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Chloroplasts and mitochondria both contain circular DNA and 70S ribosomes, supporting the endosymbiotic theory of their origin. They are also surrounded by an outer and an inner membrane (a double membrane or envelope). Nuclei contain linear DNA associated with histones and do not contain 70S ribosomes. Lysosomes are bounded by a single membrane and do not contain DNA or ribosomes.

Award 1 mark for selecting option A, which correctly identifies that only chloroplasts and mitochondria possess circular DNA, 70S ribosomes, and a double membrane.

First, calculate the value of one division of the stage micrometer: 0.1 mm = 100 \(\mu\text{m}\). Next, find the actual distance of the overlapping region: 5 divisions of the stage micrometer = \(5 \times 100\,\mu\text{m} = 500\,\mu\text{m}\). These 5 divisions align with 25 eyepiece units (epu), so 25 epu = 500 \(\mu\text{m}\). This means 1 epu = \(500\,\mu\text{m} / 25 = 20\,\mu\text{m}\). The plant cell is 12 epu wide, so its actual width is \(12 \times 20\,\mu\text{m} = 240\,\mu\text{m}\).

Award 1 mark for option C, which shows the correct calculation steps leading to 240 \(\mu\text{m}\).

Hydrogen bonds and hydrophobic interactions are weak, non-covalent interactions that help to stabilize the tertiary structure of proteins. They are easily disrupted by the increased kinetic energy associated with a moderate rise in temperature. Peptide bonds (primary structure) and disulfide bonds (covalent bonds in tertiary structure) are much stronger and require significantly higher energy to break.

Award 1 mark for option B, identifying hydrogen bonds and hydrophobic interactions as the heat-labile weak interactions.

A competitive inhibitor competes with the substrate for the active site of the enzyme. At sufficiently high substrate concentrations, the substrate outcompetes the inhibitor, allowing the reaction to reach its original maximum rate. Therefore, \(V_{\max}\) remains unchanged. However, because a higher substrate concentration is required to achieve half of the maximum rate (half \(V_{\max}\)), the Michaelis-Menten constant (\(K_m\)) increases, indicating a lower apparent affinity for the substrate.

Award 1 mark for option C, identifying that the competitive inhibitor leaves \(V_{\max}\) unchanged but increases \(K_m\).

Water always moves down a water potential gradient from a region of higher (less negative) water potential to a region of lower (more negative) water potential by osmosis. Here, water moves from the plant cells (\(-600\text{ kPa}\)) out into the surrounding sucrose solution (\(-900\text{ kPa}\)). As water leaves, the vacuole and cytoplasm shrink, and the protoplast eventually pulls away from the cell wall, causing the cells to become plasmolysed.

Award 1 mark for option B, identifying the net movement of water out of the cells by osmosis and the resulting plasmolysis.

During anaphase, the centromeres divide, allowing the sister chromatids to separate. The shortening of the spindle microtubules then pulls the separated chromatids (now individual chromosomes) towards opposite poles of the cell.

Award 1 mark for option A, identifying anaphase as the mitotic stage where sister chromatids separate.

First, find the complementary mRNA sequence transcribed from the DNA template strand: DNA \(\text{3'-TAC GTT CGA CTA-5'}\) is transcribed to mRNA \(\text{5'-AUG CAA GCU GAU-3'}\). During translation, tRNA anticodons pair complementarily with the mRNA codons. The codon \(\text{5'-AUG-3'}\) pairs with tRNA anticodon \(\text{3'-UAC-5'}\), \(\text{5'-CAA-3'}\) pairs with \(\text{3'-GUU-5'}\), \(\text{5'-GCU-3'}\) pairs with \(\text{3'-CGA-5'}\), and \(\text{5'-GAU-3'}\) pairs with \(\text{3'-CUA-5'}\). Thus, the sequence of anticodons from first to last is \(\text{UAC, GUU, CGA, CUA}\).

Award 1 mark for option B, which correctly translates the sequence into the matching tRNA anticodons.

The trachea requires continuous patency, which is provided by C-shaped rings of cartilage. It also contains smooth muscle to adjust airway volume and goblet cells to secrete mucus to trap pathogens. In contrast, bronchioles (especially terminal bronchioles) do not contain cartilage to allow for complete constriction when needed. They do contain smooth muscle to regulate airflow, but they lack goblet cells to prevent mucus accumulation from clogging these very narrow airways.

Award 1 mark for option A, which correctly identifies the presence and absence of each structural tissue type in the trachea and bronchioles.

Both chloroplasts and mitochondria contain circular DNA, 70S ribosomes, and are bounded by a double membrane. The nucleus contains linear DNA and is bounded by a double membrane (the nuclear envelope), but it does not contain 70S ribosomes (it contains nucleoli where ribosome subunits are assembled, and the ribosomes in the cytoplasm are 80S). The nucleoid is the region in prokaryotes containing circular DNA, but it is not bounded by any membrane.

1 mark for identifying that only chloroplasts and mitochondria meet all three criteria: containing DNA, 70S ribosomes, and being bounded by a double membrane.

Statement 1 is correct because both glycogen and amylopectin are branched polysaccharides and therefore contain both alpha-1,4-glycosidic bonds (in the chains) and alpha-1,6-glycosidic bonds (at the branch points). Statement 2 is incorrect because both are polymers of alpha-glucose, not beta-glucose. Statement 3 is correct because glycogen has more frequent branch points (approximately every 8 to 12 glucose units) compared to amylopectin (approximately every 24 to 30 glucose units). Statement 4 is incorrect because glycogen is a storage polysaccharide in animal and fungal cells, whereas starch (composed of amylose and amylopectin) is the storage polysaccharide in plant cells.

1 mark for identifying that statement 1 and statement 3 are correct, while statements 2 and 4 are incorrect.

A competitive inhibitor binds reversibly to the active site of the enzyme, competing directly with the substrate. At very high substrate concentrations, the substrate virtually always outcompetes the inhibitor for the active site, meaning the maximum rate of reaction (Vmax) remains unchanged. However, because more substrate is required to achieve half of this maximum rate (Vmax/2), the Michaelis-Menten constant (Km) increases, indicating a lower apparent affinity of the enzyme for its substrate in the presence of the inhibitor.

1 mark for identifying that a competitive inhibitor leaves Vmax unchanged but increases the Km value.

Water moves down a water potential gradient from areas of higher (less negative) water potential to areas of lower (more negative) water potential. The correct pathway begins with soil water, which has the highest water potential, entering the root hair cell sap. It then moves across the root cortex into the xylem vessel (xylem sap), travels up the plant to the leaf, enters the cell walls of the mesophyll cells, evaporates from the cell walls into the intercellular air spaces, and finally diffuses out through the stomata into the atmosphere, which has the lowest water potential.

1 mark for identifying the correct physiological sequence of water transport down the water potential gradient from soil to atmosphere.

To find the correct template DNA strand, we must determine the complementary sequence to the mRNA codons for each amino acid in order: 1. Cysteine: UGC is coded by the complementary DNA triplet ACG (since A pairs with U, C with G, and G with C). 2. Glycine: GGU is coded by the complementary DNA triplet CCA (since C pairs with G, C with G, and A with U). 3. Alanine: GCA is coded by the complementary DNA triplet CGT (since C pairs with G, G with C, and T with A). Thus, the sequence ACG-CCA-CGT is a valid template strand sequence. Option B is incorrect because it is the non-template (coding) DNA sequence with thymine. Option C is incorrect because the middle triplet GGT would transcribe to CCA, which codes for proline, not glycine. Option D is incorrect because it contains uracil, which is found in RNA, not DNA.

1 mark for correctly transcribing the mRNA codons back to complementary DNA template triplets.

Companion cells are living cells that contain mitochondria (to produce ATP for active loading of sucrose), plasmodesmata (connecting them to sieve tube elements), and active cytoplasm. Mature xylem vessel elements are dead, hollow tubes that completely lack cytoplasm and organelles such as mitochondria, and they do not possess plasmodesmata (instead they have pits for lateral water movement). Therefore, all three features (1, 2, and 3) are present in companion cells but absent from mature xylem vessel elements.

1 mark for identifying that mitochondria, plasmodesmata, and cytoplasm are all present in companion cells but completely absent in mature xylem vessels.

The hepatic artery carries oxygenated blood directly from the aorta, so it has high hydrostatic pressure and high oxygen concentration. The hepatic vein carries deoxygenated blood away from the liver back to the heart via the vena cava, so it has low hydrostatic pressure and low oxygen concentration. The hepatic portal vein carries blood from the small intestine to the liver; while it has low hydrostatic pressure and low oxygen concentration, it has a very high concentration of glucose and other absorbed nutrients immediately after a meal before the liver regulates and stores them.

1 mark for correctly identifying the physiological differences in pressure, oxygenation, and nutrient levels among the three main liver blood vessels.

Penicillin is an antibiotic that specifically inhibits the enzyme transpeptidase, which is responsible for catalyzing the cross-linking of peptidoglycan chains in bacterial cell walls. This weakens the cell wall, causing the bacteria to burst due to osmotic pressure (lysis). Viruses do not have a cell wall (they have a protein capsid and lack peptidoglycan), nor do they undergo binary fission or cell wall synthesis, making them entirely unaffected by penicillin.

1 mark for identifying that penicillin targets peptidoglycan cell wall synthesis in bacteria, which is absent in viruses.

Mitochondria contain their own circular DNA and 70S ribosomes, reflecting their evolutionary origin from prokaryotes. The enzymes of the Krebs cycle are located within the mitochondrial matrix. While chloroplasts also contain circular DNA and 70S ribosomes, they contain photosynthetic pigments and enzymes of the Calvin cycle, not the Krebs cycle. Amyloplasts lack the machinery for the Krebs cycle and do not typically contain ribosomes or DNA in active form like mitochondria. Lysosomes do not contain DNA or ribosomes.

1 mark for identifying that mitochondria contain circular DNA, 70S ribosomes, and Krebs cycle enzymes.

The trisaccharide is made from two glucose molecules and one fructose molecule. A bond between two \(\alpha\)-glucose molecules is an \(\alpha\)-1,4-glycosidic bond (as found in maltose). The bond between an \(\alpha\)-glucose and a fructose molecule is an \(\alpha\)-1,\(\beta\)-2-glycosidic bond (as found in sucrose). Therefore, these two types of bonds must be hydrolyzed to release the free monosaccharides.

1 mark for selecting the option with both \(\alpha\)-1,4 (between glucose units) and \(\alpha\)-1,\(\beta\)-2 (between glucose and fructose units) glycosidic bonds.

According to the induced-fit hypothesis, the enzyme's active site is not a rigid shape. When the substrate binds, it induces a conformational change in the enzyme. This change puts physical stress on the chemical bonds of the substrate, making them easier to break and thus lowering the activation energy of the reaction.

1 mark for identifying that substrate binding induces a conformational change in the enzyme, which strains substrate bonds and lowers activation energy.

First, calculate the initial water potential of the plant cells: \(\Psi_{\text{cell}} = \Psi_s + \Psi_p = -800\text{ kPa} + 200\text{ kPa} = -600\text{ kPa}\). For net water movement to occur out of the cell, the external solution must have a lower (more negative) water potential than the cell (\(\Psi_{\text{solution}} < \Psi_{\text{cell}}\)). Solutions 3 (\(-800\text{ kPa}\)) and 4 (\(-1000\text{ kPa}\)) both have water potentials lower than \(-600\text{ kPa}\), so water will move out of the cells into these solutions.

1 mark for calculating the cell's water potential as \(-600\text{ kPa}\) and correctly identifying that Solutions 3 and 4 have lower water potentials, causing net water loss.

During metaphase of mitosis, the DNA has already replicated. There are 16 chromosomes aligned at the equator. Each chromosome consists of 2 sister chromatids (making a total of \(16 \times 2 = 32\) sister chromatids) joined together at a single centromere. Thus, there are 16 centromeres in total.

1 mark for identifying 16 chromosomes, 32 sister chromatids, and 16 centromeres in metaphase.

The DNA template strand sequence is 3\('- TAC GGT CAT ATT - 5'\). Transcription produces a complementary mRNA strand: 5\('- AUG CCA GUA UAA - 3'\). During translation, tRNA molecules with complementary anticodons bind to these codons. The anticodons complementary to the mRNA codons are: for AUG, the anticodon is UAC; for CCA, it is GGU; for GUA, it is CAU; and for UAA, it is AUU. Thus, the anticodons are UAC, GGU, CAU, AUU.

1 mark for identifying the correct tRNA anticodons as UAC, GGU, CAU, AUU based on complementary base pairing.

In a healthy mammalian lung, the bronchus contains irregular plates of cartilage, goblet cells to secrete mucus, and a ciliated epithelium to sweep mucus away. As the airways narrow into bronchioles, cartilage and goblet cells are lost (absent), but a ciliated epithelium remains to continue clearing any debris.

1 mark for correctly matching the presence of cartilage plates, goblet cells, and cilia in the bronchus, and their absence (except cilia) in the bronchiole.

Individual 1 receives pre-formed antibodies (passive) via medical intervention (artificial), so this is artificial passive immunity. Individual 2 is infected naturally and produces their own immune response (active), so this is natural active immunity. Individual 3 receives an injection (artificial) to stimulate their own immune response (active) without causing full disease, which is artificial active immunity.

1 mark for correctly classifying Individual 1 as artificial passive, Individual 2 as natural active, and Individual 3 as artificial active.

Fraction W contains a double membrane, circular DNA, and 70S ribosomes, which is characteristic of mitochondria and chloroplasts. Fraction X contains hydrolytic enzymes in an acidic environment, which is characteristic of a lysosome. Fraction Y is continuous with the nuclear envelope and has ribosomes, identifying it as the rough endoplasmic reticulum (RER). Fraction Z has a double membrane with pores, which is the nucleus. Therefore, row A is the correct identification.

1 mark for the correct identification of all four cell structures in row A.

First, find the actual size of the stage micrometer divisions: 1 division = \(0.1\text{ mm} = 100\mu\text{m}\). Thus, 4 divisions = \(400\mu\text{m}\). Next, calculate the value of 1 eyepiece graticule division (epd): 50 epd = \(400\mu\text{m}\), so 1 epd = \(400 / 50 = 8\mu\text{m}\). Finally, calculate the actual width of the plant cell: \(12\text{ epd} \times 8\mu\text{m/epd} = 96\mu\text{m}\).

1 mark for the correct calculation showing \(96\mu\text{m}\).

Amylose, amylopectin, and glycogen are all polymers of \(\alpha\)-glucose. Amylose is an unbranched helical chain containing only \(\alpha\)-1,4 glycosidic bonds. Amylopectin and glycogen are branched chains containing both \(\alpha\)-1,4 glycosidic bonds in their straight sections and \(\alpha\)-1,6 glycosidic bonds at their branch points.

1 mark for identifying the correct monomer (\(\alpha\)-glucose) and correct glycosidic bonds for all three polysaccharides.

Inhibitor X is a competitive inhibitor because its effects can be overcome by increasing the substrate concentration, allowing \(V_{\max}\) to be reached. Inhibitor Y is a non-competitive inhibitor because it binds to an allosteric site and changes the shape of the active site, preventing \(V_{\max}\) from being achieved regardless of the substrate concentration.

1 mark for the correct explanation of competitive and non-competitive inhibitors based on the experimental results.

The mRNA transcribed from the template DNA 3'- T A C C G G T T T C G A A C T -5' in the 5' to 3' direction is 5'- A U G G C C A A A G C U U G A -3'. The tRNA anticodons align antiparallel (3' to 5') and complementary to the mRNA codons, resulting in: 3'- U A C -5', 3'- C G G -5', 3'- U U U -5', 3'- C G A -5', and 3'- A C U -5'. Note that the tRNA anticodons (3' to 5') have the same base sequence as the DNA template (3' to 5'), except that Uracil (U) replaces Thymine (T) in RNA.

1 mark for selecting the correct complementary tRNA anticodons in the 3' to 5' direction, ensuring that Uracil is used instead of Thymine.

Mature xylem vessel elements are dead cells that lose their protoplast (including the nucleus) and have highly lignified cellulose walls. Mature phloem sieve tube elements are living cells, but they lose their nucleus during development to facilitate translocation; they are not lignified and retain their cellulose cell walls. Therefore, both cell types lack a nucleus at maturity, xylem is lignified while phloem is not, and both contain cellulose in their cell walls.

1 mark for correctly identifying the structural features of mature xylem and phloem elements.

Fluid X is blood plasma because it contains red blood cells and a high concentration of large soluble plasma proteins that cannot pass through the capillary endothelium. Fluid Y is tissue fluid, which is formed by the ultrafiltration of blood plasma; it lacks red blood cells and contains very few proteins. Fluid Z is lymph, which drains from tissue fluid and contains a high concentration of lymphocytes produced in the lymph nodes.

1 mark for the correct identification of all three body fluids.

1. MMR vaccine introduces antigens artificially, and the host actively produces antibodies and memory cells (active artificial). 2. Antivenom provides ready-made antibodies artificially to neutralize venom without activating the host's immune response (passive artificial). 3. Colostrum naturally transfers maternal antibodies to the baby, providing temporary protection without memory cells (passive natural). 4. Influenza infection naturally introduces pathogens, prompting the host to actively produce antibodies and memory cells (active natural).

1 mark for correctly classifying all four immunity scenarios.

Centrioles are non-membrane-bound organelles consisting of a cylinder of nine triplets of microtubules arranged in a ring (9+0 arrangement). They organize the spindle fibres during cell division in animal cells and some lower plants, but are absent from most flowering plants (which can still organize spindles without centrioles). Spindle fibres themselves are composed of the protein tubulin, not actin. The nucleolus is responsible for ribosome assembly, not spindle fibre synthesis.

Award 1 mark for the correct option A. Reject all other options.

Amylose is composed of \(\alpha\)-glucose monomers linked by \(\alpha\)-1,4-glycosidic bonds, which causes the chain to coil into a helical shape without requiring monomer rotation. Cellulose is composed of \(\beta\)-glucose monomers linked by \(\beta\)-1,4-glycosidic bonds. To allow the formation of these bonds, alternating glucose monomers must rotate 180 degrees, resulting in a straight chain that can form hydrogen bonds with adjacent chains to produce microfibrils. Amylose is unbranched and does not form microfibrils.

Award 1 mark for the correct option C. Reject all other options.

Collagen is a fibrous structural protein. Its primary structure consists of a repeating triplet sequence of amino acids, where every third amino acid is glycine. The small R-group of glycine (a single hydrogen atom) allows the three polypeptide helical chains to pack very tightly together to form a triple helix. Collagen is not a globular protein, does not contain haem prosthetic groups, and is held together primarily by hydrogen bonds and covalent cross-links rather than disulfide bonds in its triple helical structure.

Award 1 mark for the correct option B. Reject all other options.

Mechanism 1 is facilitated diffusion, which relies on channel or carrier proteins but is passive (no ATP) and only moves molecules down their concentration gradient. Mechanism 2 is active transport, which requires carrier proteins (pumps) and ATP energy to move substances against their concentration gradient. Mechanism 3 is simple diffusion, which is passive and occurs directly through the phospholipid bilayer without the aid of membrane proteins.

Award 1 mark for the correct option B. Reject all other options.

Chromosomes aligning along the equator of the spindle is the defining characteristic of metaphase. At this stage, the nuclear envelope, which began breaking down during prophase, has completely disintegrated. This allows spindle microtubules originating from opposite poles of the cell to attach to the centromeres of the chromosomes.

Award 1 mark for the correct option A. Reject all other options.

1. Transcription of the template DNA strand (3'-TAC GGC TTA ACG-5') produces mRNA in an antiparallel orientation: 5'-AUG CCG AAU UGC-3'.
2. During translation, tRNA molecules pair antiparallel to the mRNA codons:
- Codon 1: 5'-AUG-3' pairs with tRNA anticodon 3'-UAC-5'
- Codon 2: 5'-CCG-3' pairs with tRNA anticodon 3'-GGC-5'
- Codon 3: 5'-AAU-3' pairs with tRNA anticodon 3'-UUA-5'
- Codon 4: 5'-UGC-3' pairs with tRNA anticodon 3'-ACG-5'
3. Putting the anticodons together in the 3' to 5' direction gives: 3'-UAC GGC UUA ACG-5'. This sequence is identical to the template DNA sequence, but with Uracil (U) replacing Thymine (T).

Award 1 mark for the correct option B. Reject all other options.

Capillaries are the sites of exchange between blood and tissues. Their walls consist of a single layer of flattened (squamous) endothelial cells, minimizing the diffusion distance for oxygen, glucose, carbon dioxide, and metabolic wastes. Arteries have thick muscular walls to withstand high pressure, but blood is not moved by peristalsis. Veins have thin muscular walls and a wide lumen to offer low resistance to blood flow, and they contain valves to prevent backflow. Arterioles have muscle fibres in their walls to constrict or dilate, controlling blood flow, but they do not contain valves.

Award 1 mark for the correct option C. Reject all other options.

Penicillin functions by acting as an inhibitor of transpeptidase, the enzyme responsible for catalysing the cross-linking of peptidoglycan chains in the bacterial cell wall. This prevents the bacteria from synthesising strong, functional cell walls. When the bacteria grow and division is attempted, the weakened cell wall cannot resist osmotic pressure, resulting in osmotic lysis. It is only effective against growing cells and does not target ribosomes, mature cell walls, or RNA polymerase.

Award 1 mark for the correct option B. Reject all other options.

Part (a) Xylem vessel elements are highly adapted to their water-transporting function: 1. Lignified cell walls prevent the vessels from collapsing under negative pressure (tension) and provide structural support. 2. They are dead, hollow tubes with no cytoplasm or organelles, minimizing resistance to water flow. 3. End walls between adjacent elements are lost, forming continuous tubes for uninterrupted water columns. 4. Pits in the lignified walls allow lateral movement of water to adjacent vessels or cells.
Part (b)(i) Sieve tube elements are adapted for translocation by having: 1. Very little cytoplasm and no nucleus, vacuole, or ribosomes, leaving a clear lumen to reduce resistance to flow. 2. Sieve plates at the end walls with sieve pores, which allow the bulk flow of organic solutes. 3. Plasmodesmata connecting them to companion cells.
Part (b)(ii) Companion cells support sieve tube elements by: 1. Carrying out metabolic processes (due to having mitochondria, ribosomes, etc.) to keep the sieve tube element alive. 2. Containing many mitochondria to produce ATP for active transport of sucrose. 3. Using proton pumps and co-transporters to load sucrose into the sieve tube element against its concentration gradient.

Part (a) (Max 4 marks):
1. Lignin / lignified walls; to prevent collapse under tension / provide support;
2. No cytoplasm / no organelles / hollow / dead cells; to reduce resistance to water flow;
3. No end walls / continuous tube; to allow uninterrupted water columns;
4. Pits in walls; to allow lateral movement of water;

Part (b)(i) (Max 3 marks):
1. Little cytoplasm / cytoplasm restricted to thin peripheral layer / no nucleus / no vacuole; to maximize space / reduce resistance to flow of sap;
2. Sieve plates / sieve pores; to allow rapid movement / bulk flow of organic solutes;
3. Plasmodesmata; to allow transport / communication with companion cells;

Part (b)(ii) (Max 3 marks):
1. Carry out metabolic reactions / provide proteins / keep sieve tube element alive;
2. Mitochondria present; to produce ATP / for aerobic respiration;
3. Active loading of sucrose / operation of proton pumps / cotransporter proteins;

Part (a) As substrate concentration increases, the rate of reaction increases because there are more substrate molecules to collide with enzyme active sites, increasing the frequency of effective collisions and the rate of enzyme-substrate complex formation. However, at high substrate concentrations, the rate levels off and reaches a maximum value (\(V_{max}\)). At this point, all active sites are saturated, and enzyme concentration becomes the limiting factor.
Part (b)(i) Competitive inhibitors have a molecular shape similar to the substrate and bind temporarily to the active site, blocking the substrate. At low substrate concentrations, the inhibitor successfully competes with the substrate, significantly reducing the reaction rate. At high substrate concentrations, the substrate molecules outnumber the inhibitor molecules, outcompeting them, so the reaction rate approaches the normal maximum rate.
Part (b)(ii) For a competitive inhibitor, the maximum rate of reaction (\(V_{max}\)) remains unchanged, but the Michaelis-Menten constant (\(K_m\)) increases (reflecting lower affinity). For a non-competitive inhibitor, the maximum rate of reaction (\(V_{max}\)) decreases, while the Michaelis-Menten constant (\(K_m\)) remains unchanged.

Part (a) (Max 4 marks):
1. Low substrate conc: rate increases as substrate conc increases; due to more successful collisions / more enzyme-substrate complexes (ESCs) formed;
2. Substrate is the limiting factor at low concentrations;
3. High substrate conc: rate plateaus / reaches maximum / \(V_{max}\);
4. All active sites are saturated; enzyme concentration is the limiting factor;

Part (b)(i) (Max 3 marks):
1. Inhibitor Y has a similar shape to substrate and binds to active site;
2. Low substrate conc: rate is reduced because inhibitor blocks active sites;
3. High substrate conc: rate is restored / unaffected because substrate outcompetes the inhibitor;

Part (b)(ii) (Max 3 marks):
1. Competitive: \(V_{max}\) unchanged AND \(K_m\) increases;
2. Non-competitive: \(V_{max}\) decreases AND \(K_m\) unchanged;
3. Correct reference to the active site vs allosteric site binding explaining the difference;

Part (a)(i) The complementary mRNA sequence is 5'-AUG CCG AAU GAU UGA-3'.
Part (a)(ii) The first three codons are 5'-AUG-3', 5'-CCG-3', and 5'-AAU-3'. The corresponding tRNA anticodons are 3'-UAC-5' (or UAC), 3'-GGC-5' (or GGC), and 3'-UUA-5' (or UUA).
Part (b) Ribosomes facilitate translation by: 1. Binding to the mRNA molecule and moving along it codon by codon in the 5' to 3' direction. 2. Holding two tRNA molecules in place at the same time (at the A and P sites). 3. Allowing complementary base pairing between mRNA codons and tRNA anticodons. 4. Containing the enzyme peptidyl transferase, which catalyses the formation of a peptide bond between adjacent amino acids.
Part (c) A substitution mutation may not change the function of the protein because: 1. The genetic code is degenerate / redundant, meaning multiple codons can code for the same amino acid. Thus, the mutation may be silent and code for the exact same amino acid. 2. If a different amino acid is coded for (missense), it may have similar chemical properties (e.g., charge, hydrophobicity) or be located away from the active/binding site, preserving the protein's overall tertiary structure and function.

Part (a)(i) (1 mark):
1. 5'-AUG CCG AAU GAU UGA-3' (allow without 5' and 3' if written in correct orientation left-to-right);

Part (a)(ii) (2 marks):
1. First anticodon: UAC (or 3'-UAC-5');
2. Second and third anticodons: GGC and UUA (both correct for the second mark);

Part (b) (Max 5 marks):
1. Ribosome binds to mRNA / moves along mRNA;
2. Reads codons / moves one codon at a time in 5' to 3' direction;
3. Has slots / sites (P and A sites) to hold two tRNAs;
4. Facilitates complementary base pairing between mRNA codon and tRNA anticodon;
5. Peptidyl transferase; catalyses peptide bond formation;
6. Between amino acids carried by adjacent tRNAs;

Part (c) (Max 2 marks):
1. Silent mutation / degenerate (redundant) genetic code;
2. Different codon still codes for the same amino acid;
3. Missense mutation codes for a chemically similar amino acid / change occurs in non-essential part of polypeptide (not affecting tertiary structure);

Part (a) Bacterial cell walls are made of peptidoglycan, which consists of parallel polysaccharide chains of alternating amino sugars cross-linked by short peptide chains. Penicillin acts as an inhibitor of the enzyme transpeptidase (glycoprotein peptidase), which forms these peptide cross-links. In growing cells, as the autolysins break down parts of the cell wall to allow growth, penicillin prevents new cross-links from forming. This weakens the cell wall, causing the bacterium to burst due to osmotic pressure (lysis).
Part (b) Penicillin is ineffective against viruses because: 1. Viruses do not have a cell wall (they have a protein coat or capsid). 2. Viruses lack metabolic pathways, ribosomes, and enzymes like transpeptidase that penicillin targets. 3. Viruses reproduce inside host eukaryotic cells, which also lack peptidoglycan cell walls.
Part (c) \(\beta\)-lactamase is an enzyme produced by resistant bacteria that hydrolyses and breaks open the \(\beta\)-lactam ring of the penicillin molecule, rendering the antibiotic inactive before it can bind to transpeptidase. Another mechanism of resistance is the use of efflux pumps, which actively pump the antibiotic out of the bacterial cell, or a mutation that alters the structure of the target enzyme (transpeptidase) so penicillin can no longer bind to it.

Part (a) (Max 4 marks):
1. Peptidoglycan contains polysaccharide chains cross-linked by peptides;
2. Penicillin inhibits transpeptidase / glycoprotein peptidase;
3. Prevents formation of peptide cross-links;
4. Cell wall is weakened;
5. Water enters by osmosis and cell bursts / lyses;

Part (b) (Max 3 marks):
1. Viruses do not have cell walls / peptidoglycan / cell membranes;
2. Viruses lack their own metabolism / do not have enzymes/ribosomes;
3. Viruses are inside host cells which are not affected by penicillin;

Part (c) (Max 3 marks):
1. \(\beta\)-lactamase breaks down / hydrolyses the \(\beta\)-lactam ring of penicillin;
2. Inactivates penicillin / prevents it binding to transpeptidase;
3. Other mechanism: Efflux pumps (to pump drug out) / mutation changing shape of transpeptidase (target site modification) / decrease in membrane permeability;

Part (a) Monoclonal antibody production involves: 1. Injecting a mouse with the target antigen to stimulate an immune response. 2. Harvesting plasma cells (B-lymphocytes) from the mouse's spleen. 3. Fusing these B-lymphocytes with cancerous myeloma cells (using a fusogen like polyethylene glycol, PEG) to produce hybridoma cells. 4. Screening and identifying hybridoma cells that produce the desired antibody. 5. Cloning the selected hybridoma cells to produce large quantities of identical antibodies.
Part (b) An IgG antibody consists of four polypeptide chains (two heavy and two light chains) held together by disulfide bonds, forming a Y-shape. At the ends of the 'Y' branches are the variable regions. These regions have a specific amino acid sequence and a unique tertiary structure that forms an antigen-binding site complementary in shape to a specific epitope of the target antigen.
Part (c) Active immunity involves the host's own immune system producing antibodies and memory cells in response to an antigen (either naturally via infection or artificially via vaccination), providing long-term protection. Passive immunity involves receiving pre-formed antibodies from an external source (e.g., breast milk or injection of antitoxins), which provides immediate but temporary protection as no memory cells are produced.

Part (a) (Max 5 marks):
1. Inject animal/mouse with antigen;
2. Isolate/harvest B-lymphocytes/plasma cells from spleen;
3. Fuse with tumor / myeloma cells;
4. Using PEG / fusogen / electrofusion;
5. Select / screen for hybridoma cells producing desired antibody;
6. Clone selected hybridoma cell (by mitosis) to produce identical cells;

Part (b) (Max 3 marks):
1. Four polypeptide chains / two heavy and two light chains connected by disulfide bonds;
2. Variable regions (at the tips);
3. Specific amino acid sequence / tertiary structure / 3D shape;
4. Complementary to a specific antigen / epitope;

Part (c) (Max 2 marks):
1. Active involves production of antibodies by the individual's own plasma cells AND memory cells are produced (long-term);
2. Passive involves introduction of external antibodies AND no memory cells are produced (short-term);

Part (a) 1. Rough endoplasmic reticulum (RER): Ribosomes on the RER surface translate the mRNA to synthesize the amylase polypeptide chain. The chain enters the lumen of the RER, where it folds into its tertiary shape. 2. Golgi body: Amylase is transported from the RER to the Golgi body in transport vesicles. The Golgi body chemically modifies the protein (e.g., glycosylation) and packages the mature amylase into secretory vesicles. 3. Secretory vesicles: These vesicles bud off from the Golgi trans-face, move through the cytoplasm, and fuse with the cell surface membrane to release amylase via exocytosis.
Part (b) Eukaryotic cells contain membrane-bound organelles, whereas prokaryotic cells do not. Specifically: 1. Eukaryotic pancreatic cells have a true nucleus enclosed by a double membrane (nuclear envelope), while prokaryotic cells have circular DNA free in the cytoplasm (nucleoid region). 2. Eukaryotes have membrane-bound organelles such as mitochondria, RER, and Golgi, which are completely absent in prokaryotic cells.
Part (c) Microtubules form an internal network of tracks (part of the cytoskeleton) along which motor proteins (like kinesin or dynein) attached to secretory vesicles can travel, using energy from ATP hydrolysis to pull the vesicles to the plasma membrane.

Part (a) (Max 6 marks):
1. RER: Ribosomes synthesize the polypeptide / protein;
2. Protein enters RER lumen where it folds;
3. Transport vesicles bud off RER and carry protein to Golgi body;
4. Golgi body: Modifies protein (e.g., adds sugars / glycosylation);
5. Golgi packages protein into secretory vesicles;
6. Secretory vesicles: Move to and fuse with cell surface membrane (exocytosis);

Part (b) (Max 2 marks):
1. Pancreatic cell has nuclear envelope / double membrane around nucleus, prokaryote has no nuclear membrane / nucleoid;
2. Pancreatic cell has membrane-bound organelles (mitochondria/RER/Golgi) whereas prokaryotes lack all membrane-bound organelles;

Part (c) (Max 2 marks):
1. Microtubules act as tracks / pathways;
2. Motor proteins attach to vesicles and move along them;
3. Uses ATP / energy to transport vesicles to membrane;

**(a)** To prepare \(10.0\text{ cm}^3\) of each concentration from \(5.0\%\) stock protease solution **P**, use the formula: \(C_1 V_1 = C_2 V_2\).
For a \(1.0\%\) solution: \(5.0 \times V_1 = 1.0 \times 10.0 \implies V_1 = 2.0\text{ cm}^3\) of **P**. The volume of water **W** needed is \(10.0 - 2.0 = 8.0\text{ cm}^3\).
Repeating this calculation for each concentration yields:
- \(1.0\%\): \(2.0\text{ cm}^3\) of **P**, \(8.0\text{ cm}^3\) of **W**
- \(2.0\%\): \(4.0\text{ cm}^3\) of **P**, \(6.0\text{ cm}^3\) of **W**
- \(3.0\%\): \(6.0\text{ cm}^3\) of **P**, \(4.0\text{ cm}^3\) of **W**
- \(4.0\%\): \(8.0\text{ cm}^3\) of **P**, \(2.0\text{ cm}^3\) of **W**
- \(5.0\%\): \(10.0\text{ cm}^3\) of **P**, \(0.0\text{ cm}^3\) of **W**

**(b)** Method:
1. Place a fixed volume (e.g., \(5.0\text{ cm}^3\)) of the \(1.0\%\) milk suspension **M** and \(2.0\text{ cm}^3\) of pH 7.5 buffer into a test tube.
2. Draw a distinct black cross on a piece of white paper and place it behind the test tube.
3. Add a fixed volume (e.g., \(2.0\text{ cm}^3\)) of the prepared protease concentration. Start a stopwatch immediately.
4. Observe the tube and stop the stopwatch when the mixture transitions from turbid/cloudy to clear, allowing the black cross to become clearly visible.
5. Keep temperature constant by placing the tube in a water bath.
6. Repeat the procedure three times at each concentration to calculate a mean.

**(c)** Graph Construction:
- **X-axis**: Protease concentration / %
- **Y-axis**: Rate of reaction / \(\text{s}^{-1}\)
- **Scale**: Linear, even, and large enough to occupy at least half of the grid (e.g. x-axis: 2 cm = 1.0%, y-axis: 2 cm = 4.0 \(\text{s}^{-1}\)).
- **Plotting**: Plot points carefully with a small 'x' or circled dot.
- **Line**: Join points with straight, ruled lines from point to point, with no extrapolation.

**(d)** Explanation:
- As the concentration of protease increases, the rate of reaction increases.
- This is because a higher enzyme concentration provides a greater number of active sites.
- There is a higher frequency of successful collisions between the protease active sites and the protein (casein) substrate molecules.
- This leads to the formation of more enzyme-substrate complexes (ESCs) per unit time, resulting in a faster rate of hydrolysis.

**(e)** Sources of Error and Modifications:
1. *Error*: Difficulty in determining the exact endpoint (when the suspension is fully 'clear' is subjective).
*Modification*: Use a colorimeter to measure the change in light absorbance or transmission over time.
2. *Error*: Fluctuations in room temperature affecting enzyme activity.
*Modification*: Use a thermostatically-controlled water bath to keep the temperature of the tubes constant.

**(a) Dilution Table [Total: 4 marks]**
- **1 mark**: Table drawn with clear column headings and correct units: Protease concentration / %, Volume of protease stock **P** / \(\text{cm}^3\), and Volume of distilled water **W** / \(\text{cm}^3\).
- **1 mark**: Shows five concentrations: 1.0, 2.0, 3.0, 4.0, and 5.0.
- **1 mark**: Calculates correct volumes of P (2.0, 4.0, 6.0, 8.0, 10.0) respectively.
- **1 mark**: Calculates correct volumes of W (8.0, 6.0, 4.0, 2.0, 0.0) respectively, showing consistent decimal places.

**(b) Practical Method [Total: 5 marks]**
- **1 mark**: Standardizes independent variable (describes how to change concentration of protease) and dependent variable (time taken for milk suspension to clear).
- **1 mark**: Details standardizing controlled variables (using a constant volume of milk suspension, and a constant volume of buffer to maintain constant pH).
- **1 mark**: Provides a clear method to detect the endpoint objectively (e.g., placing a black cross behind the tube and stopping the timer when it becomes visible).
- **1 mark**: Replicates the experiment at least three times at each concentration to obtain a mean or identify anomalies.
- **1 mark**: Details keeping temperature constant (e.g., using a water bath).

**(c) Graph Plotting [Total: 4 marks]**
- **1 mark**: Axes correctly labeled with units: x-axis: "Protease concentration / %", y-axis: "Rate of reaction / \(\text{s}^{-1}\)".
- **1 mark**: Scale is linear and occupies more than half of the grid in both directions.
- **1 mark**: All points plotted accurately to within 0.5 of a small square.
- **1 mark**: Points joined point-to-point with straight, sharp, ruled lines, with no extrapolation beyond the first or last point.

**(d) Explanation [Total: 4 marks]**
- **1 mark**: Identifies the positive correlation/proportional relationship between enzyme concentration and reaction rate.
- **1 mark**: Explains that higher protease concentration means more active sites are available.
- **1 mark**: Mentions increased frequency of successful/effective collisions between enzyme active sites and substrate molecules.
- **1 mark**: Refers to more enzyme-substrate complexes (ESCs) being formed per unit time.

**(e) Errors and Improvements [Total: 4 marks]**
- **1 mark**: Identifies subjective endpoint / human error in judging when the cross is visible.
- **1 mark**: Suggests using a colorimeter to measure absorbance or transmission of light quantitatively.
- **1 mark**: Identifies temperature fluctuation during the investigation as an error.
- **1 mark**: Suggests using a thermostatically-controlled water bath.

Part (a) Low-power plan drawing:
The drawing should show a clear transverse section of slide J1. The lines must be sharp, continuous, and drawn with a sharp pencil (no shading). It must feature:
- A thick outer cuticle represented on both upper and lower epidermis.
- Palisade mesophyll as a distinct layer beneath the upper epidermis.
- A central midrib with a well-developed vascular bundle (xylem on the upper side, phloem on the lower side).
- Sunken stomatal pits on the lower epidermis (if present on slide).
- Correct labels for 'xylem' (pointing to the vessel elements inside the bundle) and 'palisade mesophyll' (pointing to the palisade layer).

Part (b) High-power cell drawing:
- Only three adjacent cells must be drawn.
- Cell walls must be represented by double lines.
- The cells must be elongated, closely packed, and oriented vertically.
- One clear label line pointing directly to the cell wall of a cell.

Part (c) Comparison Table & Adaptation:
Example table of differences:
Anatomical FeatureSlide J1 (Xerophytic leaf)Fig. 2.1 (Aquatic leaf)Cuticle thicknessThick cuticleVery thin / absent cuticleStomata positionSunken in pits on lower surfaceOn upper surface only, not sunkenMesophyll air spacesCompact with small air spacesLarge air chambers / aerenchyma presentVascular bundle developmentLarge / well-developed xylemReduced / small vascular tissue
Physiological advantage of a feature from slide J1:
- Thick cuticle: Reduces water loss via cuticular transpiration by forming a barrier to water vapor diffusion.
- Sunken stomata: Traps moist air inside the pits, reducing the water potential gradient between the leaf interior and the atmosphere, thereby reducing the rate of transpiration.

Part (d) Calculation:
1. Calculate the value of 1 epu in micrometres (\(\mu\text{m}\)):
\(10\text{ epu} = 0.25\text{ mm} = 250\ \mu\text{m}\)
\(1\text{ epu} = \frac{250\ \mu\text{m}}{10} = 25\ \mu\text{m}\)

2. Calculate the actual thickness of 18 epu:
\(\text{Actual thickness} = 18 \times 25\ \mu\text{m} = 450\ \mu\text{m}\)

(a) Plan Drawing [6 marks]:

- MMO (Decision): Sharp, thin, continuous lines drawn with a sharp pencil. No shading or sketchy lines. [1]

- MMO (Observation): Large drawing occupying at least 50% of the available space. [1]

- MMO (Observation): Correct tissue plan showing upper epidermis, palisade mesophyll, spongy mesophyll, lower epidermis, and a distinct central vascular bundle. [1]

- PDO (Layout): No individual cells drawn anywhere in the tissue plan. [1]

- PDO (Layout): Correct proportions of layers (e.g., palisade layer is wider than the upper epidermis; vascular bundle is centralized). [1]

- ACE (Interpretation): Label lines are straight, do not cross, and point precisely to 'xylem' (inner region of vascular bundle) and 'palisade mesophyll'. [1]



(b) High-Power Drawing [4 marks]:

- MMO (Decision): Only three adjacent palisade cells drawn, showing double lines to represent cell wall thickness. [1]

- MMO (Observation): Cells are drawn columnar/elongated in shape and arranged closely side-by-side. [1]

- PDO (Layout): Clean lines, drawing is large (occupying at least 5 cm of height). [1]

- ACE (Interpretation): Correct label with a straight line pointing directly to 'cell wall'. [1]



(c) Comparison and Adaptation [6 marks]:

- PDO (Layout): Comparison presented as a table with clear headings ('Feature', 'Slide J1', 'Fig. 2.1'). [1]

- ACE (Interpretation): Identifies three distinct, contrasting anatomical differences between Slide J1 and Fig. 2.1 (e.g., Cuticle: thick vs thin; Stomata: lower surface/sunken vs upper surface/not sunken; Air spaces: small/compact vs large chambers; Vascular tissue: well-developed vs reduced). [3 marks, 1 for each correct comparison]

- ACE (Interpretation): Correctly selects one feature from slide J1 (e.g., thick cuticle or sunken stomata). [1]

- ACE (Interpretation): Explains how this feature reduces transpiration / water loss by trapping moist air or reducing water potential gradient. [1]



(d) Calibration and Calculation [3 marks]:

- Step 1: Shows calibration of 1 epu (e.g., \(0.25\text{ mm} = 250\ \mu\text{m}\) or \(1\text{ epu} = 25\ \mu\text{m}\) or \(0.025\text{ mm}\)). [1]

- Step 2: Shows multiplication of 18 by the value of 1 epu (e.g., \(18 \times 25\)). [1]

- Step 3: Correct final answer of \(450\ \mu\text{m}\) with appropriate units. (Accept \(4.5 \times 10^2\ \mu\text{m}\)). [1]