An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 (V2) Cambridge IGCSE Biology (0610) paper. Not affiliated with or reproduced from Cambridge.
Paper 22 (Extended Multiple Choice)
Answer all 40 multiple-choice questions. For each question, choose the single best answer option from A, B, C, or D.
40 Question · 40 marks
Question 1 · multiple-choice
1 marks
Which row correctly pairs a tissue in a dicotyledonous leaf with its adaptation for photosynthesis?
A.spongy mesophyll: tightly packed cells to absorb maximum sunlight
B.palisade mesophyll: high concentration of chloroplasts positioned near the upper surface
C.upper epidermis: large intercellular air spaces for rapid diffusion of gases
D.guard cells: complete absence of chloroplasts to allow light through
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Worked solution
Palisade mesophyll cells contain the highest concentration of chloroplasts and are packed closely together near the top surface of the leaf to absorb the maximum amount of light energy for photosynthesis. Spongy mesophyll cells are loosely packed to allow gas diffusion, the upper epidermis is transparent and generally lacks chloroplasts, and guard cells do contain chloroplasts.
Marking scheme
B [1]
Question 2 · multiple-choice
1 marks
What occurs in the skin of a human when the body temperature rises above normal?
A.Arterioles constrict, shunt vessels dilate, and sweat production decreases.
B.Arterioles dilate, shunt vessels constrict, and sweat production increases.
C.Arterioles dilate, shunt vessels dilate, and sweat production increases.
D.Arterioles constrict, shunt vessels constrict, and sweat production decreases.
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Worked solution
When body temperature rises, the thermoregulatory centre initiates vasodilation: arterioles supplying the surface capillary loops dilate (widen), shunt vessels constrict, and more warm blood flows close to the skin surface where heat is lost by radiation and convection. Sweat glands also increase sweat secretion to cool the skin by evaporation.
Marking scheme
B [1]
Question 3 · multiple-choice
1 marks
The statements describe energy flow within a woodland food web:
oak tree \(\rightarrow\) caterpillars \(\rightarrow\) blue tits \(\rightarrow\) sparrowhawks
Which statement about this food chain is correct?
A.Blue tits occupy the primary consumer trophic level.
B.The sparrowhawks receive more total energy than the caterpillars.
C.Energy is transferred between trophic levels by feeding, and energy is lost as heat at each level.
D.Approximately 90% of the energy in the oak trees is successfully transferred to the caterpillars.
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Worked solution
Energy enters ecosystems from sunlight via photosynthesis and is transferred along the food chain through feeding (ingestion). At each trophic level, a large proportion of energy is lost as heat from cellular respiration, excretion, and uneaten matter, meaning only approximately 10% is passed to the next level.
Marking scheme
C [1]
Question 4 · multiple-choice
1 marks
A template strand of DNA has the base sequence:
TAC GGC TTA
What is the base sequence of the mRNA strand transcribed from this DNA template?
A.AUG CCG AAU
B.ATG CCG AAT
C.UAC GGC UUA
D.ATC CCA AAT
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Worked solution
During transcription, complementary base pairing occurs between DNA and RNA nucleotides. In RNA, Uracil (U) pairs with Adenine (A), Adenine (A) pairs with Thymine (T), Guanine (G) pairs with Cytosine (C), and Cytosine (C) pairs with Guanine (G). Therefore, TAC GGC TTA transcribes into AUG CCG AAU.
Marking scheme
A [1]
Question 5 · multiple-choice
1 marks
Which row correctly identifies an enzyme, its substrate, its product, and the organ that secretes it?
A.enzyme: amylase | substrate: starch | product: glucose | organ of secretion: stomach
B.enzyme: protease | substrate: protein | product: amino acids | organ of secretion: pancreas
C.enzyme: lipase | substrate: fatty acids | product: glycerol | organ of secretion: liver
D.enzyme: maltase | substrate: maltose | product: starch | organ of secretion: small intestine
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Worked solution
Protease enzymes (such as trypsin) are secreted by the pancreas into the small intestine where they hydrolyse proteins into peptides and amino acids. Amylase digests starch into maltose (not glucose directly), lipase breaks down lipids into fatty acids and glycerol and is produced by the pancreas (not the liver), and maltase digests maltose into glucose.
Marking scheme
B [1]
Question 6 · multiple-choice
1 marks
Which layer of cells in a dicotyledonous leaf contains the highest concentration of chloroplasts per cell and is the primary site of photosynthesis?
A.Upper epidermis
B.Palisade mesophyll
C.Spongy mesophyll
D.Lower epidermis
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Worked solution
The palisade mesophyll consists of columnar cells packed closely together near the upper surface of the leaf, containing the greatest density of chloroplasts to maximise light absorption for photosynthesis.
Marking scheme
B is correct [1]; palisade mesophyll is the main photosynthetic tissue containing the highest chloroplast density.
Question 7 · multiple-choice
1 marks
What physiological responses occur in the human skin when core body temperature drops below \(37\ ^\circ\text{C}\)?
A.Arterioles near the skin surface constrict, shunt vessels dilate, and sweat secretion decreases
B.Arterioles near the skin surface dilate, shunt vessels constrict, and sweat secretion decreases
C.Arterioles near the skin surface constrict, shunt vessels constrict, and sweat secretion increases
D.Arterioles near the skin surface dilate, shunt vessels dilate, and sweat secretion increases
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Worked solution
When body temperature decreases, arterioles supplying skin surface capillaries constrict (vasoconstriction) and shunt vessels dilate to divert blood away from the skin surface, reducing heat loss by radiation. Sweat gland activity also decreases.
Marking scheme
A is correct [1]; vasoconstriction of surface arterioles, dilation of shunt vessels, and decreased sweat secretion prevent heat loss.
Question 8 · multiple-choice
1 marks
In a marine ecosystem, phytoplankton fix \(50\,000\text{ kJ}\) of energy into biomass. Zooplankton feed on the phytoplankton, and small fish feed on the zooplankton. Assuming an average energy transfer efficiency of \(10\%\) between successive trophic levels, how much energy is transferred into the biomass of the small fish?
A.\(50\text{ kJ}\)
B.\(500\text{ kJ}\)
C.\(5\,000\text{ kJ}\)
D.\(45\,000\text{ kJ}\)
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B is correct [1]; \(50\,000 \times 0.10 \times 0.10 = 500\text{ kJ}\).
Question 9 · multiple-choice
1 marks
A section of the template strand of a DNA molecule has the base sequence: \(\text{TAC-GGC-TTA}\). Which option correctly shows the complementary sequence of bases on the mRNA strand formed during transcription and the number of amino acids it codes for?
A.mRNA sequence: \(\text{AUG-CCG-AAU}\); Number of amino acids: \(3\)
B.mRNA sequence: \(\text{AUG-CCG-AAU}\); Number of amino acids: \(9\)
C.mRNA sequence: \(\text{ATG-CCG-TTA}\); Number of amino acids: \(3\)
D.mRNA sequence: \(\text{UTC-CCG-UUA}\); Number of amino acids: \(9\)
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Worked solution
During transcription, adenine (A) in DNA pairs with uracil (U) in RNA, thymine (T) pairs with adenine (A), and guanine (G) pairs with cytosine (C). Thus, \(\text{TAC-GGC-TTA}\) transcribes into \(\text{AUG-CCG-AAU}\). Each triplet of bases (codon) codes for one amino acid, so \(9\) bases / \(3\text{ bases per codon} = 3\) amino acids.
Marking scheme
A is correct [1]; transcription base pairing gives AUG-CCG-AAU and 3 codons code for 3 amino acids.
Question 10 · multiple-choice
1 marks
Which row correctly identifies an enzyme active in the human duodenum, its specific substrate, and the main product formed?
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Worked solution
Trypsin is a protease secreted by the pancreas that acts in the duodenum, breaking down proteins/polypeptides into peptides. Pepsin acts in the stomach. Amylase breaks starch into maltose. Lipase breaks lipids into fatty acids and glycerol.
Marking scheme
A is correct [1]; trypsin is active in the duodenum and hydrolyses proteins to peptides.
Question 11 · multiple-choice
1 marks
Which row correctly matches a leaf tissue with its adaptation for photosynthesis?
B.spongy mesophyll — tightly packed cells without air spaces to maximize light absorption
C.upper epidermis — dense layer of chloroplast-rich cells to absorb sunlight directly
D.cuticle — cellular layer that actively transports carbon dioxide into the leaf
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Worked solution
Palisade mesophyll cells are elongated and arranged vertically near the upper surface of the leaf, packed with numerous chloroplasts to absorb the maximum amount of light. Spongy mesophyll cells are loosely packed with large air spaces to facilitate gas diffusion. The upper epidermis is transparent and contains very few or no chloroplasts. The cuticle is a non-cellular waxy layer that prevents water loss.
Marking scheme
A is correct [1]; B is incorrect as spongy mesophyll has loose packing with air spaces; C is incorrect as epidermal cells generally lack chloroplasts; D is incorrect as the cuticle is a non-cellular layer that prevents water loss.
Question 12 · multiple-choice
1 marks
What happens in the human skin in response to a decrease in core body temperature?
A.Arterioles supplying skin capillaries dilate and sweat production increases.
B.Arterioles supplying skin capillaries constrict and shunt vessels dilate.
C.Arterioles supplying skin capillaries dilate and shunt vessels constrict.
D.Shunt vessels constrict and skeletal muscles remain completely relaxed.
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Worked solution
When core body temperature falls, vasoconstriction occurs: arterioles supplying the superficial skin capillaries constrict, while shunt vessels dilate. This redirects blood flow through deeper vessels away from the skin surface, reducing heat loss by radiation and convection.
Marking scheme
B is correct [1]; A is incorrect because sweat secretion increases during overheating; C is incorrect because arterioles constrict (not dilate) during cooling; D is incorrect because shivering involves rapid muscle contraction, not relaxation.
Question 13 · multiple-choice
1 marks
Why do food chains in natural ecosystems rarely exceed four or five trophic levels?
A.All ingested energy at lower trophic levels is stored permanently without loss.
B.The total biomass of apex predators always exceeds that of primary producers.
C.Energy is lost at each trophic level via respiration, excretion, and heat.
D.Carnivores at each successive stage convert 100% of consumed food into new biomass.
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Worked solution
Energy is lost at every trophic level due to metabolic processes such as cellular respiration (heat loss), movement, excretion, and unconsumed parts. Typically, only about 10% of energy is transferred to the next level, leaving insufficient energy to sustain viable populations beyond four or five trophic levels.
Marking scheme
C is correct [1]; A is incorrect as energy is lost, not concentrated indefinitely; B is incorrect as biomass generally decreases at higher trophic levels; D is incorrect as energy transfer efficiency is roughly 10%, not 100%.
Question 14 · multiple-choice
1 marks
Which statement correctly describes the role of mRNA during protein synthesis?
A.DNA leaves the nucleus and binds to ribosomes in the cytoplasm.
B.mRNA carries amino acids directly from the cytoplasm into the nucleus.
C.Ribosomes replicate the DNA strand within the nucleus.
D.mRNA carries a copy of the genetic code from the nucleus to the ribosome.
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Worked solution
During protein synthesis, mRNA is formed in the nucleus as a complementary copy of the gene's base sequence (transcription). It then passes through nuclear pores into the cytoplasm, where it binds to a ribosome to direct the assembly of amino acids in a specific sequence (translation).
Marking scheme
D is correct [1]; A is incorrect because DNA remains within the nucleus; B is incorrect because mRNA carries the code from DNA, not amino acids directly; C is incorrect because ribosomes do not synthesize DNA.
Question 15 · multiple-choice
1 marks
Which row correctly identifies the site of production, substrate, and main product of salivary amylase?
A.salivary glands | starch | maltose
B.stomach | protein | amino acids
C.stomach | lipids | fatty acids and glycerol
D.pancreas | lipids | simple sugars
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Worked solution
Salivary amylase is produced and secreted by the salivary glands into the mouth. It catalyzes the breakdown of starch (a complex carbohydrate) into reducing sugars, primarily maltose (a disaccharide), working near a neutral pH.
Marking scheme
A is correct [1]; B describes pepsin/protease in the stomach; C describes lipase (produced by the pancreas, not stomach); D describes trypsin/protease (acting on proteins, not lipids).
Question 16 · multiple-choice
1 marks
Which structural adaptation of a dicotyledonous leaf is correctly matched with its function in photosynthesis?
A.large intercellular air spaces in the spongy mesophyll — allow rapid diffusion of carbon dioxide to photosynthesising cells
B.tightly packed palisade mesophyll cells near the lower epidermis — maximise the absorption of incident sunlight
C.thick transparent waxy cuticle on the upper surface — stores water required for photolysis
D.high density of stomata on the upper epidermis — prevents excessive water loss while admitting light
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Worked solution
Intercellular air spaces in the spongy mesophyll layer create a large surface area and provide pathways for the rapid diffusion of carbon dioxide and oxygen between the stomata and photosynthesising cells.
Marking scheme
A is correct [1]; B is incorrect as palisade mesophyll cells are located near the upper epidermis; C is incorrect as the cuticle reduces water loss rather than storing water; D is incorrect as stomata are predominantly situated on the lower epidermis to minimise excessive transpiration.
Question 17 · multiple-choice
1 marks
A person moves from a warm environment into a cold room. Which physiological responses occur in the skin to help maintain a constant internal body temperature?
A.arterioles supplying surface capillaries dilate and shunt vessels constrict
B.arterioles supplying surface capillaries constrict and shunt vessels dilate
C.sweat glands increase secretion and hair erector muscles relax
D.sweat glands decrease secretion and arterioles supplying surface capillaries dilate
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Worked solution
In response to cold, arterioles supplying surface capillaries constrict (vasoconstriction), reducing blood flow to the skin surface to decrease heat loss by radiation. Concurrently, shunt vessels dilate to allow blood to bypass surface capillaries.
Marking scheme
B is correct [1]; A and C describe responses to warm environments or incorrect vessel actions; D is incorrect because arterioles supplying surface capillaries constrict in the cold rather than dilate.
Question 18 · multiple-choice
1 marks
The following feeding relationship exists in an aquatic ecosystem:
phytoplankton \(\rightarrow\) zooplankton \(\rightarrow\) small fish \(\rightarrow\) squid \(\rightarrow\) seal
Which row correctly identifies the trophic level of the small fish and the organism acting as the primary consumer?
A.trophic level of small fish: 2 ; primary consumer: phytoplankton
B.trophic level of small fish: 3 ; primary consumer: small fish
C.trophic level of small fish: 3 ; primary consumer: zooplankton
D.trophic level of small fish: 4 ; primary consumer: zooplankton
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Worked solution
Phytoplankton occupy trophic level 1 (producers). Zooplankton feed on producers and occupy trophic level 2 (primary consumers). Small fish feed on primary consumers and occupy trophic level 3 (secondary consumers).
Marking scheme
C is correct [1]; A, B, and D assign incorrect trophic levels or confuse the role of the primary consumer.
Question 19 · multiple-choice
1 marks
Which statement correctly describes the relationship between DNA, genes, and proteins?
A.A gene is a sequence of bases in DNA that codes for the sequence of amino acids in a specific protein.
B.A protein is a sequence of nucleotides that codes for the structure of a gene in DNA.
C.A gene is a chain of amino acids that determines the order of bases in a DNA molecule.
D.DNA is a polymer of amino acids that carries instructions for the synthesis of genes.
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Worked solution
A gene is a length of DNA that codes for a specific protein. The sequence of bases in the gene determines the sequence of amino acids assembled during protein synthesis.
Marking scheme
A is correct [1]; B incorrectly states that proteins are made of nucleotides; C incorrectly defines a gene as an amino acid chain; D incorrectly describes DNA as a polymer of amino acids.
Question 20 · multiple-choice
1 marks
An investigation was conducted to study the breakdown of egg white suspension (protein) by an enzyme extracted from the human alimentary canal. Four test-tubes were incubated at \(37\ ^\circ\text{C}\) for 2 hours:
Digestion of egg white occurred only in test-tube 1.
Which enzyme was used and from which organ was it extracted?
A.pepsin from the stomach
B.trypsin from the pancreas
C.amylase from the salivary glands
D.lipase from the pancreas
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Worked solution
Pepsin is a protease that functions in the acidic environment (\(\text{pH } 2\)) of the stomach. Boiled pepsin denatures (test-tube 2), and alkaline conditions inhibit its activity (test-tube 3).
Marking scheme
A is correct [1]; B is incorrect because trypsin operates at an alkaline pH (around pH 8) in the small intestine; C and D are incorrect because amylase digests starch and lipase digests fats.
Question 21 · multiple-choice
1 marks
Which row correctly identifies the adaptations of the palisade mesophyll layer for maximum rate of photosynthesis?
A.tightly packed cells, high density of chloroplasts, positioned near the upper surface
B.loosely arranged cells, low density of chloroplasts, positioned near the lower surface
C.tightly packed cells, low density of chloroplasts, positioned near the lower surface
D.loosely arranged cells, high density of chloroplasts, positioned near the upper surface
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Worked solution
Palisade mesophyll cells are column-shaped and tightly packed near the upper epidermis where sunlight exposure is greatest. They contain the highest concentration of chloroplasts per cell to maximize light absorption for photosynthesis.
Marking scheme
A [1]; 1 mark for identifying the correct combination of palisade mesophyll adaptations.
Question 22 · multiple-choice
1 marks
A person moves from a warm room into cold outdoor air. Which combination of physiological responses in the skin helps reduce heat loss from the body?
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Worked solution
In cold conditions, arterioles supplying skin surface capillaries constrict (vasoconstriction) and shunt vessels dilate, diverting blood away from the skin surface to minimize heat loss by radiation and convection. Sweat gland activity also decreases to minimize evaporative cooling.
Marking scheme
A [1]; 1 mark for correct physiological responses of skin vessels and sweat glands to cold.
Question 23 · multiple-choice
1 marks
The following food chain is found in a grassland habitat: grass \(\rightarrow\) grasshopper \(\rightarrow\) lizard \(\rightarrow\) hawk. If the total energy contained within the grass is \(84\,000\text{ kJ}\), and approximately \(10\%\) of energy is transferred between successive trophic levels, how much energy is transferred to the tertiary consumer?
A.\(84\text{ kJ}\)
B.\(840\text{ kJ}\)
C.\(8\,400\text{ kJ}\)
D.\(8.4\text{ kJ}\)
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A [1]; 1 mark for correct calculation of energy reaching the tertiary consumer.
Question 24 · multiple-choice
1 marks
Which statement correctly describes protein synthesis in a human cell?
A.Transcription occurs in the nucleus to produce mRNA, and translation occurs at the ribosome where amino acids are linked.
B.Transcription occurs at the ribosome to produce tRNA, and translation occurs in the nucleus where DNA is copied.
C.Transcription occurs in the cytoplasm to join amino acids, and translation occurs at the ribosome to make mRNA.
D.Transcription occurs in the nucleus to replicate DNA, and translation occurs in the cytoplasm to break down proteins.
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Worked solution
Protein synthesis consists of two main stages: transcription, which occurs in the nucleus where an mRNA copy of a gene's base sequence is made; and translation, which occurs at the ribosome where tRNA molecules bring specific amino acids that are assembled into a polypeptide chain.
Marking scheme
A [1]; 1 mark for correctly matching transcription (nucleus, mRNA formation) and translation (ribosome, amino acid assembly).
Question 25 · multiple-choice
1 marks
Which row correctly identifies a digestive enzyme, its site of action, and the reaction it catalyses in the human alimentary canal?
A.Enzyme: amylase | Site of action: mouth | Reaction: starch \(\rightarrow\) maltose
B.Enzyme: pepsin | Site of action: duodenum | Reaction: proteins \(\rightarrow\) amino acids
C.Enzyme: trypsin | Site of action: stomach | Reaction: proteins \(\rightarrow\) peptides
D.Enzyme: lipase | Site of action: stomach | Reaction: lipids \(\rightarrow\) fatty acids and glycerol
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Worked solution
Salivary amylase is secreted into the mouth (oral cavity) where it breaks down starch into maltose. Pepsin acts in the stomach (not duodenum), trypsin acts in the duodenum (not stomach), and lipase acts primarily in the small intestine (not stomach).
Marking scheme
A [1]; 1 mark for correct identification of enzyme, site of action, and chemical reaction.
Question 26 · multiple-choice
1 marks
The cross-section of a dicotyledonous leaf contains different specialized cell layers.
Which row correctly describes the adaptations of the palisade mesophyll layer and the spongy mesophyll layer?
A.Palisade mesophyll: contains few chloroplasts per cell; Spongy mesophyll: tightly packed without air spaces
B.Palisade mesophyll: tightly packed with many chloroplasts; Spongy mesophyll: loosely arranged with large air spaces
C.Palisade mesophyll: loosely arranged with large air spaces; Spongy mesophyll: tightly packed with many chloroplasts
D.Palisade mesophyll: contains no chloroplasts; Spongy mesophyll: primary site of light absorption
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Worked solution
Palisade mesophyll cells are vertically elongated, packed closely together, and contain the highest concentration of chloroplasts to maximize light absorption for photosynthesis. Spongy mesophyll cells are loosely packed with large intercellular air spaces to facilitate the diffusion and rapid exchange of carbon dioxide and oxygen between the stomata and the photosynthesizing cells.
Marking scheme
B is correct [1]
A is incorrect: Palisade mesophyll contains a high density of chloroplasts, not low. C is incorrect: Spongy mesophyll has large air spaces, not tightly packed cells. D is incorrect: Reverses the roles and structural characteristics of palisade and spongy layers.
Question 27 · multiple-choice
1 marks
A student runs a race on a hot, sunny day.
Which combination of physiological responses will occur in the skin to help decrease core body temperature?
A.Arterioles near the skin surface constrict; sweat production increases; shunt vessels dilate
B.Arterioles near the skin surface dilate; sweat production decreases; shunt vessels constrict
C.Arterioles near the skin surface dilate; sweat production increases; shunt vessels constrict
D.Arterioles near the skin surface constrict; sweat production decreases; shunt vessels dilate
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Worked solution
During overheating, arterioles supplying the skin surface capillaries dilate (vasodilation) and shunt vessels constrict, increasing blood flow to the skin surface to maximize heat loss by radiation and convection. Sweat glands increase secretion of sweat, which evaporates from the skin surface, absorbing latent heat and cooling the body.
Marking scheme
C is correct [1]
A is incorrect: Arterioles dilate (widen), not constrict, to increase blood flow to surface capillaries. B is incorrect: Sweat production increases, not decreases. D is incorrect: Shunt vessels constrict, directing more blood to surface capillaries, and arterioles dilate.
Question 28 · multiple-choice
1 marks
The diagram represents a food chain in an open ocean ecosystem:
If the phytoplankton fix \( 450\,000\text{ kJ} \) of energy, and there is an efficiency of energy transfer of \( 10\% \) between each consecutive trophic level, how much energy reaches the mackerel?
A.\( 450\text{ kJ} \)
B.\( 4\,500\text{ kJ} \)
C.\( 45\text{ kJ} \)
D.\( 45\,000\text{ kJ} \)
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Therefore, \( 450\text{ kJ} \) of energy is transferred to the mackerel.
Marking scheme
A is correct [1]
B is incorrect: \( 4\,500\text{ kJ} \) is the energy reaching the small fish (trophic level 3). C is incorrect: \( 45\text{ kJ} \) is the energy reaching the tuna (trophic level 5). D is incorrect: \( 45\,000\text{ kJ} \) is the energy reaching the zooplankton (trophic level 2).
Question 29 · multiple-choice
1 marks
Which statement correctly describes the synthesis of a protein from the genetic code stored in DNA?
A.DNA leaves the nucleus and attaches directly to ribosomes to assemble amino acids.
B.mRNA is produced at the ribosome and moves into the nucleus to copy the DNA code.
C.The sequence of bases in mRNA determines the sequence of fatty acids joined at the ribosome.
D.mRNA is transcribed in the nucleus, travels to the cytoplasm, and passes through a ribosome where amino acids are assembled.
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Worked solution
In protein synthesis, the DNA base sequence of a gene is transcribed to form a complementary mRNA molecule inside the nucleus. The mRNA then passes out of the nucleus through a nuclear pore into the cytoplasm, where it binds to a ribosome. The ribosome reads the codons on the mRNA to assemble amino acids in the specific sequence determined by the gene.
Marking scheme
D is correct [1]
A is incorrect: DNA remains in the nucleus and does not move to ribosomes. B is incorrect: mRNA is synthesized in the nucleus, not cytoplasm, and carries genetic information from the nucleus to ribosomes. C is incorrect: Ribosomes assemble amino acids into proteins, not glucose into starch/glycogen.
Question 30 · multiple-choice
1 marks
Which row correctly identifies an enzyme, its site of action in the human alimentary canal, its substrate, and the optimum pH environment for its activity?
A.Enzyme: Amylase | Site of action: Mouth | Substrate: Protein | Optimum pH: Acidic
B.Enzyme: Trypsin | Site of action: Stomach | Substrate: Protein | Optimum pH: Acidic
C.Enzyme: Pepsin | Site of action: Stomach | Substrate: Protein | Optimum pH: Acidic
D.Enzyme: Lipase | Site of action: Small intestine | Substrate: Starch | Optimum pH: Alkaline
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Worked solution
Pepsin is an endopeptidase (protease) secreted in the stomach (gastric juice) that hydrolyses proteins into shorter polypeptide chains. It functions at an acidic optimum pH of approximately pH 1.5–2.0, provided by hydrochloric acid in gastric juice. Amylase acts on starch at neutral/slightly alkaline pH, trypsin acts in the duodenum at alkaline pH (around pH 8), and lipase hydrolyses lipids/fats at alkaline pH in the small intestine.
Marking scheme
C is correct [1]
A is incorrect: Salivary amylase breaks down starch (not protein) and works at neutral pH (around pH 7). B is incorrect: Trypsin functions in the small intestine (duodenum) at an alkaline pH (around pH 8), not acidic. D is incorrect: Lipase breaks down fats into glycerol and fatty acids, not amino acids, and operates at an alkaline pH.
Question 31 · multiple-choice
1 marks
Which row correctly matches a leaf tissue layer with its adaptation for photosynthesis?
A.Upper epidermis: contains the highest concentration of chloroplasts per cell
B.Palisade mesophyll: cells are column-shaped and tightly packed near the upper surface
C.Spongy mesophyll: cells fit tightly together with no intercellular spaces to prevent gas loss
D.Lower epidermis: covered with a thick waxy cuticle to maximise carbon dioxide uptake
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Worked solution
Palisade mesophyll cells are vertically elongated (column-shaped) and closely packed near the upper surface of the leaf where light intensity is highest, and they contain the highest concentration of chloroplasts to maximize light absorption for photosynthesis.
Marking scheme
B [1]
Question 32 · multiple-choice
1 marks
A person exercises vigorously in a warm environment. Which physiological responses occur in their skin to help maintain a constant internal body temperature?
A.Arterioles constrict and sweat gland secretion decreases
B.Arterioles dilate and sweat gland secretion increases
C.Shunt vessels dilate and sweat gland secretion decreases
D.Shunt vessels constrict and sweat gland secretion decreases
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Worked solution
When body temperature rises, thermoreceptors detect the increase and the hypothalamus coordinates a cooling response: arterioles supplying surface capillaries dilate (vasodilation) to increase blood flow to the skin surface for radiant heat loss, and sweat glands secrete more sweat to remove heat energy via evaporative cooling.
Marking scheme
B [1]
Question 33 · multiple-choice
1 marks
A woodland food chain is shown: oak tree \(\rightarrow\) caterpillars \(\rightarrow\) blue tits \(\rightarrow\) sparrowhawk. Which row correctly describes the shape of the pyramid of numbers and the pyramid of biomass for this food chain?
A.Pyramid of numbers: upright; Pyramid of biomass: inverted
B.Pyramid of numbers: narrow base; Pyramid of biomass: upright
C.Pyramid of numbers: upright; Pyramid of biomass: narrow base
D.Pyramid of numbers: narrow base; Pyramid of biomass: inverted
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Worked solution
A single oak tree supports many caterpillars, which in turn support fewer blue tits and an even smaller number of sparrowhawks, giving the pyramid of numbers a narrow base. However, the oak tree has a very large dry mass, so the total biomass decreases at each successive trophic level, resulting in an upright pyramid of biomass with a broad base.
Marking scheme
B [1]
Question 34 · multiple-choice
1 marks
The sequence of bases in a section of a DNA template strand is: TAC CGA TTT GCA. What is the sequence of bases in the mRNA molecule transcribed from this strand?
A.AUG GCU AAA CGU
B.AUG CGA UUU GCA
C.ATG GCT AAA CGT
D.UAC CGA UUU GCA
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Worked solution
During transcription, complementary RNA nucleotides pair with DNA template bases (A pairs with U, T pairs with A, C pairs with G, G pairs with C). Therefore: TAC \(\rightarrow\) AUG, CGA \(\rightarrow\) GCU, TTT \(\rightarrow\) AAA, GCA \(\rightarrow\) CGU.
Marking scheme
A [1]
Question 35 · multiple-choice
1 marks
Which row correctly identifies the site of production, substrate, and optimum pH for the digestive enzyme trypsin?
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Worked solution
Trypsin is a protease secreted by the pancreas into the duodenum. It hydrolyses proteins into peptides in an alkaline environment (optimum pH around 8), maintained by hydrogencarbonate in bile and pancreatic juice.
Marking scheme
B [1]
Question 36 · multiple-choice
1 marks
Which feature of the palisade mesophyll layer is an adaptation to maximise photosynthesis?
A.Cells are loosely packed with large air spaces between them to increase gas diffusion
B.Cells contain a high concentration of chloroplasts and are arranged vertically near the upper surface
C.Cells lack chloroplasts to allow maximum penetration of light to deeper tissues
D.Cells secrete a thick waxy layer that prevents light reflection from the leaf surface
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Worked solution
Palisade mesophyll cells are situated near the upper epidermis, are arranged vertically and closely packed together, and contain a very high density of chloroplasts to absorb the maximum amount of light energy for photosynthesis. Loosely arranged cells with large air spaces is a feature of the spongy mesophyll.
Marking scheme
B ; [1]
Question 37 · multiple-choice
1 marks
What physiological changes occur in the skin when human body temperature falls below the set point?
A.Arterioles supplying skin capillaries dilate and shunt vessels constrict
B.Arterioles supplying skin capillaries constrict and shunt vessels dilate
C.Arterioles supplying skin capillaries constrict and sweat secretion increases
D.Arterioles supplying skin capillaries dilate and sweat secretion decreases
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Worked solution
When internal body temperature drops, the thermoregulatory centre in the brain triggers responses to reduce heat loss: arterioles supplying skin surface capillaries constrict (vasoconstriction) and shunt vessels dilate, redirecting blood flow away from the skin surface to retain heat.
Marking scheme
B ; [1]
Question 38 · multiple-choice
1 marks
Which statement correctly describes the role of mRNA in protein synthesis?
A.It carries a copy of the genetic code from the nucleus to ribosomes in the cytoplasm
B.It remains permanently inside the nucleus to replicate chromosomal DNA
C.It directly joins fatty acids and glycerol together at the cell membrane
D.It forms the permanent structural double helix of genes within the nucleus
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Worked solution
During protein synthesis, mRNA is transcribed from DNA in the nucleus and carries a copy of the genetic code to the ribosomes in the cytoplasm, where translation occurs to assemble amino acids into a polypeptide chain.
Marking scheme
A ; [1]
Question 39 · multiple-choice
1 marks
Which row correctly matches a digestive enzyme to its substrate and main product?
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Worked solution
Maltase is an enzyme that catalyses the breakdown of the disaccharide maltose into the monosaccharide glucose on the outer membranes of epithelial cells in the small intestine. Amylase breaks down starch into maltose, lipase breaks down lipids into fatty acids and glycerol, and protease breaks down proteins into polypeptides and amino acids.
Marking scheme
C ; [1]
Question 40 · multiple-choice
1 marks
A plant shoot is placed horizontally in complete darkness. After three days, the shoot bends upwards. Which statement explains this response?
A.Auxin accumulates on the lower side of the shoot, causing cells on that side to elongate faster
B.Auxin accumulates on the upper side of the shoot, causing cells on that side to elongate faster
C.Auxin is broken down on the lower side of the shoot, stimulating cell division on the upper side
D.Auxin moves away from the lower side, inhibiting cell elongation on the lower side
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Worked solution
Gravity causes auxin to accumulate on the lower side of a horizontally placed shoot. In shoots, a higher concentration of auxin stimulates cell elongation. Consequently, cells on the lower side elongate more rapidly than cells on the upper side, causing the shoot to bend upwards (negative gravitropism).
Marking scheme
A ; [1]
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(ii) State the balanced chemical equation for photosynthesis. .................................................................................................................................. [2]
(b) An investigation was carried out to measure the rate of photosynthesis in an aquatic plant, Elodea canadensis, placed at different distances from a light source. The volume of oxygen released per minute was recorded.
(i) Calculate the percentage decrease in the volume of \(\text{O}_2\) collected per minute when the distance from the lamp is increased from 10 cm to 30 cm.
Show your working. Give your answer to two significant figures.
(ii) Explain why the volume of oxygen collected decreases as the distance from the lamp increases. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. [3]
[Total: 11]
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Worked solution
(a)(i) Layer B is the palisade mesophyll. Adaptations include: cells are packed tightly together vertically to maximize light absorption per unit area; they contain a large number of chloroplasts; they are located near the upper surface to capture maximum light.
(a)(ii) \(6\text{CO}_2 + 6\text{H}_2\text{O} \xrightarrow{\text{light and chlorophyll}} \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\).
(b)(ii) As the plant is moved further away, light intensity decreases. Light provides the energy required for photosynthesis; when light intensity is low, it becomes the limiting factor. Therefore, fewer water molecules are split / less light energy is trapped by chlorophyll, reducing the rate at which oxygen gas is released.
Marking scheme
(a)(i) 1. palisade (mesophyll) ; 2. contain many / large number of / high density of chloroplasts / chlorophyll ; 3. packed tightly together / elongated / column-shaped / arranged vertically ; 4. situated near upper surface (to receive maximum sunlight) ; [max 3]
(b)(i) 1. correct subtraction: \(1.85 - 0.85 = 1.00\) ; 2. dividing change by 1.85: \(\frac{1.00}{1.85} \times 100\) ; 3. 54(%) (correct rounding to 2 sig figs) ; [A 54.1 for 2 marks if rounding incorrect; allow ecf for arithmetic error in step 1] [3]
(b)(ii) 1. light intensity decreases (with distance) / light is limiting factor ; 2. less light (energy) absorbed / trapped by chlorophyll ; 3. lower rate of photosynthesis ; 4. less water split / less photolysis / less \(\text{O}_2\) produced as a byproduct ; [max 3]
Question 2 · structured-theory
11 marks
2 Homeostasis maintains a constant internal environment within the human body.
(a) Define the term negative feedback. .................................................................................................................................. .................................................................................................................................. [2]
(b) Fig. 2.1 shows changes in a person's skin when moving from a warm environment to a cold environment.
(i) Describe the process of vasoconstriction in response to cold temperatures and explain how it helps reduce heat loss. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. [4]
(ii) State two other physiological responses of the body that occur when body temperature falls below the set point of 37 °C. 1. ............................................................................................................................. 2. ............................................................................................................................. [2]
(c) Blood glucose concentration is also controlled by negative feedback.
Explain how the pancreas and liver work together to lower blood glucose concentration when it rises above normal. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. [3]
[Total: 11]
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Worked solution
(a) Negative feedback is a regulatory control mechanism whereby any deviation of a variable away from a normal set point triggers a corrective mechanism that reverses the change, restoring the variable back to the set point.
(b)(i) Vasoconstriction: arterioles supplying skin surface capillaries narrow/constrict due to contraction of smooth muscle in their walls. Shunt vessels open/dilate. As a result, blood flow to surface capillaries decreases, leading to reduced heat loss via radiation and conduction to the external environment.
(b)(ii) 1. Shivering (rapid involuntary muscle contractions producing heat from increased respiration). 2. Contraction of hair erector muscles making hairs stand up to trap an insulating layer of still air.
(c) When blood glucose rises above normal, beta cells in the islets of Langerhans in the pancreas detect the increase and secrete the hormone insulin into the bloodstream. Insulin binds to receptors on liver (and muscle) cells, stimulating them to increase glucose uptake and activate enzymes that convert glucose into insoluble glycogen for storage.
Marking scheme
(a) 1. change / deviation from normal set point / optimum level (is detected) ; 2. stimulates a corrective mechanism / response to return level back to set point / normal ; [2]
(b)(i) 1. arterioles (supplying surface capillaries) constrict / narrow / reduce lumen diameter [R capillaries constrict] ; 2. shunt vessels dilate / widen ; 3. less blood flows through skin surface capillaries / blood diverted away from skin surface ; 4. less heat lost by radiation / conduction / convection ; [4]
(b)(ii) 1. shivering / involuntary muscle contraction (to release heat by respiration) ; 2. hair erector muscles contract / hairs stand on end (trapping insulating air layer) ; 3. decrease in sweat production / sweat glands inactive ; 4. increased metabolic rate / adrenaline secretion ; [max 2]
(c) 1. pancreas / islets of Langerhans detects high glucose and secretes insulin ; 2. insulin transported in blood to liver (and muscle) cells ; 3. liver converts glucose into glycogen (for storage) / increases uptake of glucose ; [3]
Question 3 · structured-theory
11 marks
3 Fig. 3.1 represents a food web in a temperate woodland ecosystem.
Oak tree \(\rightarrow\) Aphids \(\rightarrow\) Ladybirds \(\rightarrow\) Blue tits \(\rightarrow\) Sparrowhawk Oak tree \(\rightarrow\) Caterpillars \(\rightarrow\) Blue tits \(\rightarrow\) Sparrowhawk Oak tree \(\rightarrow\) Wood mice \(\rightarrow\) Owls
(a) (i) Define the term trophic level. .................................................................................................................................. [1]
(ii) State the name of the principal source of energy for this ecosystem. .................................................................................................................................. [1]
(b) In this woodland ecosystem, the total energy contained in the biomass of each trophic level was measured per \(\text{m}^2\) per year.
(c) Explain why energy is lost at each trophic level in a food chain. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. [3]
(d) Explain why food chains rarely contain more than four or five trophic levels. .................................................................................................................................. .................................................................................................................................. [2]
[Total: 11]
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Worked solution
(a)(i) A trophic level is the position an organism occupies in a food chain, food web, pyramid of numbers, or pyramid of biomass. (a)(ii) Sunlight / light energy from the Sun.
(b)(i) \(\text{Efficiency} = \left(\frac{\text{Energy in primary consumers}}{\text{Energy in primary producers}}\right) \times 100 = \left(\frac{2640}{24000}\right) \times 100 = 11\%\).
(b)(ii) \(\text{Efficiency} = \left(\frac{\text{Energy in secondary consumers}}{\text{Energy in primary consumers}}\right) \times 100 = \left(\frac{237.6}{2640}\right) \times 100 = 9.0\%\).
(c) Energy is lost between trophic levels because: organisms lose energy as heat generated during metabolic processes like respiration; not all parts of the organism are eaten (e.g. bones, woody bark, roots); some material consumed is indigestible and passed out as faeces (egestion); energy is lost in excretory products like urea.
(d) Because approximately 90% of energy is lost at each successive trophic level, very little energy reaches the top levels. After 4–5 levels, the remaining energy is too small to support the metabolic and reproductive needs of another viable breeding population.
Marking scheme
(a)(i) position of an organism in a food chain / food web / pyramid of numbers / biomass ; [1] (a)(ii) the Sun / sunlight / solar radiation ; [1]
(c) 1. lost as heat / thermal energy from cellular respiration ; 2. not all parts of the organism are ingested / consumed (e.g. roots, bones, fur) ; 3. not all ingested food is digested / absorbed / lost through egestion / faeces ; 4. lost through excretion / urine / urea / waste products ; [max 3]
(d) 1. large proportion / ~90% of energy is lost at each trophic level / only ~10% transferred ; 2. after 4–5 levels there is not enough energy remaining to support / sustain another trophic level / population ; [2]
Question 4 · structured-theory
11 marks
4 DNA carries the genetic code required to synthesise proteins within cells.
(a) Define the term gene. .................................................................................................................................. .................................................................................................................................. [2]
(b) Fig. 4.1 outlines stages of protein synthesis.
Stage 1: DNA \(\rightarrow\) mRNA (in the nucleus) Stage 2: mRNA \(\rightarrow\) Polypeptide chain (at the ribosome)
(i) State the name of the base present in RNA that replaces thymine (T). .................................................................................................................................. [1]
(ii) Describe the roles of mRNA and ribosomes in the process of protein synthesis. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. [4]
(c) A single base mutation occurs in the DNA template strand of a gene coding for an enzyme, changing the triplet code from CCT to CAT.
Explain how this mutation could result in the loss of enzyme activity. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. [4]
[Total: 11]
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Worked solution
(a) A gene is a length of DNA that codes for a specific protein.
(b)(i) Uracil (U).
(b)(ii) Role of mRNA: It is formed by complementary base pairing against the DNA template in the nucleus (transcription). It carries the genetic instructions from the nucleus through nuclear pores into the cytoplasm. Role of ribosomes: Ribosomes provide the site for translation; they read the sequence of codons on mRNA, allowing specific amino acids to be joined in the correct sequence by peptide bonds to form a polypeptide.
(c) A mutation alters the base triplet from CCT to CAT in DNA, resulting in a different codon on the mRNA strand. During translation, a different amino acid is incorporated into the growing polypeptide chain. This changes the sequence of amino acids (primary structure), which alters hydrogen/ionic/disulfide bonds and changes the 3D tertiary conformation of the enzyme. As a result, the active site is altered in shape and is no longer complementary to the substrate, preventing enzyme-substrate complexes from forming.
Marking scheme
(a) a length of DNA ; that codes for a protein ; [2]
(b)(i) uracil / U ; [1]
(b)(ii) 1. mRNA carries a copy of the gene / genetic code / sequence of bases ; 2. mRNA moves from nucleus to cytoplasm / ribosome ; 3. ribosome attaches to mRNA / reads codons / triplets ; 4. ribosome matches specific amino acids / joins amino acids together in correct order (forming peptide bonds / polypeptide) ; [4]
(c) 1. different mRNA codon produced / codon changes ; 2. different amino acid inserted into polypeptide / sequence of amino acids changes ; 3. alters the folding / tertiary structure / 3D shape of the protein ; 4. active site changes shape / no longer complementary to substrate ; 5. substrate can no longer fit / bind / no enzyme-substrate complex forms ; [max 4]
Question 5 · structured-theory
11 marks
5 Pathogens enter the body and can cause transmissible diseases.
(a) State the definition of a pathogen. .................................................................................................................................. [1]
(b) White blood cells provide defence against pathogens.
(i) Outline the function of phagocytes during an infection. .................................................................................................................................. .................................................................................................................................. [2]
(ii) Describe how lymphocytes respond to foreign antigens to provide active immunity. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. [4]
(c) Fig. 5.1 shows antibody concentration in the blood following a primary vaccination and a secondary booster injection.
[Graph shows: - Initial vaccination at day 0: antibody levels slowly rise to 10 arbitrary units by day 14 and decline by day 28. - Booster injection at day 35: antibody levels rise rapidly to 90 arbitrary units by day 42 and remain elevated.]
Using Fig. 5.1 and your biological knowledge, explain two differences between the primary and secondary immune responses.
(d) Explain why antibiotics are effective against bacterial infections but do not work against viral infections. .................................................................................................................................. .................................................................................................................................. [2]
[Total: 11]
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Worked solution
(a) A pathogen is a disease-causing organism (such as a bacterium, virus, fungus, or protoctist).
(b)(i) Phagocytes detect foreign pathogens, move towards them, engulf them (phagocytosis/endocytosis), and break them down using digestive enzymes stored inside lysosomes.
(b)(ii) When lymphocytes encounter foreign antigens on a pathogen, specific B-lymphocytes with complementary receptors bind to the antigen. They undergo clonal selection and rapid mitosis. These cells differentiate to secrete large numbers of specific antibodies that bind to and neutralise or agglutinate pathogens. Some lymphocytes become memory cells that remain in the bloodstream for long periods, providing long-term active immunity.
(c) 1. The secondary response has a much shorter lag phase / starts much faster because memory cells recognize the antigen immediately. 2. The secondary response produces a significantly higher concentration of antibodies (90 vs 10 arbitrary units) and persists longer in the blood.
(d) Antibiotics work by disrupting specific bacterial metabolic pathways or structures, such as peptidoglycan cell wall synthesis or bacterial 70S ribosomes. Viruses do not possess cell walls, cell membranes, or their own metabolic machinery; instead, they hijack host cellular machinery to replicate inside host cells, where antibiotics cannot affect them.
Marking scheme
(a) a disease-causing organism ; [1]
(b)(i) 1. engulf / ingest pathogens (by phagocytosis) ; 2. digest / destroy pathogens using enzymes (in lysosomes) ; [2]
(b)(ii) 1. recognise / bind to specific (foreign) antigens ; 2. divide by mitosis / clone (to produce plasma cells) ; 3. produce / secrete antibodies ; 4. antibodies have complementary shape to antigen / bind to pathogens ; 5. produce memory cells (for long-term immunity) ; [max 4]
(c) 1. secondary response produces higher concentration of antibodies / higher peak (90 vs 10 a.u.) ; 2. secondary response is faster / shorter delay / steeper rate of increase (due to presence of memory cells) ; 3. antibodies remain in blood for longer duration in secondary response ; [max 2]
(d) 1. antibiotics disrupt bacterial cell structures / cell wall synthesis / bacterial metabolism / ribosomes ; 2. viruses do not have cell walls / do not have metabolism / reproduce inside host cells ; [2]
Question 6 · structured-theory
11 marks
1 Leaves are the primary organs of photosynthesis in terrestrial flowering plants.
(a) State the names of two distinct tissues within a leaf that contain chloroplasts. 1 .................................................................................................................................... 2 .............................................................................................................................. [2]
(b) Describe and explain two structural adaptations of the palisade mesophyll layer that maximize light absorption for photosynthesis. .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .............................................................................................................................. [3]
(c) A student prepared a peel of the lower epidermis of a privet leaf and examined it using a light microscope.
(i) In a circular field of view with an area of \(0.20\text{ mm}^2\), the student counted 34 stomata. Calculate the stomatal density in stomata per \(\text{mm}^2\).
Space for working.
Stomatal density = ................................................. stomata per \(\text{mm}^2\) [1]
(ii) Explain the advantage to a terrestrial plant of having stomata located mainly on the lower epidermis rather than the upper epidermis. [2] .................................................................................................................................... ....................................................................................................................................
(d) Explain how guard cells control the opening of stomata in bright daylight. [3] .................................................................................................................................... .................................................................................................................................... ....................................................................................................................................
[Total: 11]
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Worked solution
(a) Any two from: palisade mesophyll, spongy mesophyll, guard cells.
(b) The palisade mesophyll is situated directly beneath the upper epidermis, positioning it close to the light source. The cells are columnar (elongated vertically) and closely packed, ensuring minimal light passes through unabsorbed. Furthermore, each cell contains a high number of chloroplasts which can migrate within the cytoplasm to optimize light capture.
(c)(i) Stomatal density = \(\frac{\text{number of stomata}}{\text{area}} = \frac{34}{0.20} = 170\text{ stomata per mm}^2\).
(c)(ii) The lower surface of the leaf is shaded from direct sunlight and experiences lower temperatures and reduced air movement compared to the upper surface. Placing stomata on the lower surface reduces the rate of transpiration and prevents excessive water loss.
(d) In daylight, guard cells accumulate solutes, lowering their water potential. Water enters the guard cells from surrounding epidermal cells by osmosis. The guard cells become turgid. Because the inner cell wall adjacent to the stomatal pore is thicker and less elastic than the outer wall, the cells curve outward as they swell, widening the stomatal pore.
Marking scheme
(a) Any two from: - palisade mesophyll ; - spongy mesophyll ; - guard cells ; [max 2]
(b) Any three from: - positioned near / directly below upper epidermis (to receive maximum sunlight) ; - columnar / elongated cells arranged vertically (increases pathway of light through cell) ; - tightly packed / no large intercellular spaces (to maximize light absorption per unit area) ; - contain many / high density of chloroplasts ; - chloroplasts can move / relocate towards light ; [max 3]
(c)(i) 170 ; [1]
(c)(ii) Any two from: - lower surface is shaded / out of direct sunlight ; - lower surface is cooler / experiences less heating ; - reduces rate of transpiration / evaporation / water loss / diffusion of water vapour ; - protects plant from wilting / water stress ; [max 2]
(d) Any three from: - water enters guard cells by osmosis ; - guard cells become turgid / swell ; - inner wall is thicker / less flexible / less elastic than outer wall ; - guard cells bend / curve / bow outwards (opening the pore) ; [max 3]
Question 7 · structured-theory
11 marks
2 Homeostasis is the maintenance of a constant internal environment in the human body.
(a) Define the term homeostasis. [2] .................................................................................................................................... ....................................................................................................................................
(b) Blood glucose concentration is maintained within narrow limits by the endocrine system.
(i) Explain how the pancreas and the liver respond to an increase in blood glucose concentration above normal levels. [4] .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... ....................................................................................................................................
(ii) A healthy volunteer had a fasting blood glucose concentration of \(4.5\text{ mmol dm}^{-3}\). Forty minutes after drinking a glucose solution, their blood glucose concentration rose to \(7.2\text{ mmol dm}^{-3}\).
Calculate the percentage increase in blood glucose concentration.
(c) Explain the importance of negative feedback in the control of blood glucose concentration. [3] .................................................................................................................................... .................................................................................................................................... ....................................................................................................................................
[Total: 11]
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Worked solution
(a) Homeostasis is defined as the maintenance of a constant internal environment within set limits.
(b)(i) When blood glucose concentration rises above normal, beta cells in the pancreas detect the increase and secrete the hormone insulin into the bloodstream. Insulin travels to target organs, predominantly the liver (and skeletal muscle). Insulin stimulates liver cells to increase uptake of glucose from the blood and convert excess glucose into the insoluble storage polysaccharide glycogen (glycogenesis). As a result, blood glucose concentration decreases back to normal.
(c) Negative feedback is a regulatory mechanism where a change away from a normal set point triggers a corrective physiological response in the opposite direction. If blood glucose becomes too high, insulin brings it down; if it becomes too low, glucagon brings it up. This prevents harmful extremes: high glucose causing water loss from cells by osmosis, or low glucose depriving brain and body cells of substrate for aerobic respiration.
Marking scheme
(a) maintenance of a constant internal environment ; within set limits / set point ; [2]
(b)(i) Any four from: - pancreas / islets / beta cells detect high blood glucose concentration ; - pancreas secretes / releases insulin ; - insulin travels in the blood / plasma ; - liver (or muscle) cells absorb / take up more glucose ; - glucose converted to glycogen ; - rate of respiration / storage of glucose increases ; - blood glucose concentration decreases / returns to set point ; [max 4]
(c) Any three from: - deviation / change from normal level / set point is detected ; - triggers corrective action / response in opposite direction ; - prevents blood glucose becoming dangerously high / hyperosmotic conditions / water drawn out of body cells by osmosis ; - prevents blood glucose becoming dangerously low / ensures constant supply of glucose for cell respiration / prevents brain damage / coma ; [max 3]
Paper 62 (Alternative to Practical)
Answer all practical-based structured tasks, including graphing, scientific drawing, data tabulation, and experimental design.
2 Question · 40 marks
Question 1 · alternative-to-practical
20 marks
1 A student investigated the effect of sodium hydrogencarbonate concentration (a source of dissolved carbon dioxide) on the rate of photosynthesis in spinach leaf discs.
The student used the following method: • A cork borer was used to cut 50 circular discs from fresh spinach leaves. • The discs were placed inside a syringe filled with sodium hydrogencarbonate solution, and the plunger was pulled while blocking the tip to create a vacuum. This removed trapped air from the spongy mesophyll, causing all discs to sink to the bottom. • Five test-tubes were set up containing \(20\text{ cm}^3\) of different concentrations of sodium hydrogencarbonate (\(\text{NaHCO}_3\)) solution: \(0.2\%\), \(0.4\%\), \(0.6\%\), \(0.8\%\), and \(1.0\%\). • 10 sunken leaf discs were placed into each test-tube. • A lamp was positioned \(15\text{ cm}\) away from the test-tubes to illuminate the discs. • A stopwatch was started, and the time taken for 5 leaf discs (50%) to float to the surface was recorded in seconds. • The experiment was repeated two more times for each concentration.
Table 1.1 shows the student's results.
Table 1.1 concentration of \(\text{NaHCO}_3\) / %time taken for 5 discs to float in trial 1 / stime taken for 5 discs to float in trial 2 / stime taken for 5 discs to float in trial 3 / smean time taken / s0.22152282172200.41641581581600.61231181191200.8949092............1.0787473............ (a) (i) Complete Table 1.1 by calculating the mean time taken for \(0.8\%\) and \(1.0\%\) \(\text{NaHCO}_3\) solutions. [2]
(ii) State the independent variable and the dependent variable in this investigation. Independent variable: .................................................................................................... Dependent variable: ...................................................................................................... [2]
(iii) State two variables that were kept constant in this investigation. 1 ..................................................................................................................................... 2 ..................................................................................................................................... [2]
(iv) Explain why the leaf discs float to the surface as photosynthesis occurs. [1]
(v) Identify one source of error in this experimental method and suggest an improvement to overcome it. Source of error: ............................................................................................................. Improvement: ............................................................................................................... [2]
(b) The rate of photosynthesis can be estimated using the equation: \[\text{rate of photosynthesis} = \frac{1000}{\text{mean time taken / s}}\] Calculate the rate of photosynthesis for the \(0.4\%\) \(\text{NaHCO}_3\) solution. Give your answer to two decimal places.
Space for working.
................................................................................... arbitrary units [2]
(c) Aquatic plants, such as Cabomba, produce bubbles of gas during photosynthesis.
Plan an investigation to determine the effect of light intensity on the rate of photosynthesis in an aquatic plant.
In your answer, you should include: • how you will change the independent variable • how you will measure the dependent variable • key variables you will keep constant and how you will do this • a suitable control experiment • how you will ensure the reliability and accuracy of your results • any relevant safety precautions. [6]
(d) Describe the steps required to safely test a leaf for the presence of starch and state the result for a leaf that has been photosynthesising. [3]
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(a) (ii) Independent variable: concentration of sodium hydrogencarbonate (\(\text{NaHCO}_3\)) solution. Dependent variable: time taken for 5 leaf discs to float to the surface.
(a) (iii) Any two from: distance of the lamp from the tubes (light intensity), temperature, volume of solution (\(20\text{ cm}^3\)), number of leaf discs per tube (10), diameter/surface area/type of leaf discs.
(a) (iv) Oxygen is produced as a byproduct of photosynthesis, which fills the air spaces (spongy mesophyll) of the leaf disc, making it less dense than the surrounding solution and causing it to float.
(a) (v) Source of error: Leaf discs may bump into each other or stick to the tube walls / heat from lamp may raise temperature / difficulty observing the exact moment 5 discs float. Improvement: Space discs out or use wider tubes / use an LED lamp or heat shield / use video recording with a timer to review exact floating time.
(c) • Independent variable: Change the distance between the lamp and the plant (at least 5 distances, e.g. 10, 20, 30, 40, 50 cm). • Dependent variable: Count the number of bubbles released per minute / collect gas in an inverted measuring cylinder or gas syringe over a fixed time period. • Controlled variables: Use a water bath / heat shield to keep temperature constant; add a fixed concentration of sodium hydrogencarbonate solution (\(0.5\%\)) to ensure carbon dioxide is not limiting; use the same piece/length of aquatic plant. • Reliability: Repeat the measurement 3 times at each distance and calculate a mean. • Safety: Use dry hands when operating electrical equipment / ensure no water spills near the lamp.
(d) 1. Boil leaf in water for 1 minute to break cell membranes. 2. Place leaf in a boiling tube of ethanol and heat in a hot water bath (do not use a Bunsen flame directly as ethanol is flammable) to extract chlorophyll. 3. Dip leaf in warm water to soften it. 4. Spread on a white tile and add iodine solution. 5. A colour change from yellow-brown / orange to blue-black confirms the presence of starch.
Marking scheme
(a) (i) 92 ; 75 ; [2]
(a) (ii) concentration of sodium hydrogencarbonate / \(\text{NaHCO}_3\) (solution) ; time taken (for 5 leaf discs) to float ; [2]
(a) (iii) Any two from: • distance of lamp / light intensity ; • temperature (of solution/water) ; • volume of solution / \(20\text{ cm}^3\) ; • number of leaf discs / 10 ; • size / area / diameter / species of leaf discs ; [2]
(a) (iv) oxygen (gas) produced (by photosynthesis) / gas fills air spaces / decreases density (of leaf disc) ; [1]
(a) (v) error: discs sticking together / sticking to glass wall / heat from lamp affecting rate / difficulty timing 5 discs simultaneously ; improvement: tap tube gently / use wider container / use water shield or LED lamp / record with video and playback ; [2]
(b) \(1000 / 160\) ; \(6.25\) ; [2]
(c) Marking points (max 6): 1. IV: change distance of light source (minimum 5 specified distances) / use bulbs of different power (wattage) ; 2. DV: count number of bubbles in a stated time period (e.g. per minute) / measure volume of gas collected in syringe / measuring cylinder ; 3. CV: keep temperature constant using a water bath / heat filter / LED lamp ; 4. CV: constant concentration of sodium hydrogencarbonate / constant length/mass/species of plant ; 5. Reliability: repeat at each light intensity / distance at least 3 times and calculate a mean ; 6. Equilibration: allow plant to acclimatise for 2–5 minutes before recording ; 7. Safety: keep water away from electrical lamp / handle hot water bath with tongs / care with glassware ; [max 6]
(d) 1. boil leaf in water (to kill cells / soften tissue) ; 2. boil in ethanol using a water bath (to remove chlorophyll / decolourise leaf) ; 3. safety: use a water bath / turn off Bunsen burner because ethanol is flammable ; 4. add iodine solution AND positive result is blue-black / negative is yellow-brown ; [max 3]
Question 2 · alternative-to-practical
20 marks
2Fig. 2.1 is a photomicrograph of a cross-section of a dicotyledonous leaf.
Fig. 2.1 [Photomicrograph showing upper cuticle, upper epidermis, tightly packed palisade mesophyll cells, loosely arranged spongy mesophyll cells with air spaces, vascular bundle with xylem and phloem, lower epidermis with stomata, and a line XY drawn vertically through the whole leaf blade] (a) (i) Make a large, clear biological drawing of the tissue layers shown in Fig. 2.1. Do not draw individual cells. Label the palisade mesophyll and the vascular bundle. [5]
(ii) Line XY on Fig. 2.1 represents the thickness of the leaf blade. The length of line XY on Fig. 2.1 is \(68\text{ mm}\). The actual thickness of the leaf blade is \(0.40\text{ mm}\).
Calculate the magnification of Fig. 2.1 using the formula: \[\text{magnification} = \frac{\text{length of line } XY}{\text{actual thickness of leaf blade}}\] Give your answer to the nearest whole number.
(b) A student investigated the rate of water loss from leaves taken from the same plant. Four identical leaves, A, B, C, and D, were treated as follows: • Leaf A: no petroleum jelly applied (control) • Leaf B: petroleum jelly applied to the lower surface only • Leaf C: petroleum jelly applied to the upper surface only • Leaf D: petroleum jelly applied to both upper and lower surfaces.
Petroleum jelly is impermeable to water. The stalk of each leaf was sealed with petroleum jelly. The mass of each leaf was measured every hour for 5 hours. The total mass lost by each leaf was calculated.
Table 2.1 shows the results.
Table 2.1 time / hoursmass lost by Leaf A / gmass lost by Leaf B / gmass lost by Leaf C / gmass lost by Leaf D / g00.000.000.000.0010.320.080.260.0220.610.150.500.0430.880.220.720.0541.140.280.930.0651.380.341.120.07 (i) On the grid, plot a line graph of the mass lost against time for Leaf A and Leaf B. Use a ruler to join the points with straight lines. Label both lines clearly or include a key. [4]
(ii) Describe the trend shown by the data for Leaf A. [2]
(iii) Compare the total mass lost by Leaf B with that lost by Leaf C after 5 hours and explain the difference in terms of leaf structure. [2]
(c) (i) Plant leaves store carbohydrates produced during photosynthesis as starch, but also contain reducing sugars. Describe how you would test a liquid extract prepared from leaf tissue for the presence of reducing sugars. Include the expected positive colour change. [3]
(ii) State one safety precaution needed when carrying out the test for reducing sugars. [1]
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Worked solution
(a) (i) Drawing assessment points: 1. Line quality: Single, unbroken, sharp pencil lines without feathering or overlapping. 2. Size: Drawing fills more than \(50\%\) of the provided space. 3. Detail & layers: Accurate representation of all tissue layers in plan view (upper epidermis, palisade mesophyll layer, spongy mesophyll layer, lower epidermis, vascular bundle/vein). No individual cells drawn. 4. Proportions: Palisade layer depth proportional to total leaf blade thickness. 5. Labels: Palisade mesophyll and vascular bundle correctly identified with straight, non-intersecting label lines.
(a) (ii) \[\text{Magnification} = \frac{\text{length of line } XY}{\text{actual thickness}} = \frac{68\text{ mm}}{0.40\text{ mm}} = 170\] Magnification = \(\times 170\).
(b) (i) Graph plotting: • Axes: \(x\)-axis: time / hours; \(y\)-axis: mass lost / g (or mass lost by leaf / g). • Scale: Linear, suitable intervals (e.g. \(x\)-axis: 1 unit = 2 cm, \(y\)-axis: 0.2 g = 2 cm), covering more than half the grid area. • Plotting: All 6 points for Leaf A \((0,0), (1,0.32), (2,0.61), (3,0.88), (4,1.14), (5,1.38)\) and 6 points for Leaf B \((0,0), (1,0.08), (2,0.15), (3,0.22), (4,0.28), (5,0.34)\) plotted accurately to within half a small square. • Lines: Sharp, straight lines connecting points with a ruler (point-to-point), and lines clearly labelled 'Leaf A' and 'Leaf B' (or key provided).
(b) (ii) As time increases, the mass lost increases continually throughout the 5 hours. The rate of mass loss is highest in the first hour (\(0.32\text{ g/h}\)) and gradually decreases slightly over time.
(b) (iii) Leaf B lost significantly less mass (\(0.34\text{ g}\)) than Leaf C (\(1.12\text{ g}\)) because stomata are primarily located on the lower surface/epidermis of the leaf. Covering the lower surface with petroleum jelly blocks most stomata and prevents water vapour from diffusing out.
(c) (i) • Add an equal volume of Benedict's solution/reagent to the leaf extract. • Heat the mixture in a thermostatically controlled hot water bath (above \(80\text{ }^\circ\text{C}\)) for 3–5 minutes. • An initial blue colour changing to green, yellow, orange, or brick-red precipitate indicates the presence of reducing sugars.
(c) (ii) Wear safety goggles to protect eyes from hot liquid spitting / use a test-tube holder or rack to handle hot glassware.
Marking scheme
(a) (i) 1. O (outline): single, clear, continuous lines with no shading / sketchy lines ; 2. S (size): occupies at least 50% of the available space ; 3. D (detail): correct tissue layers drawn (upper epidermis, palisade mesophyll, spongy mesophyll, lower epidermis, vascular bundle) AND no single cells drawn ; 4. P (proportion): palisade layer drawn deeper than upper epidermis and roughly 1/3 to 1/2 of leaf thickness ; 5. L (labels): palisade mesophyll AND vascular bundle correctly labelled with ruled lines ; [5]
(a) (ii) measurement \(68\text{ mm}\) (\(\pm 1\text{ mm}\)) / \(6.8\text{ cm}\) ; correct formula: \(68 / 0.40\) (or equivalent with consistent units) ; \(\times 170\) (accept \(\times 168\) to \(\times 173\) if measurement was \(67\)–\(69\text{ mm}\)) ; [3]
(b) (i) 1. A (axes): correct labels with units (time / hours AND mass lost / g) ; 2. S (scale): linear scales, filling \(>50\%\) of grid in both directions ; 3. P (plotting): all points for Leaf A and Leaf B plotted accurately within \(\pm 0.5\) small square ; 4. L (lines/key): points connected with neat ruled straight lines point-to-point AND lines labelled / key given ; [4]
(b) (ii) mass lost increases as time increases / direct relationship ; rate of water loss is constant / highest in first hour / decreases slightly over time ; [2]
(b) (iii) Leaf B lost less mass than Leaf C / data quoted (\(0.34\text{ g}\) vs \(1.12\text{ g}\)) ; more stomata on lower surface (epidermis) than upper surface / lower surface has higher stomatal density ; [2]
(c) (i) add Benedict's solution / reagent ; heat / place in a hot water bath (at least \(80\text{ }^\circ\text{C}\) / boiling water) ; colour change from blue to green / yellow / orange / brick-red (precipitate) ; [3]
(c) (ii) use a water bath (not naked flame) / wear safety goggles / handle hot tubes with test-tube holder / tongs ; [1]
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