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Question 1

Topic: 12.1

(a) Glycolysis is a biochemical pathway that occurs in the cytoplasm of cells.
In glycolysis, a molecule of glucose is metabolised to two molecules of pyruvate. The process is outlined in Fig. 1.1.

(i) Explain why glucose is phosphorylated at the beginning of glycolysis.

(ii) Suggest one use of the reduced NAD that is produced in glycolysis.

(iii) Name the type of phosphorylation reaction by which ATP is made during glycolysis.
(b) Pyruvate can enter the mitochondrion by active transport.
Describe the main conditions that are required for pyruvate to enter the mitochondrion by active transport.
(c) Pyruvate is involved in the link reaction in the matrix of the mitochondrion.
Describe the link reaction.

▶️ Answer/Explanation

Solution

(a)(i) Glucose is phosphorylated to make it more reactive and unstable, preventing it from leaving the cell. This also maintains a concentration gradient for further metabolic processes.

Explanation: Phosphorylation adds phosphate groups to glucose, increasing its energy level and ensuring it remains in the cell for glycolysis.

(a)(ii) Reduced NAD is used to carry hydrogen atoms/electrons for redox reactions, such as reducing pyruvate to lactate in anaerobic respiration.

Explanation: NADH acts as an electron carrier, donating electrons in later stages (e.g., the electron transport chain) to generate ATP.

(a)(iii) Substrate-level phosphorylation.

Explanation: ATP is directly synthesized from the transfer of a phosphate group from a substrate (e.g., phosphoenolpyruvate) to ADP.

(b) Pyruvate enters the mitochondrion via active transport, requiring: (1) a carrier protein, (2) ATP/energy, (3) low pyruvate concentration outside the mitochondrion, and (4) oxygen availability.

Explanation: Active transport moves pyruvate against its concentration gradient, necessitating energy and specific transport proteins.

(c) The link reaction involves: (1) decarboxylation of pyruvate, (2) oxidation (dehydrogenation) forming NADH, and (3) combination with coenzyme A to produce acetyl CoA.

Explanation: Pyruvate loses a carbon as CO₂, and the remaining 2-carbon compound binds to CoA, entering the Krebs cycle.

Question 2

Topic: 16.1

(a) In sexual reproduction, meiosis occurs to produce haploid gametes from a diploid cell.
Explain the meaning of the terms haploid and diploid.

(b) Describe the behaviour of chromosomes during the main stages of meiosis I in animal cells.
prophase I
metaphase I
anaphase I
telophase I

(c) The production of haploid gametes by meiosis also involves division of the cytoplasm.
State the term used to describe this division of the cytoplasm.

▶️ Answer/Explanation

Solution

(a)

Final Answer: Haploid is one (complete) set of chromosomes; diploid is two (complete) sets of chromosomes.

Explanation: A haploid cell (n) contains a single set of chromosomes, while a diploid cell (2n) contains two sets—one from each parent. Gametes are haploid to ensure the correct chromosome number is restored during fertilization.

(b)

Final Answer:
Prophase I: Chromosomes condense; homologous pairs form bivalents; crossing over occurs.
Metaphase I: Homologous pairs align at the equator; independent assortment occurs.
Anaphase I: Homologous chromosomes separate and move to opposite poles.
Telophase I: Chromosomes decondense; cytokinesis may begin.

Explanation: In meiosis I, homologous chromosomes undergo pairing, recombination, and segregation, reducing the chromosome number by half. Each stage ensures genetic diversity and proper distribution of chromosomes.

(c)

Final Answer: Cytokinesis.

Explanation: Cytokinesis is the division of the cytoplasm following nuclear division (meiosis or mitosis), resulting in two distinct daughter cells.

Question 3

Topic: 17.2

(a) The sea blush, Plectritis congesta, is a flowering plant that grows on the west coast of North America.
Individual sea blush plants produce fruit that is either winged or wingless. Investigations have
shown that this characteristic is controlled by a single gene with two alleles:
• a dominant winged fruit allele
• a recessive wingless fruit allele.
Fig. 3.1 shows the difference in structure between winged fruit and wingless fruit phenotypes.

(i) A large sample of sea blush fruits was collected and their fruit-wing characteristic was recorded.

Name the type of variation that is shown for the fruit-wing characteristic of the sea blush.

(ii) Different sea blush plants can have fruit with different colours.
Suggest two ways in which a new fruit colour could naturally occur in a sea blush population.

(iii) Early taxonomists classified sea blush plants with winged fruits as a different species to sea blush plants with wingless fruits.

Since this time, evidence from observations and experiments has confirmed that the plants belong to the same species.

Suggest three examples of the evidence obtained that helped to confirm that these sea blush plants belong to the same species.

(b) Natural selection occurs in populations, such as in populations of sea blush plants.
Explain why natural selection occurs in populations.

▶️ Answer/Explanation

Solution

(a)(i) Discontinuous variation.

Explanation: The fruit-wing characteristic is controlled by a single gene with two distinct alleles (winged and wingless), resulting in clear-cut phenotypes with no intermediates, which is typical of discontinuous variation.

(a)(ii) Two possible ways:

Mutation: A new allele for fruit colour could arise due to a random mutation in the DNA.
Cross-pollination: Introduction of new alleles through hybridisation with other sea blush plants.

Explanation: New traits like fruit colour can emerge from genetic mutations or recombination during sexual reproduction, introducing variability in the population.

(a)(iii) Three pieces of evidence:

Interbreeding: Winged and wingless plants can produce fertile offspring, confirming they are the same species.
Genetic similarity: DNA analysis shows identical or highly similar gene sequences.
Shared niche: Both types occupy the same ecological habitat and exhibit similar physiological traits.

Explanation: Reproductive compatibility, genetic analysis, and ecological observations are key evidence for classifying organisms as the same species.

(b) Natural selection occurs because:

Populations exhibit genetic variation due to mutations and sexual reproduction.
Competition for limited resources leads to differential survival.
Individuals with advantageous traits (e.g., better fruit dispersal) survive and reproduce, passing on their alleles.

Explanation: Natural selection acts on heritable traits, favoring those that enhance survival and reproduction, thereby increasing the frequency of beneficial alleles in the population.

Final Answer:

3(a)(i) discontinuous ;
3(a)(ii) any two from:
1 mutation A description
or
(formation of a) new allele ;
2 change in, environment / selection pressure ; A description
3 cross pollination / cross breeding / hybridisation ;
4 new allele combination / codominant alleles together ;

3(a)(iii) any three from:
both types
1 (inter)breed to produce fertile offspring / not reproductively isolated ;
2 occupy the same niche ;
3 same / similar, morphology / physiology / biochemistry ;
A description I fruit shape
4 have, similar / same, genes / DNA sequences / genomes ;
5 similar / same, proteins / amino acid sequences ;

3(b) any five from:
1 (populations) can produce many offspring / AW ;
2 genetic variation occurs (in populations) / described ;
3 ref. mutations / sexual reproduction / gamete formation / random fertilisation ;
4 ref. competition qualified ; e.g. mates / food / resources
5 ref. selection pressure ;
6 some individuals have advantageous, adaptations / phenotype / traits / AW ;
7 (so) they, survive / reproduce / have a selective advantage ;
8 (so) pass on their advantageous allele(s) ;
9 (so) the (advantageous) allele frequency increases ;

Question 4

Topic: 19.2

Genetic technology uses many different enzymes and techniques.
(a) Restriction endonucleases are used in genetic modification. These enzymes occur naturally
in prokaryotic cells. More than 3500 different restriction endonucleases have been identified
and it is thought there are many more to discover.
Name two domains that are a source of restriction endonucleases.
(b) Originally, the method used to obtain a restriction endonuclease was to:
• grow large numbers of the specific prokaryotic cells that are the source of the enzyme
• break open the cells and extract and purify the restriction endonuclease.
This original method produced only a small quantity of restriction endonuclease and was not economical.
The newer method for large-scale production is to:
• obtain the gene coding for a specific restriction endonuclease
• introduce the gene into Escherichia coli, with a promoter that allows the gene to be expressed continuously.

The newer method increases the quantity of specific restriction endonuclease produced.
Suggest and explain the steps needed to carry out the newer method for large-scale
production of a specific restriction endonuclease.

(c) Describe the advantages of databases for the study and use of restriction endonucleases.
(d) Electrophoresis is a technique used in genetic technology.
Paper chromatography is a technique used to investigate the photosynthetic pigments found in chloroplasts.
Compare the similarities and differences between electrophoresis and chromatography.

▶️ Answer/Explanation

Solution

(a) Bacteria and Archaea

Explanation: Restriction endonucleases are naturally found in prokaryotic cells, which belong to the domains Bacteria and Archaea. These enzymes serve as a defense mechanism against viral DNA.

(b)

Steps for large-scale production:
1. Obtain the gene: Extract the restriction endonuclease gene from the prokaryotic source or synthesize it.
2. Amplify the gene: Use PCR to create multiple copies of the gene.
3. Insert into a vector: Cut a plasmid with a restriction enzyme and ligate the gene into it.
4. Transform E. coli: Introduce the recombinant plasmid into E. coli for expression.
5. Culture and extract: Grow the bacteria in a fermenter and purify the enzyme.

Explanation: This method leverages recombinant DNA technology to produce large quantities of the enzyme efficiently by using E. coli as a host.

(c)

Advantages of databases:
1. Large data storage: Stores extensive sequences of DNA and enzymes.
2. Quick access: Enables rapid sharing and retrieval of genetic information.
3. Comparative studies: Helps in analyzing and comparing enzyme structures and functions.
4. Discovery: Facilitates identification of new restriction enzymes.

Explanation: Databases streamline research by providing centralized, searchable repositories of genetic data.

(d)

Comparison of Electrophoresis and Chromatography

Explanation: Both techniques separate molecules based on size/charge (electrophoresis) or solubility (chromatography). Electrophoresis uses an electric field, while chromatography relies on a mobile and stationary phase.

Question 5

Topic: 17.3

(a) Warfarin is a poison used to kill rats. Some rats are resistant to warfarin and can survive the effects of the poison. Warfarin resistance is due to a mutant dominant allele at a single gene locus. Researchers investigated the population of brown rats on a large island where there are no predators of rats. On this island, warfarin is used to try to control the rat population. The researchers found that 40% of the population of rats was resistant to warfarin.

(i) Use Hardy–Weinberg equation 1 and equation 2 to calculate the percentage of the population of rats on the island that is heterozygous for warfarin resistance.

equation 1 p + q = 1
equation 2 p$^{2}$ + 2pq + q$^{2}$ = 1
Show your working.

(ii) In natural populations, it is usual to find that only some of the conditions of the Hardy–Weinberg principle are met. Suggest and explain which of the conditions of the Hardy–Weinberg principle are most likely to be met for the island population of brown rats.

(b) Dominant advantageous alleles and recessive advantageous alleles both naturally occur in populations. Explain why, when a new dominant advantageous allele occurs, its frequency increases more quickly in the population than when a new recessive advantageous allele occurs.

▶️ Answer/Explanation

Solution

(a)(i) 35%

Explanation: Given that 40% of the population is resistant (either homozygous dominant or heterozygous), we have $ p^2 + 2pq = 0.4 $. Since resistance is dominant, the remaining 60% are homozygous recessive ($ q^2 = 0.6 $). Solving for $ q $, we get $ q = \sqrt{0.6} \approx 0.775 $, and $ p = 1 – q \approx 0.225 $. The heterozygous frequency is $ 2pq = 2 \times 0.225 \times 0.775 \approx 0.35 $ or 35%.

(a)(ii)

Explanation: The conditions most likely met are:

No migration (isolated island population).
Random mating (no restrictions mentioned).
Large population (no genetic drift effects).
No selection pressure (warfarin resistance is advantageous, but predators are absent).

(b)

Explanation: A dominant advantageous allele is expressed even in heterozygotes, so selection acts on it immediately. In contrast, a recessive advantageous allele is only expressed in homozygotes, so its frequency increases slowly as it must first reach a sufficient frequency for homozygotes to appear.

Question 6

Topic: 14.2

(a) Fig. 6.1 is a photomicrograph of a section through a Bowman’s capsule and a glomerulus.

On Fig. 6.1, use label lines and letters to label:
F – the location of the glomerular filtrate
P – the location of the blood plasma.

(b) The glomerular filtration rate (GFR) is the rate at which fluid filters from the blood in the glomerulus into the Bowman’s capsule.
Fig. 6.2 shows the mean GFR values for healthy males and females of different age groups.

(i) Use Fig. 6.2 to calculate the yearly rate of decrease in mean GFR from the 20–29 age group to the 80–89 age group for females.

Show your working.
Give your answer to one decimal place and include units.

(ii) The age of a person and whether they are male or female can affect GFR.
Suggest and explain other factors that can affect GFR.

(c) The glomerular filtrate passes from the Bowman’s capsule to the proximal convoluted tubule, where selective reabsorption takes place.
Describe the process of selective reabsorption.

▶️ Answer/Explanation

Solution

(a)

Explanation: F is labeled in the Bowman’s capsule where the glomerular filtrate collects, and P is labeled in the glomerulus where blood plasma is located.

(b)(i) \frac{118-52}{60} = 1.1 \text{ cm}^3 \text{ min}^{-1} \text{ year}^{-1}

Explanation: The mean GFR for females decreases from 118 cm³/min (20–29 age group) to 52 cm³/min (80–89 age group) over 60 years. The yearly rate of decrease is calculated as (118 – 52)/60 = 1.1 cm³/min/year.

(b)(ii)

Explanation: Other factors affecting GFR include:

High blood pressure increases GFR due to greater hydrostatic pressure.
Kidney diseases (e.g., damage to nephrons) decrease GFR.
Dehydration lowers blood water potential, reducing GFR.
Drugs like diuretics or vasodilators can alter GFR.

(c)

Explanation: Selective reabsorption involves:

Active transport of Na⁺ out of tubule cells into blood, creating a gradient.
Na⁺, glucose, and amino acids are reabsorbed via cotransport or diffusion.
Water follows by osmosis due to the lowered water potential in cells.
Useful substances (e.g., glucose) are returned to the blood, while waste remains in the filtrate.

Question 7

Topic: 15.1

(a) A striated muscle cell (muscle fibre) will contract when stimulated by a motor neurone at a neuromuscular junction.
Air temperature can affect the temperature of striated muscle cells.
Investigations have shown that the efficiency of contraction of striated muscle cells decreases when the air temperature decreases.
Suggest reasons why a reduction in temperature can decrease the efficiency of contraction of striated muscle cells.

(b) The greater blue-ringed octopus, Hapalochlaena lunulata, produces tetrodotoxin (TTX). TTX is a neurotoxin. If a mammal is bitten by this octopus, the effect of TTX can cause the death of the mammal.
Fig. 7.1 shows a greater blue-ringed octopus.

TTX binds to voltage-gated sodium ion channels in the axon of a neurone and changes the tertiary structure of the channel protein.
Suggest how TTX may affect the functioning of a motor neurone.

▶️ Answer/Explanation

Solution

(a)

Explanation: A decrease in temperature reduces the efficiency of muscle contraction due to:

Slower enzyme activity (e.g., ATPase, acetylcholinesterase), reducing ATP production and neurotransmitter breakdown.
Reduced ion movement (Ca²⁺, Na⁺), impairing action potentials and calcium binding to troponin.
Fewer cross-bridge formations between actin and myosin, decreasing power strokes.
Lower metabolic rates, limiting energy availability for contraction.

(b)

Explanation: TTX disrupts motor neurone function by:

Blocking voltage-gated Na⁺ channels, preventing depolarization and action potentials.
Inhibiting neurotransmitter (ACh) release at neuromuscular junctions, stopping muscle stimulation.
Causing paralysis due to failed signal transmission to muscles.

Final Answer:

7(a) Respiration slows, enzyme activity decreases, ion movement reduces, and cross-bridge formation declines.

7(b) TTX blocks Na⁺ channels, preventing depolarization and causing paralysis.

Question 8

Topic: 13.2

(a) In plants such as rice, Oryza sativa, rubisco has a low rate of activity, which in turn affects the rate of photosynthesis. The cereal crop sorghum, Sorghum bicolor, has a high rate of activity of rubisco. A genetically modified (GM) variety of rice was produced. Parts of the quaternary structure of rubisco in rice were altered to be the same as the rubisco in sorghum. The rate of activity of rubisco in non-GM rice and GM rice was measured at different concentrations of atmospheric carbon dioxide (CO₂). The results are shown in Fig. 8.1.

Fig. 8.1: Rubisco activity in GM and non-GM rice

(i) Compare the two curves shown in Fig. 8.1 and explain why the curve for non-GM rice levels off.

(ii) Suggest which part of the rubisco molecule was altered to produce the GM variety of rice.

(b) Fig. 8.2 shows an outline of the Calvin cycle.

(i) Name the process that involves the enzyme rubisco.

(ii) State the type of reaction that occurs when GP is converted to TP.

(iii) RuBP is regenerated from molecules of TP. State how many molecules of RuBP are produced from 10 molecules of TP.

(iv) TP molecules that are not involved in the regeneration of RuBP can be used in the synthesis of other molecules. State two molecules that can be produced from these TP molecules.

▶️ Answer/Explanation

Solution

(a)(i)

Comparison: The GM rice shows a higher rubisco activity rate than non-GM rice at all CO₂ concentrations. Both curves increase initially, but the non-GM rice levels off at higher CO₂ concentrations.

Explanation for leveling off: The non-GM rice curve plateaus because CO₂ is no longer the limiting factor—other factors like light intensity, temperature, or enzyme saturation become limiting.

(a)(ii) Active site or allosteric site.

Explanation: The active site of rubisco was likely altered to enhance its catalytic efficiency, mimicking sorghum’s rubisco.

(b)(i) Carbon fixation (or carboxylation).

(b)(ii) Reduction (or hydrogenation).

(b)(iii) 6 molecules of RuBP.

Explanation: The Calvin cycle requires 10 TP molecules to regenerate 6 RuBP molecules (5 TP → 3 RuBP, but scaling for 10 TP).

(b)(iv) Any two of: glucose, fructose, sucrose, starch, cellulose, amino acids, glycerol, or fatty acids.

Question 9

Topic: 14.1

(a) The passage outlines the endocrine system.
Complete the passage by using the most appropriate scientific terms.
The endocrine system consists of tissues and ……………………………………………………….. ,
which secrete hormones such as insulin, glucagon and antidiuretic hormone (ADH).
Glucagon only affects target cells that have complementary receptors, which are located
on the ……………………………………………………….. . Glucagon binds to these receptors
and this leads to the production of a ……………………………………………………….. , which
transfers the signal throughout the cytoplasm. The target cells for insulin are in the liver,
……………………………………………………….. and adipose tissue. The target cells for ADH are
those of the distal convoluted tubule and the ……………………………………………………….. of the
kidney nephron. These hormones are involved in ……………………………………………………….. ,
which is the maintenance of a constant internal environment.

(b) The blood glucose concentration of a person was measured at regular intervals after the ingestion of a meal rich in glucose.
Fig. 9.1 shows the results of this investigation.

(i) Explain how the blood glucose concentration is reduced during phase A of the curve.

(ii) Suggest why the blood glucose concentration increases again during phase B.

▶️ Answer/Explanation

Solution

(a)

Answer: glands ;
cell surface membrane / plasma membrane ;
second messenger / cAMP ;
(striated) muscles ;
collecting duct ;
homeostasis ;

Explanation: The endocrine system consists of glands that secrete hormones. Glucagon binds to receptors on the cell surface membrane, triggering a second messenger (cAMP) to relay the signal. Insulin targets the liver, muscles, and adipose tissue, while ADH acts on the collecting duct of nephrons. These processes maintain homeostasis.

(b)(i)

Answer:
1. Insulin is produced/released ;
2. Increase in uptake of glucose into cells ;
3. Detail of uptake ; e.g., increased permeability of cell surface membrane to glucose / GLUT added to membrane ;
4. Increase in respiration of glucose / phosphorylation of glucose ;
5. Stimulates glycogenesis / conversion of glucose to lipids ;

Explanation: During phase A, insulin secretion enhances glucose uptake by cells, increasing membrane permeability (via GLUT transporters). Glucose is either respired or stored as glycogen (glycogenesis) and lipids, lowering blood glucose levels.

(b)(ii)

Answer: Reference to glucagon ;

Explanation: In phase B, glucagon stimulates glycogenolysis (breakdown of glycogen into glucose) in the liver, releasing glucose back into the bloodstream, causing the observed rise.

Question 10

Topic: 18.3

(a) Many zoos around the world are involved in captive breeding programmes for endangered species.
Outline the role of zoos in the conservation of endangered species, other than captive breeding programmes.

(b) Chester Zoo, in the United Kingdom, controls the European captive breeding programme for the eastern black rhinoceros, Diceros bicornis michaeli.
Fig. 10.1 shows an eastern black rhinoceros.

In the captive breeding programme for the eastern black rhinoceros, 15.8% of females breed each year while in the wild 23.7% of females breed each year.
Suggest reasons why fewer females breed in captivity than in the wild.

(c) Sometimes assisted reproduction is used in captive breeding programmes for the eastern black rhinoceros.
Fig. 10.2 outlines the process of IVF that is used in these programmes.

Outline:
• the event A that occurs in the production of the embryos
• the two other possible outcomes, B and C, for the embryos that have been produced.

▶️ Answer/Explanation

Solution

(a)

Explanation: Zoos contribute to conservation by providing a safe location (protected from predators/poachers), conducting research (disease control, diet, behavior), raising funds, collaborating with local communities/governments, educating the public, and offering medical care to endangered species.

(b)

Explanation: Fewer females breed in captivity due to an unnatural environment, reduced fertility, refusal to mate, limited availability of suitable mates, stress, and controlled breeding policies by zoos to manage population capacity.

(c)

Explanation: Event A: Fertilization occurs when sperm is added to the egg.
Outcome B: Embryos can be implanted into a surrogate mother for gestation.
Outcome C: Embryos can be cryopreserved (frozen) for future use in breeding programs.

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