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Question 1

Topic: 14.1 (Homeostasis in mammals)

(a) Fig. 1.1 is a diagram of a nephron.

(b) The cells of the proximal convoluted tubule are adapted to carry out selective reabsorption.
Describe and explain how these cells are adapted to carry out selective reabsorption.

▶️ Answer/Explanation

Solution

(a)

1(a) A – label line to collecting duct ;
B – label line to inner part of Bowman’s capsule ;
C – label line to, loop of Henle /
part of collecting duct below level of convoluted tubules ;
D – label line to wider blood vessel entering Bowman’s capsule ;

Explanation: The labels A-D are identified based on the nephron structure. A is the collecting duct, B is the inner part of Bowman’s capsule, C is the loop of Henle, and D is the afferent arteriole (wider blood vessel entering Bowman’s capsule).

(b)

any four from:
1 microvilli, for large surface area / increases surface area (for reabsorption) ;
2 cotransporter (proteins) for movement of, glucose / amino acids
(with sodium ions) ;
3 tight junctions / described, to, stop substances passing in between cells /
cause substances to pass though cells ;
4 many mitochondria to provide ATP for, active transport / pumping of Na+;
5 folded basal membrane for many sodium (potassium) pumps ;

Explanation: The proximal convoluted tubule cells are adapted for selective reabsorption through structural and functional features. Microvilli increase surface area, cotransporters facilitate glucose/amino acid movement, tight junctions ensure substances pass through cells, mitochondria provide ATP for active transport, and folded basal membranes house sodium pumps.

Question 2

Topic: 16.2 (The role of genes in determining the phenotypes)

The scientist Gregor Mendel investigated differences in the length of the stem in the pea plant, Pisum sativum. In 1866, he published the results of his investigation into this trait (characteristic). Fig. 2.1 shows a diagram of a pea plant.

Mendel observed that the pea plants he grew either had tall stems or dwarf (short) stems. In his investigation, Mendel carried out crosses using pea plants with these two phenotypes.

(a) From the results of these crosses, Mendel demonstrated that tall stems were dominant to dwarf stems in pea plants. It is now known that the stem length trait in pea plants is controlled by one gene that has two alleles: • a dominant allele, Le • a recessive allele, le.

Describe a cross that could be carried out and how the results of the cross could be analysed to determine the genotype of a pea plant with a tall stem.

(b) The scientists P W Brian and H G Hemming identified that the difference in the length of the stem in pea plants was associated with the presence or absence of gibberellin. They published their findings in 1955.

(i) Gibberellin leads to a response in plant cells by binding to specific receptor molecules. State the term used to describe a molecule, such as gibberellin, that binds to specific receptor molecules and leads to a response in cells.

(ii) Suggest the response of the cells in the internode region of the stem, as labelled in Fig. 2.1, to the presence of gibberellin and describe how this response affects the trait investigated by Mendel.

▶️ Answer/Explanation

Solution

(a)

Explanation: To determine the genotype of a tall pea plant (either homozygous dominant Le Le or heterozygous Le le), a test cross is performed by crossing the tall plant with a dwarf plant (le le). If all offspring are tall, the parent is homozygous dominant. If a 1:1 ratio of tall to dwarf offspring is observed, the parent is heterozygous.

Steps:

Perform a test cross between the tall plant and a homozygous recessive dwarf plant (le le).
Observe the offspring phenotypes.
If all offspring are tall, the tall parent is Le Le.
If any dwarf offspring appear, the tall parent is Le le.

(b)(i) Hormone / Ligand.

Explanation: Gibberellin is a plant hormone (ligand) that binds to specific receptors, triggering cellular responses such as growth regulation.

(b)(ii)

Explanation: Gibberellin stimulates cell elongation in the internode region, leading to increased stem height. This explains why pea plants with functional gibberellin pathways (tall phenotype) grow taller than those without (dwarf phenotype).

Steps:

Gibberellin binds to receptors in internode cells.
This triggers cell elongation and/or division.
Resulting in taller stems, as observed in Mendel’s tall phenotype.

Question 3

Topic: 16.2 (The role of genes in determining the phenotypes)

Cystic fibrosis is an autosomal recessive genetic disease. People with cystic fibrosis have a homozygous recessive genotype.

(a) Explain the meaning of the terms homozygous and recessive.

(b) (i) In 2020:

• there were 10800 people with cystic fibrosis in the UK
• the UK population was estimated to be 67100000 people.

A proportion of people in the UK population are heterozygous for the gene that causes cystic fibrosis and do not have symptoms of the disease.

Use the Hardy–Weinberg principle to calculate the number of people in the UK population who are expected to be heterozygous for the gene that causes cystic fibrosis.

The two equations for the Hardy–Weinberg principle are provided.

number of people in the UK expected to be heterozygous for the gene =……………………………………….
(ii) The Hardy–Weinberg principle provides a useful estimate of the number of people in the UK who are heterozygous for cystic fibrosis. However, the estimate is lower than the actual number. This underestimation occurs because not all the conditions of the Hardy–Weinberg principle apply.

In the UK in 2020, the mean life expectancy of:

• people with cystic fibrosis was approximately 50 years
• all people was approximately 80 years.

Explain how this information accounts for the underestimation of the number of people in the UK that are heterozygous for cystic fibrosis.
(c) A screening programme for cystic fibrosis was introduced in 2007 for all children born in the UK. Children are tested within seven days of their birth. Children identified from the screening programme as being at high risk of having cystic fibrosis can have a genetic test to confirm whether they have the disease.
(i) Table 3.1 shows the median predicted life expectancy for people born in the UK who have cystic fibrosis. Predictions are shown for people born in 2008, 2012, 2016 and 2020.

Describe the trend shown in Table 3.1 and outline how early screening for cystic fibrosis may have contributed to this trend.
(ii) In many countries, a genetic test for cystic fibrosis is available to adults who do not have cystic fibrosis but have a family member who either has cystic fibrosis or is heterozygous for the gene that causes cystic fibrosis.

These adults include partners, parents, offspring, brothers and sisters of the family member. The aim is to find out if any of these adults are heterozygous for the gene that causes cystic fibrosis.

Discuss the ethical and social considerations of making a genetic test for cystic fibrosis available to these adults.

▶️ Answer/Explanation

Solution

(a)

Homozygous: Both alleles of a gene are the same.

Recessive: The phenotype of the allele is masked by a dominant allele, requiring two copies for expression.

Explanation: Homozygous means identical alleles (e.g., aa), while recessive alleles only show their effect when homozygous.

(b)(i)

Number of heterozygous people = 1,680,000

Explanation: Using Hardy-Weinberg, \( q = \sqrt{\frac{10,800}{67,100,000}} = 0.0127 \), \( p = 1 – q = 0.9873 \), and \( 2pq = 0.0251 \). Multiplying by population: \( 67,100,000 \times 0.0251 = 1,680,000 \).

(b)(ii)

Explanation: The Hardy-Weinberg principle assumes no selection, but cystic fibrosis reduces life expectancy, leading to underestimation of heterozygotes due to natural selection against affected individuals.

(c)(i)

Trend: Median life expectancy increased from 38.8 (2008) to 50.6 years (2020).

Explanation: Early screening enables prompt treatment, improving survival rates over time.

(c)(ii)

Ethical considerations: Informed reproductive decisions, cost, stigma, and test accuracy are key factors.

Explanation: Testing helps plan care but raises issues like discrimination and confidentiality.

Question 4

Topic: 17.2 (Natural and artificial selection)

Holstein Friesian cattle are a breed of cattle used by dairy farmers in many countries of the world for the high milk yield of their cows. Fig. 4.1 shows Holstein Friesian cattle.

Milk yield in Holstein Friesian cattle is affected by heat stress. Heat stress occurs when homeostatic mechanisms are not enough to keep the body temperature down to normal levels. One of the factors that contributes to heat stress is air temperature. Fig. 4.2 shows the mean daily air temperature in Central Europe and the mean monthly milk yield per cow of Holstein Friesian cattle in Central Europe.

(a) With reference to Fig. 4.2, describe the trends in air temperature and milk yield from April to August.

(b) Many dairy farmers in tropical regions use cattle breeds that are tolerant to heat stress (heat-tolerant cattle). These heat-tolerant cattle can tolerate higher air temperatures than Holstein Friesian cattle before heat stress occurs and have milder symptoms of heat stress than Holstein Friesian cattle for the same high air temperatures. Where heat stress does not occur, heat-tolerant cattle produce a lower milk yield than Holstein Friesian cattle under the same conditions. Scientists compared DNA sequences of Holstein Friesian cattle and heat-tolerant cattle for a number of genes known to have an effect on body temperature. Twenty genes were found that had alleles associated only with heat-tolerant cattle. With reference to the information provided, including the data in Fig. 4.2:

• state the type (pattern) of phenotypic variation shown by milk yield in cattle
• identify factors that cause phenotypic variation in milk yield in cattle.

In each case, give a reason for your choice.

(c) The scientists found that one of the genes studied, PRLR, has a dominant allele known as SLICK. The SLICK allele was identified in Senepol cattle, a heat-tolerant breed, and is not found in Holstein Friesian cattle. Cattle with the SLICK allele have short hair due to reduced hair growth. Scientists have used selective breeding to introduce the SLICK allele into Holstein Friesian cattle. The milk yields of normal Holstein Friesian cattle and Holstein Friesian cattle with the SLICK allele are shown in Fig. 4.3, during March (5°C) and September (14°C).

With reference to Fig. 4.3, describe the effect of the SLICK allele on milk yield in Holstein Friesian cattle.

(d) The SLICK allele differs from the recessive allele by a single nucleotide deletion. This results in a frameshift mutation and introduces a premature stop codon in the PRLR gene. Scientists can use gene editing to replicate this mutation in Holstein Friesian cattle. This provides a way to introduce the SLICK allele into Holstein Friesian cattle without selective breeding. Compare gene editing and selective breeding for introducing the SLICK allele into Holstein Friesian cattle. Include similarities and differences in your answer.

▶️ Answer/Explanation

Solution

(a)

Answer: Temperature increases from April to August, while milk yield decreases. For example, in April, the temperature is ~10°C with a milk yield of ~795 kg/cow, whereas in August, the temperature rises to ~18°C, and the milk yield drops to ~715 kg/cow.

Explanation: The graph shows a clear inverse relationship between air temperature and milk yield. As temperature rises, the cows experience heat stress, reducing their milk production efficiency.

text

(b)

Answer: The phenotypic variation in milk yield is continuous because it shows a range of values without distinct categories. Factors causing variation include:

Genetic factors: Differences in alleles (e.g., 20 genes identified) affect heat tolerance and milk yield.
Environmental factors: Air temperature directly impacts milk yield due to heat stress.

Explanation: Milk yield is influenced by both genetic traits (polygenic inheritance) and environmental conditions, leading to a continuous distribution of phenotypes.

(c)

Answer: The SLICK allele increases milk yield, especially at higher temperatures. In March (5°C), the difference is small (~20 kg), but in September (14°C), the difference is larger (~50 kg). Cattle with the SLICK allele maintain higher milk yields under both conditions.

Explanation: The allele reduces heat stress by promoting short hair growth, allowing cows to regulate body temperature better and sustain milk production.

(d)

Answer:

Aspect	Selective Breeding	Gene Editing
Time Required	Slow (multiple generations)	Fast (single generation)
Precision	Less precise (relies on natural recombination)	Highly precise (targets specific genes)
Unintended Traits	May introduce unwanted traits	Minimizes unintended changes
Ethical Concerns	Generally accepted	Raises ethical debates

Explanation: Gene editing offers faster and more precise results but involves ethical considerations, while selective breeding is slower but more traditional.

Question 5

Topic: 17.3 (Evolution)

Complete the following paragraphs using the most appropriate word or words.

The theory of evolution describes a process that can lead to the formation of new species from pre-existing species over ………………………………………. .

DNA sequence data of different species can be compared to show evolutionary relationships. Two species that have a more recent common ancestor share more ……………………………………… in the DNA nucleotide sequences of their genomes than two species that are more distantly related.

Mitochondrial DNA can also be used in the study of evolutionary relationships. Mitochondrial DNA is inherited only from the female gamete, and its nucleotide sequence is unaffected by ………………………………………. during the production of gametes.

DNA sequence data can be stored in large biological………………………………………. , allowing faster comparison of the nucleotide sequences of genomes using computer software. DNA sequence data can also be used to predict the……………………………….. sequences of proteins produced by a species.

A……………………………………………….. can be used to detect many different mRNA molecules at the same time in studies that compare gene expression between different species.

▶️ Answer/Explanation

Solution

Final Answer:

(many) generations / (many) years / (a long) time / AW ;
mutations / similarities ;
meiosis / crossing over / recombination ;
databases ;
amino acid ;
microarray ;

Explanation:

1. Evolution occurs over many generations or a long period of time, leading to speciation.

2. Closely related species share more similarities in DNA sequences due to a recent common ancestor.

3. Mitochondrial DNA remains unchanged because it bypasses recombination during gamete formation.

4. DNA data is stored in databases for computational analysis, and it helps predict amino acid sequences in proteins.

5. A microarray detects multiple mRNA molecules simultaneously, aiding in gene expression studies.

Question 6

Topic: 12.1 (Respirometer)

A respirometer is a piece of apparatus that can be used to measure the rate of respiration of living tissue such as germinating peas.
A simple respirometer is shown in Fig. 6.1.

A student carried out an investigation to determine the effect of temperature on the rate of respiration of germinating peas.
• The student set up the respirometer as shown in Fig. 6.1 and placed the respirometer in a water-bath at 10°C.
• After five minutes, the student used the syringe to adjust the position of the coloured liquid in the right-hand side of the U-shaped tube so that it lined up with 0cm on the ruler. The student immediately started a timer.
• The germinating peas used up oxygen, causing the coloured liquid in the U-shaped tube to move.
• The student measured the distance moved by the coloured liquid after 20 minutes.
• The student repeated the experiment at temperatures of 20°C, 30°C, 40°C and 50°C.

(a) State the function of the potassium hydroxide solution used in the investigation.
(b) Suggest how the validity of the results could be assessed.
(c) Explain why the respirometer was left in the water-bath for five minutes before starting the experiment.
(d) The rate of movement of the coloured liquid in the U-shaped tube, calculated from the results, is shown in Table 6.1.

Plot a graph of the results shown in Table 6.1 on the grid in Fig. 6.2. Draw a curved line of best fit.

(e) The rate of movement of the coloured liquid is related to the rate of respiration.
Explain the effect of temperature on the rate of respiration shown in Table 6.1 and Fig. 6.2.

▶️ Answer/Explanation

Solution

(a) Absorb carbon dioxide (produced by peas).

Explanation: KOH reacts with CO₂ to form K₂CO₃, ensuring only oxygen consumption affects liquid movement.

(b) Repeat experiments and calculate standard deviation.

Explanation: Consistency across repeats and statistical analysis (e.g., low SD) confirm reliability. A control (e.g., glass beads) validates that liquid movement is due to respiration.

(c) Allow temperature equilibration.

Explanation: The 5-minute delay ensures the respirometer and peas reach the water-bath temperature, eliminating thermal expansion artifacts.

(d)

Explanation: Points plotted at (10°C, 0.2), (20°C, 0.4), (30°C, 0.6), (40°C, 0.5), (50°C, 0.3). A smooth curve peaks at 30°C, showing optimal temperature.

(e) Rate increases up to 30°C due to enhanced enzyme activity, then declines as enzymes denature.

Explanation: 1. Increase (10–30°C): Higher kinetic energy boosts enzyme-substrate collisions (Q₁₀ effect).
2. Decline (>30°C): Enzymes (e.g., cytochrome oxidase) denature, reducing active site efficiency.
3. Optimum at 30°C: Peak respiration rate aligns with maximum enzyme stability.

Question 7

Topic: 13.1 (Photosynthesis as an energy transfer process)

(a) The light-dependent stage of photosynthesis occurs within chloroplasts. In this stage, electrons are emitted from the chlorophyll a molecules and passed to electron acceptors. If a redox indicator, such as DCPIP, is added to a suspension of illuminated chloroplasts, electrons will be transferred to DCPIP, causing the colour of the DCPIP to change from blue to colourless.

A student investigated the effect of the wavelength of light (colour of light) on the rate of photosynthesis.

DCPIP was added to three colorimeter tubes, each containing a suspension of chloroplasts. The chloroplast suspensions were kept in the dark until required.
The colorimeter tubes were each exposed to light of a different colour: red, blue or green. The intensity of light was the same for all tubes, and each was exposed to light for four minutes. All other conditions were kept the same.
The absorbance of each chloroplast suspension was measured at one-minute intervals using a colorimeter.

The results are shown in Fig. 7.1.

(i) Explain why the chloroplast suspensions were kept in the dark until required.

(ii) Describe the results shown in Fig. 7.1.

(iii) With reference to the light-dependent stage of photosynthesis, explain the differences between the results shown in Fig. 7.1 for red light and for green light.

(b) Changes in the atmospheric carbon dioxide concentration, light intensity and temperature can affect the rate of photosynthesis. These three factors directly affect different processes of photosynthesis.

Complete Table 7.1 using a tick (✓) to identify the processes that can be directly affected by each factor or a cross (✗) to identify the processes that are not directly affected by each factor. Indirect effects where a change in the rate of one process affects the rate of a different process should not be considered. A tick or a cross must be placed in the final column of every row.

▶️ Answer/Explanation

Solution

(a)(i) To prevent any premature electron excitation or photosynthesis initiation before the experiment begins, ensuring all measurements start from the same baseline.

(a)(ii)

All three colors show decreasing absorbance over time, indicating DCPIP reduction.
Blue light shows the fastest reduction (steepest slope), followed by red, while green shows minimal change.
Final absorbance values: Blue (lowest), Red (intermediate), Green (highest).

(a)(iii)

Red light is strongly absorbed by chlorophyll, exciting many electrons that reduce DCPIP rapidly (steep slope).
Green light is poorly absorbed (reflected), resulting in fewer excited electrons and minimal DCPIP reduction (flat curve).
The difference demonstrates chlorophyll’s absorption spectrum peaks in red/blue wavelengths.

(b)

Explanation:

Light intensity directly affects the light-dependent stage (✓) but not the Calvin cycle (✗).
CO₂ concentration directly affects the Calvin cycle (✓) but not the light-dependent stage (✗).
Temperature affects enzyme-dependent processes (both stages), but only the Calvin cycle is marked (✓) as it’s more temperature-sensitive due to Rubisco activity.

Question 8

Topic: 15.1 (Control and coordination in mammals)

(a) Approximately 2 × 10⁹ people in the world are currently infected with the bacterial disease tuberculosis (TB) caused by Mycobacterium tuberculosis. Early diagnosis is important so that treatment can begin.

APOPO is a non-profit organisation that has trained African giant pouched rats, Cricetomys gambianus, to use their sense of smell to detect M. tuberculosis. They do this by sniffing a sample of thick mucus from the lungs of people who may have TB. The African giant pouched rats are able to detect the presence of M. tuberculosis with an accuracy of 87-93%.

(i) The type of receptor cell used by African giant pouched rats to detect M. tuberculosis is the same as that used in human taste buds.

Name this type of receptor cell.

(ii) Suggest why African giant pouched rats trained to detect M. tuberculosis may also be able to detect other species of Mycobacterium that cause TB.

(b) The African giant pouched rat belongs to the kingdom Animalia in the domain Eukarya.

Complete Table 8.1 to show the full classification of the African giant pouched rat.

(c) Differences between members of the domain Eukarya and members of the domain Bacteria include the presence or absence of particular membrane-bound cell structures.

Outline other differences in the characteristic features of members of the domain Eukarya and members of the domain Bacteria.

(d) Describe, with reference to the structure of viruses, how viruses are classified.

▶️ Answer/Explanation

Solution

(a)(i)

Chemoreceptor

Explanation: Both rats and humans use chemoreceptors to detect chemical signals – olfactory receptors in smell and taste receptors in taste buds.

(a)(ii)

1. Different Mycobacterium species produce similar volatile organic compounds
2. They share common ancestry and genetic similarities
3. The same chemoreceptors respond to related chemical signatures

Explanation: Mycobacterium species have conserved metabolic pathways that produce similar detectable compounds.

(b)

Explanation: Full classification: Domain Eukarya → Kingdom Animalia → Phylum Chordata → Class Mammalia → Order Rodentia → Family Nesomyidae → Genus Cricetomys → Species gambianus

(c)

Explanation: These fundamental differences reflect their evolutionary divergence about 2 billion years ago.

(d)

1. By nucleic acid type (DNA/RNA, single/double stranded)
2. Presence/absence of lipid envelope
3. Capsid structure (icosahedral, helical, complex)
4. Host specificity (animal, plant, bacterial)
5. Disease symptoms caused

Explanation: The Baltimore classification system categorizes viruses primarily by their genome type and replication strategy.

Question 9

Topic: 15.2 (control and coordination in plants)

(a) An investigation was carried out to study the effect of different intensities of blue light on the percentage germination of barley seeds. Barley seeds were exposed to blue light for a period of seven days. All other variables were kept constant. The results are shown in Table 9.1.

The effect of blue light on the concentration of abscisic acid (ABA) was also investigated. ABA concentration was measured at intervals over seven days in barley seeds exposed to blue light at an intensity of 57 arbitrary units. The results are shown in Table 9.2.

For comparison, in the dark the concentration of ABA in barley seeds fell from 100au at the start (day 0) to 45au on day 1 and did not increase from day 1 to day 7. ABA is thought to affect gibberellin synthesis or activity. Using the information in Table 9.1 and Table 9.2, describe the effect of blue light on the germination of barley seeds and suggest an explanation for this effect.

(b) After germination, auxin is important in the growth of barley plants. Describe and explain the role of auxin in cell elongation.

▶️ Answer/Explanation

Solution

(a)

Explanation:

Blue light decreases the percentage germination of barley seeds.
As the intensity of blue light increases, the percentage germination decreases (e.g., 100% in dark vs. 20% at 57 units).
Blue light increases the concentration of ABA (from 100au to 150au by day 7).
ABA inhibits gibberellin synthesis/activity, which is essential for seed germination.
ABA promotes dormancy by preventing the breakdown of DELLA proteins, which suppress germination.

(b)

Explanation: Auxin promotes cell elongation through the following steps:

Auxin binds to receptors on the cell surface membrane.
Stimulates proton pumps to move H⁺ ions into the cell wall.
Lowers the pH of the cell wall, activating expansins.
Expansins break hydrogen bonds between cellulose microfibrils, loosening the cell wall.
K⁺ channels open, increasing osmotic potential.
Water enters by osmosis, increasing turgor pressure and cell volume.

Question 10

Topic: 15.1 (control and coordination in mammals)

(a) Striated muscle is composed of myofibrils. Myofibrils contain several structural proteins including troponin, tropomyosin, actin and myosin. Outline the roles of these four structural proteins in the contraction of a sarcomere.

(b) Drugs that cause muscle paralysis (paralytic drugs) are used during surgery to stop the patient moving. One commonly used paralytic drug is succinylcholine, which works by preventing contraction of muscles. Succinylcholine is able to prevent muscles contracting because it has a similar shape to acetylcholine. Suggest how succinylcholine is able to prevent muscles contracting.

(c) Duchenne muscular dystrophy (DMD) is a genetic disease that is caused by a single gene. DMD affects striated muscle, and symptoms of the disease first appear at an early age. A fibrous protein, dystrophin, stabilises muscle fibres during contraction. A person with DMD produces non-functioning dystrophin or no dystrophin at all. The disease occurs in about four in 100,000 people and mainly affects boys. Suggest and explain why boys are more likely to have DMD than girls.

▶️ Answer/Explanation

Solution

(a) Roles of structural proteins in sarcomere contraction:

Explanation:

Troponin binds calcium ions (Ca²⁺), causing a conformational change.
This change moves tropomyosin away from actin’s binding sites, exposing them.
Myosin heads bind to actin, forming cross-bridges.
Myosin performs a power stroke, pulling actin filaments inward, shortening the sarcomere.
ATP binding to myosin detaches the cross-bridges, resetting the cycle.

(b) Mechanism of succinylcholine:

Explanation:

Succinylcholine mimics acetylcholine (ACh) and binds to ACh receptors at the neuromuscular junction.
It acts as a competitive inhibitor, blocking ACh from binding.
Without ACh binding, sodium channels remain closed, preventing depolarization.
No action potential is generated, so muscle contraction cannot occur.

(c) Why DMD primarily affects boys:

Explanation:

DMD is X-linked recessive; the dystrophin gene is on the X chromosome.
Males (XY) have only one X chromosome, so a single recessive allele causes the disease.
Females (XX) must inherit two recessive alleles to express DMD (rare).
Heterozygous females are carriers but typically asymptomatic.

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