Biology 5090/21 — May/June 2025

Walkthrough

In any food web, the energy originates from the sun. Producers (plants and algae) capture this light energy and convert it into chemical energy via photosynthesis. All subsequent consumers obtain their energy by eating the producers or other consumers. Therefore, the ultimate source of energy for this pond food web is the sun.

Key Takeaways

All food chains and webs are powered by solar energy captured by autotrophs (producers). Without the sun, there would be no energy input to sustain the ecosystem.

Common Mistakes

Candidates often write 'food' or 'plants' as the energy source. While plants contain energy, they do not create it; they convert solar energy. The question asks for the source of energy for the web, which is external to the biological components: the sun.

Things to Be Careful About

The command word is 'State'. A single word or short phrase is required. 'Sunlight' or 'Sun' are both acceptable. Do not write 'photosynthesis' as that is the process, not the source.

Walkthrough

We need to classify the organisms in Fig. 2.1.

1. Producers (Trophic Level 1): Water weed and alga. (Given as 2).
2. Primary Consumers (Herbivores, Trophic Level 2): Freshwater snail (eats water weed and alga) and insect larva (eats alga).
3. Secondary Consumers (Carnivores/Omnivores eating primary consumers, Trophic Level 3):
   • Water shrew: Eats freshwater snail and insect larva (both primary consumers). So, water shrew is a secondary consumer.
   • Frog: Eats insect larva (primary consumer). So, frog is a secondary consumer.
   • Note on Snake: The snake eats the frog and water shrew. Since the frog and water shrew are at level 3, the snake is at level 4 (tertiary consumer). There is no direct arrow from insect larva to snake, so the snake is not a secondary consumer here.
   • Count: Water shrew and Frog = 2.
4. Carnivores (Organisms that eat other animals):
   • Water shrew: Eats snail and insect larva. (Yes)
   • Frog: Eats insect larva. (Yes)
   • Snake: Eats water shrew and frog. (Yes)
   • Hawk: Eats water shrew and snake. (Yes)
   • Note: Freshwater snail and insect larva eat plants/algae, so they are herbivores, not carnivores.
   • Count: Water shrew, Frog, Snake, Hawk = 4.

Key Takeaways

• Secondary consumers feed directly on primary consumers (herbivores). In this web, they are at the third trophic level.
• Carnivores are organisms that eat meat (other animals). An organism can be a secondary, tertiary, or quaternary consumer and still be a carnivore.
• Always trace the arrows: arrows point from the food to the feeder.

Common Mistakes

• Counting the snake as a secondary consumer. The snake eats the frog (level 3) and water shrew (level 3), making the snake a tertiary consumer (level 4). It only becomes a secondary consumer if it ate the insect larva directly, which the diagram does not show.
• Missing the hawk as a carnivore. The hawk eats the water shrew and the snake, both animals.
• Confusing 'secondary consumer' with 'carnivore'. All secondary consumers in this web are carnivores, but not all carnivores are secondary consumers (snake and hawk are higher up).

Things to Be Careful About

• The table asks for the number of organisms. Ensure you count unique species/organisms, not the number of arrows or paths.
• 'Carnivores' includes any animal that eats other animals, regardless of whether it is a secondary, tertiary, or quaternary consumer.

Walkthrough

To determine trophic levels, we count the steps from the producer.

Path 1 (Hawk eats Water Shrew):

1. Producer: Alga or Water weed
2. Primary Consumer: Insect larva or Freshwater snail
3. Secondary Consumer: Water shrew
4. Tertiary Consumer: Hawk

In this chain, the hawk is at the fourth trophic level.

Path 2 (Hawk eats Snake):
The snake eats the frog (or water shrew, but let's look at the longer chain to get the 5th level).

1. Producer: Alga
2. Primary Consumer: Insect larva
3. Secondary Consumer: Frog
4. Tertiary Consumer: Snake
5. Quaternary Consumer: Hawk

In this chain, the hawk is at the fifth trophic level.

Since the hawk eats both water shrews (making it a tertiary consumer) and snakes (making it a quaternary consumer), it feeds at two different trophic levels.

Key Takeaways

• An organism can occupy multiple trophic levels in a food web if it has multiple food sources at different levels.
• Trophic level = number of steps from the producer + 1 (Producer is level 1).
• Alternatively, count the number of organisms the food has passed through. If food passes through 3 organisms (Producer → Primary → Secondary → Hawk), Hawk is 4th level.

Common Mistakes

• Only finding one food chain. Look for all paths leading to the hawk.
• Miscounting the levels. Remember producers are level 1.
• Not explaining why. The question asks to 'explain', so you must mention the two different prey (snakes vs. water shrews) and the resulting trophic levels.

Things to Be Careful About

• The mark scheme accepts 'fifth trophic level / quaternary consumer' and 'fourth trophic level / tertiary consumer'. Use precise terminology.
• Ensure you mention both scenarios (feeding on snakes and feeding on water shrews).

Walkthrough

The question asks why shrews need more oxygen in cold water.

1. Identify the organism type: Shrews are mammals. Mammals are endotherms (homeotherms), meaning they maintain a constant internal body temperature regardless of the environment.
2. Effect of cold environment: Cold water conducts heat away from the body much faster than air. The shrew will lose body heat rapidly.
3. Response: To maintain constant body temperature, the shrew must increase heat production. Heat is a by-product of respiration (chemical energy released as heat).
4. Link to oxygen: Aerobic respiration equation: Glucose + Oxygen → Carbon dioxide + Water + Energy. To increase the rate of respiration (and thus heat production), the shrew must consume oxygen at a faster rate.
5. Additional factor: Diving involves muscle activity. Muscle contraction requires ATP from respiration. The mark scheme also accepts 'more energy/respiration needed for increased muscle contraction', but the thermoregulation point is the primary biological reason for 'cold water' specifically.

Key Takeaways

• Mammals maintain constant body temperature.
• In cold environments, metabolic rate (respiration rate) increases to generate heat.
• Respiration is an aerobic process requiring oxygen; higher rate = higher oxygen demand.

Common Mistakes

• Saying 'shrews get cold'. They don't; they regulate their temperature. They lose heat faster.
• Forgetting the link to respiration. Cold -> Heat loss -> Need heat -> Respiration produces heat -> Respiration needs oxygen.
• Not mentioning that shrews are mammals/endotherms. This justifies why they regulate temperature.

Things to Be Careful About

• The mark scheme gives max 2 marks. Two clear points are needed: (1) Mammals regulate body temperature / need to stay warm, (2) Respiration releases heat / needs oxygen.
• Avoid saying 'shrews need oxygen to stay warm'. Be precise: 'Respiration releases heat, and respiration requires oxygen'.

Walkthrough

This question asks for an explanation of evolution by natural selection. We must use the standard Darwinian steps: Variation, Selection, Survival/Reproduction, Inheritance.

1. Variation (Mutation): In any population, there is genetic variation. A random mutation occurred (or exists) in some water shrews that produces this special oxygen-storing protein. (Mark scheme: 'some water shrews have more protein / a mutation occurred').
2. Selection Pressure (Competition): Food is limited. Shrews need to dive to get food. The ability to stay underwater longer is advantageous. (Mark scheme: 'water shrews are competing for food').
3. Differential Survival: Shrews with the protein can store more oxygen, stay down longer, and collect more food. Shrews without it might run out of oxygen and starve or fail to get enough energy. Thus, those with the protein are more likely to survive. (Mark scheme: 'shrews that have the protein can collect more food... survive').
4. Reproduction: Survivors live long enough to reproduce. (Mark scheme: 'reproduce').
5. Inheritance: They pass the alleles (genetic code) for the protein to their offspring. (Mark scheme: 'pass on the alleles').
6. Result: Over time, the next generation will have a higher proportion of individuals with the protein. (Mark scheme implies this with 'max 4' points covering the chain).

Key Takeaways

• Natural selection requires pre-existing variation (mutation).
• The environment creates a selection pressure (need for food/oxygen).
• Individuals with the advantageous trait survive and reproduce better (fitness).
• The trait is genetic (alleles) and is passed on.

Common Mistakes

• Lamarckian evolution: Saying 'shrews developed the protein because they needed it' or 'shrews used the protein so it grew'. Evolution does not work by need or use; it works by selection of existing variation. Use 'mutation' and 'variation'.
• Forgetting inheritance: Stopping at 'they survive'. Must mention passing genes/alleles to offspring.
• Vague language: 'They are better'. Say 'more likely to survive/reproduce'.

Things to Be Careful About

• The mark scheme has 5 potential points but max 4 marks. Cover the main chain: Mutation/Variation → Advantage (more food) → Survival → Reproduction → Inheritance.
• Use the word 'alleles' or 'genes'. 'Hereditary material' is okay, but 'alleles' is precise 5090 terminology.
• Do not say 'the shrews evolved the protein'. Say 'the protein arose via mutation and was selected for'.

Walkthrough

The question asks for the starting points of starch and protein digestion. We look at Fig. 3.1 and recall the digestive pathway.

1. Starch digestion: Chemical digestion of starch begins in the mouth (label A). Salivary glands secrete saliva containing the enzyme salivary amylase, which breaks down starch into maltose. Although digestion continues in the small intestine, it starts here.
2. Protein digestion: Chemical digestion of proteins begins in the stomach (label G). The stomach lining secretes pepsinogen, which is activated to pepsin by hydrochloric acid. Pepsin breaks proteins into polypeptides.

The oesophagus (not labelled, between A and G) only transports food. The small intestine (E) is where most digestion is completed, but not where it starts for these macronutrients.

Key Takeaways

• Starch digestion starts in the mouth (salivary amylase).
• Protein digestion starts in the stomach (pepsin in acidic conditions).
• Always distinguish between where digestion starts and where it is completed.

Common Mistakes

• Saying digestion starts in the small intestine for everything. While the small intestine is the main site, starch and protein digestion have earlier starting points.
• Confusing the mouth (A) with the salivary glands (also A, but the question asks for the part where digestion starts, which is the oral cavity/mouth).
• Writing "stomach" instead of the letter G when the question asks for the letter.

Things to Be Careful About

• The question asks for letters, not names. Write A and G.
• Starch digestion begins in the mouth, not the stomach. Protein digestion begins in the stomach, not the mouth or small intestine.

Walkthrough

We need to describe the roles of organs B and C (liver and gall bladder) specifically in lipid (fat) digestion.

1. Organ B is the liver. The liver is a large organ that synthesises bile. Bile is not an enzyme; it is a mixture of salts and pigments.
2. Organ C is the gall bladder. The liver produces bile continuously, but the body doesn't always need it. The gall bladder acts as a reservoir, storing and concentrating bile between meals.
3. Mechanism: When fatty food enters the small intestine, bile is released. Bile contains bile salts which act as detergents. They break large fat droplets into smaller ones. This process is called emulsification. Emulsification does not chemically break bonds; it increases the surface area of the fat, allowing the enzyme lipase (from the pancreas) to work much faster.

Key Takeaways

• Liver = production of bile.
• Gall bladder = storage of bile.
• Bile = emulsification (physical change, not chemical), increasing surface area for lipase.

Common Mistakes

• Saying the liver or gall bladder produces digestive enzymes. Bile is not an enzyme; it has no active site.
• Saying bile digests fat. Bile only emulsifies (physically breaks up) fat. Lipase does the chemical digestion.
• Forgetting to mention surface area. Emulsification's main purpose is to increase surface area for enzymes.

Things to Be Careful About

• Use the correct organ names: liver (B) and gall bladder (C).
• The mark scheme specifically looks for the word emulsifies or emulsification.
• Ensure you link emulsification to a faster rate of digestion (surface area increase).

Walkthrough

The question asks to explain reduced growth in untreated coeliac disease using Fig. 3.2. We must build a logical chain from structure to function to outcome.

1. Observe Fig. 3.2: The left diagram (healthy) shows tall, finger-like villi. The right diagram (coeliac) shows villi that are flattened, blunted, and shorter.
2. Relate structure to function: The function of villi in the small intestine (organ E) is absorption of nutrients. The tall, numerous villi provide a large surface area. Flattened villi mean a lower surface area.
3. Consequence of lower surface area: With less surface area, the rate and total amount of nutrient absorption (into the blood) decreases. Less glucose, amino acids, fatty acids, etc., enter the blood.
4. Link to growth: The question mentions a child. Children are growing, which requires building new cells and tissues. This process (anabolism) needs amino acids to make new proteins. If absorption is poor, the child is starved of amino acids, so growth is reduced (stunted growth or failure to thrive).

Key Takeaways

• Villi increase surface area for absorption.
• Damage to villi (as in coeliac disease) reduces absorption.
• Growth requires building blocks (amino acids) from digested food.

Common Mistakes

• Saying "less food is digested". Digestion may still happen, but absorption is the problem. The villi are for absorption, not digestion (though brush border enzymes are on them, the primary issue here is surface area for uptake).
• Forgetting to mention amino acids or proteins in the final step. Growth is specifically linked to protein synthesis.
• Not referring to Fig 3.2. You must mention that the villi are smaller/shorter.

Things to Be Careful About

• The mark scheme awards marks for: (1) villi smaller/shorter/lower surface area, (2) less absorption of nutrients, (3) less amino acids/proteins for growth. Ensure all three are covered.
• Do not say "the child does not eat enough". The problem is absorption, not intake.

Walkthrough

The scientists want to deliver an enzyme to organ E (small intestine) to digest gluten. The enzyme is in a tablet. The coating dissolves in alkali but not acid. Why?

1. Path of the tablet: Swallowed tablet goes to the stomach (organ G) first, then to the small intestine (organ E).
2. Stomach conditions: The stomach lining secretes hydrochloric acid (HCl). The pH is very low (around 1-2), making it highly acidic.
3. Enzyme sensitivity: Enzymes are proteins. Their function depends on the shape of their active site (lock-and-key model). Extreme pH levels (too acidic or too alkaline) can break the bonds holding the protein shape together, causing denaturation. If the enzyme is denatured, gluten cannot be digested.
4. Why resist acid? If the coating dissolved in the stomach (acidic), the enzyme would be exposed to HCl. Unless this specific enzyme is adapted to work in acid (which the question implies it isn't, as it needs to be protected), it would be denatured or damaged before reaching its target.
5. Small intestine conditions: Organ E (small intestine) receives bile from the liver/gall bladder and pancreatic juice, which are alkaline (basic). The pH is around 7-8.
6. Why dissolve in alkali? The coating is designed to break down in alkaline conditions. Once the tablet reaches the small intestine (organ E), the alkaline environment dissolves the coating, releasing the enzyme. The enzyme is now in its optimal pH environment and can digest the gluten.

Key Takeaways

• Stomach = acidic (HCl). Small intestine = alkaline (bile/pancreatic juice).
• Enzymes can be denatured by incorrect pH.
• A coating can protect an enzyme until it reaches the correct pH environment.

Common Mistakes

• Saying "the acid would kill the enzyme". Enzymes are not alive; they are denatured or damaged.
• Saying "the coating protects the stomach from the enzyme". The goal is to protect the enzyme from the stomach.
• Not mentioning that the small intestine is alkaline. The question states the coating dissolves in alkaline conditions, so you must link this to the environment of organ E.

Things to Be Careful About

• Use the word denatured or denaturation.
• Refer to hydrochloric acid or acid in the stomach.
• Refer to alkaline conditions in organ E (small intestine).
• The explanation must cover both parts: why it doesn't dissolve in acid (protection) and why it does dissolve in alkali (release).

Walkthrough

The cholera bacterium does not damage the intestine directly; it secretes a toxin that makes the cells lining the small intestine secrete chloride ions (Cl⁻) into the lumen. Adding dissolved ions lowers the water potential of the gut contents, so water moves by osmosis from the blood and tissues (higher water potential) into the intestine (lower water potential). This is why cholera causes severe watery diarrhoea and dehydration. Each gap is one mark, and the mark scheme's exact words are 'toxin', 'chloride' and 'decrease / reduce'.

Key Takeaways

• The cholera toxin → chloride ion secretion → lowered water potential → osmotic water loss chain is a standard 5090 sequence.
• Water always moves from the higher water potential to the lower water potential across a partially permeable membrane.

Common Mistakes

• Writing 'poison' or 'acid' instead of toxin — the precise term is required.
• Writing 'sodium' instead of chloride ions.
• Writing 'increase' for the water potential — adding ions decreases water potential; confusing this reverses the direction of osmosis.
• Saying 'water moves in by diffusion' — it moves by osmosis, through a partially permeable membrane.

Things to Be Careful About

• The mark scheme accepts 'decrease / reduce' for the third gap — use one of those exact verbs.
• Keep the causal chain in order; each gap is marked independently, so a wrong word in one gap does not affect the others.

Walkthrough

The project aims to reduce deaths by 95%, so 5% of the original number would remain — that is the maximum if the project succeeds. 5% of 95 000 is 95 000 ÷ 100 × 5 = 4750. The double semicolon in the mark scheme means the correct answer alone carries both marks, so the arithmetic must be right.

Key Takeaways

• 'Reduce by 95%' leaves 5% remaining — a classic percentage trap.
• Percentage of a quantity = (percentage ÷ 100) × quantity.

Common Mistakes

• Calculating 95% of 95 000 (= 90 250) instead of the 5% that remains — this is the number saved, not the number of deaths.
• Subtracting 95 from 95 000, or other misreads of the percentage.

Things to Be Careful About

• Read carefully: the question asks for the number of deaths after the reduction, not the number prevented.
• No unit is needed on the answer line — it is a number of people.

Walkthrough

Vaccination produces active immunity — the body makes its own antibodies. The sequence is: (1) the vaccine contains a weakened or harmless form of the pathogen, so it cannot cause the disease but still carries antigens on its surface; (2) the antigens are recognised as foreign and stimulate lymphocytes (white blood cells) to divide and secrete antibodies with a shape complementary to the antigen — the lock-and-key idea; (3) some lymphocytes become memory cells, which persist long after the antibodies have gone; (4) on a real infection the memory cells respond immediately, producing antibodies so quickly that the pathogen is destroyed before it multiplies enough to cause symptoms. That rapid secondary response is what 'being immune' means. Each of the four mark-scheme points is one mark.

Key Takeaways

• Active immunity: antigen → lymphocytes → antibodies + memory cells → faster, stronger secondary response.
• Antibodies are specific — each fits only one antigen.
• Memory cells are the reason immunity lasts and the reason the second encounter is fought off quickly.

Common Mistakes

• Saying the vaccine 'kills the pathogens in the body' — it prevents disease; it is not a cure for an existing infection.
• Writing 'white blood cells' alone without naming lymphocytes — the precise cell type is the marking point.
• Omitting memory cells — a very commonly dropped mark.
• Confusing antibodies (proteins made by lymphocytes) with antigens (molecules on the pathogen surface).
• Saying antibodies 'eat' pathogens — antibodies bind to antigens; phagocytes engulf pathogens.

Things to Be Careful About

• The command word is describe, so give the sequence of events; the final 'rapidly produced' idea is a separate mark, so make the speed explicit.
• Four marks = four distinct points; write them as separate statements so the examiner can award each one.

Walkthrough

'Discuss the practical problems' means use each row of Table 4.1 and ask what could go wrong in the real world. Row 1: two doses one to six weeks apart — people must come back a second time; in remote areas, or with mobile populations, many will not return, so they are not fully protected. Row 2: vaccine added to a glass of water — the water must be clean; in regions without a safe water supply the water itself could contain pathogens (ironically, including cholera) or make people ill. Row 3: storage at 2–8 °C — the 'cold chain' needs refrigerators and a reliable electricity supply; without them the vaccine may spoil and be ineffective. Each of these is one mark, and each problem must be tied to the specific feature of the vaccine, not a generic complaint.

Key Takeaways

• A good evaluation answer always links the problem back to the specific data given (doses, route, storage temperature).
• Vaccination programmes fail practically, not just biologically — logistics, infrastructure and compliance matter.

Common Mistakes

• Giving generic answers like 'it is expensive' or 'people may refuse it' — the marks are tied to the three features in the table.
• Saying 'the water may be dirty' without linking it to the oral route of this vaccine.
• Saying 'it needs to be kept cold' but not mentioning refrigeration / power supply — the mark wants the practical consequence.
• Writing about how the vaccine works immunologically — the question asks about practical problems, not mechanism.

Things to Be Careful About

• Three marks = three distinct problems; take one from each row of the table so you do not repeat yourself.
• Use the data: quote 'two doses', 'glass of water' and '2–8 °C' in your reasoning to anchor each point.

Walkthrough

Photosynthesis has two raw materials — carbon dioxide and water — and two products — glucose and oxygen. The stem already gives $6\text{CO}_2$, so the first blank is water, $\text{H}_2\text{O}$, and the product blanks are glucose, $\text{C}_6\text{H}_{12}\text{O}_6$, and oxygen, $\text{O}_2$.

Balancing: glucose contains 12 hydrogens, so six waters are needed on the left ($6 \times \text{H}_2\text{O}$ gives 12 H). Those six waters also supply six oxygens, which together with the twelve oxygens from $6\text{CO}_2$ make eighteen oxygens in total; glucose holds six, leaving twelve to form $6\text{O}_2$. Hence the fully balanced equation:

Key Takeaways

• Learn the photosynthesis equation as a fixed item: $6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$.
• One mark is for the correct formulae, one for correct balancing — both are needed.

Common Mistakes

• Writing $\text{H}_2\text{O}_2$ instead of $\text{H}_2\text{O}$ (hydrogen peroxide is a completely different substance).
• Forgetting the coefficient 6 on water or oxygen — the formulae mark may still score but the balancing mark is lost.
• Writing the word equation when the symbol equation is asked for.
• Reversing the equation (respiration direction).

Things to Be Careful About

• Subscripts must be written correctly: $\text{C}_6\text{H}_{12}\text{O}_6$, not C6H12O6 typed loosely if precision is expected.
• Check every atom balances before moving on — count C, H and O on each side.

Walkthrough

The seaweed doubles every 10 days. Over 40 days there are $40 \div 10 = 4$ doubling periods. Starting at 2 tonnes:

• after 10 days: 4 tonnes
• after 20 days: 8 tonnes
• after 30 days: 16 tonnes
• after 40 days: 32 tonnes

Equivalently, $2 \times 2^4 = 32$. The common error is treating 40 days as 40 doublings, or halving instead of doubling, or stopping one interval early.

Key Takeaways

• Doubling problems: first find the number of doubling intervals, then apply the doubling that many times.
• $2 \times 2^n$ grows fast — exponential growth is exactly why growing seaweed is attractive for removing carbon dioxide.

Common Mistakes

• Dividing 40 by 10 incorrectly or counting only three doublings (giving 16 tonnes).
• Multiplying 2 by 40 (giving 80 tonnes) — linear instead of exponential thinking.
• Omitting the unit "tonnes" on the answer line.

Things to Be Careful About

• Write out each doubling step so an arithmetic slip is visible and may earn error-carried-forward credit.
• The answer line asks for biomass in tonnes — include the unit.

Walkthrough

Plants absorb nitrate ions ($\text{NO}_3^{-}$) from the soil (or, here, supplied to the seaweed in water). Nitrogen is an essential element in amino acids, and amino acids are joined together to make proteins. Proteins are needed for growth and for making enzymes, so without nitrates the plant cannot build new tissue.

The two marks map onto the two steps: (1) nitrates → amino acids, (2) amino acids → proteins. Both links must be present.

Key Takeaways

• Nitrate ions supply nitrogen for making amino acids and then proteins.
• Magnesium ions, the other standard mineral, are needed for chlorophyll — do not mix the two up.

Common Mistakes

• Saying nitrates are "used for photosynthesis" or "give energy" — they are building blocks for proteins, not fuel.
• Stopping at "amino acids" without linking on to proteins — that loses the second mark.
• Confusing nitrate with magnesium (chlorophyll).

Things to Be Careful About

• Give the full chain: nitrate → amino acid → protein. Each arrow is a marking point.

Walkthrough

Normally, when an organism dies, decomposers (bacteria and fungi) break down its body. Decomposers respire aerobically — they need oxygen — and respiration releases the carbon in the dead material back to the atmosphere as carbon dioxide:

carbon compounds in dead seaweed + oxygen → carbon dioxide + water (+ energy)

At the bottom of the ocean there is very little dissolved oxygen. Without oxygen the decomposers cannot carry out aerobic respiration, so decomposition effectively stops. The carbon compounds in the dead seaweed are therefore not broken down and no carbon dioxide is released — the carbon stays trapped in the bundles on the ocean floor. This is why sinking the seaweed removes carbon from the atmosphere long-term.

Both halves are needed for the marks: (1) no decomposition because decomposers require oxygen, (2) therefore no aerobic respiration and the carbon is not released.

Key Takeaways

• Decomposers respire aerobically; their respiration returns carbon from dead organic matter to the atmosphere as carbon dioxide.
• Removing oxygen blocks decomposition, locking carbon away — the principle behind carbon sequestration schemes like this one.

Common Mistakes

• Writing only "the seaweed cannot decay" without explaining why (no oxygen for decomposers' respiration) — an "explain" question needs the reason.
• Saying the seaweed itself needs oxygen — the point is about the decomposers' respiration.
• Suggesting anaerobic decomposition releases lots of carbon dioxide quickly — anaerobic breakdown is slow and incomplete here.
• Confusing respiration with photosynthesis in the explanation.

Things to Be Careful About

• Name the decomposers (bacteria and fungi) and name aerobic respiration explicitly — these precise terms carry the marks.
• Make the final link explicit: no respiration by decomposers → no carbon dioxide released → carbon stays trapped.

Walkthrough

The question gives three features and asks you to explain why each suggests wind OR insect pollination. The trick is that sycamore shows features of BOTH mechanisms, so you must pair each feature with the mechanism it serves.

1. Small and green coloured — showy, brightly coloured petals exist purely to advertise the flower to insects. A small green flower has no such advertisement, which is exactly what you expect if the wind is doing the pollinating: there is no point spending energy on petals no insect needs to see.
2. Large quantities of small, light pollen grains — wind pollination is wasteful; most grains never reach another flower. So wind-pollinated plants compensate by making enormous numbers of very light grains that stay airborne and drift far. Heavy sticky pollen would fall straight down and be useless on the wind.
3. Contains nectar — nectar is a sugary reward that lures insects into the flower. While feeding, the insect brushes against anthers and stigma, carrying pollen between flowers. This is the classic insect-pollination feature.

So two points score here: one linking the small/green flowers or the abundant light pollen to wind pollination, and one linking nectar to insect attraction.

Key Takeaways

• Wind-pollinated flowers: small, dull, no nectar, huge quantities of light pollen, feathery stigmas.
• Insect-pollinated flowers: large, colourful, scented, nectaries, smaller amounts of heavier sticky pollen.
• Every structural feature of a flower can be argued as an adaptation for its pollination method.

Common Mistakes

• Writing "the flowers are pollinated by both" without explaining WHICH feature suits WHICH agent — the marks sit in the pairing plus reason.
• Saying nectar "attracts pollinators" vaguely instead of naming insects.
• Saying light pollen "travels further" without mentioning it being carried by the wind.

Things to Be Careful About

• This is an "Explain" question: every point needs its because-clause, not just the observation.
• The mark scheme accepts either the small/green feature OR the large/light pollen for the wind mark — one is enough, but give the clearest.

Walkthrough

Self-pollination is essentially inbreeding: the offspring receive very similar alleles from both parents, so variation is low and harmful recessive alleles can be made homozygous. Cross-pollination mixes alleles from two different plants, so the offspring are genetically more varied. Variation matters because if the environment changes — a new disease, a drought — some individuals in a varied population are likely to carry combinations of alleles that let them survive and reproduce.

The mark scheme wants the key word variation: 'increased variation' or the equivalent idea of a better capacity to respond to environmental change.

Key Takeaways

• Cross-pollination → gametes from different plants → more genetic variation in offspring.
• Variation is the raw material that lets a population survive environmental change (and underpins natural selection).

Common Mistakes

• Writing vague answers like 'healthier plants' or 'stronger offspring' — the scheme wants variation specifically.
• Confusing cross-pollination with cross-fertilisation terminology or drifting into hybrid vigour arguments beyond the syllabus wording.

Things to Be Careful About

• One mark only — one clean sentence is enough; do not pad.

Walkthrough

The question asks for the blood group (phenotype) of a person with the genotype $I^A I^O$. In the ABO blood group system, the $I^A$ and $I^B$ alleles are codominant, meaning that if both are present, both A and B antigens are expressed (blood group AB). The $I^O$ allele is recessive to both $I^A$ and $I^B$. Therefore, in the genotype $I^A I^O$, the dominant $I^A$ allele determines the phenotype, and the individual will have blood group A.

Key Takeaways

• $I^A$ and $I^B$ are codominant alleles.
• $I^O$ is a recessive allele.
• Genotype $I^A I^O$ results in blood group A; genotype $I^B I^O$ results in blood group B.

Common Mistakes

• Writing "A" instead of "blood group A". The question asks for the blood group, so the full phrase is safer.
• Confusing $I^A I^O$ with $I^A I^B$ and answering blood group AB.

Things to Be Careful About

• Ensure you use the exact terminology "blood group A" rather than just "A" to match the mark scheme.

Walkthrough

Blood group O is the phenotype expressed only when neither the $I^A$ nor the $I^B$ allele is present. Since $I^O$ is recessive to both $I^A$ and $I^B$, an individual must inherit an $I^O$ allele from both parents to show blood group O. Therefore, the only possible genotype for blood group O is homozygous recessive: $I^O I^O$.

Key Takeaways

• Blood group O is always homozygous recessive ($I^O I^O$).
• You cannot have a heterozygous genotype that results in blood group O.

Common Mistakes

• Writing $I^o I^o$ with a lowercase 'o'. The standard notation uses $I^O$ (capital I, capital O) or $I^o$ (capital I, lowercase o) consistently, but $I^O I^O$ is the most widely accepted form in 5090.
• Suggesting $I^A I^O$ or $I^B I^O$ results in blood group O.

Things to Be Careful About

• The mark scheme accepts $I^o I^o$, but writing $I^O I^O$ is standard. Ensure the superscripts are clear.

Walkthrough

First, determine the genotypes of parents (5) and (6) from the family tree (Fig. 7.1).

• Parent (6) has blood group O, so their genotype must be $I^O I^O$. The gametes they produce will all carry the $I^O$ allele. This is already provided in the top row headers of the Punnett square.
• Parent (5) has blood group A. They could theoretically be $I^A I^A$ or $I^A I^O$. However, parent (5) and (6) have a child (7) with blood group O ($I^O I^O$). Child (7) must have inherited one $I^O$ from each parent. Therefore, parent (5) must carry an $I^O$ allele, making their genotype $I^A I^O$. The row headers in the Punnett square confirm this, showing gametes $I^A$ and $I^O$.

Now, complete the Punnett square by combining the gametes:

• Top-left: $I^A$ (from parent 5) + $I^O$ (from parent 6) = $I^A I^O$
• Top-right: $I^A$ (from parent 5) + $I^O$ (from parent 6) = $I^A I^O$
• Bottom-left: $I^O$ (from parent 5) + $I^O$ (from parent 6) = $I^O I^O$
• Bottom-right: $I^O$ (from parent 5) + $I^O$ (from parent 6) = $I^O I^O$

Key Takeaways

• A child with blood group O ($I^O I^O$) proves that both parents carry at least one $I^O$ allele.
• Punnett squares for ABO blood groups use the same 2x2 grid format as monohybrid crosses, but with three alleles.

Common Mistakes

• Forgetting to fill in the top row headers with $I^O$ and $I^O$ (though the question provides the row headers, candidates sometimes leave the column headers blank or write the wrong allele).
• Writing $I^A I^O$ and $I^O I^A$ as different genotypes; order does not matter in genotype notation.

Things to Be Careful About

• The mark scheme specifically shows the completed table. Ensure all four inner cells are filled.
• Use the exact allele notation ($I^A$, $I^O$) as given in the question.

Walkthrough

The question asks for the probability that the child is male AND has blood group A. These are two independent events.

1. Probability of blood group A: Look at the completed Punnett square. Two of the four boxes contain $I^A I^O$, which results in blood group A. So, the probability of blood group A is $\frac{2}{4} = \frac{1}{2}$.
2. Probability of being male: In humans, the sex of the child is determined by the sperm from the father. Half the sperm carry an X chromosome and half carry a Y chromosome, giving a $\frac{1}{2}$ probability of a male child.
3. Combined probability: Multiply the two independent probabilities: $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.

Key Takeaways

• Blood group and sex are inherited independently.
• To find the probability of two independent events both occurring, multiply their individual probabilities.

Common Mistakes

• Forgetting to include the $\frac{1}{2}$ probability for the child being male. Many candidates just read $\frac{1}{2}$ from the Punnett square and stop.
• Calculating the probability as $\frac{1}{4}$ of the children rather than the probability for this specific child. (Though numerically the same, the reasoning must be clear).

Things to Be Careful About

• The mark scheme accepts $\frac{1}{4}$, 25%, 0.25, "one in four", or 1:3 (odds). Any of these is correct, but $\frac{1}{4}$ or 25% is the standard way to express probability.

Walkthrough

We need to determine the genotype of each person in the family tree and count how many are homozygous (have two identical alleles).

• (1) Female, blood group AB: Genotype must be $I^A I^B$. (Heterozygous)
• (2) Male, blood group A: Genotype is $I^A I^?$ . They have a child (3) with blood group B. Child (3) must be $I^B I^O$ (since parent (1) is $I^A I^B$, child (3) got $I^B$ from (1), so must get $I^O$ from (2)). Thus, parent (2) must carry $I^O$. Genotype of (2) is $I^A I^O$. (Heterozygous)
• (3) Female, blood group B: Got $I^B$ from (1) and $I^O$ from (2). Genotype is $I^B I^O$. (Heterozygous)
• (4) Male, blood group AB: Genotype must be $I^A I^B$. (Heterozygous)
• (5) Female, blood group A: Got $I^A$ from (2) (since (1) is $I^A I^B$, she could get $I^A$ or $I^B$; but she has an O child, so she must have $I^O$). Genotype is $I^A I^O$. (Heterozygous)
• (6) Male, blood group O: Genotype must be $I^O I^O$. (Homozygous)
• (7) Female, blood group O: Genotype must be $I^O I^O$. (Homozygous)
• (8) Male, blood group A: Got $I^O$ from parent (6). Genotype is $I^A I^O$. (Heterozygous)

Counting the homozygous individuals: (6) and (7). Total = 2.

Key Takeaways

• Blood group O is always homozygous ($I^O I^O$).
• You can deduce hidden alleles by looking at the phenotypes of offspring. If a child is homozygous recessive (O), both parents must carry at least one recessive allele.

Common Mistakes

• Assuming blood group A or B individuals are homozygous. Blood group A can be $I^A I^A$ or $I^A I^O$; blood group B can be $I^B I^B$ or $I^B I^O$. You must use the pedigree to decide.
• Forgetting that blood group AB ($I^A I^B$) is heterozygous.

Things to Be Careful About

• The question asks for the number of people, not the genotypes. Ensure you count correctly.
• Double-check your deductions. For example, person (5) is blood group A, but because she has a blood group O child, she must be heterozygous ($I^A I^O$).