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Palisade mesophyll cells are located near the upper epidermis. They are columnar (cylindrical) in shape, closely packed together with minimal intercellular space, and contain a high density of chloroplasts to maximize light absorption for photosynthesis. In contrast, spongy mesophyll cells are loosely packed, more spherical in shape, and have fewer chloroplasts.

1 mark for choosing option B.

When body temperature rises, the body aims to lose excess heat through radiation and convection. To achieve this, the arterioles supplying the capillaries near the skin surface vasodilate (widen) to increase blood flow near the surface. Simultaneously, shunt vessels constrict (narrow) to redirect more blood into these surface capillaries rather than bypassing them.

1 mark for choosing option B.

Let's trace the food chains to determine the trophic levels:
1. Phytoplankton (Producer) \(\rightarrow\) Zooplankton (Primary consumer) \(\rightarrow\) Large fish (Secondary consumer).
2. Phytoplankton (Producer) \(\rightarrow\) Zooplankton (Primary consumer) \(\rightarrow\) Small fish (Secondary consumer) \(\rightarrow\) Large fish (Tertiary consumer).
Therefore, the large fish acts as both a secondary and a tertiary consumer.

1 mark for choosing option C.

The correct sequence is:
1. Transcription occurs first in the nucleus, where mRNA is synthesized from the DNA template (Event 2).
2. The mRNA molecule then leaves the nucleus through a nuclear pore (Event 4).
3. In the cytoplasm, the mRNA binds to a ribosome (Event 1).
4. Translation occurs as tRNA molecules carry specific amino acids to the ribosome, where they are linked together in the correct sequence to form a polypeptide (Event 3).
Thus, the correct sequence is 2 \(\rightarrow\) 4 \(\rightarrow\) 1 \(\rightarrow\) 3.

1 mark for choosing option B.

Let's review each option:
- Amylase breaks down starch into maltose (not glucose directly).
- Pepsin breaks down protein into smaller peptides (not immediately into free amino acids).
- Lipase acts in the duodenum (small intestine) to digest lipids (fats and oils) into fatty acids and glycerol. This is correct.
- Maltase breaks down maltose into glucose (not sucrose).

1 mark for choosing option C.

Passive immunity is a short-term, temporary defence against a pathogen using antibodies that were produced by another individual (e.g., from mother to baby via breast milk or placenta, or via an injection of antitoxins/antibodies). Memory cells are not produced, and the host's own lymphocytes do not actively synthesize these antibodies.

1 mark for choosing option C.

Auxin is synthesized in the shoot tip and diffuses downwards. When exposed to unilateral light, auxin moves away from the light and accumulates on the shaded side of the shoot. In shoots, a higher concentration of auxin stimulates cell elongation. Consequently, the cells on the shaded side elongate more rapidly than those on the illuminated side, causing the shoot to bend towards the light source (positive phototropism).

1 mark for choosing option A.

During anaerobic respiration:
- In yeast (alcoholic fermentation), glucose is broken down to release ethanol and carbon dioxide (no water is formed).
- In human muscle cells (lactic acid fermentation), glucose is broken down to produce lactic acid only (no carbon dioxide or water is formed).
Therefore, Row B is correct.

1 mark for choosing option B.

Guard cells have a characteristic structure where the inner cell wall (adjacent to the pore) is thicker and less elastic than the outer wall. This asymmetry causes the cells to curve outward when they absorb water and become turgid, opening the stoma. Palisade mesophyll cells are closely packed and column-shaped to absorb light, whereas spongy mesophyll cells are loosely packed with air spaces to facilitate gas diffusion. Upper epidermal cells do not contain chloroplasts (except for guard cells), as their primary role is protection and allowing light to pass through.

Correct option selected: 1 mark. Other options are incorrect due to mismatched structural adaptations and functions.

When the body is exposed to a hot environment, it needs to increase heat loss. Arterioles supplying the skin capillaries dilate (vasodilation) to increase blood flow near the skin surface, while shunt vessels constrict to redirect blood into these superficial capillaries. Simultaneously, sweat glands increase sweat production; evaporation of this sweat removes heat from the skin surface.

Correct option selected: 1 mark. Other options describe incorrect vascular or glandular responses to a hot environment.

The pyramid of numbers represents the absolute number of individual organisms at each trophic level. Since there is only 1 oak tree, but thousands of caterpillars, the base is narrow while the primary consumer level is very wide. Thus, the pyramid of numbers is spindle-shaped (narrow at the base, widening in the middle, and narrowing at the top). The pyramid of biomass represents the dry mass of living material at each level; because energy is lost at each transfer, biomass always decreases up the food chain, resulting in a standard pyramid shape (broadest at the base, tapering to the top).

Correct option selected: 1 mark. Other options incorrectly represent either the pyramid of numbers or the pyramid of biomass.

Protein synthesis begins in the nucleus where the DNA of a gene unwinds and is transcribed to make mRNA (2). The mRNA then leaves the nucleus through a nuclear pore into the cytoplasm (4). In the cytoplasm, the mRNA attaches to and passes through a ribosome (1). Finally, the ribosome translates the mRNA sequence, assembling amino acids in the correct order to form a protein (3).

Correct option selected: 1 mark. Other sequences represent incorrect chronological steps in transcription and translation.

Lipase is synthesized by the pancreas and secreted into the small intestine (specifically the duodenum), where it breaks down lipids into fatty acids and glycerol. Amylase produced by the pancreas acts in the small intestine, not the mouth. Maltase is found on the membranes of the epithelium of the small intestine, not salivary glands. Pepsin is produced by the stomach wall and acts in the stomach, not the pancreas.

Correct option selected: 1 mark. Other options misidentify either the source, site of action, or products.

Active immunity occurs when the body produces its own antibodies and memory cells. It is 'artificial' when it is induced medically rather than naturally. Vaccination with a weakened pathogen represents artificial active immunity (C). A and B describe passive immunity (where antibodies are received from an external source). D describes natural active immunity.

Correct option selected: 1 mark. Other options represent natural passive (A), artificial passive (B), or natural active (D) immunity.

Auxin has opposite effects on shoots and roots. In shoots, a higher concentration of auxin stimulates cell elongation; the lower side grows faster, causing the shoot to bend upwards (negative gravitropism). In roots, a higher concentration of auxin inhibits cell elongation; the upper side grows faster than the lower side, causing the root to bend downwards (positive gravitropism).

Correct option selected: 1 mark. Other options represent incorrect effects of auxin on cell elongation in roots or shoots.

During ventricular systole, the pressure in the ventricles increases above the atrial pressure. This forces the atrioventricular valves to close, preventing blood from flowing back into the atria. At the same time, this high pressure forces the semilunar (aortic and pulmonary) valves to open, allowing blood to flow into the arteries.

Correct option selected: 1 mark. Other combinations of valve states are incorrect for ventricular contraction.

Palisade mesophyll cells are located near the upper surface of the leaf, are columnar in shape, are tightly packed, and contain many chloroplasts for photosynthesis. They also possess a cell wall. Spongy mesophyll cells also contain chloroplasts, guard cells contain chloroplasts and have cell walls, and upper epidermal cells typically lack chloroplasts.

1 mark for the correct option A.
- A is correct because palisade mesophyll cells have a cell wall, contain many chloroplasts, and are closely packed columnar cells.
- B is incorrect because spongy mesophyll cells contain chloroplasts.
- C is incorrect because guard cells possess a cell wall.
- D is incorrect because upper epidermal cells lack chloroplasts.

In a hot environment, body temperature must be regulated. The arterioles supplying the capillaries near the skin surface dilate (vasodilation). This increases blood flow near the skin surface, allowing more heat to be lost by radiation and convection. Thus, skin arterioles dilate, and the rate of heat loss from the skin surface increases.

1 mark for the correct option D.
- D is correct because moving into a warm environment triggers vasodilation (dilation of skin arterioles) to increase blood flow near the surface, thereby increasing the rate of heat loss.
- A, B and C are incorrect because they specify either constriction (which decreases heat loss in the cold) or a mismatch between vessel diameter and heat loss.

In food chain 1, the small fish is a secondary consumer (third trophic level) because it eats Daphnia (a primary consumer). In food chain 2, the small fish eats dragonfly nymphs (secondary consumers), making it a tertiary consumer (fourth trophic level). Therefore, the small fish occupies both the third and fourth trophic levels.

1 mark for the correct option C.
- C is correct because in food chain 1, small fish are secondary consumers (3rd trophic level), and in food chain 2, they are tertiary consumers (4th trophic level).
- A is incorrect because algae are producers.
- B is incorrect because dragonfly nymphs are secondary consumers (3rd trophic level).
- D is incorrect because kingfishers are tertiary/quaternary consumers.

Transcription is the process by which an mRNA molecule is made in the nucleus from a DNA template. Translation is the process where ribosomes in the cytoplasm use the mRNA sequence to assemble amino acids into a protein.

1 mark for the correct option A.
- A is correct because transcription occurs in the nucleus and synthesizes mRNA from a DNA template, whereas translation occurs at the ribosomes to build a protein from the mRNA template.
- B, C, and D are incorrect because they swap the locations, products, or definitions of transcription and translation.

Pepsin is a protease that digests protein (albumin) into smaller peptides. It requires acidic conditions (provided by hydrochloric acid) to function efficiently. In Test-tube 1, active pepsin and hydrochloric acid are present, leading to rapid digestion of the albumin, turning the cloudy suspension clear. In Test-tube 2, water is used instead of acid, meaning the pH is not optimal for pepsin. In Test-tube 3, the boiled pepsin is denatured and inactive. In Test-tube 4, no enzyme is present.

1 mark for the correct option A.
- A is correct because pepsin is active only in acidic conditions. Test-tube 1 contains both active pepsin and hydrochloric acid.
- Test-tube 2 lacks acid (water is neutral).
- Test-tube 3 has boiled (denatured) pepsin.
- Test-tube 4 has no enzyme.

Active immunity involves the production of antibodies by the person's own immune system, resulting in long-term protection and the formation of memory cells. Passive immunity involves receiving antibodies from an external source (such as across the placenta or through breast milk), providing temporary, short-term protection with no memory cell production.

1 mark for the correct option A.
- A is correct because active immunity involves the host body producing its own antibodies, while passive immunity involves acquiring antibodies from an external source.
- B is incorrect because active immunity produces memory cells, not passive.
- C is incorrect because active immunity is long-term, not passive.
- D is incorrect because passive immunity is acquired via transferred antibodies (e.g. placenta) and active via infection or vaccine.

In a shoot exposed to unidirectional light, auxin diffuses away from the light and accumulates on the shaded side. Auxin stimulates cell elongation in shoots. The increased elongation on the shaded side causes the shoot to bend towards the light source.

1 mark for the correct option B.
- B is correct because auxin accumulates on the shaded side of a plant shoot and stimulates cell elongation there, causing unequal growth and bending towards the light.
- A, C, and D are incorrect because they fail to identify the shaded side as having higher auxin concentration or incorrectly describe the effect of auxin on shoot cell growth.

Osmosis is the net movement of water molecules from a region of higher water potential to a region of lower water potential down a water potential gradient through a partially permeable membrane. It is a passive transport process and does not require energy released by respiration. Active transport, cell division, protein synthesis, and nerve impulse transmission all require energy from respiration.

1 mark for the correct option A.
- A is correct because osmosis is a passive process that relies on a water potential gradient and does not require energy from respiration.
- B, C, and D are incorrect because active transport, transmission of nerve impulses, and chemical synthesis (polypeptide assembly) are active biological processes requiring ATP/respiration energy.

The palisade mesophyll contains the highest concentration of chloroplasts to absorb maximum light energy near the upper surface of the leaf. Sugars (specifically sucrose) are transported throughout the plant via the phloem tissue.

Award 1 mark for selecting the correct option (A).
- Reject other options because xylem transports water/mineral ions, and spongy mesophyll contains fewer chloroplasts than palisade mesophyll.

To lose heat and cool down, the arterioles supplying skin capillaries dilate (vasodilation) to bring more warm blood to the surface. Sweat glands increase secretion, cooling the skin as sweat evaporates. Hair erector muscles relax, allowing hairs to lie flat and avoiding the trapping of an insulating layer of warm air.

Award 1 mark for the correct combination of homeostasis responses (B).
- Incorrect options confuse vasodilation with vasoconstriction, or incorrectly describe sweating/hair erector muscle states during overheating.

Zooplankton are preyed upon by both beetle larvae and small fish. If beetle larvae are removed from the food web, there is a reduction in predation pressure on the zooplankton, which leads to an initial increase in their population size.

Award 1 mark for the correct prediction of population change based on predator-prey dynamics (A).
- B is incorrect because small fish lost a food source, so their population would not increase.
- C is incorrect because beetle larvae do not eat small fish.
- D is incorrect because an increase in zooplankton would lead to a decrease, not an increase, in phytoplankton.

In protein synthesis, mRNA is first transcribed from DNA in the nucleus. The mRNA then moves out of the nucleus to a ribosome in the cytoplasm. The ribosome reads the sequence of codons on the mRNA, and tRNA molecules bring the corresponding amino acids to the ribosome to build the polypeptide chain.

Award 1 mark for the correct transcription and translation sequence (B).
- Reject A because DNA replication is not a step in normal protein synthesis.
- Reject C because proteins are not synthesized in the nucleus and carried by mRNA.
- Reject D because tRNA does not copy DNA.

Amylase is secreted by salivary glands into the mouth, and by the pancreas into the duodenum of the small intestine. Amylase breaks down starch into the disaccharide maltose. Maltose is later broken down into glucose by maltase.

Award 1 mark for selecting B.
- A and C are incorrect because amylase is denatured in the acidic conditions of the stomach.
- D is incorrect because amylase is active in both the mouth and small intestine, and its direct product is maltose, not glucose.

Active immunity is acquired through exposure to a pathogen (or its antigens in a vaccine), causing the body to produce its own antibodies. Memory cells persist in the lymphatic system and rapidly produce large quantities of antibodies if the pathogen is encountered again.

Award 1 mark for B.
- Reject A because injecting antibodies provides passive immunity, and memory cells do not produce antigens.
- Reject C because passive immunity is acquired through breast milk, and memory cells do not perform phagocytosis.
- Reject D because memory cells do not release digestive enzymes to destroy pathogens directly; they produce antibodies.

Gravity causes auxin to accumulate on the lower side of the horizontally placed shoot tip. Because high concentration of auxin stimulates cell elongation in shoots, cells on the lower side elongate more than those on the upper side, causing the shoot to bend and grow upwards (negative gravitropism).

Award 1 mark for A.
- Reject B because auxin accumulates on the lower side due to gravity, and shoots show negative gravitropism.
- Reject C because shoots grow upwards, not downwards.
- Reject D because gravity causes an unequal distribution of auxin even in uniform light.

Aerobic respiration completely breaks down glucose, releasing a relatively large amount of energy. Anaerobic respiration (alcohol fermentation) in yeast only partially breaks down glucose, releasing a much smaller amount of energy per glucose molecule.

Award 1 mark for C.
- Reject A because anaerobic respiration in yeast (fermentation) does produce carbon dioxide along with ethanol.
- Reject B because anaerobic respiration in yeast produces alcohol, but aerobic respiration does not produce lactic acid.
- Reject D because aerobic respiration requires oxygen, whereas anaerobic respiration occurs in its absence.

Palisade mesophyll cells are column-shaped and closely packed together near the upper surface of the leaf. They contain many chloroplasts to absorb as much light as possible for photosynthesis. Air spaces are characteristic of the spongy mesophyll, the waxy layer is the cuticle, and xylem vessels (not palisade cells) are specialised for water transport and lack cell walls and nuclei.

1 mark for selecting B, which correctly links the physical layout and abundance of chloroplasts in palisade cells with light absorption.

In a cold environment, the body acts to conserve heat. Vasoconstriction occurs, where arterioles supplying skin capillaries constrict to reduce blood flow near the skin surface, reducing heat loss by radiation. Additionally, sweat glands secrete less sweat to minimise heat loss through evaporation.

1 mark for selecting A, identifying both vasoconstriction and decreased sweat secretion as responses to cold.

To calculate the percentage efficiency of energy transfer from grasshopper to frog: \(\frac{\text{Energy in frog}}{\text{Energy in grasshopper}} \times 100 = \frac{80}{1000} \times 100 = 8.0\%\).

1 mark for the correct calculation leading to choice B.

A gene is defined as a length of DNA that codes for a specific protein. DNA itself is double-stranded and made of nucleotides, whereas proteins are macromolecules made of amino acid chains.

1 mark for selecting C, which contains the correct biological definition of a gene in relation to DNA and proteins.

Amylase is produced by the salivary glands (and pancreas). Its substrate is starch, which it breaks down into maltose. Maltase is the enzyme that subsequently breaks maltose down into glucose, while pepsin acts in the stomach on proteins, and lipase acts on lipids.

1 mark for selecting A, identifying the correct digestive enzyme properties.

Active immunity is the defense against a pathogen by antibody production inside the individual's own body, which leads to the formation of memory cells and long-term protection. Passive immunity involves receiving pre-made antibodies from another individual or source, providing only temporary, short-term protection with no memory cell production.

1 mark for selecting B, which accurately highlights how antibodies are sourced in active versus passive immunity.

Auxin is synthesized in the shoot tip and diffuses away from light, leading to a higher concentration of auxin on the shaded side of the shoot. In stems, auxin stimulates cell elongation, meaning cells on the shaded side grow longer than those on the illuminated side, causing the shoot to bend towards the light source.

1 mark for selecting B, correctly identifying auxin accumulation on the shaded side and its stimulatory effect on cell elongation in shoots.

Plant palisade mesophyll cells have a cellulose cell wall, chloroplasts, and a large central vacuole, which are absent in human red blood cells. Red blood cells also lack a nucleus (unlike the mesophyll cell), but cell membrane and cytoplasm are present in both cell types.

1 mark for choosing A, identifying the trio of structures unique to the plant mesophyll cell relative to the mammalian red blood cell.

(a) Palisade mesophyll cells are adapted by being elongated (columnar) and packed tightly together near the upper surface to maximize light capture. They contain a very high concentration of chloroplasts, which are located near the edges of the cell to minimize the diffusion distance for carbon dioxide and optimize light absorption.

(b) The spongy mesophyll has loosely packed cells with large air spaces between them. This allows carbon dioxide to diffuse rapidly from the stomata to all mesophyll cells and oxygen to diffuse out. Additionally, the cells have thin, moist cell walls, allowing gases to dissolve before diffusing into or out of the cells.

(c) Restricting stomata to the lower epidermis reduces water loss via transpiration. The lower surface is shaded from direct sunlight and is cooler than the upper surface, which decreases the rate of evaporation of water from the spongy mesophyll.

(d) Phloem tissue transports sugars (sucrose) away from the leaf. Sieve tube elements are adapted by having sieve plates with pores to allow continuous flow of cell sap, and very little cytoplasm with no nucleus to minimize resistance to flow. Companion cells, located adjacent to the sieve tubes, contain many mitochondria to provide the ATP required for active loading of sugars into the phloem.

(a) [Max 3 marks]
- Elongated / columnar cell shape [1]
- Cells packed closely/tightly together [1]
- Arranged vertically/at right angles to the upper surface [1]
- Large numbers of / many chloroplasts [1]
- Chloroplasts move / are located near the cell wall / periphery [1]

(b) [Max 3 marks]
- Cells are loosely packed / irregular in shape [1]
- Large intercellular air spaces / air gaps [1]
- Facilitates rapid diffusion of carbon dioxide and oxygen [1]
- Thin / moist cell walls allow gases to dissolve [1]

(c) [Max 2 marks]
- Reduces water loss / rate of transpiration [1]
- Lower surface is cooler / shaded from direct sunlight / less air movement [1]
- Less evaporation of water occurs [1]

(d) [Max 3 marks]
- Phloem (sieve tubes / companion cells) [1]
- Sieve tube elements have little cytoplasm / no nucleus / no ribosomes to allow free flow [1]
- Sieve plates with pores allow transport of organic solutes [1]
- Companion cells contain many mitochondria to provide energy / ATP for active loading [1]

(a) Homeostasis is defined as the maintenance of a constant internal environment within restricted limits.

(b) (i) Vasoconstriction involves the narrowing (constriction) of arterioles supplying skin capillaries. This reduces the volume of blood flowing through the capillaries near the surface of the skin, meaning less heat is lost to the environment by radiation, conduction, or convection.
(ii) Shivering is the rapid, involuntary contraction of skeletal muscles. This muscle activity requires a high rate of aerobic respiration, which releases heat energy as a metabolic by-product, raising the core body temperature.

(c) (i) The hypothalamus contains thermoreceptors that directly monitor the temperature of the blood flowing through it. It also receives electrical impulses via sensory neurones from temperature receptors located in the skin.
(ii) Negative feedback is a control system where a change in a parameter (e.g., a drop in temperature) triggers actions that counteract or reverse the change, bringing the parameter back to its optimum set point. Once the set point is reached, the corrective mechanisms are switched off.

(a) [Max 2 marks]
- Maintenance of a constant / stable internal environment [1]
- Within narrow limits / set point [1]

(b) (i) [Max 3 marks]
- Arterioles in the skin constrict / narrow [1] (reject: capillaries constrict)
- Shunt vessels dilate [1]
- Less blood flows through capillaries near the skin surface [1]
- Reduces heat loss by radiation / conduction / convection [1]

(b) (ii) [Max 2 marks]
- Rapid / involuntary contraction of skeletal muscles [1]
- Increases rate of respiration [1]
- Respiration releases heat energy [1]

(c) (i) [Max 2 marks]
- Thermoreceptors in the hypothalamus monitor blood temperature [1]
- Receives nerve impulses from temperature receptors in the skin [1]

(c) (ii) [Max 2 marks]
- A deviation from the set point / normal temperature triggers a corrective mechanism [1]
- The response opposes / reverses the direction of the change [1]
- Returns temperature to normal and switches off the corrective action [1]

(a) A producer is an organism that makes its own organic nutrients, usually using energy from sunlight through photosynthesis.

(b) (i) Atlantic puffins occupy the fourth trophic level (tertiary consumers).
(ii) Marine phytoplankton → zooplankton → herring → Atlantic puffins (or harbor seals).

(c) Energy transfer is inefficient because energy is lost as heat to the environment during respiration, through movement, and via excretion or egested waste (faeces). Also, not all parts of the zooplankton are eaten or digested by the herring. This continuous loss of energy means that less energy is available at each successive trophic level. Eventually, after 4 or 5 levels, there is insufficient energy remaining to support a viable population at a higher level.

(d) (i) The zooplankton population will increase because there are fewer herring to prey on them, leading to a decreased mortality rate.
(ii) The harbor seal population will decrease because herring is a major food source. With fewer prey available, seals will experience food shortages, leading to increased starvation or forced emigration, unless they can find alternative prey.

(a) [Max 1 mark]
- An organism that makes its own organic nutrients / food using light energy / through photosynthesis [1]

(b) (i) [Max 1 mark]
- Tertiary consumer / fourth trophic level [1]

(b) (ii) [Max 1 mark]
- Marine phytoplankton → zooplankton → herring → Atlantic puffins / harbor seals [1]
(Accept: correct written chain with arrows representing energy transfer)

(c) [Max 4 marks]
- Energy lost as heat from respiration [1]
- Energy lost through movement / metabolic processes [1]
- Energy lost in undigested material / faeces / excretion [1]
- Not all parts of organisms are eaten [1]
- Only about 10% of energy is transferred to the next level [1]
- Insufficient energy remains at the top of the food chain to support another level [1]

(d) (i) [Max 2 marks]
- Zooplankton population increases [1]
- Due to reduced predation / fewer herring eating them [1]

(d) (ii) [Max 2 marks]
- Harbor seal population decreases [1]
- Due to loss of a food source / starvation / competition for other prey [1]

(a) (i) A gene is a length of DNA that codes for a specific protein.
(ii) An allele is an alternative / different version of a gene.

(b) In the nucleus, the DNA double helix unwinds, and a complementary copy of the gene's sequence is made in the form of mRNA (transcription). The mRNA molecule is small enough to leave the nucleus via a nuclear pore and enters the cytoplasm, where it binds to a ribosome (the site of protein synthesis). The ribosome reads the mRNA bases in triplets (codons). Transfer RNA (tRNA) molecules carry specific amino acids to the ribosome, matching their anticodons to the mRNA codons. The amino acids are linked together in a specific sequence by peptide bonds to form a polypeptide chain, which then folds into a functional protein.

(c) A change in the DNA base sequence (a mutation) alters the triplet code on the gene. This leads to a different codon sequence on the mRNA, which can cause a different amino acid to be inserted into the polypeptide chain. Changing even one amino acid can alter the final 3D shape / folding of the protein. If the protein is an enzyme, this change may alter the shape of its active site so that the substrate can no longer bind to it, rendering it non-functional.

(a) (i) [Max 2 marks]
- Length of DNA [1]
- That codes for a specific protein [1]

(a) (ii) [Max 1 mark]
- Alternative / different version of a gene [1]

(b) [Max 5 marks]
- Complementary copy of gene / DNA is made as mRNA [1]
- mRNA molecule leaves the nucleus [1]
- mRNA travels to / binds to a ribosome in cytoplasm [1]
- Ribosome is the site of protein synthesis [1]
- Ribosome reads mRNA sequence in triplets / codons [1]
- tRNA brings specific amino acids to the ribosome [1]
- Amino acids join together to form a protein / polypeptide [1]

(c) [Max 3 marks]
- Mutation / change in DNA base sequence changes the mRNA codons [1]
- Results in a different amino acid sequence in the polypeptide [1]
- Alters the folding / 3D shape of the protein [1]
- Active site of enzyme changes shape / substrate can no longer fit [1]

(a) (i) The protease enzyme in the stomach is pepsin, and its optimum pH is acidic (usually around pH 1.5 to 2.0).
(ii) The hydrochloric acid in the stomach provides the optimum acidic pH for pepsin to function efficiently. Additionally, the acid kills ingested bacteria and other pathogens by denaturing their enzymes, acting as a critical non-specific body defense.

(b) (i) Bile contains bile salts which emulsify fats. This means they break down large fat globules into many tiny droplets, significantly increasing the surface area for lipase enzymes to act upon. Bile is also alkaline, so it neutralizes the acidic mixture (chyme) coming from the stomach, creating the optimum alkaline conditions for pancreatic lipase to work.
(ii) The products of lipid digestion are fatty acids and glycerol.

(c) Chemical digestion is essential because food consists of large, insoluble molecules that cannot pass through cell membranes. Chemical digestion uses enzymes to break these down into small, soluble molecules that can be absorbed through the wall of the small intestine (villi) into the bloodstream.

(a) (i) [Max 2 marks]
- Pepsin [1]
- pH 1.5 to 2.5 / acidic [1]

(a) (ii) [Max 3 marks]
- Hydrochloric acid lowers pH / creates acidic conditions [1]
- Denatures proteins in pathogens / kills bacteria [1]
- Activates pepsinogen / provides optimum pH for pepsin [1]

(b) (i) [Max 3 marks]
- Bile emulsifies lipids / fats [1]
- Breaks large globules into small droplets [1]
- Increases surface area for lipase action [1]
- Neutralizes stomach acid / chyme [1]
- Provides optimum pH for pancreatic enzymes [1]

(b) (ii) [Max 1 mark]
- Fatty acids and glycerol [1] (both required)

(c) [Max 2 marks]
- Large, insoluble molecules cannot be absorbed [1]
- Must be broken down into small, soluble molecules [1]
- To cross the lining of the gut / enter capillaries [1]

(a) A pathogen is a disease-causing organism (such as a bacterium or virus), whereas a transmissible disease is an illness/disease in which the pathogen can be passed from one host to another.

(b) A vaccine contains weakened, harmless, or dead forms of a pathogen (or its isolated antigens). When injected or ingested, the antigens trigger an immune response. Lymphocytes with complementary receptors recognize these antigens and produce specific antibodies to destroy them. Crucially, some of these lymphocytes divide to become memory cells. If the active pathogen enters the body in the future, these memory cells recognize it immediately and rapidly produce massive quantities of antibodies, destroying the pathogen before symptoms develop.

(c) (i) Passive immunity is acquired when the infant receives pre-formed, ready-made antibodies from another individual, in this case, directly from the mother through colostrum or breast milk.
(ii) Passive immunity is temporary because the infant's own lymphocytes are not activated by antigens, meaning no memory cells are produced. Over time, the maternal antibodies are naturally broken down, metabolized, and cleared from the infant's circulation, leaving them without long-term protection.

(a) [Max 2 marks]
- Pathogen: a disease-causing organism [1]
- Transmissible disease: a disease whose pathogen can be passed/transmitted from one host to another [1]

(b) [Max 4 marks]
- Vaccine contains harmless / dead / weakened pathogens / antigens [1]
- Antigens trigger an immune response [1]
- Lymphocytes produce specific antibodies [1]
- Memory cells are produced [1]
- Future infection triggers rapid / large-scale antibody production [1]

(c) (i) [Max 2 marks]
- Direct transfer of antibodies [1]
- From mother to infant via breast milk / placenta [1]
- No contact with pathogen / antigen required [1]

(c) (ii) [Max 3 marks]
- No active immune response in the infant [1]
- No memory cells are produced [1]
- Maternal antibodies are foreign and are eventually broken down / cleared [1]

(a) Phototropism is a growth response in which the direction of plant growth is determined by the direction of light.

(b) Auxin is synthesized in the tip of the plant shoot and diffuses down the stem. When the shoot is exposed to unilateral light (light from one side), auxin migrates horizontally away from the light, accumulating on the shaded side. In shoots, a higher concentration of auxin stimulates cell elongation. Consequently, the cells on the shaded side elongate much more than those on the illuminated side, causing the shoot to bend and grow towards the light source.

(c) (i) Plant roots are positively gravitropic, meaning they grow downwards in the direction of gravity. In contrast, plant shoots are negatively gravitropic and grow upwards, away from the force of gravity.
(ii) Growing downwards ensures that the roots penetrate deep into the soil. This provides robust anchorage for the plant, keeping it stable. It also places the roots in closer contact with deeper soil layers containing water and dissolved mineral ions, which are vital for plant survival.

(a) [Max 1 mark]
- A growth response where direction of growth is determined by the direction of light [1]

(b) [Max 5 marks]
- Auxin is made in the shoot tip [1]
- Auxin diffuses downwards [1]
- Unidirectional light causes auxin to move to the shaded side [1]
- Higher concentration of auxin on the shaded side [1]
- Auxin stimulates cell elongation (in shoots) [1]
- Shaded side grows / elongates faster than the lit side [1]
- Causes the shoot to bend towards the light [1]

(c) (i) [Max 2 marks]
- Roots grow downwards / towards gravity / are positively gravitropic [1]
- Shoots grow upwards / away from gravity / are negatively gravitropic [1]

(c) (ii) [Max 3 marks]
- Anchors the plant firmly in the soil [1]
- Allows roots to reach water deep underground [1]
- Allows roots to absorb mineral ions / nutrients [1]

PART (a)
(i) Table of results:
| Leaf | Count 1 | Count 2 | Count 3 | Count 4 | Count 5 | Mean Count (per \(0.25 \text{ mm}^2\)) |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| A | 18 | 19 | 21 | 17 | 20 | 19.0 |
| B | 22 | 19 | 20 | 21 | 18 | 20.0 |
| C | 17 | 18 | 19 | 16 | 20 | 18.0 |
Overall Mean Count = 19.0

(ii) Mean stomata per \(\text{mm}^2\):
\(\text{Mean count in } 0.25 \text{ mm}^2 = 19.0\)
\(\text{Mean count per } 1 \text{ mm}^2 = \frac{19.0}{0.25} = 76\) stomata per \(\text{mm}^2\).

(iii) The student repeated the counts five times for each leaf and used three different leaves (replicates).

PART (b)
(i) Prevents / reduces excessive water loss by transpiration since the upper surface is exposed to direct sunlight and higher temperatures.
(ii) Drawing criteria: Large size, clear single lines without shading, two bean-shaped guard cells showing a central opening (stoma) with four surrounding epidermal cells correctly drawn. Label lines to 'guard cell' and 'stoma'.

PART (c)
- Independent variable: The surface of the leaf that is sealed/exposed (e.g., Upper surface sealed with Vaseline vs Lower surface sealed with Vaseline vs Both sealed vs Neither sealed).
- Dependent variable: Change in mass of the leafy shoot (or percentage mass loss) after a set period of time.
- Control variables: Temperature, light intensity, wind speed, relative humidity, starting mass/surface area of the shoots, duration of the experiment.
- Procedure: Select four similar leafy shoots from the same plant. Treat them as follows: Shoot 1: coat upper surface of all leaves with Vaseline; Shoot 2: coat lower surface with Vaseline; Shoot 3: coat both surfaces (control); Shoot 4: coat neither surface (control). Place each shoot in a small flask of water, seal the top of the flask with plasticine to prevent evaporation of water from the flask itself. Measure the initial mass of each setup using a balance. Leave all four setups under the same environmental conditions (e.g., constant temperature and light) for 24 hours. Re-weigh each setup and calculate the percentage mass loss: \\frac{\\text{initial mass} - \\text{final mass}}{\\text{initial mass}} \\times 100\). Repeat the entire experiment three times to calculate average values and ensure reliability.

PART (a)(i) [5 marks]
1. Table drawn with clear borders and appropriate headers with units (e.g., Leaf, Count 1 to 5, Mean count per \(0.25 \text{ mm}^2\)).
2. All raw counts correctly entered into the table.
3. Mean calculated correctly for Leaf A (19.0), B (20.0), and C (18.0).
4. Overall mean count for the lower epidermis calculated correctly (19.0).
5. All calculated mean values presented to 1 decimal place consistently.

PART (a)(ii) [2 marks]
1. Correct working shown: \(19.0 / 0.25\) or \(19.0 \times 4\).
2. Correct final answer: \(76\) (accept answer based on candidate's mean count from (a)(i)).

PART (a)(iii) [1 mark]
1. Multiple counts (five replicates) per leaf OR using three different leaves.

PART (b)(i) [1 mark]
1. Reduces water loss / reduces rate of transpiration (do not accept 'prevents water loss completely').

PART (b)(ii) [5 marks]
1. Drawing size: occupies at least half of the provided space.
2. Line quality: clean, continuous lines drawn with a sharp pencil, absolutely no shading or sketchy lines.
3. Accuracy of cells: two distinct bean-shaped guard cells forming a stoma, surrounded by four adjacent epidermal cells of appropriate shape.
4. Labelling: straight, uncrossed label lines pointing directly to a guard cell and the stoma.
5. Correct labels: 'guard cell' and 'stoma' / 'stomatal pore'.

PART (c) [6 marks]
1. Independent variable: covering different surfaces (upper surface vs lower surface) with waterproof barrier / petroleum jelly / Vaseline.
2. Dependent variable: measuring mass loss / percentage mass loss over time using an electronic balance (or timing color change of cobalt chloride paper).
3. Control variables (at least two): same temperature, same humidity, same light intensity, same wind speed, same leaf species / surface area.
4. Experimental setup description: seal cut stems to prevent water loss from stem / use oil/plasticine on water reservoir to prevent evaporation from container.
5. Duration and process: leave setups for a specified, realistic time period (e.g., 24 hours) under identical conditions, then re-weigh.
6. Replication: repeat the experiment at least three times to obtain mean values and ensure reliability.

PART (a)
(i) Rates of reaction:
- pH 1.5: \(1000 / 85 = 11.76\)
- pH 2.5: \(1000 / 140 = 7.14\)
- pH 3.5: \(1000 / 320 = 3.13\)
- pH 4.5: \(1000 / 680 = 1.47\)

(ii) Graph plotting:
- X-axis label: pH
- Y-axis label: Rate of reaction / arbitrary units
- Scale: Linear, covering more than 50% of the grid (e.g., x-axis from 0 to 6; y-axis from 0 to 12).
- Points: Plotted accurately as small crosses at (1.5, 11.76), (2.5, 7.14), (3.5, 3.13), and (4.5, 1.47).
- Line: Points connected with thin, clean, ruled straight lines or a smooth line of best fit. No extrapolation past 4.5 or below 1.5.

(iii) Estimation: At pH 2.0, the rate of reaction is approximately \(9.45\) (accept range from \(9.0\) to \(9.8\) depending on candidate's plotted line).
Description: Draw a vertical line from pH 2.0 on the x-axis to meet the curve/line, then read horizontally across to the y-axis to find the rate of reaction value.

(iv) Pepsin is a stomach enzyme with an optimum pH in highly acidic conditions (pH 1.5–2.0). At pH 5.5, the environment is too alkaline, causing the pepsin protein to denature. The specific shape of its active site is altered, so it is no longer complementary to the albumin substrate. As a result, enzyme-substrate complexes cannot form, and the albumin remains undigested (suspension remains cloudy).

PART (b)
(i) Independent variable: pH
Dependent variable: Time taken for the suspension to clear (or rate of reaction).

(ii) Control variables:
1. Temperature of the reaction mixture: control by placing all tubes in a thermostatically controlled water bath set to \(37^\circ\text{C}\).
2. Concentration and volume of pepsin: control by measuring exactly the same volume (e.g., \(2\text{ cm}^3\)) of the same stock pepsin solution using a graduated pipette or syringe.

PART (c)
- Source of error: Judging the \"completely clear\" end-point is subjective and can vary between runs, causing inconsistency.
- Improvement: Use a colorimeter to measure the light transmission through the test tubes at fixed intervals (e.g., every 10 seconds), or place a black card with a bold white cross behind the test tube and stop the timer when the cross becomes clearly visible.

PART (a)(i) [3 marks]
1. Correct formula application shown for at least one calculation.
2. All four rates calculated correctly: pH 1.5 = 11.76, pH 2.5 = 7.14, pH 3.5 = 3.13, pH 4.5 = 1.47.
3. All values rounded correctly to 2 decimal places.

PART (a)(ii) [5 marks]
1. Axes labelled correctly with units where appropriate (x-axis: 'pH'; y-axis: 'Rate of reaction / arbitrary units' or 'Rate of reaction / \(\text{s}^{-1}\)').
2. Linear scale chosen so that the plotted points occupy more than half of the grid in both directions.
3. All four calculated points plotted accurately within half a small square.
4. Plotted points joined with a thin, sharp, ruled line or smooth curve.
5. No extrapolation beyond the plotted points (no line drawn to 0 or 5.5).

PART (a)(iii) [2 marks]
1. Correct rate value read from candidate's graph at pH 2.0 (approx. \(9.0 - 9.8\)).
2. Description of drawing a vertical line from 2.0 on x-axis to the plotted line and reading across to the y-axis.

PART (a)(iv) [2 marks]
1. Enzyme / pepsin denatured at pH 5.5.
2. Shape of active site is altered / no longer complementary to substrate (egg white/albumin) so no enzyme-substrate complexes can form.

PART (b)(i) [2 marks]
1. Independent variable: pH.
2. Dependent variable: Time taken for egg white to clear / rate of reaction.

PART (b)(ii) [4 marks]
1. Control variable 1: Temperature AND controlled using a water bath / thermostatically controlled water bath.
2. Control variable 2: Volume/concentration of pepsin OR volume/concentration of egg white suspension AND controlled using a syringe / pipette / measuring cylinder.
(Award 2 marks for each complete pair of variable + control method)

PART (c) [2 marks]
1. Source of error: Subjective determination of the end point / difficult to judge when completely clear.
2. Improvement: Use of a colorimeter to measure light transmittance / use a black cross behind the tube and time until it is visible.