DNA using restriction enzymes.
Denature DNA into single strands and label with fluorescent dyes.

Explanation: The DNA must be isolated, cut into manageable pieces, and tagged for hybridization with microarray probes.

(b)(i) 2,400,000 or 2,500,000 nucleotides.

Explanation: The fluorescence drops to ~50% between ~18.5M and ~21M nucleotides, indicating a deletion of ~2.5M nucleotides.

(b)(ii)

DiGeorge and control DNA are labeled with different fluorescent dyes.
DNA hybridizes to complementary probes on the microarray.
Deletions reduce fluorescence (50%) compared to control (100%).

Explanation: The microarray detects deletions by comparing fluorescence intensities between test and control DNA at specific probe locations.

(b)(iii)

Variable deletion sizes (800k–3.1M nucleotides) affect different genes.
Heterozygosity: Remaining alleles may be dominant/recessive.
Environmental factors influence phenotypic expression.

Explanation: Phenotypic variability arises from differences in deleted gene content, allelic interactions, and external factors.

Question 5

Topic: 16.1

Meiosis is described as a reduction division because the number of chromosomes in the daughter cells is reduced by half.

(a) Table 5.1 describes some of the events that take place during four of the different stages of meiosis in an animal cell.

Complete Table 5.1 by:
• outlining the behaviour of the spindle fibres during anaphase I
• identifying the stage of meiosis in which spindle fibres re-form the spindle in daughter cells
• drawing a diagram to show telophase II.
You do not need to add labels to your diagram showing telophase II.

(b) Explain the need for a reduction division during meiosis.

▶️ Answer/Explanation

Solution

(a)

Explanation:

Anaphase I: Spindle fibres shorten, pulling homologous chromosomes to opposite poles.
Spindle re-formation: Occurs during prophase II in daughter cells.
Telophase II diagram: Shows four haploid cells with unpaired chromosomes (no labels required).

(b)

Answer:

Meiosis produces haploid gametes (n) for sexual reproduction.
Fertilization combines two gametes (n + n) to form a diploid zygote (2n).
Maintains species’ chromosome number across generations.
Prevents polyploidy (excessive chromosome doubling) in offspring.

Explanation: Reduction division ensures genetic stability by halving chromosome number in gametes, allowing restoration of diploidy upon fertilization. This prevents chromosomal aberrations and maintains biodiversity.

Question 6

Topic: 12.2

(a) Fig. 6.1 is a diagram of a section through a mitochondrion.

The four arrows, A, B, C and D, show the movement of molecules and ions. Use the letters to identify all the arrows (one or more) that show:
(i) active transport of protons
(ii) diffusion of carbon dioxide
(b) Outline the role of the mitochondrial matrix in respiration

▶️ Answer/Explanation

Solution

(a)(i) D

Explanation: Arrow D represents the active transport of protons (\(H^+\)) from the mitochondrial matrix into the intermembrane space. This process requires energy (ATP) and is crucial for establishing the proton gradient used in chemiosmosis.

(a)(ii) C and D

Explanation: Carbon dioxide (\(CO_2\)), a byproduct of respiration, diffuses out of the mitochondrion. Arrow C shows \(CO_2\) diffusing from the matrix to the intermembrane space, while arrow D (in reverse) indicates \(CO_2\) exiting the mitochondrion entirely.

(b) Role of the Mitochondrial Matrix:

Explanation: The mitochondrial matrix is the site of key metabolic processes:
1. Link Reaction & Krebs Cycle: Pyruvate is decarboxylated and converted into acetyl-CoA, which enters the Krebs cycle to generate reduced NAD and FAD.
2. Protein Synthesis: Contains DNA and ribosomes to produce enzymes (e.g., dehydrogenases) essential for respiration.
3. Substrate-Level Phosphorylation: Direct ATP production occurs during the Krebs cycle (e.g., via succinyl-CoA synthetase).

(c) Effect of Oxygen Lack on Oxidative Phosphorylation:

Explanation: Oxygen is the final electron acceptor in the electron transport chain (ETC). Without it:
1. ETC Stops: Electrons cannot be passed to oxygen, halting electron flow.
2. No Proton Gradient: \(H^+\) ions are not pumped into the intermembrane space, preventing chemiosmosis.
3. ATP Production Ceases: No ATP is synthesized via ATP synthase.
4. NAD/FAD Accumulation: Reduced NAD and FAD cannot be recycled, inhibiting earlier respiration stages.

Question 7

Topic: 14.1

(a) Fig. 7.1 is a diagram representing a synapse between a chemoreceptor cell from a human taste bud and a dendrite of a sensory neurone.

In an experiment, different concentrations of sodium chloride solution were applied to the microvilli of the chemoreceptor cell. The membrane potential of the chemoreceptor cell and the membrane potential of the dendrite of the sensory neurone were recorded for each concentration. The resting potential of this chemoreceptor cell is –50mV and the resting potential of the dendrite of this sensory neurone is –70mV. The results are shown in Table 7.1. Table 7.1

Explain the results shown in Table 7.1.
(b) Describe the differences in structure and function between sensory neurones and motor neurones.

▶️ Answer/Explanation

Solution

(a)

Explanation: At \(10.1 \, \mathrm{g \, dm}^{-3}\), the chemoreceptor membrane does not depolarize (stays at \(-50 \, \mathrm{mV}\)), so no neurotransmitter is released, and the sensory neurone remains at resting potential (\(-70 \, \mathrm{mV}\)). At higher concentrations (\(31.0 \, \mathrm{g \, dm}^{-3}\)), the chemoreceptor depolarizes, releasing neurotransmitter, which triggers depolarization (action potential) in the sensory neurone. The response follows the all-or-nothing law, where a threshold must be crossed for an impulse to be generated.

(b)

Differences between sensory and motor neurones

Explanation: Sensory neurones transmit impulses from receptors to the CNS, while motor neurones carry impulses from the CNS to effectors. Structurally, sensory neurones have long dendrites and short axons, whereas motor neurones have short dendrites and long axons. Additionally, sensory neurones are unipolar, while motor neurones are multipolar.

Question 8

Topic: 13.1

(a) Describe the functions of the internal membranes of the chloroplast in photosynthesis.

(b) Rubisco activase (RA) is an enzyme that has an effect on the activity of rubisco.
An investigation was carried out on the effect of RA on the activity of rubisco.

Solutions of rubisco and RuBP were added to two tubes, A and B.
RA was added to tube \(\mathbf{A}\).
Both tubes were incubated at \(25^{\circ} \mathrm{C}\) for 6 minutes.
The activity of rubisco was measured every 30 seconds.

All conditions were kept the same, except for the addition of RA to tube \(\mathbf{A}\). The results are shown in Fig. 8.1.

Describe the results shown in Fig. 8.1 and suggest an explanation for the effect of RA on the activity of rubisco.

▶️ Answer/Explanation

Solution

(a)

Explanation: The internal membranes (thylakoids) of the chloroplast play key roles in photosynthesis:

They contain photosynthetic pigments (e.g., chlorophyll) to absorb light energy.
They house photosystems I & II and the electron transport chain for light-dependent reactions.
They facilitate photolysis (splitting of water) and ATP synthesis via chemiosmosis.
Their stacked structure (grana) increases surface area for efficient light absorption.

(b)

Description of Results:

Tube A (with RA) shows a steady increase in rubisco activity over time (e.g., from ~3.0 to ~11.5 units).
Tube B (without RA) maintains constant low activity (~0.4 units).

Explanation: RA activates rubisco by modifying its active site, enabling better binding with RuBP and faster product release. This explains the rising activity in Tube A, while Tube B lacks this activation mechanism.

Question 9

Topic: 14.2

(a) Fig. 9.1 is a diagram of a relaxed sarcomere in striated muscle

(i) On Fig. 9.1, use label lines and letters to label:
• an actin filament with the letter P
• a myosin filament with the letter R. [2]
(ii) State what happens to the A-band and the I-band when the sarcomere contracts.
(b) The plant Strychnos toxifera produces the toxin curare, which can cause muscle paralysis in mammals.
The toxin acts by binding to receptors on the cell surface membranes (sarcolemma) of muscle cells at neuromuscular junctions.
(i) Suggest how binding of curare to receptors may cause muscle paralysis.

(ii) Suggest why the action of curare may lead to the death of a mammal

▶️ Answer/Explanation

Solution

(a)(i) P pointing to thin filament (actin); R pointing to thick filament (myosin).

Explanation: Actin filaments (P) are the thinner filaments, while myosin filaments (R) are the thicker filaments in the sarcomere.

(a)(ii) A-band – stays the same; I-band – gets narrower.

Explanation: The A-band remains constant as it represents the length of the myosin filament, while the I-band shortens during contraction as actin filaments slide over myosin.

(b)(i) Curare competes with acetylcholine, blocking receptors and preventing depolarization. This inhibits \( \mathrm{Na}^+ \) channel opening, \( \mathrm{Ca}^{2+} \) release, and muscle contraction.

Explanation: Curare acts as a competitive inhibitor at neuromuscular junctions, disrupting the signal for muscle activation.

(b)(ii) Curare can cause death by paralyzing respiratory muscles (ribs/diaphragm) or cardiac muscles, leading to suffocation or heart failure.

Explanation: Without muscle function, essential processes like breathing and circulation cease, resulting in fatality.

Question 10

Topic: 18.1

(a) The passage in Fig. 10.1 is about biodiversity.
Complete the passage by using the most appropriate scientific terms.

(b) The International Union for Conservation of Nature (IUCN) Red List of Threatened Species is updated regularly.
Table 10.1 shows the numbers of endangered animal species counted every three years between 2007 and 2019.

(i) Calculate the rate of increase in the number of endangered species between 2007 and 2019.
Show your working.
Give your answer to the nearest whole number.

(ii) More species of fish were listed as endangered in 2019 than species of mammals. Suggest reasons why more fish species than mammal species are endangered.

▶️ Answer/Explanation

Solution

(a)

abundance / numbers / population (size);
Simpson’s;
genes;
alleles;
adapt / evolve;
habitats / niches;

Explanation: The passage describes biodiversity, which includes the abundance of species, Simpson’s diversity index, genetic diversity (genes and alleles), and the ability of species to adapt to different habitats.

(b)(i)

\[ \begin{aligned} & \text{Rate of increase} = \frac{14234 – 7851}{12} \\ & = \frac{6383}{12} \\ & = 532 \text{ per year (to the nearest whole number).} \end{aligned} \]

Explanation: The total increase in endangered species from 2007 to 2019 is \(14234 – 7851 = 6383\). Dividing this by 12 years gives an annual rate of \(532\) species per year.

(b)(ii)

More fish species are endangered due to overfishing, habitat destruction (e.g., coral reef degradation), pollution (e.g., plastic waste), and climate change affecting aquatic ecosystems. Mammals may have better conservation efforts or slower reproductive rates, making fish more vulnerable.

Explanation: Fish face greater anthropogenic pressures, such as unsustainable fishing practices and marine pollution, while mammals often benefit from targeted conservation programs and public awareness.

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