Biology 9700/42 — February/March 2024

Background Concept

A nephron is the functional unit of the kidney, and the printed diagram (Fig. 1.1) shows its main regions: Bowman's capsule enclosing the glomerulus, the proximal convoluted tubule (PCT), the loop of Henle, the distal convoluted tubule (DCT), and the collecting duct. The cortex contains Bowman's capsule, the PCT and the DCT, while the medulla contains the loop of Henle and most of the collecting duct.

Bowman's capsule has two epithelial layers. The outer (parietal) layer is a simple squamous epithelium; the inner (visceral) layer wraps around the glomerular capillaries and is made of specialised cells called podocytes. Their interdigitating foot processes form filtration slits that contribute to the filtration barrier during ultrafiltration.

The glomerulus is supplied by two arterioles of unequal diameter. The afferent arteriole is wider than the efferent arteriole. Because blood enters a wider vessel and leaves through a narrower one, pressure builds up inside the glomerular capillary tuft, which is the driving force for ultrafiltration.

Urine is finally produced once fluid has been modified along the nephron and collecting duct. The collecting duct is the last region of the nephron that any given drop of filtrate passes through before entering the renal pelvis as urine, so this is the region that contains urine (or, more precisely, the most concentrated urine that the nephron produces).

Understanding the Question

The question gives you a printed diagram of a nephron and asks you to add four labelled lines (A, B, C and D), each pointing to a specific structure:

• A – a region that contains urine
• B – a region that contains podocytes
• C – a region of the nephron that lies within the medulla
• D – the afferent arteriole

Because these are four discrete structural identifications, you must place each label line on the correct feature of the diagram, not just the general area. The mark scheme is precise: B must point to the inner part of Bowman's capsule, and D must point to the wider of the two vessels entering the capsule (not the narrower efferent arteriole, which also enters/leaves Bowman's capsule region).

Approach

Work through the labels one at a time, identifying each feature by its position and what is drawn there:

1. Urine is the fluid that leaves the nephron — find the most distal tubule shown, the collecting duct.
2. Podocytes sit on the inner (visceral) layer of Bowman's capsule that hugs the glomerular capillary tuft.
3. The medulla of the kidney contains the loop of Henle and the lower portion of the collecting duct (anything that drops below the level of the convoluted tubules).
4. The afferent arteriole is the vessel that brings blood into the glomerulus and is wider than the efferent arteriole that carries blood away.

Step-by-Step Reasoning

• Label A – The collecting duct is the right-hand tube that joins onto the distal convoluted tubule and runs down the side of the nephron. By this point the fluid has been adjusted by selective reabsorption and is the closest thing to urine still within the nephron. Add line A to this duct.
• Label B – Look inside Bowman's capsule where the glomerular capillaries are drawn. The inner wall of the capsule (the one in contact with the capillaries) is made of podocytes. The outer wall is simple squamous epithelium with no podocytes. Line B must touch the inner wall.
• Label C – The medulla is the inner part of the kidney. Two nephron structures dip down into it: the loop of Henle (the U-shaped hairpin) and the lower part of the collecting duct (below the cortex). Either is acceptable, but the most obvious medullary target is the loop of Henle.
• Label D – At the glomerulus, two arterioles are drawn. The vessel entering the glomerulus is the afferent arteriole; the one leaving is the efferent arteriole. The afferent is the wider of the two, so line D must go to the wider vessel. (Wider lumen on entry is also why glomerular pressure is high enough for ultrafiltration.)

Key Takeaways

• The cortex contains Bowman's capsule, the PCT and the DCT; the medulla contains the loop of Henle and the lower collecting duct.
• Podocytes are in the visceral (inner) layer of Bowman's capsule, not the parietal (outer) layer.
• The afferent arteriole is wider than the efferent arteriole — the diameter difference is what generates the high hydrostatic pressure required for ultrafiltration.
• The collecting duct is the final part of the nephron and the region that contains the most processed fluid (essentially urine) before it leaves for the renal pelvis.

Common Mistakes

• Pointing B to the outer wall of Bowman's capsule. Podocytes are on the inner (visceral) layer that hugs the capillaries, not the outer parietal layer.
• Pointing D to the narrower vessel (the efferent arteriole). The afferent is the wider of the two.
• Pointing A to the bladder or ureter. The diagram shows only a single nephron — the largest tube on the right is the collecting duct, which is the closest region on the figure that contains urine.
• Pointing C at the proximal convoluted tubule or distal convoluted tubule. These are in the cortex, not the medulla.

Things to Be Careful About

• Place each label line directly on the named structure, not on an adjacent tubule or vessel.
• For D, do not just point at "a vessel" — the mark requires the wider entering vessel specifically.
• For C, the mark scheme accepts either the loop of Henle or the lower collecting duct; choose whichever is clearer in the printed figure.

Background Concept

The proximal convoluted tubule (PCT) is the first coiled region of the nephron after Bowman's capsule. Its cells are responsible for selective reabsorption: they take back useful substances (glucose, amino acids, vitamins, many ions and about 85% of the water) from the filtrate and return them to the blood, while leaving behind wastes such as urea.

Reabsorption in the PCT is mainly by active transport (often co-transport) rather than simple diffusion, because the useful molecules are being moved against a concentration gradient and the cell must be selective about what it takes back. To do this fast enough, PCT cells need:

• a large surface area exposed to the filtrate (so molecules can be picked up quickly);
• specific membrane proteins that recognise and transport the molecules to be reabsorbed;
• tight control over the route fluid takes (so nothing slips past between cells);
• a steady supply of ATP to power the active transport.

This combination of features is an excellent example of the biological principle that structure follows function.

Understanding the Question

The command words are describe and explain, so each point must do two things: name the structural feature of the PCT cell and explain how that feature helps selective reabsorption. Marks will not be awarded for structure alone ("it has microvilli") or for function alone ("reabsorption happens") — you need the link.

The question specifies the PCT, so answers about other regions of the nephron (e.g. features of the loop of Henle or distal tubule) are not creditworthy here.

Approach

Think about the journey of a glucose molecule (or amino acid) from the filtrate back into the blood. It must:

1. Enter the apical (lumen-facing) surface of the PCT cell — so the cell needs a large surface area and specific transporters.
2. Cross the cell — so the cell needs ATP and metabolic machinery.
3. Leave through the basal surface into the tissue fluid and capillaries — so the basal membrane needs to be adapted to pump solutes outwards.

Work through the features of the cell in that order: microvilli, co-transporters, mitochondria, basal infoldings, plus the tight junctions that ensure the fluid actually goes through the cell rather than slipping between them.

Step-by-Step Reasoning

• Microvilli (1 mark). The apical surface of each PCT cell is covered in microvilli, forming a brush border. This dramatically increases the surface area in contact with the filtrate, so more transporters can be present and more molecules can be reabsorbed per unit time.
• Co-transporter proteins (1 mark). Specific carrier proteins in the apical membrane bind $\text{Na}^{+}$ and a useful solute (e.g. glucose or an amino acid) together and move them into the cell. The $\text{Na}^{+}$ moves down its electrochemical gradient (which was set up by the basal $\text{Na}^{+}$/$\text{K}^{+}$ pumps); the solute is "dragged" with it. This is the key mechanism of selective reabsorption for glucose and amino acids.
• Tight junctions (1 mark). Adjacent PCT cells are joined by tight junctions near their apical surfaces. These seal the gaps between cells, so filtrate cannot leak through the spaces between them. Everything that is reabsorbed must therefore pass through the cell, where the transporters can act on it selectively. This is what makes the reabsorption selective.
• Many mitochondria (1 mark). Active transport (especially the $\text{Na}^{+}$/$\text{K}^{+}$ pump) requires large amounts of ATP. PCT cells are packed with mitochondria to supply this ATP by aerobic respiration. The high mitochondrial density is a direct indicator of the high metabolic demand of the tissue.
• Folded basal membrane (1 mark). The basal surface is highly infolded, giving extra surface area in contact with the tissue fluid. This accommodates very large numbers of $\text{Na}^{+}$/$\text{K}^{+}$ pumps, which keep the intracellular $\text{Na}^{+}$ concentration low. The low internal $\text{Na}^{+}$ maintains the gradient that drives co-transport at the apical surface — so the basal infoldings are essential for the apical co-transport to keep working.

Key Takeaways

• Selective reabsorption in the PCT depends on the $\text{Na}^{+}$ gradient set up by basal $\text{Na}^{+}$/$\text{K}^{+}$ pumps (using ATP from mitochondria).
• Microvilli and basal infoldings both increase surface area — but for different purposes (apical uptake and basal pumping respectively).
• Tight junctions make the reabsorption selective by forcing the filtrate through cells, where the appropriate transporters are located.
• This is a classic structure–function question: each structure has a specific role in the reabsorption mechanism.

Common Mistakes

• Stating the structure without the function ("it has microvilli") — earns nothing. You must say what the microvilli do (increase surface area for reabsorption).
• Stating "many mitochondria for respiration" without linking it to ATP use for active transport / the $\text{Na}^{+}$/$\text{K}^{+}$ pump.
• Saying tight junctions "hold the cells together" — this is true but not what earns the mark; the mark requires that tight junctions prevent substances passing between cells so that reabsorption must occur through the cells (which is what makes it selective).
• Confusing the apical and basal surfaces, or omitting the $\text{Na}^{+}$ ion in descriptions of co-transport.
• Describing features of the loop of Henle (water-permeable walls, counter-current multiplier) — these are adaptations of a different region and do not earn credit here.

Things to Be Careful About

• The mark scheme is strict: "co-transporter proteins" must be linked to movement of glucose or amino acids with $\text{Na}^{+}$ — not just "transport proteins".
• The mark scheme uses the phrasing "folded basal membrane" — saying "folded cell surface" is too vague.
• "Many" mitochondria — the qualifier matters; a single mitochondrion per cell is not enough to support the high rate of active transport.
• You only need to make four mark-worthy points; do not waste time on a fifth if you have clearly written the four best ones.

Background Concept

Mendel showed that each inherited trait is controlled by discrete factors (now called genes), and that each gene has two versions — alleles — one inherited from each parent. In pea plants, stem length is controlled by one gene with two alleles:

• $Le$ — dominant allele (tall stem)
• $le$ — recessive allele (dwarf stem)

A pea plant with a tall stem therefore has one of two possible genotypes: homozygous dominant ($Le\,Le$) or heterozygous ($Le\,le$). Both genotypes produce the same tall phenotype, because the dominant $Le$ allele masks the recessive $le$ allele. The problem is that, just by looking at a tall plant, you cannot tell which of these two genotypes it carries.

A test cross solves this. A test cross is a controlled genetic cross in which an organism showing the dominant phenotype is mated with an organism that is homozygous recessive for the same gene. Because the recessive parent can only contribute a recessive $le$ allele to its offspring, the phenotypes of the offspring reveal the alleles carried by the dominant parent.

Understanding the Question

Part (a) asks the candidate to design a cross that distinguishes a $Le\,Le$ tall plant from a $Le\,le$ tall plant, and to describe how the offspring results would be interpreted.

The command word is "describe", so the answer must include both the procedure (what cross to perform) and the analysis (how the offspring allow you to deduce the parent's genotype). Four marks are available, which corresponds to four distinct creditable points in the mark scheme.

Approach

The strategy is to:

1. Identify the appropriate cross — a test cross with a homozygous recessive partner.
2. Predict the two possible sets of offspring, depending on the unknown parent's genotype.
3. Match the observed offspring ratio to the prediction to deduce the parent's genotype.

A Punnett square is not required, but the underlying logic is the same as for one. Both the procedure and the two possible outcomes must be stated to score full marks.

Step-by-Step Reasoning

Step 1 — Choose the cross partner.
A dwarf plant must be $le\,le$ (homozygous recessive), because the dwarf phenotype can only appear when both alleles are recessive. This partner is guaranteed to pass a single $le$ allele to every offspring.

Step 2 — Predict the offspring if the tall parent is $Le\,Le$ (homozygous dominant).

Every offspring receives $Le$ from the tall parent and $le$ from the dwarf parent, giving genotype $Le\,le$ in all cases. All offspring are tall.

Step 3 — Predict the offspring if the tall parent is $Le\,le$ (heterozygous).

Half the offspring inherit $Le$ (tall, $Le\,le$) and half inherit only $le$ (dwarf, $le\,le$). The expected ratio is 1 tall : 1 dwarf.

Step 4 — Match observation to prediction.

• If every offspring is tall → the tall parent must have been $Le\,Le$.
• If approximately half the offspring are dwarf → the tall parent must have been $Le\,le$.

Step 5 — Practical detail (AVP).
The mark scheme accepts an additional point for any sensible practical detail. In Mendel's actual experiments this included transferring pollen by hand (pea flowers normally self-pollinate, so the anthers must be removed from the female parent before crossing), harvesting the resulting seeds, and growing them on to score the phenotypes. Any one of these is sufficient to earn the extra point.

Key Takeaways

• A test cross uses a homozygous recessive partner to expose the alleles hidden in a dominant-phenotype organism.
• The two possible offspring outcomes — all dominant vs. a 1:1 mix of dominant and recessive — are diagnostic of the parent's genotype.
• This logic applies to any gene with a clear dominant/recessive relationship, not just to stem length in peas.

Common Mistakes

• Crossing two tall plants. This tells you nothing, because any offspring ratio could be produced by either $Le\,Le$ or $Le\,le$ parents, depending on which tall plant you happened to use.
• Using a heterozygous partner. A non-recessive partner does not act as a test. The recessive parent must be guaranteed homozygous.
• Stopping at "1:1 ratio" without the conditional. You must state the two possible outcomes and link each one to a parent genotype, not just give one outcome.
• Omitting the $Le$/$le$ notation. The mark scheme expects the precise allele symbols; writing "homozygous dominant" without the genotype loses credit in contexts where notation is required.

Things to Be Careful About

• The cross is described, not necessarily performed — but the offspring analysis must be stated, not implied.
• The Punnett square grid itself is not a required mark, but writing out the gametes (e.g. "$Le$ or $le$ from the tall parent; $le$ from the dwarf parent") makes the reasoning explicit and is safer in a written answer.
• "Test cross" is a single technical term that earns a mark on its own — do not bury it inside a paragraph.

Background Concept

In diploid organisms, every gene exists in two copies — one on each of the pair of homologous chromosomes inherited from the two parents. These two versions of the gene are called alleles, and they may be identical or different. The complete set of alleles an organism carries is its genotype; the observable characteristics that result (also influenced by the environment) make up its phenotype.

When the two alleles at a particular gene locus are identical, the genotype is said to be homozygous at that locus. It can be homozygous dominant (e.g. AA) or homozygous recessive (e.g. aa). When the two alleles differ (Aa), the genotype is heterozygous.

Whether an allele's effect is seen in the phenotype depends on the dominance relationship between the alleles. A dominant allele is one whose effect appears in the phenotype whenever the allele is present (so both AA and Aa show the dominant phenotype). A recessive allele is one whose effect is only seen in the phenotype when no dominant allele is present — that is, only in the homozygous recessive genotype aa. In a heterozygote Aa, the dominant allele's product is sufficient to produce the dominant phenotype, and the recessive allele's effect is hidden or "masked".

Understanding the Question

This is a 2-mark definitions question. The candidate must write a clear explanation of two terms: homozygous and recessive. The question stem notes that cystic fibrosis is autosomal recessive and that affected people are homozygous recessive, providing context for why the terms matter, although the definitions themselves are general biology.

The command word is "Explain the meaning" — so the answer must give the precise biological meaning of each term, not just an example. The mark scheme credits:

• For homozygous: a statement that both alleles of a gene are the same.
• For recessive: a statement about the phenotype being masked by a dominant allele, OR that two copies of the recessive allele are needed for the phenotype to be displayed.

Approach

Write a one-sentence definition for each term, using the precise vocabulary the mark scheme rewards: alleles, gene / genotype, the same, phenotype, masked, dominant, recessive.

Step-by-Step Reasoning

Homozygous:

• A diploid organism has two alleles for every gene.
• "Homozygous" describes the situation where those two alleles are identical — e.g. AA or aa.
• The key statement is that both alleles of a gene are the same.

Recessive:

• An allele is recessive if its phenotypic effect is only visible when there is no dominant allele present.
• A dominant allele (when present) masks the effect of a recessive allele.
• Two copies of the recessive allele (the homozygous recessive genotype) are required for the recessive phenotype to appear.

Key Takeaways

• Homozygous = an identical pair of alleles at a particular gene locus.
• Recessive = an allele whose phenotype is only expressed when the genotype is homozygous recessive (no dominant allele is present to mask it).
• These definitions are foundational for any later work on inheritance patterns, genetic crosses, and population genetics.

Common Mistakes

• Defining homozygous as simply "having two alleles" — too vague. The alleles must be the same.
• Defining recessive as "rare" or "weak" — recessive refers to the dominance relationship, not frequency or strength.
• Saying that a recessive allele "does nothing" — it codes for a product, but the effect is only visible when no dominant allele is present.
• Confusing homozygous with heterozygous — homozygous = identical alleles; heterozygous = different alleles.

Things to Be Careful About

• The mark scheme requires reference to phenotype in the definition of recessive — saying "the allele is masked" without specifying what is being masked (the phenotype) does not earn the mark.
• "Two copies of the allele" or "no dominant allele" both work for the recessive definition — either captures the essence.
• Use precise CIE terminology: gene, allele, genotype, phenotype, dominant, recessive, homozygous, heterozygous.

Background Concept

Holstein Friesian cattle are a high-yielding dairy breed whose milk production is sensitive to environmental conditions, particularly heat stress. Heat stress occurs when an animal's homeostatic mechanisms (sweating, panting, reducing metabolic rate) cannot dissipate heat fast enough, so core body temperature rises. Sustained high temperatures reduce feed intake, alter metabolism, and divert energy away from milk synthesis, lowering yield.

Fig. 4.2 is a paired bar chart in which the grey bars (left axis) give mean daily air temperature and the white bars (right axis) give mean monthly milk yield per cow, with standard-error bars. Reading across the months from April to August, the temperature bars rise while the yield bars fall — a clear inverse relationship over this part of the year.

Understanding the Question

The command word is describe, and the mark scheme requires two points: a qualitative statement of the trends, and a quantitative comparison supported by figures taken from the graph. You must mention BOTH variables and say how each changes between the two named months.

Approach

Look at the grey (temperature) bars and the white (yield) bars for April and August. The temperature bar grows taller between these two months; the yield bar shrinks. To earn the second mark, take a numerical read-off from each bar and quote the values with their units.

Step-by-Step Reasoning

1. April: the grey bar reaches 10 °C on the left-hand axis.
2. August: the grey bar reaches 18 °C on the left-hand axis — temperature has risen.
3. April: the white bar reaches the right-hand axis level of approximately 792–798 $\text{kg cow}^{-1}$.
4. August: the white bar reaches approximately 711–716 $\text{kg cow}^{-1}$ — yield has fallen.
5. Therefore, as temperature increased from 10 °C to 18 °C, mean monthly milk yield per cow fell from roughly 795 to roughly 714 $\text{kg cow}^{-1}$, consistent with heat stress suppressing production.

Key Takeaways

• Heat stress in cattle is associated with reduced milk yield.
• Dual-axis bar charts must be read against the correct axis; always check the legend and key.
• Describe requires both the trend (qualitative) and supporting data (quantitative).

Common Mistakes

• Quoting only one variable (e.g. "temperature increases") without mentioning milk yield.
• Giving a comparison without any numerical values.
• Reading off the wrong axis (the right-hand axis is for yield, not temperature).

Things to Be Careful About

• The mark scheme accepts a small range because you are reading off a graph by eye; values such as 10 °C and 795 $\text{kg cow}^{-1}$ in April, 18 °C and 714 $\text{kg cow}^{-1}$ in August are all acceptable.
• Quote the units (°C and $\text{kg cow}^{-1}$).

Background Concept

Phenotypic variation within a species can be:

• Discontinuous — falls into discrete categories with no intermediates (e.g. ABO blood groups, Mendel's pea seed shape).
• Continuous — shows a full range of values with no clear categories (e.g. height, mass, milk yield); typically produced by the combined action of many genes (polygenic) plus environmental effects.

The causes of phenotypic variation are therefore genetic (alleles at one or many loci) and environmental (diet, temperature, light, etc.). Where many genes contribute, the trait is described as polygenic.

Understanding the Question

The stem explicitly supplies the information you need:

• Fig. 4.2 shows monthly milk yields that change gradually with temperature — a smooth range, not distinct categories.
• Twenty different genes were found to be associated with heat tolerance and therefore influence milk yield.
• Heat-tolerant and Holstein Friesian cattle produce different milk yields under the same conditions — a genetic difference independent of environment.

You must (1) name the type of variation with a reason, and (2) identify causes of variation with reasons.

Approach

Apply the textbook definitions to the evidence given:

• Smooth range of values → continuous variation.
• Yield changes with air temperature → environmental cause.
• 20 different genes have an effect, and the two breeds differ → polygenic / genetic cause.

Step-by-Step Reasoning

1. Type of variation — continuous: Fig. 4.2 shows milk yields that fall gradually from month to month rather than into sharply separated groups. There are intermediate values, so the variation is continuous, not discontinuous.
2. Environmental cause — air temperature: The graph demonstrates that yield falls as temperature rises. Air temperature is an external (environmental) factor that influences the phenotype.
3. Genetic cause — polygenic / many genes: Scientists identified 20 genes associated with body temperature and heat tolerance. In addition, under identical conditions, heat-tolerant cattle produce less milk than Holstein Friesian cattle, indicating that the genetic constitution of the two breeds influences yield. A trait influenced by many genes is polygenic, producing continuous variation when combined with environmental effects.

Key Takeaways

• Continuous variation arises from the combined action of many genes and environmental influences.
• A graph showing a smooth range of values, not discrete classes, is the classic signature of continuous variation.
• Identifying a cause of variation requires a piece of evidence from the question that points to it.

Common Mistakes

• Saying "discontinuous" because different breeds have different yields — breeds overlap in their ranges, so the trait is continuous.
• Naming only one cause (genetic OR environmental) when both apply.
• Giving a reason that does not actually use the data provided (e.g. citing "diet" without any evidence in the question).

Things to Be Careful About

• The mark scheme requires a reason for each choice — not just the label.
• "Polygenic" must be tied to the evidence of many genes, not asserted on its own.
• "Continuous" must be tied to the absence of distinct categories, not just to the existence of a graph.

Background Concept

The SLICK allele is a dominant allele of the PRLR (prolactin receptor) gene. It arose in Senepol cattle and produces a short-haired phenotype that helps the animal lose heat more readily, reducing the impact of heat stress. Because heat stress depresses milk yield, an allele that mitigates heat stress should improve milk yield — but only in conditions warm enough to cause heat stress.

Fig. 4.3 compares the same breed (Holstein Friesian) with and without SLICK at two different mean daily air temperatures. This is a controlled comparison: the only genetic difference is the PRLR allele, so any difference in yield can be attributed to it.

Understanding the Question

The command word is describe. You need to use Fig. 4.3 to explain what the SLICK allele does to milk yield. The mark scheme rewards:

• a general statement that SLICK increases yield,
• a small difference in March (cool month) and a larger difference in September (warmer month),
• the observation that SLICK cattle maintain their yield across the two months while normal cattle do not,
• supporting figures from the graph.

Approach

Read the four bars in Fig. 4.3 and compare them: the white bars are normal Holstein Friesian cattle, the grey bars are SLICK carriers. Note the size of the gap between the two bars in March versus September, and note whether the bars change height between the two months.

Step-by-Step Reasoning

1. Overall effect — In both March and September the SLICK bar is taller than the normal bar, so the SLICK allele increases milk yield in both months (mp1).
2. March, 5 °C — The two bars are close in height: normal ≈ 780–785 $\text{kg cow}^{-1}$ and SLICK ≈ 795–800 $\text{kg cow}^{-1}$, a difference of about 15 $\text{kg cow}^{-1}$ — small (mp2).
3. September, 14 °C — The gap is much wider: normal ≈ 690–700 $\text{kg cow}^{-1}$ and SLICK ≈ 770–780 $\text{kg cow}^{-1}$, a difference of about 80 $\text{kg cow}^{-1}$ — much larger (mp3).
4. Stability of yield — Comparing the SLICK bars between the two months, the height barely changes (≈ 800 versus ≈ 775 $\text{kg cow}^{-1}$), while the normal bars fall markedly (≈ 785 → ≈ 695 $\text{kg cow}^{-1}$). SLICK cattle maintain their milk yield across both temperatures (mp4).
5. Biological interpretation — At 5 °C, neither animal is heat-stressed, so the SLICK phenotype offers little extra benefit. At 14 °C, normal Holsteins begin to suffer heat stress and their yield falls, whereas the short hair of SLICK cattle allows better heat loss and the yield is preserved.

Key Takeaways

• The effect of an allele can depend on the environment — SLICK is most beneficial at warmer temperatures.
• When describing a graph comparison, give the direction of the effect, the size of the difference, and supporting figures.
• Maintain the distinction between "the allele increases yield" (always true in this dataset) and "the allele maintains yield" (true relative to the normal comparison across the two months).

Common Mistakes

• Stating only the direction of the effect without comparing March and September.
• Omitting numerical values from the graph.
• Saying SLICK "decreases" milk yield because one of the months shows a small fall — read both bars; the SLICK bar is always at least as high as the normal one.
• Confusing the alleles of the gene with the alleles affecting the trait — the SLICK allele is a single allele whose effect is described, not a comparison between two alleles within the same cow.

Things to Be Careful About

• Quote the figures from the graph, not invented values.
• Use the units $\text{kg cow}^{-1}$ and the temperatures °C as in the question.
• A small difference means a small gap between the bars, not a small absolute yield.

Background Concept

Selective breeding is the traditional method of improving livestock: animals with the desired trait (here, heat tolerance in Senepol cattle) are mated with the breed to be improved (Holstein Friesian) and the offspring carrying the desired allele are selected and back-crossed over many generations. The whole genome of the donor parent is being transferred, so other (often unwanted) alleles come along too.

Gene editing uses molecular tools (e.g. CRISPR–Cas9) to make a precise change at a specific locus. Here, a single nucleotide deletion is introduced into the PRLR gene of a Holstein Friesian zygote, replicating the SLICK mutation. Only the targeted gene is altered; the rest of the genome remains that of the recipient breed.

The two methods therefore differ in speed, precision, side-effects on other genes, impact on genetic diversity, and regulatory and ethical considerations.

Understanding the Question

The command word is compare, and the mark scheme requires both similarities and differences for six marks. The list of 12 possible marking points in the mark scheme shows the breadth expected: similarity, effect, confirmation, agency, gene number, proportion of offspring, generations, location of procedure, technical demands, genetic variation, regulation, and outbreeding.

Approach

For each row of the mark scheme, give a one-clause statement for the gene editing side and, where the mark scheme asks for a contrast, a one-clause statement for the selective breeding side. Aim for at least six distinct points.

Step-by-Step Reasoning

1. Similarity — both are artificial / performed by humans. (mp4) Whether the allele is introduced by gene editing in a lab or by selective breeding on a farm, in both cases humans choose the parents and decide which animals to keep.
2. Similarity — both produce first-generation offspring that can be screened. (mp1 + mp3) In both methods, the first (or near-first) generation can be checked for the SLICK allele by genetic testing (DNA) or by phenotypic observation (short hair).
3. Difference — gene number. (mp5) Gene editing alters only the PRLR gene; selective breeding transfers the whole genome of the donor (Senepol), so many other genes are brought in alongside.
4. Difference — proportion of offspring with the allele. (mp6) Because gene editing acts on the zygote, all offspring are homozygous or heterozygous for the edited allele. With selective breeding, the allele segregates and only some offspring inherit it.
5. Difference — speed. (mp7) Gene editing produces SLICK cattle in one generation; selective breeding takes many generations of crossing and back-crossing.
6. Difference — procedure / location. (mp8 + mp9) Gene editing is performed in a laboratory on embryos, zygotes or cells using molecular methods and requires training and equipment. Selective breeding is performed on the farm by mating (or artificial insemination) — a physiological method that needs no molecular equipment.
7. Difference — outbreeding. (mp12) Gene editing can be done within the Holstein Friesian breed, so no outbreeding is required. Selective breeding requires crossing with a different breed (Senepol), introducing genes from outside the breed.
8. Difference — genetic diversity. (mp10) Because gene editing changes only one locus, it maintains the rest of the genome and therefore the heterozygosity and desirable characteristics of Holstein Friesian cattle. Selective breeding, especially with repeated back-crossing and selection, reduces genetic variation and risks losing desirable traits.
9. Difference — regulation. (mp11) Gene-edited animals are produced by a novel technique and typically require regulatory approval (in many countries they are treated like GMOs). Selectively bred animals are produced by conventional breeding and do not require such approval.
10. Difference — known vs unknown effects. (mp2) Because SLICK is a naturally occurring allele, its effects are well established; selective breeding is therefore relatively predictable. Gene editing is a newer technique and may have negative or unknown long-term effects on the animal.

Key Takeaways

• Gene editing and selective breeding can achieve the same outcome (introducing SLICK) but at very different speeds, with very different impacts on the rest of the genome.
• Selective breeding moves a whole block of DNA from donor to recipient, so linked alleles come along too — this is why it is slow and reduces variation.
• Gene editing is precise but raises regulatory and ethical questions because it is a new molecular technology.

Common Mistakes

• Listing only differences and no similarities (or vice versa).
• Repeating the same idea twice (e.g. "selective breeding takes longer" and "selective breeding takes many generations").
• Saying gene editing "does not change any genes" — it does change one specific gene.
• Saying selective breeding "does not reduce genetic variation" — repeated selection with a limited breeding pool does reduce variation.
• Omitting the practical contrast (lab vs farm) and the regulatory contrast.

Things to Be Careful About

• A compare question usually needs the two methods side by side in each point; a one-sided list does not satisfy the question.
• Do not invent advantages for either method that are not supported by the question (e.g. do not claim gene editing is "cheaper").
• Keep the technical terms precise: allele, gene, genome, outbreeding, heterozygosity.

Background Concept

A respirometer is a closed apparatus that measures the net change in gas volume when a living tissue respires. Aerobic respiration consumes $\text{O}_2$ and releases $\text{CO}_2$ in roughly equal volumes, so the gas volume inside a sealed tube would stay almost constant if both gases were present. To turn the respirometer into a true measure of oxygen uptake, the $\text{CO}_2$ must be removed. Potassium hydroxide ($\text{KOH}$) reacts with $\text{CO}_2$:

so the only net change in gas volume is the loss of oxygen. This causes the pressure inside the sealed tube to fall, drawing the coloured liquid up the U-shaped tube towards the test-tube side. The distance the liquid moves is therefore a direct measure of oxygen uptake by the peas.

Understanding the Question

The command word is "state", so a single concise sentence is all that is needed. The mark scheme rewards the idea that KOH removes the $\text{CO}_2$ released by the peas.

Approach

Recall that respirometers rely on absorbing the $\text{CO}_2$ so that oxygen uptake alone causes the liquid to move.

Step-by-Step Reasoning

1. Germinating peas respire aerobically, taking in $\text{O}_2$ and giving out $\text{CO}_2$.
2. Inside the sealed respirometer, $\text{O}_2$ is consumed and $\text{CO}_2$ is released — equal volumes, so the gas volume would not change.
3. KOH absorbs the $\text{CO}_2$, so the volume of gas in the tube decreases in proportion to $\text{O}_2$ uptake.
4. The coloured liquid moves to compensate for the pressure drop, and this movement is the measurement.

Key Takeaways

The KOH in a respirometer converts what is essentially a closed-volume system into an effective $\text{O}_2$ meter by selectively scrubbing out the $\text{CO}_2$ of respiration.

Common Mistakes

• Saying KOH "absorbs oxygen" or "absorbs gas" — the role is specifically to absorb $\text{CO}_2$.
• Saying KOH "measures" respiration — it does not measure anything, it only conditions the gas mixture.

Things to Be Careful About

The mark scheme requires "carbon dioxide (produced by peas)" — be explicit about which gas is being absorbed and that it comes from the peas, not from the air.

Background Concept

"Validity" in experimental biology means the extent to which the results actually measure what the experiment claims to measure — here, the effect of temperature on respiration rate. Two broad classes of issue threaten validity:

• Physical/environmental artefacts: changes in gas volume that are not caused by respiration (e.g. expansion of air as the respirometer warms up in the water-bath, or changes in atmospheric pressure).
• Confounding variables: other things that vary between trials alongside the independent variable (e.g. different volumes of peas at each temperature).

Validity is distinct from reliability (whether the result can be repeated); valid results are usually also reliable, but reliable results are not automatically valid.

Understanding the Question

The command word is "suggest", so the candidate is free to propose a range of methods. Two marks are available and the mark scheme offers four alternative marking points, any two of which earn the marks. The strongest answers combine a control with a repeat-and-statistics approach.

Approach

To improve validity: (1) remove or account for any movement not caused by respiration, and (2) confirm that the independent variable is the only thing that differs between trials. A control apparatus addresses (1); repeats, statistics and standardised variables address (2).

Step-by-Step Reasoning

1. Idea of repeatability: perform the experiment more than once at each temperature. Compare the readings: the more similar the repeats, the more confidence in the result.
2. Statistical measure: quantify the spread of the repeats — calculate the standard deviation, standard error, or carry out a named test (t-test, correlation). A small spread means the data are consistent.
3. Control respirometer: set up a second respirometer identical to the first, but with glass beads (or an equal volume of dead/boiled peas) instead of living peas. This control should show little or no liquid movement, but any small movement reveals changes due to temperature or pressure. Subtract the control reading from the experimental reading.
4. Checking other variables: confirm that volume of peas, mass of peas, volume and concentration of $\text{KOH}$, time, and the orientation of the apparatus are the same at every temperature — only the temperature should vary.

Key Takeaways

Validity of a respirometer experiment is improved by (i) a control to remove non-respiratory movement and (ii) repeats with statistics to confirm consistency, plus (iii) standardising every variable other than the one being tested.

Common Mistakes

• Confusing validity with reliability and only suggesting "do more repeats" — repeats alone improve reliability, not validity.
• Suggesting a "control with no peas" rather than specifying an equal volume of glass beads or dead peas, which would not have the same heat capacity / mass / gas space.
• Vague answers such as "improve accuracy" or "reduce human error" — the mark scheme rejects these.

Things to Be Careful About

The control must be as close to the experimental set-up as possible, differing only in whether the peas are alive. Volume of material is critical because it determines the gas space inside the tube.

Background Concept

Gas volume is highly sensitive to temperature (Charles's law: at constant pressure, volume is proportional to absolute temperature). Inside a sealed respirometer any temperature change will expand or contract the trapped air, moving the coloured liquid independently of respiration. The same principle means enzyme activity — and therefore respiration rate — only becomes stable once the peas themselves are at the target temperature.

Understanding the Question

The procedure leaves the respirometer in the water-bath for five minutes before adjusting the meniscus to zero and starting the timer. The command word is "explain", and one mark is available. The mark scheme accepts the single word "acclimatisation" or "equilibration".

Approach

Identify what is being allowed to equilibrate: the peas, the air inside, and the KOH solution all need to reach the water-bath temperature so that the initial reading represents a true zero rather than a transient thermal-expansion effect.

Step-by-Step Reasoning

1. At the moment the respirometer is placed in the water-bath, the peas and the air inside the tube are at room temperature, not at the bath temperature.
2. Over the first few minutes, heat transfers from the bath into the tube, warming the air (which expands) and warming the peas (which alters enzyme activity).
3. If timing started immediately, the early part of the run would be dominated by these thermal-transient effects and the measured distance would not reflect respiration at the intended temperature.
4. The five-minute wait gives the system time to come to thermal equilibrium, so that the trial starts with both peas and gas already at the target temperature.

Key Takeaways

Any experiment that measures a rate at a set temperature must allow the apparatus and its contents to reach that temperature before timing begins; otherwise thermal expansion or contraction of the air contaminates the result.

Common Mistakes

• "To let the peas germinate" — they are already germinating.
• "To let the KOH absorb the $\text{CO}_2$ from the previous trial" — irrelevant; this is the start of the experiment.
• "To let the syringe warm up" — too narrow; the main reason is to equilibrate the peas and the trapped air.

Things to Be Careful About

The mark scheme accepts just the single word "acclimatisation" or "equilibration". A full sentence is fine but unnecessary.

Background Concept

A good biological graph conveys both the data and the underlying trend. CIE examiners reward:

• correct axis labels with units;
• sensible scales that use at least half the printed grid;
• accurate plotting (a small cross or dot in a clear circle);
• a single smooth curve (or straight line) of best fit, not dot-to-dot joining;
• no extrapolation beyond the range of the data.

Understanding the Question

We are given five pairs of numbers in Table 6.1 and an empty graph grid (Fig. 6.2) whose axes are already labelled and scaled. Two marks are available: one for the five correctly placed points and one for an appropriate smooth curve, and the mark scheme explicitly rejects extrapolation.

Approach

Place each point at the intersection of its x- and y-values, then sweep a single smooth curve through them, taking care that the line is a curve (because the data peak and fall) and is not extrapolated beyond $x = 10$ or $x = 50$.

Step-by-Step Reasoning

1. Read the axes carefully: x is temperature in °C, with major lines every $10\,^\circ\text{C}$ from $0$ to $50$ (and minor lines every $2\,^\circ\text{C}$). y is rate of movement in $\text{mm min}^{-1}$, with major lines every $0.10$ from $0.00$ to $1.40$ (and minor lines every $0.02$).
2. Plot each point at the correct intersection; a small neat cross ("×") in a circle is the cleanest convention.
   • $x = 10$, $y = 0.40$
   • $x = 20$, $y = 0.70$
   • $x = 30$, $y = 1.30$
   • $x = 40$, $y = 1.15$
   • $x = 50$, $y = 0.60$
3. Inspect the shape: the points rise from $10$ to $30\,^\circ\text{C}$, then fall — so a single smooth curve, not two straight lines, is the right choice. The peak is at $30\,^\circ\text{C}$.
4. Draw a smooth curve of best fit in one continuous sweep. The line should pass as close as possible to all five points; small deviations (a point lying a fraction of a mm off the line) are acceptable because biological data have natural scatter.
5. Stop the curve at the data range — do not continue it to $x = 0$ or beyond $x = 50$, because that would be extrapolation (the mark scheme rejects this).

Key Takeaways

• A curve of best fit is a single smooth line representing the trend; do not join points with a zig-zag of straight segments.
• Always check whether the data rise and fall — if they do, a smooth curve (not two lines) is the appropriate model.
• Never extrapolate a best-fit curve beyond the data range unless the question asks you to.

Common Mistakes

• Joining the points with a series of straight line segments — this scores zero for "smooth curve".
• Extrapolating the curve to $0\,^\circ\text{C}$ or beyond $50\,^\circ\text{C}$ — explicitly rejected by the mark scheme.
• Mis-reading the y-scale (e.g. placing $1.30$ at the $1.20$ or $1.40$ line).
• Drawing two separate lines (one up, one down) instead of one smooth curve.

Things to Be Careful About

• Use a sharp pencil and a ruler-free hand for the curve.
• Do not extend the curve through the y-axis or past the last data point.
• Each plotted point should be unambiguous; if you use a cross, make the centre lie exactly on the data value.

Background Concept

Respiration is a series of enzyme-catalysed reactions (glycolysis, the link reaction, Krebs cycle, oxidative phosphorylation). Like any enzyme-controlled process, its rate is highly sensitive to temperature, and the familiar bell-shaped curve seen in this question is the same pattern observed for almost all enzymes:

• Below the optimum: the rate is limited by kinetic energy. At low temperature, molecules move slowly, collide infrequently, and a small fraction of collisions have the activation energy needed to form an enzyme–substrate complex.
• At the optimum: kinetic energy is high enough for most collisions to be successful, and the enzyme's tertiary structure is still intact.
• Above the optimum: thermal motion begins to break the weak interactions (hydrogen bonds, hydrophobic interactions, ionic bonds) that hold the tertiary structure together. The active site loses its specific shape, the substrate can no longer bind, and the enzyme is denatured. The effect is largely irreversible.

Understanding the Question

The question refers explicitly to Table 6.1 and Fig. 6.2, so the explanation must follow the shape of the plotted data — a rise from $10$ to $30\,^\circ\text{C}$, a peak at $30\,^\circ\text{C}$ ($1.30\,\text{mm min}^{-1}$), and a fall to $50\,^\circ\text{C}$ ($0.60\,\text{mm min}^{-1}$). Three marks are available, and the mark scheme lists five candidate points: the role of enzymes, the kinetic-energy rise, the collision/complex reason for the rise, denaturation as the cause of the fall, and an optional further detail such as active-site shape change or naming the optimum.

Approach

Use enzyme kinetics to explain the rise, and enzyme denaturation to explain the fall. Anchor the explanation to the specific data values so the answer is clearly tied to Table 6.1.

Step-by-Step Reasoning

1. Anchor in the biology: respiration is catalysed by enzymes, so its rate is governed by the same temperature effects that govern any enzyme-catalysed reaction.
2. Rising phase (10 → 30 °C): the data increase from $0.40$ to $1.30\,\text{mm min}^{-1}$. At higher temperature the enzyme and substrate molecules have more kinetic energy, so:
   • they collide more frequently;
   • a greater proportion of those collisions have the activation energy needed to form an enzyme–substrate complex.
     The net effect is more successful collisions / more enzyme–substrate complexes per unit time, so the rate of respiration rises.
3. Falling phase (30 → 50 °C): the data fall from $1.30$ to $0.60\,\text{mm min}^{-1}$. Above the optimum, the high temperature disrupts the hydrogen bonds and other weak interactions that maintain the enzyme's tertiary structure. The active site loses its specific shape and can no longer bind the substrate — the enzyme is denatured. With fewer functional active sites, the rate of respiration falls.
4. Optimum: the maximum rate at $30\,^\circ\text{C}$ is the optimum temperature for the enzymes in these peas; below it kinetic energy limits the rate, above it denaturation limits the rate.

Key Takeaways

• An enzyme-controlled rate–vs–temperature curve has a single peak at the optimum temperature.
• Below the optimum: more kinetic energy → more successful collisions → more enzyme–substrate complexes → higher rate.
• Above the optimum: enzymes denature, the active site shape changes, and the rate falls sharply.
• Respiration in peas behaves exactly like any other enzyme-controlled process on a temperature curve.

Common Mistakes

• Explaining the rise purely as "molecules move faster" without linking it to successful collisions or to enzyme–substrate complexes — the mark scheme insists on this link.
• Saying the enzymes "die" or "are killed" — the correct term is denature.
• Stating that all enzymes are denatured at $40\,^\circ\text{C}$ (or some other fixed value) — denaturation starts above the optimum and increases with temperature; the degree of denaturation, not its presence, is what changes.
• Failing to identify the optimum temperature explicitly from the data ($30\,^\circ\text{C}$).
• Giving a general explanation of temperature effects without tying it to the specific shape of the curve in Fig. 6.2.

Things to Be Careful About

• Use the precise terms "kinetic energy", "successful/effective collisions", "enzyme–substrate complex" and "denaturation".
• The optimum temperature is not a property of "all enzymes" — it is the optimum for the enzymes in the peas in this experiment, and it is the temperature at which the rate in Table 6.1 is highest.
• Be careful with the direction of the fall: the rate decreases because the enzyme is denatured, not because of any other limitation such as substrate concentration.

Background Concept

Photosynthesis has two linked stages:

• The light-dependent stage (photophosphorylation) on the thylakoid membranes, in which light energy drives electron transport and ATP synthesis.
• The light-independent stage (Calvin cycle) in the stroma, in which CO2 is fixed by the enzyme rubisco into glycerate 3-phosphate (GP), reduced by ATP and reduced NADP from the light-dependent stage, and used to regenerate ribulose bisphosphate (RuBP).

Three common limiting factors are:

• Carbon dioxide concentration — supplies the substrate for the Calvin cycle.
• Light intensity — supplies the energy for photophosphorylation.
• Temperature — affects the rate of enzyme-catalysed reactions and the behaviour of membrane proteins.

Understanding the Question

The question asks which combinations of these three factors and two stages show a DIRECT effect. Indirect effects — for example, lower light intensity reduces ATP and reduced NADP supply and so eventually slows the Calvin cycle — are explicitly excluded.

Approach

For each (factor, stage) pair ask: "If I change the factor on its own, will the rate of this stage change because of a direct action on that stage?" If yes → ✓; if the stage is only affected because something else is affected first → ✗.

Step-by-Step Reasoning

• CO2 + Calvin cycle: CO2 is the substrate for carbon fixation by rubisco. A change in CO2 directly alters the rate of the Calvin cycle. ✓
• CO2 + photophosphorylation: Photophosphorylation does not use CO2. Changing CO2 has no direct effect on it. ✗
• Light intensity + Calvin cycle: The Calvin cycle does not require light; it requires the ATP and reduced NADP produced by photophosphorylation, but that is an INDIRECT effect and so is excluded. ✗
• Light intensity + photophosphorylation: Light provides the energy that drives electron transport and ATP synthesis. Changing light intensity directly alters the rate of photophosphorylation. ✓
• Temperature + Calvin cycle: The Calvin cycle is catalysed by enzymes (notably rubisco), so its rate is directly affected by temperature. ✓
• Temperature + photophosphorylation: Photophosphorylation involves protein components — the electron carriers in the electron transport chain and ATP synthase — embedded in the thylakoid membrane. Temperature directly affects the activity of these proteins. ✓

Key Takeaways

• CO2 and light intensity each directly affect only one stage of photosynthesis.
• Temperature is unusual in directly affecting BOTH stages.
• "Direct" means acting on the named stage itself, not via an intermediate effect on the other stage.

Common Mistakes

• Ticking the light intensity–Calvin cycle cell (the Calvin cycle does not use light directly).
• Crossing the temperature–photophosphorylation cell (photophosphorylation involves proteins whose activity depends on temperature).
• Treating the ATP/reduced NADP link between the two stages as a "direct" effect of light on the Calvin cycle.

Things to Be Careful About

• The question requires a tick or a cross in EVERY row — leaving a row blank loses marks.
• Read the rubric — INDIRECT effects must NOT be considered.

Background Concept

Living organisms are classified into a hierarchy of nested ranks. The principal ranks, from broadest to most specific, are:

Domain → Kingdom → Phylum → Class → Order → Family → Genus → Species

The binomial name of a species has two parts: the genus (capitalised) and the specific epithet (lower-case), both italicised. Together the two words uniquely identify the species within the hierarchy.

Understanding the Question

The table has four ranks already completed (kingdom, class, family, species) and four blanks. The blanks correspond to the ranks phylum, order and genus (×2 — the rank label and the answer). The right-hand column already reveals the answers for phylum (Chordata) and order (Rodentia); the candidate must:

1. Name the three missing ranks.
2. State the genus of Cricetomys gambianus (the species name in the question stem).

Marking: 4 correct = 2 marks; 1–3 correct = 1 mark.

Approach

• Recall the sequence of ranks so you can name them in the correct order.
• Recognise that the first word of a binomial name is the genus.

Step-by-Step Reasoning

1. The rank between kingdom and class is phylum (Animalia → Chordata).
2. The rank between class and family is order (Mammalia → Rodentia).
3. The rank between family and species is genus.
4. The binomial in the question is Cricetomys gambianus. By the convention Genus species, the genus is Cricetomys.
5. Writing all four items correctly earns 2 marks; 1–3 correct earns 1 mark.

Key Takeaways

• The Linnaean hierarchy has a fixed order of ranks; memorise it as King Philip Came Over For Good Soup (Kingdom, Phylum, Class, Order, Family, Genus, Species).
• In a binomial, the capitalised first word = genus; the lower-case second word = specific epithet.

Common Mistakes

• Writing "Division" instead of "Phylum" — Division is the botanical equivalent, not used in zoology.
• Writing the genus in lower case or omitting italics on the exam paper.
• Reversing genus and specific epithet (e.g. writing gambianus Cricetomys).

Things to Be Careful About

The genus is the missing entry on the right-hand side of the table; the missing entries on the left are the rank names. Both must be supplied for full marks. The binomial is already italicised in the question stem — keep the convention.

Background Concept

The three domains of life — Eukarya, Bacteria and Archaea — are distinguished by fundamental differences in cell biology. The question has already told you the headline difference (presence/absence of membrane-bound organelles such as the nucleus). Many other features also separate the two domains:

• DNA organisation: eukaryotes package DNA with histone proteins into linear chromosomes inside a nucleus; bacteria carry a single circular chromosome, often with small circular plasmids, and lack histones.
• Cell wall chemistry: where present, eukaryote cell walls are made of cellulose (plants, algae) or chitin (fungi); bacterial cell walls are made of peptidoglycan (murein).
• Ribosomes: eukaryote cytoplasmic ribosomes are 80S (organelle ribosomes are 70S); bacterial ribosomes are 70S — exploited by antibiotics such as streptomycin.
• Reproduction: eukaryotes divide by mitosis (and many reproduce sexually via meiosis); bacteria divide by binary fission and reproduce asexually.
• Size: typical eukaryotic cells are > 5 µm in diameter; typical bacterial cells are 1–5 µm.

Understanding the Question

The stem already acknowledges that the presence/absence of membrane-bound organelles (e.g. a true nucleus) differs between Eukarya and Bacteria. You are asked for other differences. Each mark requires a paired comparison: stating the Eukarya side AND the Bacteria side of the same feature.

Approach

Choose four reliable comparative features and write them as parallel statements: "Eukarya have X, whereas Bacteria have Y." Do not repeat the membrane-bound organelle difference.

Step-by-Step Reasoning

1. DNA structure — eukaryotes have linear DNA associated with histones; bacteria have circular DNA (and often plasmids), no histones. ✓
2. Cell wall — only some eukaryotes have walls (cellulose/chitin); all bacteria have walls (peptidoglycan). ✓
3. Cell division — eukaryotes use mitosis; bacteria use binary fission. ✓
4. Reproduction — many eukaryotes reproduce sexually; bacteria reproduce asexually. ✓
5. Ribosomes — eukaryotes have 80S ribosomes (and 70S in organelles); bacteria have 70S only. ✓
6. Cell size — eukaryotic cells > 5 µm; bacterial cells 1–5 µm. ✓

Any four of these gives full marks. A common mark-scheme trap is to award the mark only if both sides of the comparison appear, so a one-sided statement (e.g. "eukaryotes have linear DNA") does not earn the mark.

Key Takeaways

• Domains are separated by many features, not just membrane-bound organelles.
• Each mark requires a paired (Eukarya vs Bacteria) statement.
• The size distinction and ribosome size are the most distinctive single features and are worth memorising.

Common Mistakes

• Repeating the membrane-bound organelle difference (already given in the stem).
• Giving only one side of the comparison — e.g. "eukaryotes have histones" without "bacteria lack histones".
• Writing "eukaryotes have a cell wall" — wrong, because only some do.
• Confusing the peptidoglycan/murein cell wall with chitin or cellulose.

Things to Be Careful About

Use precise vocabulary: peptidoglycan (not "sugar"), histones (not "proteins"), binary fission (not "splitting"). Quantities (80S vs 70S) are easy marks — learn them.

Background Concept

Viruses are acellular particles: a nucleic-acid genome (DNA or RNA) enclosed in a protein capsid, sometimes wrapped in a host-derived phospholipid envelope. They do not fit into the Linnaean hierarchy because they are not living cells. Instead, virologists classify them by their structural and molecular features.

Key classification features:

• Nucleic acid type: DNA viruses (e.g. adenovirus, herpesvirus) vs RNA viruses (e.g. influenza, coronavirus, HIV).
• Strandedness: single-stranded (ss) or double-stranded (ds).
• Capsid symmetry: helical, icosahedral or complex.
• Envelope: enveloped vs non-enveloped (this affects transmission and resistance to environmental conditions).
• Host type: animal, plant, bacterial (bacteriophage), archaeal.
• Mode of replication and disease caused are also used.

Understanding the Question

The question says "with reference to the structure of viruses" — so you should focus on features visible or molecularly detectable in the virion. Two marks are available; the mark scheme accepts any two structural features.

Approach

Pick the two most reliable structural/molecular features: the type of nucleic acid and whether it is single- or double-stranded. These are unambiguous and easy to remember.

Step-by-Step Reasoning

1. The genetic material of a virion is either DNA or RNA — never both. This is the first axis of virus classification.
2. The nucleic acid may be single-stranded (e.g. HIV, influenza) or double-stranded (e.g. adenovirus, herpesvirus). This subdivides the groups further.
3. Other structural features the mark scheme allows: enveloped vs non-enveloped, with or without a tail sheath (bacteriophages), and the type of host they infect.

Any two of these give the two marks.

Key Takeaways

• Viruses are classified by their features, not by Linnaean ranks.
• The most fundamental feature is the type of nucleic acid in the virion.
• Enveloped vs non-enveloped is a key determinant of how a virus is transmitted and how it is inactivated (enveloped viruses are usually more sensitive to heat, detergents and desiccation).

Common Mistakes

• Saying viruses are classified into "kingdoms" or by their "size alone" — viruses do not have a Linnaean kingdom, and size is too crude a feature on its own.
• Forgetting that a virus contains either DNA or RNA, not both.
• Confusing the capsid (protein coat) with the envelope (host-derived phospholipid bilayer).

Things to Be Careful About

The question asks for classification with reference to structure — so keep your points structural (nucleic acid, strand-ness, envelope, capsid, tail). Avoid drifting into non-structural features (host range, disease) unless you have already given the structural ones.

Background Concept

Seeds require specific environmental cues to germinate, including appropriate temperature, water and oxygen. Light is also a powerful environmental signal that influences whether seeds remain dormant or germinate. Many small seeds require light to germinate, but in some species (such as certain cereal grains like barley) light actually inhibits germination — this prevents seeds buried too deeply from germinating before they have access to sufficient light for photosynthesis.

Two key plant hormones interact to control germination:

• Gibberellin: promotes germination by stimulating the synthesis of enzymes (e.g. alpha-amylase) that mobilise stored food reserves in the endosperm. Gibberellin works by causing the breakdown of DELLA repressor proteins, which otherwise inhibit growth-promoting gene expression.
• Abscisic acid (ABA): inhibits germination and maintains seed dormancy, allowing seeds to survive unfavourable conditions.

The balance between these two hormones determines whether a seed germinates.

Understanding the Question

The question presents two pieces of experimental data:

• Table 9.1: percentage germination of barley seeds at four different blue light intensities after 7 days.
• Table 9.2: ABA concentration over 7 days in seeds exposed to 57 au blue light, plus a comparison with seeds kept in the dark.

The question has two parts:

1. Describe the effect of blue light on germination (this must use the data).
2. Suggest an explanation for this effect (this must use the ABA data and the hint that ABA affects gibberellin).

The command word 'describe' requires data-based statements, while 'suggest an explanation' allows mechanistic reasoning.

Approach

1. Identify the trend in Table 9.1: how does germination change with light intensity?
2. Quote supporting data points to back up the trend.
3. Use Table 9.2 to identify what blue light does to ABA concentration over time.
4. Combine this with the hint that ABA affects gibberellin synthesis or activity to construct a mechanistic explanation.

Step-by-Step Reasoning

Describing the effect:

From Table 9.1, as blue light intensity increases from 0 to 57 au, the percentage germination falls from 98.0% to 14.7%. This shows that blue light inhibits germination of barley seeds, and the higher the intensity, the greater the inhibition.

Explaining the effect:

From Table 9.2, in blue light at 57 au, ABA concentration rises from 100 au at day 0 to 381 au by day 7. By contrast, in the dark, ABA falls to 45 au on day 1 and stays low. Therefore blue light causes ABA concentration to increase.

The question states that ABA affects gibberellin synthesis or activity. Since gibberellin is needed to mobilise stored food for germination, increased ABA must be inhibiting this gibberellin action, thereby preventing germination.

Further points that earn credit: ABA could prevent the breakdown of DELLA proteins (which would otherwise allow growth-promoting genes to be expressed), or ABA could promote dormancy directly by inhibiting gene expression needed for germination.

Key Takeaways

• Barley seeds show an inverse relationship between blue light intensity and germination.
• Blue light increases ABA concentration.
• ABA inhibits gibberellin action, preventing the mobilisation of food reserves needed for germination.
• Hormones often work in antagonistic pairs (here ABA versus gibberellin) to fine-tune plant responses to the environment.

Common Mistakes

• Simply stating 'light decreases germination' without quoting data — candidates must support the trend with numbers.
• Describing only the effect without suggesting an explanation.
• Confusing the direction of the effect (some candidates may say blue light increases germination because they expect light to be positive for plants in general).
• Failing to link ABA to gibberellin, missing the mechanism.
• Saying 'ABA decreases germination' without explaining that it does so by inhibiting gibberellin.

Things to Be Careful About

• The data must be quoted correctly: 98.0% at 0 au versus 14.7% at 57 au.
• Use the precise term 'concentration of ABA' rather than vague 'amount of ABA'.
• The question gives a specific hint that ABA affects gibberellin — this MUST be incorporated into the explanation.
• Be precise about the difference between 'synthesis' and 'activity' of gibberellin — either is acceptable.
• ABA's other mechanisms (DELLA breakdown, dormancy, gene expression) are valid additional points to gain extra marks.

Background Concept

Auxin (indole-3-acetic acid, IAA) is a plant growth hormone produced mainly at the shoot apex and transported downwards. It promotes cell elongation in young shoots and is involved in phototropism, gravitropism and apical dominance.

The acid growth hypothesis explains how auxin causes cell elongation. The mechanism has two coupled phases:

1. Wall-loosening phase: auxin triggers the secretion of H+ ions into the cell wall, lowering its pH. The acidic environment activates expansin proteins, which break hydrogen bonds holding cellulose microfibrils together, making the wall more extensible.
2. Osmotic water-uptake phase: as the wall loosens, the cell is poised to take up water. Auxin also causes K+ channels to open, allowing K+ to enter the cell. This lowers the water potential (Psi) of the cytoplasm, so water enters by osmosis, increasing turgor pressure and stretching the loosened wall — the cell elongates.

For elongation to continue, the cell must also deposit new cell wall material (cellulose, hemicellulose) to maintain wall thickness.

Understanding the Question

This question asks for both description (what happens) and explanation (why/how it happens) of auxin's role in cell elongation. The mark scheme rewards a step-by-step account of the acid growth mechanism. Candidates must give at least 7 distinct points covering binding, proton pumping, wall acidification, expansin activation, microfibril loosening, K+ uptake, and water entry.

Approach

Work through the mechanism in logical sequence:

1. How auxin enters signalling — receptor binding.
2. The wall-loosening pathway — proton pumps to H+ secretion to expansins to cellulose microfibrils.
3. The osmotic pathway — K+ channels to K+ uptake to lower water potential to osmotic water entry to turgor-driven expansion.

Each step has a clear cause-and-effect relationship that must be explicit. The two phases are interdependent: wall loosening permits expansion, and osmotic uptake drives expansion.

Step-by-Step Reasoning

1. Auxin binds to a receptor on the cell surface membrane. The receptor detects auxin and initiates intracellular signalling.

2. Auxin stimulates proton pumps. The proton pumps are membrane proteins (H+-ATPases) that actively transport H+ using ATP.

3. Protons move from the cytoplasm into the cell wall. This is active transport against the electrochemical gradient; the proton pumps use ATP.

4. The pH of the cell wall decreases. The wall becomes acidic (pH around 4.5–5).

5. Expansins are activated. These proteins are inactive at neutral pH but become active in acidic conditions.

6. Expansins loosen the linkage between cellulose microfibrils. They disrupt the non-covalent bonding in the cell wall matrix.

7. By breaking hydrogen bonds. This is the precise mechanism — expansins do not break covalent bonds; they disrupt hydrogen bonds between cellulose and hemicellulose (xyloglucan). Note the mark scheme explicitly says 'Ignore weakening bonds' — candidates must specify HOW the wall loosens.

8. K+ channels open. Auxin signalling also activates potassium channels in the plasma membrane.

9. K+ ions diffuse into the cytoplasm. They move down their electrochemical gradient.

10. The water potential of the cytoplasm decreases. The accumulation of K+ (a solute) lowers Psi, making the cytoplasm more negative.

11. Water enters the cell by osmosis. Water moves from the higher water potential outside to the lower water potential inside.

12. Turgor pressure increases / the cell volume increases. The incoming water pushes against the loosened cell wall, causing the cell to elongate.

Key Takeaways

• The acid growth hypothesis has two coupled phases: wall loosening and osmotic water uptake.
• Auxin acts through receptor-mediated signalling at the plasma membrane.
• H+ secretion activates expansins, which specifically break hydrogen bonds (not covalent bonds).
• Solute accumulation (K+) lowers water potential, driving osmotic water entry.
• Turgor pressure against a loosened wall causes cell elongation.
• The mechanism is ATP-dependent at the proton-pump step.

Common Mistakes

• Saying 'expansins break bonds' without specifying hydrogen bonds — the mark scheme requires 'hydrogen bonds'.
• Saying auxin 'weakens' the wall without explaining the mechanism.
• Confusing the direction of proton movement (some candidates say H+ enters the cell instead of leaving it).
• Skipping the K+ / osmotic phase entirely.
• Forgetting to mention the role of turgor pressure in cell elongation.
• Stating that auxin directly causes water uptake (it does so indirectly via K+ accumulation and wall loosening).
• Saying the cell wall becomes 'stronger' instead of 'loosened'.

Things to Be Careful About

• Use precise terminology: 'proton pumps', not 'pumps' alone; 'cellulose microfibrils', not 'cell wall'; 'water potential', not 'water concentration'.
• The two phases are sequential and interdependent — wall loosening makes elongation possible, and osmotic uptake drives expansion.
• Turgor pressure increases because water enters, not because of any active pumping of water.
• The mechanism applies to young, growing cells — mature cells with rigid walls do not elongate this way.
• The mark scheme explicitly says 'Ignore weakening bonds'; simply saying 'weakens the wall' is insufficient, they must specify HOW (hydrogen bonds).

Background Concept

A sarcomere is the contractile unit of a myofibril, bounded by two Z lines, with thin (actin) filaments anchored at the Z lines and thick (myosin) filaments lying between them. Contraction occurs when the thin filaments slide over the thick filaments, shortening the sarcomere — the sliding filament model. The four proteins in the question play distinct, tightly coupled roles.

In a relaxed muscle, the myosin-binding sites on actin are physically blocked by the long, rod-shaped tropomyosin molecule lying along the actin groove. Troponin is a small globular complex attached to tropomyosin at regular intervals; it is the calcium sensor of the system. When the sarcoplasmic reticulum releases $\text{Ca}^{2+}$ into the sarcoplasm, $\text{Ca}^{2+}$ binds to troponin, troponin undergoes a conformational change, and this drag-moves tropomyosin away from the myosin-binding sites on actin.

Myosin is a motor protein. Each myosin head carries an actin-binding site and an ATPase site. Once the binding sites on actin are exposed, a myosin head (already energised by hydrolysing ATP to ADP + $\text{P}_i$) binds actin — this is the cross-bridge. The head then pivots (the power stroke), pulling the actin filament towards the M line, which shortens the sarcomere. After the power stroke, a new ATP molecule binds to the myosin head; this reduces the head's affinity for actin, so it detaches. ATP is then hydrolysed again, re-cocking the head. The cycle repeats as long as $\text{Ca}^{2+}$ and ATP are available.

Understanding the Question

Part (a) asks for an outline — a brief, ordered account, not an essay — of what each of the four named proteins does during contraction of a single sarcomere. Four marks are available, so four distinct points are needed. The mark scheme offers five possible points, so the candidate can omit one. Each protein must contribute something to the answer, and the role of each should be tied to contraction, not to muscle structure in general.

Approach

The trick with this kind of question is to keep the proteins in the order in which they act during the contraction cycle: $\text{Ca}^{2+}$-binding trigger (troponin) → exposure of binding sites (tropomyosin / actin) → cross-bridge formation and power stroke (myosin) → detachment using ATP (myosin). Linking each step to the next makes the answer read as a single coherent mechanism rather than four disjoint facts, which is what examiners reward.

Step-by-Step Reasoning

Mark point 1 — troponin and $\text{Ca}^{2+}$
When the motor neurone fires, an action potential sweeps down the T-tubules and triggers $\text{Ca}^{2+}$ release from the sarcoplasmic reticulum. This $\text{Ca}^{2+}$ binds to troponin, causing it to change shape. This single event is the on-switch of contraction.

Mark point 2 — tropomyosin moves
Because troponin is physically attached to tropomyosin, the shape change drags tropomyosin away from the myosin-binding sites on the actin filament, exposing them.

Mark point 3 — myosin binds actin (cross-bridge formation)
Energised myosin heads (ADP + $\text{P}_i$ still bound after the previous ATP hydrolysis) now bind to the newly exposed sites on actin.

Mark point 4 — power stroke
The myosin head pivots, releasing ADP and $\text{P}_i$. This pulls the actin filament towards the centre of the sarcomere. The mark scheme accepts either "shortens the sarcomere" or "Z lines move closer together / towards the M line" as evidence that the candidate understands the geometry. The latter is a more precise way of saying the same thing and is what CIE mark schemes tend to prefer.

Mark point 5 — detachment using ATP
A fresh ATP molecule binds to the myosin head. Binding ATP (not yet its hydrolysis) reduces the myosin's affinity for actin, so the head detaches. The myosin head's intrinsic ATPase activity then hydrolyses the ATP, re-cocking the head for the next cycle. Note that the mark scheme's wording is "myosin head is ATPase / binds to ATP, to allow detachment" — both halves of this point are needed: ATP binding causes detachment, and the myosin head has ATPase activity to re-energise it.

Key Takeaways

• The four proteins act in a fixed sequence: troponin (Ca²⁺ sensor) → tropomyosin (gatekeeper on actin) → actin (filament that gets pulled) → myosin (motor that pulls and detaches via ATP).
• The sarcomere shortens because actin filaments slide over myosin filaments; the filaments themselves do not contract.
• ATP has two distinct roles: it is hydrolysed to energise the power stroke, and its binding (before hydrolysis) is what releases the myosin head from actin. A cadaver becomes rigid (rigor mortis) precisely because no fresh ATP is made to allow detachment.

Common Mistakes

• Saying "calcium binds to tropomyosin" — it binds to troponin.
• Describing the power stroke as "myosin shortening" or "myosin contracting" — myosin does not get shorter; it pivots, and it is the actin filament that moves.
• Omitting ATP and the detachment step — without it the cycle is not a cycle and the muscle could not relax or contract again.
• Saying "myosin uses energy from ATP to contract" — vague; the mark scheme wants "myosin head is an ATPase / binds ATP to allow detachment".

Things to Be Careful About

• The mark scheme uses "actin" and "myosin" as bold key terms — answers that omit these in favour of vague phrases like "the filaments" can lose marks.
• Either wording for the power stroke is accepted ("shorten sarcomere" or "Z lines move closer together / towards M line"); the second is more rigorous and is the wording the examiner is hoping for.
• "Troponin changes shape" is the precise phrase — do not say "troponin activates" or "troponin moves"; troponin is the trigger, tropomyosin is the part that physically moves.