2023 Cambridge IGCSE Biology (0610) Practice Paper with Answers

Stomach protease (pepsin) is adapted to the highly acidic conditions of gastric juice, meaning its optimum pH is around 2.0. Additionally, since it is a human enzyme, its optimum temperature is body temperature, which is approximately 37 °C. Temperatures as high as 80 °C will denature the enzyme, rendering it inactive.

1 mark for selecting pH 2.0 and 37 °C as the optimum conditions for a human gastric enzyme.

Passive immunity involves receiving antibodies from an external source rather than producing them. Natural passive immunity occurs when these antibodies are transferred naturally, such as from mother to fetus across the placenta or through colostrum/breast milk. Option B is artificial active immunity, C is natural active immunity, and D is artificial passive immunity.

1 mark for identifying the transfer of maternal antibodies via the placenta as passive natural immunity.

According to the concept of limiting factors, the rate of photosynthesis is limited by the factor that is in the shortest supply. Since temperature, water, and carbon dioxide levels are maintained at optimum or high levels, the reduced sunlight on a cloudy day makes light intensity the limiting factor.

1 mark for identifying light intensity as the limiting factor under the described conditions.

Xerophytes have specific adaptations to conserve water: sunken stomata trap moist air near the leaf surface to reduce diffusion; rolled leaves limit the exposure of stomata to air currents; a thick waxy cuticle prevents cuticular transpiration.

1 mark for recognizing sunken stomata, rolled leaves, and a thick cuticle as xerophytic adaptations.

The electrical impulse in a reflex arc begins at the receptor (which detects the stimulus). It travels along the sensory neurone to the central nervous system (spinal cord), crosses a synapse to the relay neurone, then crosses another synapse to the motor neurone, which carries the impulse to the effector (muscle) to carry out the response.

1 mark for the correct order of components in a reflex arc.

A concentrated salt solution has a lower water potential than the cytoplasm of the plant cells. Consequently, water moves out of the cells by osmosis down the water potential gradient, causing the cytoplasm to pull away from the cell wall, resulting in plasmolysis.

1 mark for identifying plasmolysis and the outward net movement of water.

Restriction endonucleases are used to cut the DNA of the human chromosome and the bacterial plasmid, creating complementary sticky ends. DNA ligase is then used to join the sugar-phosphate backbones of the cut gene and plasmid together to form recombinant DNA.

1 mark for identifying DNA ligase as the joining enzyme in genetic modification.

Fertilizer runoff causes an algal bloom at the surface of the pond. This bloom blocks sunlight from reaching deeper aquatic plants, which then die because they cannot photosynthesise. Decomposers (bacteria) multiply rapidly as they feed on the dead plant matter, and their aerobic respiration consumes the dissolved oxygen in the water, causing fish to suffocate and die.

1 mark for the correct chronological sequence of eutrophication events.

An increase in pH from 2.0 to 7.0 (away from its optimum pH) will denature the stomach enzyme (pepsin), which decreases its activity. Decreasing the temperature from \(37\ ^\circ\text{C}\) to \(20\ ^\circ\text{C}\) reduces the kinetic energy of the enzyme and substrate molecules, leading to fewer effective collisions and a slower reaction rate. An increase in enzyme concentration increases the rate of reaction (assuming substrate is available). Thus, only changes 1 and 2 cause the rate of reaction to decrease.

A is the correct answer. 1 mark for identifying that both denaturation due to pH change and decreased kinetic energy due to cooling reduce the reaction rate.

Enzymes are proteins that act as biological catalysts. They speed up chemical reactions and remain chemically unchanged at the end of the reaction, meaning they can be reused.

C is the correct answer. 1 mark for recognizing that enzymes are unchanged biological catalysts.

Vaccination containing weakened pathogens introduces antigens into the body, stimulating the child's own immune system to produce antibodies and memory cells. This is a form of active immunity. The cells responsible for synthesizing and releasing these antibodies are lymphocytes.

A is the correct answer. 1 mark for correctly pairing active immunity with lymphocyte antibody production.

At high light intensity and constant temperature, carbon dioxide concentration is the most likely limiting factor. Increasing the concentration of carbon dioxide in the container will therefore increase the rate of photosynthesis. Decreasing the temperature would slow down the enzyme-controlled reactions, reducing the rate.

B is the correct answer. 1 mark for identifying carbon dioxide concentration as the limiting factor to be increased.

Stomata sunken in pits trap moist, humid air next to the leaf surface. This local accumulation of water vapour decreases the difference in water vapour concentration (the concentration gradient) between the internal air spaces of the leaf and the air directly outside the stomata, which reduces transpiration.

A is the correct answer. 1 mark for identifying that sunken stomata trap humid air, reducing the concentration gradient.

In a reflex arc, a stimulus is first detected by a receptor. The receptor initiates an electrical impulse that travels along a sensory neurone to the central nervous system. Inside the central nervous system, the impulse is passed across a synapse to a relay neurone, and then across another synapse to a motor neurone. The motor neurone carries the impulse to the effector (such as a muscle) to produce the response.

B is the correct answer. 1 mark for the correct chronological order of elements in a reflex arc pathway.

A concentrated sucrose solution has a lower water potential than the cell sap inside the plant cells. Consequently, water moves out of the cells by osmosis down the water potential gradient. As the cytoplasm and vacuole shrink, the cell membrane pulls away from the cell wall, leaving the cells plasmolysed.

D is the correct answer. 1 mark for matching water loss by osmosis with the plasmolysed state.

Yeast respires anaerobically to produce ethanol and carbon dioxide gas. In bread making, the bubbles of carbon dioxide gas get trapped within the gluten network of the dough, expanding and causing the bread dough to rise. The ethanol evaporates during the baking process.

A is the correct answer. 1 mark for identifying carbon dioxide as the gas that causes bread dough to rise.

An increase in temperature increases the kinetic energy of both the enzyme and substrate molecules. This causes them to move faster, leading to more frequent collisions per unit time and a higher rate of effective collisions, increasing the formation of enzyme-substrate complexes.

A - Incorrect: Kinetic energy increases, not decreases.
B - Incorrect: Denaturation only occurs at temperatures significantly above the optimum temperature.
C - Correct: Higher temperature increases kinetic energy and the frequency of effective collisions.
D - Incorrect: The rate of enzyme-substrate complex formation increases.

When the substrate concentration is very high, all active sites of the enzyme molecules are fully occupied (saturated). At this point, adding more substrate cannot increase the rate of reaction, and the enzyme concentration becomes the limiting factor.

A - Correct: All active sites are occupied.
B - Incorrect: Substrate concentration does not denature enzymes.
C - Incorrect: Enzymes do not alter the activation energy in a way that limits reaction rate as concentration changes.
D - Incorrect: Substrates do not act as competitive inhibitors to their own enzymes in this manner.

Passive immunity is temporary because the antibodies are not produced by the individual's own lymphocytes, but are instead acquired from another source (e.g., breast milk or injection). Consequently, no memory cells are produced, and the protection is short-term.

A - Incorrect: This describes active immunity.
B - Incorrect: Active immunity produces memory cells.
C - Incorrect: Passive immunity does not produce memory cells.
D - Correct: Passive immunity features antibodies from another source, no memory cells, and short-term protection.

Since increasing the light intensity does not increase the rate of photosynthesis, light intensity is no longer the limiting factor. The limiting factor must be one of the other environmental conditions held at a very low level, which in this case are temperature and carbon dioxide concentration.

A - Incorrect: Light intensity is no longer limiting.
B - Correct: Temperature and carbon dioxide concentration are limiting the rate.
C - Incorrect: Oxygen is a product of photosynthesis, not a limiting reactant.
D - Incorrect: Water availability is not a limiting factor for an aquatic plant.

Xerophytes are adapted to survive in environments with little liquid water. Features such as stomata sunk in pits, a thick waxy cuticle, and leaves reduced to spines are adaptations to reduce the rate of transpiration. Having many stomata on the upper epidermis is an adaptation of floating hydrophytes, not xerophytes.

A - Incorrect: Feature 3 is not a xerophytic adaptation.
B - Correct: Features 1, 2, and 4 reduce water loss.
C - Incorrect: Feature 3 is not a xerophytic adaptation.
D - Incorrect: Feature 2 is also a key xerophytic adaptation.

In a spinal reflex arc, the pathway starts at the receptor, which detects the stimulus. The nerve impulse is transmitted along the sensory neurone to the relay neurone in the spinal cord, then to the motor neurone, and finally to the effector (the muscle that contracts to withdraw the hand).

A - Incorrect: Order of neurones is wrong.
B - Correct: Receptor -> sensory neurone -> relay neurone -> motor neurone -> effector.
C - Incorrect: Pathway starts at the receptor, not the effector.
D - Incorrect: The relay neurone sits between the sensory and motor neurones.

Water moves from a region of higher water potential to a region of lower water potential by osmosis. The solution with the lowest water potential will draw the most water out of the potato cells, resulting in the greatest loss in mass. The potato cylinder in D decreased in mass the most (from 10.0 g to 8.1 g).

A - Incorrect: This cylinder gained mass, meaning the solution had a higher water potential than the potato.
B - Incorrect: There was no net water movement.
C - Incorrect: This cylinder lost some water, but less than D.
D - Correct: This cylinder lost the most water, showing the surrounding solution had the lowest water potential.

Restriction enzymes are used to cut DNA at specific base sequences (leaving sticky ends). DNA ligase is the enzyme used to join the sticky ends of the target gene and the plasmid vector together to form recombinant DNA.

A - Correct: Restriction enzymes cut DNA, and ligase joins DNA.
B - Incorrect: The roles are reversed.
C - Incorrect: These are not the functions of these enzymes in cloning vectors.
D - Incorrect: Messenger RNA isolation and insertion are not described correctly by these terms.

Amylase is a protein enzyme. At high temperatures such as \( 60\text{ }^\circ\text{C} \), the thermal energy breaks chemical bonds holding the enzyme's specific 3D shape together. This changes the shape of the active site (denaturation), meaning the substrate (starch) can no longer fit into it to react.

1 mark for the correct option B.
0 marks for any other option.

Active immunity is defined as defense against a pathogen by antibody production in the body. It can be gained after an infection or through vaccination. During this process, the body's own lymphocytes synthesize specific antibodies and create long-lived memory cells, providing long-term immunity.

1 mark for the correct option A.
0 marks for any other option.

At high light intensity, light is no longer the factor that limits the rate of photosynthesis. Since the rate for Curve 2 (0.03% CO2) is lower than the rate for Curve 1 (0.1% CO2), increasing the concentration of carbon dioxide increases the rate. Therefore, carbon dioxide concentration is the limiting factor for Curve 2 under these conditions.

1 mark for the correct option A.
0 marks for any other option.

A thick waxy cuticle reduces evaporation from the epidermal cells of the leaf, conserving precious water in arid environments. Sunken stomata decrease (rather than increase) transpiration; large air spaces are adaptations for floating in hydrophytes; and reducing leaves to spines decreases surface area to minimize transpiration.

1 mark for the correct option B.
0 marks for any other option.

Synaptic transmission begins when an impulse triggers the release of neurotransmitters from vesicles (4). These molecules diffuse across the gap (1), bind to receptors on the postsynaptic side (2), and trigger a new nerve impulse (3).

1 mark for the correct option A.
0 marks for any other option.

The lock and key hypothesis proposes that the enzyme (lock) and substrate (key) have complementary shapes. Only a substrate with the exact fitting shape can bind to the active site to produce a reaction.

1 mark for the correct option A.
0 marks for any other option.

Phagocytes carry out phagocytosis, engulfing pathogens and digesting them using enzymes. Lymphocytes produce specific protein molecules called antibodies that target and neutralize pathogens.

1 mark for the correct option A.
0 marks for any other option.

Glucose is soluble and highly reactive. For safe transport around the plant within the phloem sieve tubes, it is converted into sucrose. For long-term energy storage, it is condensed into starch, which is insoluble and does not alter the osmotic potential of cells.

1 mark for the correct option A.
0 marks for any other option.

The rate of oxygen production is highest at pH 7.0 (9.6 cm³/min), which is between pH 5.0 and pH 9.0. Therefore, the optimum pH lies within this range.

1 mark for correct option B.

Active immunity involves the body producing its own antibodies in response to an antigen, which leads to the production of memory cells and long-term protection.

1 mark for correct option A.

Since the carbon dioxide concentration is increased beyond the point where the rate of photosynthesis stops increasing (0.2%), carbon dioxide is no longer the limiting factor. Therefore, some other factor such as light intensity or temperature is limiting the rate.

1 mark for correct option B.

Closing stomata during the hottest part of the day reduces the rate of transpiration, conserving water in arid environments. Spines decrease surface area; sunken stomata decrease the concentration gradient of water vapor; and a thick waxy cuticle acts as a barrier to reduce water loss.

1 mark for correct option B.

Adrenaline prepares the body for vigorous action ('fight or flight') by increasing the pulse rate (to pump more blood), dilating the pupils (to let in more light), and increasing blood glucose concentration (to provide more substrate for respiration).

1 mark for correct option A.

Magnification is calculated using the formula: Magnification = Image size / Actual size. Both values must be in the same units.
Magnification = 60 mm / 0.15 mm = 400.

1 mark for correct option D.

Active transport is the movement of particles through a cell membrane from a region of lower concentration to a region of higher concentration (against a concentration gradient) using energy from respiration.

1 mark for correct option C.

Wind-pollinated flowers have small, dull petals (as they do not need to attract insects), anthers hanging outside the flower (to easily release pollen into the wind), and light, smooth pollen grains (so they can be carried long distances by wind currents).

1 mark for correct option C.

(a) The term used is **optimum pH**.

(b) (i)
- Step 1: Calculate the increase in volume of oxygen produced: $24.0 \text{ cm}^3 - 7.0 \text{ cm}^3 = 17.0 \text{ cm}^3$
- Step 2: Calculate the percentage increase: $\frac{17.0}{7.0} \times 100 = 242.857...\%$
- Step 3: Round to two significant figures: $240\%$

(ii) Between pH 4.0 and 7.0, the volume of oxygen produced increases because the pH is approaching the optimum pH (pH 7.0). At pH 7.0, the active site of the catalase is in its most effective/complementary shape, allowing more frequent successful collisions with hydrogen peroxide and a higher rate of enzyme-substrate complex formation.

(iii) At pH 9.0, the enzyme catalase is denatured. The extreme pH changes the ionic and hydrogen bonds holding the enzyme's tertiary structure, permanently altering the shape of the active site. Consequently, the substrate (hydrogen peroxide) can no longer fit into the active site, meaning no enzyme-substrate complexes can form and the reaction rate drops severely.

(c) Two controlled variables:
1. Concentration of hydrogen peroxide solution.
2. Volume of yeast suspension (enzyme concentration).

(a) optimum (pH); [1]

(b) (i)
- $24.0 - 7.0 = 17.0$ ; (Award 1 mark for difference)
- $\frac{17.0}{7.0} \times 100$ or $242.86...$ ; (Award 1 mark for correct fraction/percentage before rounding)
- $240$ (%); (Award 1 mark for correct rounding to 2 sig figs) [3]

(ii) Any three from:
- volume of oxygen / rate of reaction increases as pH increases (from 4.0 to 7.0) ;
- pH is moving closer to the optimum pH of the enzyme / catalase ;
- active site is in the correct/best shape to bind to the substrate / hydrogen peroxide ;
- leading to more frequent successful collisions / more enzyme-substrate complexes forming ; [max 3]

(iii) Any four from:
- catalase/enzyme is denatured ;
- high pH alters the shape of the active site ;
- hydrogen peroxide/substrate is no longer complementary / can no longer bind to the active site ;
- fewer / no enzyme-substrate complexes can form ;
- very few / no reactions occur (so minimal oxygen is released) ; [max 4]

(c) Any two from:
- concentration of hydrogen peroxide / substrate ;
- volume of hydrogen peroxide / substrate ;
- volume of yeast suspension / catalase extract ;
- concentration of yeast suspension ;
- duration of reaction / time of measurement (if not fixed at 2 mins) ; [max 2]

(a) (i) **Light intensity** is the limiting factor (since increasing the carbon dioxide concentration from 0.02% to 0.10% does not change the rate of bubbles produced, but increasing light intensity from 10 to 20 a.u. does increase the rate).
(ii) **Carbon dioxide concentration** is the limiting factor (since increasing light intensity from 40 to 50 a.u. at 0.02% does not change the rate of 20 bubbles/min, but increasing carbon dioxide concentration to 0.10% increases the rate to 45 bubbles/min).

(b) At 0.02% $\text{CO}_2$, the rate is limited at 20 bubbles per minute because carbon dioxide is in short supply. Carbon dioxide is a raw material required in the light-independent stage of photosynthesis to produce glucose. When the concentration is increased to 0.10% $\text{CO}_2$, this limitation is removed, allowing more carbon dioxide molecules to react per unit time, resulting in more oxygen being produced as a waste product (45 bubbles/min).

(c) (i) $\text{carbon dioxide} + \text{water} \rightarrow \text{glucose} + \text{oxygen}$
(ii) Carbon dioxide is dissolved in the surrounding water. It enters the leaf cells of submerged aquatic plants by **diffusion** directly across the cell wall and cell membrane, moving down its concentration gradient (from a high concentration in the water to a lower concentration inside the cells).

(d) Increasing temperature from $20^\circ\text{C}$ to $40^\circ\text{C}$ increases the kinetic energy of the water molecules, enzymes, and substrates. This leads to faster diffusion and more frequent successful collisions between photosynthetic enzymes (such as Rubisco) and their substrates, which increases the overall rate of photosynthesis (up to the enzyme's optimum temperature).

(a) (i) light (intensity) ; [1]
(ii) carbon dioxide (concentration) ; [1]

(b) Any four from:
- at 0.02% $\text{CO}_2$, carbon dioxide is the limiting factor / rate is plateaued / constant at 20 bubbles/min ;
- at 0.10% $\text{CO}_2$, the rate of photosynthesis is higher (45 bubbles/min) because more $\text{CO}_2$ is available ;
- carbon dioxide is a reactant / raw material in photosynthesis ;
- $\text{CO}_2$ is used to make glucose/sugar/starch / used in the light-independent stage ;
- more oxygen (by-product) is released per unit time ; [max 4]

(c) (i)
- carbon dioxide + water $\rightarrow$ glucose + oxygen ;;
- Award 1 mark for correct reactants, 1 mark for correct products.
- Reject chemical formulas unless fully balanced: $6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$. No mixed word-formula equations allowed. [2]
(ii)
- by diffusion ;
- down a concentration gradient / from high concentration in water to low concentration in leaf cells ;
- dissolves / moves across the cell wall and cell membrane ; [max 2]

(d) Any three from:
- increases kinetic energy of molecules / enzymes / substrates ;
- more frequent successful collisions / more enzyme-substrate complexes formed ;
- rate of chemical reactions / photosynthesis increases ;
- note: if $40^\circ\text{C}$ exceeds the optimum temperature of the plant's enzymes, the enzymes may begin to denature, causing the rate to decrease ; [max 3]

(a) A pathogen is a disease-causing organism.

(b) (i) Lymphocyte (or B-lymphocyte / B-cell).
(ii) Phagocyte.

(c) After the first exposure, the antibody response is much slower and weaker. There is a lag of several days, and the peak concentration is only 12 a.u. (reached on Day 14), followed by a rapid drop back to 2 a.u. After the second exposure, the antibody response is much faster and stronger. It rises in just 3 days, reaches a far higher peak of 95 a.u. (Day 57), and the antibody level remains very high (70 a.u. on Day 70) compared to the primary response.

(d) During the primary infection, some of the activated lymphocytes differentiate into memory cells. These cells survive in the blood/body for a long time. Upon a secondary exposure, these memory cells immediately recognise the specific antigens on the pathogen. They divide rapidly by mitosis and quickly differentiate into antibody-producing plasma cells, generating a massive quantity of antibodies before the pathogen can multiply enough to cause disease symptoms.

(e) Active immunity involves the production of antibodies by the individual's own body/immune system in response to an antigen (either through infection or vaccination), providing long-term protection and generating memory cells. Passive immunity involves receiving pre-formed antibodies from another organism (e.g., from mother to baby via breast milk or placenta, or via antibody injection), providing immediate but short-term protection without producing memory cells.

(a) disease-causing organism ; [1]

(b) (i) lymphocyte / B-lymphocyte / B-cell ; [1]
(ii) phagocyte ; [1]

(c) Any four from:
- lag time / delay before antibodies appear is shorter in the second exposure / secondary response is faster ;
- the peak antibody concentration is much higher in the second response (95 a.u. vs 12 a.u.) / secondary response is stronger ;
- the secondary peak is reached sooner after exposure (7 days after day 50 vs 14 days after day 0) ;
- antibody levels remain high for much longer after the second exposure (70 a.u. on Day 70 vs 2 a.u. on Day 28) ;
- use of comparative data with units (a.u. / days) from both responses ; [max 4]

(d) Any four from:
- memory cells are produced during the first exposure / primary infection ;
- they remain / persist in the bloodstream / lymphoid tissue for a long time / years ;
- they recognize the same antigen / pathogen immediately upon re-infection ;
- they clone/divide rapidly by mitosis ;
- they differentiate into antibody-producing cells (plasma cells/lymphocytes) quickly ;
- they produce a very large quantity of antibodies very quickly (before symptoms develop) ; [max 4]

(e) Any three from:
- active immunity involves the body producing its own antibodies, passive immunity involves receiving antibodies from an outside source / another organism ;
- active immunity is long-term / permanent, passive immunity is short-term / temporary ;
- active immunity produces memory cells, passive immunity does not produce memory cells ;
- active immunity is triggered by exposure to antigen / pathogen / vaccine, passive immunity is triggered by receiving antibodies (via placenta / breast milk / injection) ; [max 3]

(a) An adaptive feature is an inherited (structural, functional, or behavioural) feature of an organism that increases its fitness/chances of survival and reproduction in its environment.

(b) (i) The hairs trap water vapour (moist air) escaping from the stomata, which reduces the water potential gradient between the inside of the leaf and the outside air, thereby reducing the rate of transpiration.
(ii) Sunken stomata are protected from wind currents. They trap a layer of humid air inside the pit, which slows down the diffusion of water vapour out of the stomata.
(iii) Rolling the leaf encloses the stomata on the inner surface, hiding them from direct sunlight and wind. This creates a highly humid microclimate inside the rolled leaf, minimizing water loss by transpiration.

(c) Submerged hydrophytes have adaptations such as:
1. A very thin or absent cuticle, as water conservation is not necessary.
2. No stomata in their leaves (or stomata only on the upper surface of floating leaves), as carbon dioxide is absorbed directly from the water.
3. Large air spaces (aerenchyma) in stems and leaves to provide buoyancy and allow gas exchange/oxygen storage.

(a)
- inherited (structural / behavioural / physiological) feature ;
- increases fitness / chances of survival and reproduction (in its environment) ; [2]

(b) (i)
- traps water vapour / moist air (near stomata) ;
- reduces the water potential gradient (between inside and outside of leaf) / reduces transpiration / reduces water loss ; [2]
(ii)
- traps humid / moist air inside the pits ;
- protects stomata from wind / air currents (which would blow away humid air) / reduces diffusion of water vapour ; [2]
(iii)
- encloses / protects the stomata on the inner surface (from direct sunlight/wind) ;
- creates a humid microclimate inside the roll / reduces the exposed surface area of the leaf ; [2]

(c) Any three from:
- very thin / absent cuticle ;
- stomata absent (or present only on the upper surface for floating leaves) ;
- large air spaces / aerenchyma (for buoyancy / gas storage) ;
- reduced / poorly developed root system (water/minerals absorbed directly through leaves) ;
- highly divided / thin leaves (to increase surface area for absorption of dissolved gases / light) ;
- reduced xylem / vascular bundle (as support is provided by water) ; [max 3]

(a) (i) Shoot A will grow upwards (away from gravity / show negative gravitropism).
(ii) Auxin is produced in the shoot tip and moves downwards. Under the influence of gravity, auxin accumulates on the lower side of the horizontal shoot. In shoots, a high concentration of auxin stimulates cell elongation. Therefore, the cells on the lower side of Shoot A elongate more rapidly than the cells on the upper side, causing the shoot to bend upwards.

(b) (i) Shoot B will not bend upwards / will stop growing.
(ii) The growing tip is the site of auxin production. Removing the tip removes the source of auxin, so no cell elongation or growth can occur on either side of the shoot.

(c) Shoot C will bend upwards in the exact same manner as Shoot A. Gravitropism is a growth response to gravity and does not require light. The light-proof cap only blocks light, which has no effect on the gravity-sensing mechanism or the redistribution of auxin due to gravity in the dark.

(d) Three differences:
1. Nervous coordination involves electrical impulses transmitted along neurones, while endocrine coordination involves chemical hormones transported in the blood.
2. Nervous responses are very rapid (milliseconds), while endocrine responses are generally slower (seconds to days).
3. Nervous responses are short-lived, while endocrine responses can be long-lasting.

(a) (i) grows upwards / curves upwards / away from gravity / negative gravitropism ; [1]
(ii)
- auxin is produced in the shoot tip (and moves downwards) ;
- gravity causes auxin to accumulate on the lower side of the horizontal shoot ;
- high concentration of auxin in shoots stimulates cell elongation ;
- cells on the lower side elongate more/faster than cells on the upper side (causing upward bending) ; [4]

(b) (i) does not grow / does not bend / remains horizontal ; [1]
(ii)
- the tip of the shoot is the source of auxin / where auxin is made ;
- without the tip, there is no auxin to stimulate cell elongation / growth ; [2]

(c)
- Shoot C will bend upwards in the same way as Shoot A ;
- gravitropism is a response to gravity, not light ;
- the light-proof cap only blocks light / does not affect gravity sensing / does not affect auxin redistribution caused by gravity ; [3]

(d) Any three from:
- nervous system uses electrical impulses, endocrine system uses chemical hormones ;
- nervous transmission is through neurones, endocrine transmission is via the blood system ;
- nervous response is very rapid, endocrine response is generally slower ;
- nervous response is short-lived, endocrine response is often long-lasting ;
- nervous response is highly localized, endocrine response is widespread / affects target organs ; [max 3]

(a) (i) **Amino acids**.
(ii) **Carbon, hydrogen, oxygen, and nitrogen** (sulfur is also present in some proteins).

(b) (i) In Test-tube 1, hydrochloric acid provides a highly acidic environment / low pH, which is the optimum pH for the stomach enzyme pepsin to function efficiently, resulting in rapid digestion (15 minutes). In Test-tube 2, the pH is neutral due to water, which is far from pepsin's optimum pH, making the enzyme inactive or extremely slow (no change after 120 minutes).
(ii) Boiling the pepsin in Test-tube 4 denatured the enzyme. The extreme heat disrupted the chemical bonds holding the enzyme's specific three-dimensional shape, permanently altering the active site. The protein substrate (egg white) could no longer bind to the active site, so no enzyme-substrate complexes formed and no digestion occurred.
(iii) The egg white was digested much faster at $37^\circ\text{C}$ (Test-tube 1, 15 minutes) than at $5^\circ\text{C}$ (Test-tube 3, 110 minutes). At $37^\circ\text{C}$, the pepsin and protein molecules have much higher kinetic energy and move faster. This leads to more frequent successful collisions per unit time, resulting in a higher rate of enzyme-substrate complex formation compared to the cold $5^\circ\text{C}$ temperature.

(c) Another protease acting in the small intestine is **trypsin**, which works best in alkaline conditions (typically **pH 8**).

(a) (i) amino acids ; [1]
(ii) carbon, hydrogen, oxygen, nitrogen ;;
- Award 2 marks for all four elements.
- Award 1 mark for any three elements. (Accept sulfur as an additional element, but the main four are required) [2]

(b) (i) Any three from:
- pepsin requires acidic conditions / a low pH to function ;
- hydrochloric acid (in Tube 1) provides the optimum pH for pepsin ;
- water (in Tube 2) provides a neutral pH / incorrect pH, at which pepsin is inactive ;
- (hence) enzyme-substrate complexes form quickly in Tube 1 but not in Tube 2 ; [max 3]
(ii) Any three from:
- boiling denatures the pepsin/enzyme ;
- the shape of the active site is permanently altered / destroyed ;
- the protein substrate can no longer fit into the active site / no longer complementary ;
- no enzyme-substrate complexes can form / no reaction can occur ; [max 3]
(iii) Any four from:
- digestion is much faster in Test-tube 1 than in Test-tube 3 / ORA ;
- $37^\circ\text{C}$ is a higher temperature / closer to body temperature than $5^\circ\text{C}$ ;
- at higher temperatures, molecules have more kinetic energy / move faster ;
- leading to more frequent collisions between enzyme (pepsin) and substrate (egg white) ;
- resulting in more successful collisions / more enzyme-substrate complexes formed per unit time ; [max 4]

(c)
- trypsin ;
- alkaline / pH 8 / pH 7.5 to 8.5 ; [2]

(a) (i) The student added the three bubble counts together (36 + 34 + 35 = 105) and then divided the sum by the number of trials (3) to get 35.

(ii) The graph should have 'Distance of lamp from plant / cm' on the x-axis (with a linear scale from 0 to 50 or 60) and 'Mean bubble count' on the y-axis (with a linear scale from 0 to 50). All points from Table 1.1 should be plotted accurately with a small 'x' or a circled dot. The points should be connected with straight ruled lines or a smooth curve.

(iii) As the distance of the lamp from the plant increases, the rate of photosynthesis decreases.

(b) (i) The distance of the lamp from the plant.

(ii) Temperature of the water, concentration of carbon dioxide (sodium hydrogencarbonate solution concentration), species/size of the *Cabomba* plant.

(iii) Bubbles may be of different sizes, or bubbles may be released too fast to count accurately.

(iv) Collect the gas in a gas syringe or a graduated capillary tube to measure the actual volume.

(c) 1. Place the leaf in a beaker of boiling water for about 30 seconds to kill the cells.
2. Turn off the Bunsen burner (or use an electric water bath) and place the leaf in a tube of ethanol, then place this tube in the hot water to extract the chlorophyll.
3. Dip the leaf in warm water to soften it.
4. Spread the leaf on a white tile and add a few drops of iodine solution. A blue-black colour indicates starch is present.

### Part (a) [6 marks]
- **(i)** Add the three trials together and divide by 3 [1 mark].
- **(ii)** Axes labelled with units: x-axis: 'Distance of lamp from plant / cm', y-axis: 'Mean bubble count' [1 mark].
- Suitable linear scales where plotted points occupy more than half the grid [1 mark].
- All five points plotted accurately (within half a small square) [1 mark].
- Clean line connecting the points (smooth curve or straight lines between points) [1 mark].
- **(iii)** Inverse relationship described (as distance increases, bubble count/photosynthesis decreases) [1 mark].

### Part (b) [3 marks]
- **(i)** Distance of the lamp / light intensity [1 mark].
- **(ii)** Any two of: temperature of water, species/mass of plant, carbon dioxide concentration / sodium hydrogencarbonate concentration [2 marks].

### Part (c) [2 marks]
- **(i)** Bubbles are different sizes / human error in counting [1 mark].
- **(ii)** Collect gas using a gas syringe / graduated pipette [1 mark].

### Part (d) [3 marks]
- Heat leaf in boiling water [1 mark].
- Boil in ethanol using a water bath / no direct flame [1 mark].
- Add iodine solution [1 mark].

(a) Investigation Plan:
1. Prepare at least 5 different temperatures using water baths (e.g., 10 °C, 20 °C, 30 °C, 40 °C, and 50 °C).
2. Keep the volume (e.g., 10 cm³) and concentration (e.g., 1%) of hydrogen peroxide solution constant in each tube.
3. Cut potato cylinders of equal length (e.g., 2 cm) and diameter to keep the surface area and mass of enzyme tissue constant.
4. Equilibrate the potato cylinders and hydrogen peroxide separately in the respective water baths for 5 minutes before mixing.
5. Mix them and measure the volume of oxygen gas produced using a delivery tube and gas syringe over a fixed time (e.g., 3 minutes).
6. Repeat the experiment at each temperature at least three times to calculate a mean and identify anomalies.
7. Wear safety goggles and gloves when handling hydrogen peroxide as it is an irritant.

(b) (i) 32.0 mm
(ii) Magnification = 32.0 mm / 8.0 mm = x 4.0

(c) Reducing Sugar Test:
1. Crush/grind a sample of potato tissue using a pestle and mortar with a small amount of distilled water.
2. Decant the liquid into a test-tube.
3. Add an equal volume of Benedict's reagent to the test-tube.
4. Heat the mixture in a hot water bath (above 80 °C) for 5 minutes.
5. A colour change from blue to green, yellow, orange, or brick-red indicates the presence of reducing sugars.

### Part (a) [6 marks]
- **Independent variable:** at least 5 different temperatures used [1 mark].
- **Dependent variable:** volume of oxygen gas collected in a set time / height of foam produced [1 mark].
- **Controlled variable:** concentration/volume of hydrogen peroxide [1 mark].
- **Controlled variable:** mass/dimensions/surface area of potato pieces [1 mark].
- **Reliability:** repeat at least 3 times at each temperature and calculate a mean [1 mark].
- **Safety:** use goggles/gloves when handling hydrogen peroxide [1 mark].

### Part (b) [3 marks]
- **(i)** 32 mm (accept 31 to 33 mm) [1 mark].
- **(ii)** Correct substitution of candidate's value from (i) divided by 8.0 [1 mark].
- Correct calculation rounded to one decimal place (e.g. x 4.0 or 4.0) [1 mark].

### Part (c) [4 marks]
- Grind potato tissue with water [1 mark].
- Add Benedict's reagent [1 mark].
- Heat in a hot water bath / temperature above 80 °C [1 mark].
- Correct color change described (blue to green/yellow/orange/brick-red) [1 mark].

(a) (i) Decrease in mass of Leaf B = 4.50 - 3.78 = 0.72 g.
Percentage decrease = (0.72 / 4.50) * 100 = 16.0% (or 16%).

(ii) Decrease in mass of Leaf C = 4.00 - 3.84 = 0.16 g.
Percentage decrease = (0.16 / 4.00) * 100 = 4.0% (or 4%).

(b) (i) Leaf B had a much higher percentage decrease in mass (16.0%) compared to Leaf C (4.0%). This is because the lower surface of the leaf contains many more stomata than the upper surface. In Leaf C, the lower surface was covered with Vaseline, blocking the stomata and preventing water loss via transpiration. In Leaf B, only the upper surface was covered, leaving the stomata on the lower surface open to lose water.

(ii) Leaf D acts as a control to show that water loss occurs primarily through stomata and not across other parts of the leaf (such as the waxy cuticle).

(c) The drawing must be large (at least 10 cm in width), drawn with a sharp pencil, and feature single, clear, continuous outlines without any shading or sketching. The curved bean-shape of the two guard cells and the open central stoma pore must be clearly represented, including internal chloroplasts drawn as small distinct shapes.

(d) 1. Bean-like / curved shape.
2. Presence of chloroplasts.
3. Thicker cell wall on the inner side (facing the pore).

### Part (a) [2 marks]
- **(i)** 16.0% (or 16) [1 mark].
- **(ii)** 4.0% (or 4) [1 mark].

### Part (b) [4 marks]
- **(i)** Leaf B lost more water / mass than Leaf C [1 mark].
- Stomata are mainly located on the lower surface [1 mark].
- Vaseline on Leaf C blocked stomata on the lower surface, reducing transpiration [1 mark].
- **(ii)** Control / to show that water loss occurs via stomata and not the cuticle [1 mark].

### Part (c) [4 marks]
- Large drawing (occupies more than half the available space) [1 mark].
- Single clear lines, no sketching or shading [1 mark].
- Correct shape: two curved guard cells with an open central pore [1 mark].
- Detail: chloroplasts drawn inside guard cells and thicker inner cell walls shown [1 mark].

### Part (d) [3 marks]
- Curve / bean shape of cells [1 mark].
- Presence of chloroplasts [1 mark].
- Thicker inner cell wall [1 mark].