Cambridge IAS-Level · Thinka-original Practice Paper

2023 Cambridge IAS-Level Biology (9700) Practice Paper with Answers

Thinka Nov 2023 (V3) Cambridge International A Level-Style Mock — Biology (9700)

140 marks270 mins2023
An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V3) Cambridge International A Level Biology (9700) paper. Not affiliated with or reproduced from Cambridge.

Paper 1 (Multiple Choice)

Answer all forty multiple-choice questions. For each question, choose the correct option from A, B, C, or D.
40 Question · 40 marks
Question 1 · multiple choice
1 marks
An immunoglobulin G (IgG) antibody molecule is treated with a chemical reagent that selectively breaks only the disulfide bonds linking the light chains to the heavy chains, while leaving the heavy-to-heavy chain disulfide bonds fully intact. Which of the following describes the fragments that would be produced after this treatment?
  1. A.Two separate light chains and one dimer of two heavy chains
  2. B.Two separate light chains and two separate heavy chains
  3. C.Two fragments, each consisting of one light chain bound to one heavy chain
  4. D.Four separate polypeptide chains and no intact disulfide bonds
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Worked solution

An IgG antibody is a Y-shaped molecule composed of two identical heavy chains and two identical light chains. The light chains are connected to the heavy chains by disulfide bonds. The two heavy chains are connected to each other by disulfide bonds located in the hinge region. If a chemical selectively breaks only the disulfide bonds between the light and heavy chains while leaving the heavy-heavy disulfide bonds intact: 1. The two light chains will be completely released as two separate, free light chains. 2. The two heavy chains will remain joined together by their intact hinge disulfide bonds, forming a single dimer of heavy chains. Therefore, the resulting mixture contains two separate light chains and one dimer of two heavy chains.

Marking scheme

1 mark for identifying the correct products: two separate light chains and one dimer of heavy chains (Option A). Award 0 marks for options that incorrectly suggest heavy chains separate (B), light and heavy chains remain paired (C), or all four chains separate (D).
Question 2 · multiple choice
1 marks
An enzyme-catalysed reaction was studied under different conditions. The Michaelis-Menten constant (\(K_m\)) and maximum velocity (\(V_{\max}\)) were determined under control conditions and in the presence of two chemical substances, X and Y:

* Control: \(K_m = 2.0 \text{ mmol dm}^{-3}\), \(V_{\max} = 100 \mu\text{mol dm}^{-3} \text{ s}^{-1}\)
* With substance X: \(K_m = 5.5 \text{ mmol dm}^{-3}\), \(V_{\max} = 100 \mu\text{mol dm}^{-3} \text{ s}^{-1}\)
* With substance Y: \(K_m = 2.0 \text{ mmol dm}^{-3}\), \(V_{\max} = 45 \mu\text{mol dm}^{-3} \text{ s}^{-1}\)

Which statement correctly describes the binding sites of substances X and Y on the enzyme?
  1. A.X binds to the active site; Y binds to a site other than the active site.
  2. B.X binds to a site other than the active site; Y binds to the active site.
  3. C.X and Y both bind to the active site of the enzyme.
  4. D.X and Y both bind to sites other than the active site of the enzyme.
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Worked solution

1. Substance X increases the \(K_m\) of the enzyme from \(2.0\) to \(5.5 \text{ mmol dm}^{-3}\) but leaves the \(V_{\max}\) unchanged at \(100 \mu\text{mol dm}^{-3} \text{ s}^{-1}\). This is characteristic of a competitive inhibitor, which competes with the substrate by binding directly to the active site.
2. Substance Y decreases the \(V_{\max}\) from \(100\) to \(45 \mu\text{mol dm}^{-3} \text{ s}^{-1}\) but leaves the \(K_m\) unchanged at \(2.0 \text{ mmol dm}^{-3}\). This is characteristic of a non-competitive inhibitor, which binds to an allosteric site (a site other than the active site) and reduces catalytic activity without affecting substrate binding affinity.
Therefore, X binds to the active site, and Y binds to a site other than the active site.

Marking scheme

1 mark for selecting Option A. Accept only A. Reject options B, C, and D as they confuse the binding sites of competitive and non-competitive inhibitors based on the kinetic parameters provided.
Question 3 · multiple choice
1 marks
Which row correctly describes structural features of collagen and haemoglobin?
  1. A.Collagen: Every third amino acid is glycine, which allows the three polypeptide chains to pack closely together. Haemoglobin: Hydrophobic R-groups are oriented towards the center of the molecule, leaving hydrophilic R-groups on the outside.
  2. B.Collagen: Three alpha-helices are held together by disulfide bonds to form a strong, stretchable fibre. Haemoglobin: Four polypeptide chains are linked by covalent peptide bonds between their R-groups.
  3. C.Collagen: It is a globular protein that is highly soluble in water because of its outer hydrophilic R-groups. Haemoglobin: It is a fibrous protein with high tensile strength due to the presence of hydrophobic prosthetic groups.
  4. D.Collagen: Individual triple helices are joined by hydrogen bonds to form myofibrils. Haemoglobin: The four polypeptide chains are identical, and each contains an iron ion (\(Fe^{3+}\)) in its haem group.
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Worked solution

In collagen, every third amino acid is glycine. Glycine is the smallest amino acid (its R-group is a single hydrogen atom), which allows the three polypeptide chains to pack very closely together to form a tight, stable triple helix. In haemoglobin (a soluble globular protein), hydrophobic R-groups are oriented towards the center of the molecule to form a hydrophobic core, while hydrophilic R-groups are on the outside, facilitating solubility in aqueous media. Option B is incorrect because collagen consists of three left-handed helices (not alpha-helices) and haemoglobin chains are not linked by covalent peptide bonds between R-groups. Option C is incorrect because collagen is insoluble and fibrous, not soluble and globular. Option D is incorrect because collagen is not organized into myofibrils (which are muscle structures) and haemoglobin contains \(Fe^{2+}\) ions, not \(Fe^{3+}\), and has two different types of chains (two alpha and two beta).

Marking scheme

1 mark for identifying the correct comparative features of collagen and haemoglobin (Option A). Reject all other options due to incorrect scientific assertions about protein structure.
Question 4 · multiple choice
1 marks
A student calibrated an eyepiece graticule using a stage micrometer. The stage micrometer had scale divisions spaced exactly \(0.1 \text{ mm}\) apart. At a magnification of \(\times 100\), 20 divisions of the eyepiece graticule coincided exactly with 4 divisions of the stage micrometer. The student then replaced the stage micrometer with a slide of plant cells and viewed it at \(\times 400\) magnification. An individual chloroplast was measured to be 3 eyepiece graticule divisions in length. What is the actual length of the chloroplast in micrometres (\(\mu\text{m}\))?
  1. A.3.75 \(\mu\text{m}\)
  2. B.15.0 \(\mu\text{m}\)
  3. C.60.0 \(\mu\text{m}\)
  4. D.240.0 \(\mu\text{m}\)
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Worked solution

1. Under \(\times 100\) magnification, 4 divisions of the stage micrometer correspond to: \(4 \times 0.1 \text{ mm} = 0.4 \text{ mm} = 400 \mu\text{m}\).
2. These 4 divisions coincide with 20 eyepiece graticule divisions (epd). Therefore: \(1 \text{ epd} = 400 \mu\text{m} / 20 = 20 \mu\text{m}\) at \(\times 100\).
3. When the magnification is increased from \(\times 100\) to \(\times 400\) (an increase by a factor of 4), the physical size represented by each eyepiece division decreases by the same factor: \(1 \text{ epd at } \times 400 = 20 \mu\text{m} / 4 = 5 \mu\text{m}\).
4. The chloroplast measures 3 epd under \(\times 400\) magnification. Its actual length is: \(3 \times 5 \mu\text{m} = 15 \mu\text{m}\).

Marking scheme

1 mark for calculating the correct actual chloroplast length of \(15.0 \mu\text{m}\) (Option B). Award 0 marks for Option C (which fails to adjust for the change in magnification) or options A and D (which apply the magnification factor incorrectly).
Question 5 · multiple choice
1 marks
Which statement correctly describes the mechanism of action of penicillin on bacteria?
  1. A.It acts as a competitive inhibitor of the enzyme transpeptidase, preventing the cross-linking of peptidoglycan chains in growing cell walls.
  2. B.It binds to the small ribosomal subunit (30S), preventing the binding of tRNA and halting bacterial protein synthesis.
  3. C.It disrupts the phospholipid bilayer of the outer membrane of Gram-negative bacteria, causing leakage of essential ions.
  4. D.It inhibits bacterial DNA gyrase, preventing the replication and transcription of the bacterial circular chromosome.
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Worked solution

Penicillin works by acting as a competitive inhibitor of the enzyme transpeptidase (also known as glycopeptide transpeptidase). This enzyme is responsible for forming the cross-links between peptidoglycan chains in the bacterial cell wall. By preventing this cross-linking, the cell wall becomes structurally weak. When water subsequently enters the bacterium by osmosis, the cell cannot withstand the internal turgor pressure and bursts (lysis). Options B, C, and D describe the mechanisms of action of other antibiotics (such as tetracyclines, polymyxins, and quinolones, respectively).

Marking scheme

1 mark for selecting the correct mechanism of action of penicillin (Option A). All other options represent incorrect modes of action for penicillin.
Question 6 · multiple choice
1 marks
Which of the following correctly describes the movement of water and dissolved mineral ions through the apoplast pathway in the root of a herbaceous plant?
  1. A.Water moves through the non-living parts of the cells, including cell walls and intercellular spaces, until it reaches the endodermis where it is forced into the symplast pathway by the Casparian strip.
  2. B.Water moves through the cytoplasm and plasmodesmata of adjacent cells, driven by a concentration gradient of mineral ions established by active transport.
  3. C.Water and solutes are actively pumped across the cell surface membrane into the vacuole and then pass from vacuole to vacuole via specialized transport proteins.
  4. D.Water moves through the xylem vessel elements against gravity, driven entirely by positive hydrostatic pressure generated by the Casparian strip.
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Worked solution

The apoplast pathway refers to the movement of water and solutes through the non-living parts of the plant tissue, specifically the cellulose cell walls and intercellular spaces. When water moving via the apoplast reaches the endodermis, its progress is blocked by the waterproof Casparian strip, which contains suberin. This forces the water and dissolved mineral ions to cross the selectively permeable cell surface membrane of the endodermal cells to enter the cytoplasm, effectively switching them into the symplast pathway. This allows the plant to regulate which solutes enter the vascular bundle.

Marking scheme

1 mark for the correct description of the apoplast pathway and the role of the Casparian strip at the endodermis (Option A). Reject B (describes symplast pathway), C (describes vacuolar pathway), and D (incorrect description of xylem transport driver).
Question 7 · multiple choice
1 marks
Carbon dioxide produced by actively respiring tissues diffuses into capillaries and enters red blood cells. Which sequence of events correctly describes what happens to this carbon dioxide inside a red blood cell in a systemic capillary?
  1. A.Carbon dioxide reacts with water, catalysed by carbonic anhydrase, to form carbonic acid, which dissociates into hydrogen ions and hydrogencarbonate ions. The hydrogen ions bind to oxyhaemoglobin, releasing oxygen, and hydrogencarbonate ions diffuse out into the plasma in exchange for chloride ions.
  2. B.Carbon dioxide binds directly to the haem group of haemoglobin to form carboxyhaemoglobin, displacing oxygen. Hydrogencarbonate ions are actively pumped into the red blood cells, causing chloride ions to leave.
  3. C.Carbon dioxide is converted to carbonic acid by carbonic anhydrase in the blood plasma. Carbonic acid then enters the red blood cell to dissociate into hydrogen ions and hydrogencarbonate ions.
  4. D.Carbon dioxide reacts with water slowly without an enzyme to form carbonic acid. This dissociates, and the resulting hydrogen ions increase the affinity of haemoglobin for oxygen, holding it tighter.
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Worked solution

The sequence of events is as follows:
1. Carbon dioxide diffuses into the red blood cell, where the enzyme carbonic anhydrase rapidly catalyses its reaction with water to form carbonic acid (\(H_2CO_3\)).
2. Carbonic acid spontaneously dissociates into hydrogen ions (\(H^+\)) and hydrogencarbonate ions (\(HCO_3^-\)).
3. The hydrogen ions bind to oxyhaemoglobin, promoting the release of oxygen to the respiring tissues due to a conformational change that lowers haemoglobin's oxygen affinity (the Bohr effect).
4. Hydrogencarbonate ions diffuse out of the red blood cell down their concentration gradient into the plasma, while chloride ions enter the cell to maintain electrical neutrality (the chloride shift).

Marking scheme

1 mark for identifying the correct, sequential physiological process of carbon dioxide transport in red blood cells (Option A). Reject option B (carbon dioxide does not bind to the haem group, and chloride shift direction is reversed), C (carbonic anhydrase is inside the RBC, not plasma), and D (carbonic anhydrase is highly active, and low pH decreases oxygen affinity).
Question 8 · multiple choice
1 marks
A diploid animal cell has a chromosome number of \(2n = 16\). How many chromosomes, sister chromatids, and telomeres are present in this cell during metaphase of mitosis?
  1. A.16 chromosomes, 32 sister chromatids, 64 telomeres
  2. B.16 chromosomes, 16 sister chromatids, 32 telomeres
  3. C.32 chromosomes, 32 sister chromatids, 64 telomeres
  4. D.32 chromosomes, 64 sister chromatids, 128 telomeres
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Worked solution

1. Chromosomes: During metaphase, chromosomes are counted by the number of centromeres. There are 16 chromosomes aligned at the metaphase plate.
2. Sister chromatids: Since DNA replication occurred during S phase of interphase, each of the 16 chromosomes consists of 2 sister chromatids. Therefore, there are \(16 \times 2 = 32\) sister chromatids.
3. Telomeres: A telomere is a region of repetitive nucleotide sequences at each end of a chromatid. Each sister chromatid is a linear DNA molecule and thus has 2 telomeres (one at each end). Since there are 32 chromatids in the cell, the total number of telomeres is \(32 \times 2 = 64\) telomeres.

Marking scheme

1 mark for the correct combination of 16 chromosomes, 32 sister chromatids, and 64 telomeres (Option A). Reject other options as they miscalculate the number of chromatids or telomeres per chromatid during metaphase.
Question 9 · multiple_choice
1 marks
A student calibrated an eyepiece graticule using a stage micrometer. The stage micrometer had scale lines spaced at intervals of \(0.1\text{ mm}\). The student observed that \(50\) eyepiece graticule units coincided with \(15\) divisions on the stage micrometer. The student then replaced the stage micrometer with a slide of plant tissue and observed a xylem vessel element. At the same magnification, the diameter of the lumen of the xylem vessel element was measured as \(12\) eyepiece graticule units. What is the actual diameter of the lumen of the xylem vessel element?
  1. A.\(36\ \mu\text{m}\)
  2. B.\(180\ \mu\text{m}\)
  3. C.\(360\ \mu\text{m}\)
  4. D.\(1500\ \mu\text{m}\)
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Worked solution

First, calculate the value of one division of the stage micrometer: \(0.1\text{ mm} = 100\ \mu\text{m}\). Next, find the total distance represented by \(15\) stage micrometer divisions: \(15 \times 100\ \mu\text{m} = 1500\ \mu\text{m}\). Since these \(15\) divisions correspond to \(50\) eyepiece units, each eyepiece unit equals \(1500\ \mu\text{m} / 50 = 30\ \mu\text{m}\). Finally, multiply the measured width of the xylem lumen (\(12\) eyepiece units) by this calibration factor: \(12 \times 30\ \mu\text{m} = 360\ \mu\text{m}\).

Marking scheme

Award 1 mark for the correct calculation of actual diameter (C). Correct option is C. Incorrect options: A represents a factor of 10 decimal error; B is the result of incorrect division; D represents the total length of the 15 micrometer divisions without dividing by 50.
Question 10 · multiple_choice
1 marks
Which description of the bonds that maintain the tertiary structure of a globular protein is correct?
  1. A.Disulfide bonds are formed between the carboxyl groups of two adjacent cysteine residues.
  2. B.Hydrogen bonds are formed only between the oxygen of a carbonyl group and the hydrogen of an amine group in the peptide backbone.
  3. C.Hydrophobic interactions occur between non-polar R-groups which tend to cluster together in the interior of the protein.
  4. D.Ionic bonds are weak electrostatic attractions between positively charged R-groups and hydrophobic R-groups.
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Worked solution

Tertiary structure is stabilized by various interactions between the R-groups of amino acids. Hydrophobic interactions occur between non-polar R-groups, causing them to group together inside the protein, away from the surrounding aqueous environment. Disulfide bonds occur between sulfhydryl groups of cysteine residues, not carboxyl groups (eliminating A). Hydrogen bonds in tertiary structure occur between polar R-groups, whereas hydrogen bonds in the peptide backbone maintain secondary structures like alpha-helices and beta-pleated sheets (eliminating B). Ionic bonds are between oppositely charged polar groups, not hydrophobic groups (eliminating D).

Marking scheme

Award 1 mark for identifying the correct description of hydrophobic interactions stabilizing tertiary structure (C).
Question 11 · multiple_choice
1 marks
In the presence of a certain inhibitor, the Michaelis-Menten constant (\(K_{\text{m}}\)) of an enzyme increases, but the maximum velocity (\(V_{\max}\)) of the reaction remains unchanged. Which statement is correct for this type of inhibition?
  1. A.The inhibitor binds reversibly to an allosteric site on the enzyme, altering the shape of the active site.
  2. B.The inhibitor has a similar molecular shape to the substrate and competes with the substrate for the active site.
  3. C.The inhibitor forms strong, covalent bonds with the active site, irreversibly inactivating the enzyme.
  4. D.Increasing the substrate concentration has no effect on the degree of inhibition by this inhibitor.
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Worked solution

An increase in the Michaelis-Menten constant (\(K_{\text{m}}\)) with an unchanged maximum velocity (\(V_{\max}\)) is characteristic of competitive inhibition. In competitive inhibition, the inhibitor has a similar molecular shape to the substrate and competes with it for binding to the active site. Increasing the substrate concentration can overcome this competition, allowing the reaction to reach the original maximum velocity (\(V_{\max}\)).

Marking scheme

Award 1 mark for identifying that an increase in \(K_{\text{m}}\), while \(V_{\max}\) remains unchanged, is caused by a competitive inhibitor which competes with the substrate for the active site (B).
Question 12 · multiple_choice
1 marks
Which molecules in a cell surface membrane can act as receptors for cell signaling, and which can help to stabilize the membrane structure at high temperatures?
  1. A.act as receptors: glycoproteins and glycolipids; help to stabilize membrane: cholesterol
  2. B.act as receptors: phospholipids and cholesterol; help to stabilize membrane: glycoproteins
  3. C.act as receptors: glycoproteins only; help to stabilize membrane: glycolipids and cholesterol
  4. D.act as receptors: glycolipids and cholesterol; help to stabilize membrane: phospholipids
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Worked solution

Glycoproteins and glycolipids contain carbohydrate chains that extend from the outer surface of the cell membrane, enabling them to act as receptors for cell-signaling molecules like hormones. Cholesterol is a hydrophobic molecule embedded within the fatty acid tail region of the phospholipid bilayer. At high temperatures, cholesterol restricts the movement of fatty acid tails, stabilizing the membrane structure and reducing fluidity.

Marking scheme

Award 1 mark for identifying that glycoproteins and glycolipids act as cell-signaling receptors and cholesterol stabilizes the membrane structure at high temperatures (A).
Question 13 · multiple_choice
1 marks
During DNA replication, both strands of the double helix act as templates. If a segment of the original DNA template strand has the base sequence \(3'\text{-ATCGGTACG-5'}\), which base sequence will be found in the newly synthesised complementary strand?
  1. A.\(3'\text{-ATCGGTACG-5'}\)
  2. B.\(3'\text{-TAGCCATGC-5'}\)
  3. C.\(5'\text{-TAGCCATGC-3'}\)
  4. D.\(5'\text{-UAGCCAUGC-3'}\)
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Worked solution

DNA replication occurs in an antiparallel fashion. The complementary strand must run \(5'\) to \(3'\) when matching a \(3'\) to \(5'\) template strand. Base-pairing rules dictate that Adenine (A) pairs with Thymine (T), and Cytosine (C) pairs with Guanine (G). Thus, the template sequence \(3'\text{-ATCGGTACG-5'}\) matches with \(5'\text{-TAGCCATGC-3'}\) on the newly synthesised DNA strand. Option D contains Uracil (U), which is found in RNA, not DNA.

Marking scheme

Award 1 mark for identifying the correct complementary antiparallel DNA sequence (C).
Question 14 · multiple_choice
1 marks
Which row correctly identifies features of a mature phloem sieve tube element compared to a xylem vessel element?
  1. A.sieve tube: cytoplasm present, non-lignified walls, companion cells present; xylem: cytoplasm absent, lignified walls, companion cells absent
  2. B.sieve tube: cytoplasm absent, lignified walls, companion cells present; xylem: cytoplasm present, non-lignified walls, companion cells absent
  3. C.sieve tube: cytoplasm present, non-lignified walls, companion cells absent; xylem: cytoplasm present, non-lignified walls, companion cells present
  4. D.sieve tube: cytoplasm absent, lignified walls, companion cells absent; xylem: cytoplasm absent, lignified walls, companion cells absent
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Worked solution

Mature phloem sieve tube elements are living cells containing peripheral cytoplasm and lack a nucleus; they have non-lignified cellulose cell walls, and are associated with companion cells. Xylem vessel elements are dead cells that completely lack cytoplasm, have heavily lignified cell walls, and do not have companion cells. Therefore, row A is correct.

Marking scheme

Award 1 mark for correctly comparing the structural features of sieve tube elements and xylem vessel elements (A).
Question 15 · multiple_choice
1 marks
In red blood cells, carbon dioxide reacts with water to form carbonic acid, which dissociates into hydrogen ions and hydrogencarbonate ions. Reaction 1 is catalysed by enzyme W. Hydrogencarbonate ions diffuse out of the red blood cell, and ions X enter the cell to maintain electrical neutrality. Hydrogen ions bind to oxyhaemoglobin, causing the release of oxygen (Y). Some carbon dioxide binds directly to the terminal amine groups of haemoglobin to form Z. Which row correctly identifies W, X, Y and Z?
  1. A.W: carbonic anhydrase, X: chloride ions, Y: oxygen, Z: carbaminohaemoglobin
  2. B.W: carbonic anhydrase, X: potassium ions, Y: carbon dioxide, Z: carboxyhaemoglobin
  3. C.W: catalase, X: chloride ions, Y: oxygen, Z: carboxyhaemoglobin
  4. D.W: hydrogenase, X: hydrogen carbonate ions, Y: oxygen, Z: carbaminohaemoglobin
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Worked solution

In red blood cells, carbon dioxide and water form carbonic acid via the enzyme carbonic anhydrase (W). Carbonic acid dissociates into hydrogen ions and hydrogencarbonate ions; hydrogencarbonate ions diffuse out of the cell, while chloride ions (X) diffuse in (the chloride shift) to maintain electrical neutrality. The hydrogen ions bind to oxyhaemoglobin, triggering the release of oxygen (Y). Some carbon dioxide binds directly to the terminal amine groups of haemoglobin to form carbaminohaemoglobin (Z).

Marking scheme

Award 1 mark for identifying W as carbonic anhydrase, X as chloride ions, Y as oxygen, and Z as carbaminohaemoglobin (A).
Question 16 · multiple_choice
1 marks
At day 0, an individual is injected with an antigen for the first time. At day 30, the individual is injected with the same antigen for a second time. Which statement explains the differences between the primary immune response (after day 0) and the secondary immune response (after day 30)?
  1. A.The secondary response has a shorter delay because memory B-lymphocytes immediately start producing large amounts of antibodies.
  2. B.The secondary response is faster and produces a higher concentration of antibodies because memory B-lymphocytes rapidly divide and differentiate into plasma cells.
  3. C.The primary response is slower because T-helper cells must mutate to produce the correct antibodies.
  4. D.The secondary response lasts longer because the antibodies produced are more stable and resist breakdown for a longer period of time.
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Worked solution

In a primary immune response, the response is slow because there are few specific B-lymphocytes that must undergo clonal selection, expansion, and differentiation into antibody-producing plasma cells. In a secondary immune response, pre-existing memory cells rapidly detect the antigen, divide, and differentiate into a large number of plasma cells, leading to a much faster response and a higher concentration of antibodies. Memory B-cells themselves do not directly secrete antibodies, but differentiate into plasma cells to do so (eliminating A).

Marking scheme

Award 1 mark for explaining that the secondary response is faster and greater due to memory B-lymphocytes rapidly dividing and differentiating into plasma cells (B).
Question 17 · multiple choice
1 marks
Four statements about protein structure are listed. 1. The secondary structure is stabilized solely by hydrogen bonds between peptide group atoms. 2. The tertiary structure is stabilized by interactions between R-groups. 3. Disulfide bonds can form between any two sulfur-containing amino acids (methionine and cysteine). 4. Quaternary structure is only present in proteins composed of more than one polypeptide chain. Which statements are correct?
  1. A.1, 2 and 3 only
  2. B.1, 2 and 4 only
  3. C.2 and 4 only
  4. D.1, 3 and 4 only
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Worked solution

Statement 1 is correct because the alpha-helix and beta-pleated sheet of the secondary structure are held together by hydrogen bonds between the peptide group atoms of the polypeptide backbone. Statement 2 is correct because the tertiary structure is held in place by interactions between the R-groups of amino acids. Statement 3 is incorrect because only cysteine has a free sulfhydryl group (-SH) capable of forming a covalent disulfide bond; methionine contains a thioether group and cannot form disulfide bonds. Statement 4 is correct because quaternary structure requires the association of two or more polypeptide chains.

Marking scheme

1 mark for selecting the correct option (B) which correctly identifies statements 1, 2, and 4 as correct, and 3 as incorrect.
Question 18 · multiple choice
1 marks
If a competitive inhibitor is added to an enzyme-catalysed reaction, how do the Michaelis-Menten constant (\(K_m\)) and the maximum velocity (\(V_{\max}\)) of the reaction change?
  1. A.\(V_{\max}\) decreases and \(K_m\) remains the same
  2. B.\(V_{\max}\) remains the same and \(K_m\) increases
  3. C.\(V_{\max}\) decreases and \(K_m\) increases
  4. D.\(V_{\max}\) remains the same and \(K_m\) remains the same
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Worked solution

A competitive inhibitor competes with the substrate for the active site. This decreases the affinity of the enzyme for its substrate, which increases the \(K_m\) value (since more substrate is needed to reach half \(V_{\max}\)). However, because the inhibitor can be fully outcompeted by extremely high concentrations of substrate, the maximum rate of reaction (\(V_{\max}\)) remains unchanged.

Marking scheme

1 mark for identifying that competitive inhibition increases \(K_m\) while leaving \(V_{\max}\) unchanged (B).
Question 19 · multiple choice
1 marks
A specimen is viewed under a light microscope. An eyepiece graticule has scale divisions calibrated to \(1.5\ \mu\text{m}\) at a magnification of \(\times 400\). A mitochondrion is measured to be \(4\) eyepiece graticule units in length. If the student changes the objective lens to increase the magnification to \(\times 1000\), what is the physical length of the mitochondrion and how many eyepiece graticule units long will it appear?
  1. A.Physical length is \(6.0\ \mu\text{m}\); it will appear \(10\) graticule units long.
  2. B.Physical length is \(6.0\ \mu\text{m}\); it will appear \(1.6\) graticule units long.
  3. C.Physical length is \(1.5\ \mu\text{m}\); it will appear \(4\) graticule units long.
  4. D.Physical length is \(2.4\ \mu\text{m}\); it will appear \(10\) graticule units long.
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Worked solution

At \(\times 400\), the physical length is calculated as \(4\text{ units} \times 1.5\ \mu\text{m/unit} = 6.0\ \mu\text{m}\). When magnification increases by a factor of \(2.5\) (\(1000 / 400\)), the image of the specimen appears larger by that factor. Since the eyepiece graticule divisions themselves do not change physical size, the mitochondrion will cover \(2.5\) times more units on the graticule: \(4 \times 2.5 = 10\) units. (Alternatively, the calibration of each unit at \(\times 1000\) is \(1.5\ \mu\text{m} / 2.5 = 0.6\ \mu\text{m}\). Thus, the mitochondrion of \(6.0\ \mu\text{m}\) spans \(6.0\ \mu\text{m} / 0.6\ \mu\text{m/unit} = 10\) units).

Marking scheme

1 mark for calculating the correct physical length (6.0 micrometers) and the corresponding number of graticule units (10) under the higher magnification (A).
Question 20 · multiple choice
1 marks
Which row correctly matches the mitotic stage with the cellular events occurring during that stage in a dividing plant cell?
  1. A.Prophase: chromatin condenses, spindle fibres begin to form, and centrioles migrate to opposite poles of the cell.
  2. B.Metaphase: chromosomes align along the equator of the cell and attach to spindle fibres via their centromeres.
  3. C.Anaphase: sister chromatids are pulled apart by the shortening of spindle microtubules towards the centrioles.
  4. D.Telophase: chromosomes decondense, nuclear envelope reforms, and a cleavage furrow pinches the cell membrane.
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Worked solution

Plant cells do not possess centrioles, making statements A and C incorrect. Additionally, plant cells undergo cytokinesis by forming a cell plate rather than a cleavage furrow, making statement D incorrect. Statement B is correct for both plant and animal cells.

Marking scheme

1 mark for identifying the row describing metaphase events that can occur in plant cells, avoiding structures or processes unique to animal cells (such as centrioles or cleavage furrows) (B).
Question 21 · multiple choice
1 marks
A child is bitten by a venomous snake and immediately receives an injection of antivenom (containing specific antibodies). Three months later, the child is bitten by the same species of snake but is not immune and requires another injection of antivenom. Which row correctly describes the type of immunity provided and the reason why there was no long-term protection?
  1. A.Type of immunity: Artificial passive | Reason: No memory cells were produced from the first injection because no antigen was introduced.
  2. B.Type of immunity: Natural passive | Reason: The injected antibodies were rapidly broken down by the child's immune system.
  3. C.Type of immunity: Artificial active | Reason: The memory cells produced had a short lifespan of less than three months.
  4. D.Type of immunity: Artificial passive | Reason: The second bite contained a different antigen that could not be recognized.
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Worked solution

An injection of ready-made antibodies (antivenom) provides artificial passive immunity. Because the child's own immune system did not undergo clonal selection and expansion in response to an antigen, no memory cells were formed. As a result, once the injected antibodies were naturally cleared from the body, no immunological memory remained.

Marking scheme

1 mark for correctly matching artificial passive immunity with the absence of memory cells due to lack of antigen introduction (A).
Question 22 · multiple choice
1 marks
Which component of the cell surface membrane is responsible for maintaining membrane fluidity at low temperatures, and which component acts as a receptor for cell signalling molecules?
  1. A.Fluidity: cholesterol | Receptor: glycoprotein
  2. B.Fluidity: glycolipid | Receptor: protein channel
  3. C.Fluidity: cholesterol | Receptor: phospholipid
  4. D.Fluidity: glycoprotein | Receptor: glycolipid
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Worked solution

Cholesterol prevents fatty acid tails from packing too tightly together at low temperatures, thereby maintaining membrane fluidity. Glycoproteins act as receptors on the outer surface of the cell membrane to bind signaling molecules such as hormones.

Marking scheme

1 mark for matching cholesterol to fluidity maintenance and glycoprotein to the receptor function (A).
Question 23 · multiple choice
1 marks
The sequence of a section of the DNA template strand is \(3'\text{-TACGGCATC}-5'\). What is the sequence of the complementary mRNA strand transcribed from this DNA sequence?
  1. A.\(5'\text{-AUGCCGUAG}-3'\)
  2. B.\(5'\text{-UACGGCAUC}-3'\)
  3. C.\(3'\text{-AUGCCGUAG}-5'\)
  4. D.\(5'\text{-ATGCCGTAG}-3'\)
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Worked solution

During transcription, RNA polymerase synthesizes mRNA in the \(5'\) to \(3'\) direction by matching complementary bases to the \(3'\) to \(5'\) DNA template strand. Adenine (A) pairs with Uracil (U) in RNA, Thymine (T) pairs with Adenine (A), Cytosine (C) pairs with Guanine (G), and Guanine (G) pairs with Cytosine (C). Therefore, the DNA template sequence \(3'\text{-TACGGCATC}-5'\) produces \(5'\text{-AUGCCGUAG}-3'\).

Marking scheme

1 mark for identifying the correct mRNA sequence and directional polarity (5' to 3') (A).
Question 24 · multiple choice
1 marks
Carbon dioxide is transported in the blood in several forms. Which row correctly identifies the form in which the largest percentage of carbon dioxide is transported, and the enzyme that catalyses its conversion inside red blood cells?
  1. A.Largest percentage: hydrogencarbonate ions (\(\text{HCO}_3^-\)) | Enzyme: carbonic anhydrase
  2. B.Largest percentage: carbaminohaemoglobin | Enzyme: carbonic anhydrase
  3. C.Largest percentage: dissolved carbon dioxide in plasma | Enzyme: haemoglobinase
  4. D.Largest percentage: hydrogencarbonate ions (\(\text{HCO}_3^-\)) | Enzyme: carbonic decarboxylase
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Worked solution

Approximately 85% of carbon dioxide is transported in the blood plasma as hydrogencarbonate ions (\(\text{HCO}_3^-\)). Inside red blood cells, carbon dioxide reacts with water to form carbonic acid, a reaction catalyzed by the enzyme carbonic anhydrase. Carbonic acid then dissociates into hydrogencarbonate and hydrogen ions.

Marking scheme

1 mark for identifying hydrogencarbonate ions as the main transport form and carbonic anhydrase as the catalyzing enzyme (A).
Question 25 · multiple choice
1 marks
Which components of cell membranes are involved in maintaining the stability of the membrane at low temperatures and in cell-to-cell recognition respectively?
  1. A.cholesterol and glycoproteins
  2. B.glycolipids and phospholipids
  3. C.protein channels and cholesterol
  4. D.glycoproteins and intrinsic proteins mohs scale value close to standard definition of channel activity but wrong context in recognition context.
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Worked solution

Cholesterol maintains membrane stability and fluidity by preventing phospholipid tails from packing too tightly together at lower temperatures. Glycoproteins (and glycolipids) project from the outer membrane surface and act as cell-recognition sites or receptors.

Marking scheme

1 mark for identifying cholesterol as the stabilizer at low temperatures and glycoproteins as the recognition molecules.
Question 26 · multiple choice
1 marks
A micrograph of a plant cell shows a chloroplast with a length of 40 mm. The actual length of the chloroplast is 5 \(\mu\)m. What is the magnification of this micrograph?
  1. A.\(\times 8\)
  2. B.\(\times 80\)
  3. C.\(\times 800\)
  4. D.\(\times 8000\)
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Worked solution

First, convert the image length to the same unit as the actual length: 40 mm = 40,000 \(\mu\)m. Magnification = Image size / Actual size = 40,000 \(\mu\)m / 5 \(\mu\)m = 8000.

Marking scheme

1 mark for the correct conversion of units and calculation leading to \(\times 8000\).
Question 27 · multiple choice
1 marks
An enzyme-controlled reaction was carried out in the presence and absence of an inhibitor. The rate of reaction was plotted against substrate concentration. It was found that at very high substrate concentrations, the rate of reaction with the inhibitor was identical to the rate of reaction without the inhibitor. Which statement about the inhibitor is correct?
  1. A.It binds to an allosteric site and changes the shape of the active site.
  2. B.It binds reversibly to the active site of the enzyme.
  3. C.It forms permanent covalent bonds with the active site.
  4. D.It decreases the \(V_{max}\) of the enzyme-controlled reaction.
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Worked solution

Because the maximum rate is achieved at very high substrate concentrations, the inhibitor is competitive. Competitive inhibitors compete with substrate molecules for the active site, meaning they must bind reversibly to the active site so that increasing substrate concentration can outcompete them.

Marking scheme

1 mark for identifying that competitive inhibition is reversible and occurs at the active site.
Question 28 · multiple choice
1 marks
A portion of a template strand of DNA has the sequence: 3'- TAC GGC TTA CTA -5'. During transcription, this sequence is copied to produce mRNA, which is then translated. What is the correct sequence of anticodons on the tRNA molecules that bind to the codons of this mRNA sequence?
  1. A.5'- AUG CCG AAU GAU -3'
  2. B.3'- AUG CCG AAU GAU -5'
  3. C.3'- UAC GGC UUA CUA -5'
  4. D.5'- UAC GGC UUA CUA -3'
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Worked solution

The DNA template is 3'- TAC GGC TTA CTA -5'. The mRNA is transcribed in a complementary, antiparallel fashion to produce: 5'- AUG CCG AAU GAU -3'. During translation, the tRNA anticodons align antiparallel to the mRNA codons. The complementary sequence of anticodons to 5'- AUG CCG AAU GAU -3' in the 3' to 5' direction is 3'- UAC GGC UUA CUA -5'.

Marking scheme

1 mark for correctly determining the complementary and antiparallel tRNA anticodon sequence.
Question 29 · multiple choice
1 marks
Which of the following describes the effect of an increase in carbon dioxide concentration on the oxygen dissociation curve of haemoglobin and its physiological consequence in respiring tissues?
  1. A.The curve shifts to the left, decreasing the affinity of haemoglobin for oxygen, making oxygen more readily available to tissues.
  2. B.The curve shifts to the right, decreasing the affinity of haemoglobin for oxygen, making oxygen more readily available to tissues.
  3. C.The curve shifts to the right, increasing the affinity of haemoglobin for oxygen, reducing the release of oxygen to tissues.
  4. D.The curve shifts to the left, increasing the affinity of haemoglobin for oxygen, reducing the release of oxygen to tissues.
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Worked solution

An increase in carbon dioxide concentration shifts the oxygen dissociation curve to the right (the Bohr effect). This shift represents a decreased affinity of haemoglobin for oxygen at any given partial pressure of oxygen, enabling haemoglobin to unload oxygen more easily to actively respiring tissues.

Marking scheme

1 mark for correctly identifying that high carbon dioxide shifts the curve to the right, lowering oxygen affinity, and enhancing oxygen unloading.
Question 30 · multiple choice
1 marks
Which structural features of xylem vessel elements adapt them for the efficient transport of water? 1. Lignified cell walls to prevent collapse under tension. 2. Absence of protoplasm to reduce resistance to water flow. 3. Presence of sieve plates to allow rapid flow. 4. Pits in the walls to allow lateral movement of water.
  1. A.1, 2, 3 and 4
  2. B.1, 2 and 4 only
  3. C.1 and 3 only
  4. D.2 and 4 only
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Worked solution

Sieve plates are a feature of phloem sieve tube elements, not xylem vessels. Xylem vessel elements are dead cells lacking a protoplasm, have lignified walls to withstand tension, and contain pits for the lateral transport of water. Thus, statements 1, 2, and 4 are correct.

Marking scheme

1 mark for identifying statement 3 as false (phloem feature) and 1, 2, 4 as correct adaptations of xylem.
Question 31 · multiple choice
1 marks
A person is bitten by a venomous snake and receives an immediate injection of antivenom containing specific antibodies. Two months later, the person is bitten by the same species of snake but does not have immunity and requires another injection. What type of immunity is provided by the antivenom, and why is it not long-lasting?
  1. A.Artificial active immunity, because no memory cells are produced.
  2. B.Artificial passive immunity, because the injected antibodies are broken down and no memory cells are produced.
  3. C.Natural passive immunity, because the antibodies are biological molecules.
  4. D.Artificial passive immunity, because the immune system produces its own antibodies but they decay rapidly.
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Worked solution

Antivenom provides artificial passive immunity because pre-formed antibodies are injected directly into the body. This immunity is temporary (not long-lasting) because the recipient's immune system is not stimulated to produce its own antibodies or memory cells, and the injected foreign antibodies are eventually broken down by the body.

Marking scheme

1 mark for identifying that antivenom represents artificial passive immunity and lacks long-lasting protection due to lack of memory cells.
Question 32 · multiple choice
1 marks
During which stage of the mitotic cell cycle do centromeres divide to separate sister chromatids, and which stage immediately follows this event?
  1. A.Anaphase followed by Telophase
  2. B.Metaphase followed by Anaphase
  3. C.Anaphase followed by Metaphase
  4. D.Telophase followed by Cytokinesis
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Worked solution

Centromeres divide during anaphase, causing the sister chromatids to separate and be pulled to opposite poles of the spindle. The stage that immediately follows anaphase is telophase, where the nuclear envelopes reform around the two groups of chromosomes.

Marking scheme

1 mark for correctly linking the division of centromeres to anaphase and identifying telophase as the subsequent stage.
Question 33 · multiple choice
1 marks
A student calibrated an eyepiece graticule using a stage micrometer. The stage micrometer had divisions that were \(0.1\text{ mm}\) apart. At a magnification of \(\times 100\), 50 divisions of the eyepiece graticule aligned with 10 divisions of the stage micrometer. Under this same magnification, a plant cell was measured to be 3.5 eyepiece graticule units in length. What is the actual length of the plant cell?
  1. A.\(7\ \mu\text{m}\)
  2. B.\(35\ \mu\text{m}\)
  3. C.\(70\ \mu\text{m}\)
  4. D.\(700\ \mu\text{m}\)
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Worked solution

1. Calculate the distance represented by 10 divisions of the stage micrometer: \(10 \times 0.1\text{ mm} = 1.0\text{ mm} = 1000\ \mu\text{m}\). 2. Determine the value of 1 eyepiece graticule unit (epu): \(50\text{ epu} = 1000\ \mu\text{m}\), which means \(1\text{ epu} = 1000 / 50 = 20\ \mu\text{m}\). 3. Calculate the actual length of the plant cell: \(3.5\text{ epu} \times 20\ \mu\text{m/epu} = 70\ \mu\text{m}\).

Marking scheme

1 mark for the correct option (C). [1] Correctly calculating the calibration value of 1 epu as 20 micrometres, and multiplying by 3.5 to obtain 70 micrometres.
Question 34 · multiple choice
1 marks
An enzyme-controlled reaction is carried out in the presence of a competitive inhibitor. Which statement correctly describes the effect of this inhibitor on the maximum rate of reaction (\(V_{max}\)) and the Michaelis-Menten constant (\(K_m\)) of the enzyme?
  1. A.\(V_{max}\) is unchanged; \(K_m\) is increased.
  2. B.\(V_{max}\) is decreased; \(K_m\) is increased.
  3. C.\(V_{max}\) is unchanged; \(K_m\) is decreased.
  4. D.\(V_{max}\) is decreased; \(K_m\) is unchanged.
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Worked solution

A competitive inhibitor binds reversibly to the active site of the enzyme, competing directly with the substrate. At extremely high substrate concentrations, the substrate molecules outcompete the inhibitor molecules, allowing the reaction rate to reach the original maximum rate (\(V_{max}\) is unchanged). However, because a higher substrate concentration is required to achieve half-maximum velocity, the Michaelis-Menten constant (\(K_m\)) is increased.

Marking scheme

1 mark for the correct option (A). [1] Recognition that competitive inhibitors do not alter the maximum velocity of the enzyme but increase the apparent Michaelis-Menten constant.
Question 35 · multiple choice
1 marks
Which statements about the structure of a collagen molecule are correct? 1. It has a primary structure in which glycine occurs as every third amino acid. 2. Hydrogen bonds stabilize the triple helix structure formed by three polypeptide chains. 3. Covalent bonds form cross-links between adjacent triple helices to form a collagen fibril.
  1. A.1, 2 and 3
  2. B.1 and 2 only
  3. C.1 and 3 only
  4. D.2 and 3 only
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Worked solution

Statement 1 is correct: Collagen has a repeating primary structure where glycine is every third amino acid, allowing tight packing of the chains. Statement 2 is correct: The three helical polypeptide chains are held together by hydrogen bonds to form the triple helix molecule. Statement 3 is correct: Covalent cross-links form between adjacent collagen triple helices to form stable, high-tensile strength collagen fibrils. Therefore, all three statements are correct.

Marking scheme

1 mark for option A. [1] Correctly recognizing that statements 1, 2, and 3 are all correct statements about collagen structure and its assembly into fibrils.
Question 36 · multiple choice
1 marks
A child is bitten by a venomous snake and is immediately given an injection of antivenom containing specific antibodies. Six months later, the child is bitten by the same species of snake and again requires antivenom treatment. Which statement explains why the child did not have immunity to the snake venom during the second bite?
  1. A.The antibodies from the first injection were recognized as foreign and destroyed by the child's immune system.
  2. B.The first injection triggered a primary immune response, but the memory cells produced had died.
  3. C.The injection of antivenom provided artificial passive immunity, which does not result in the production of memory cells.
  4. D.The snake venom acts as an antigen but is too small to stimulate B-lymphocytes to undergo clonal expansion.
Show answer & marking scheme

Worked solution

An injection of antivenom containing ready-made antibodies provides artificial passive immunity. Because the child's own immune system was not stimulated to undergo clonal selection, clonal expansion, and differentiation of B-lymphocytes, no plasma cells or memory cells were produced. Consequently, no long-term active immunological memory was established, and the child remained non-immune to future bites.

Marking scheme

1 mark for option C. [1] Correctly identifying that passive immunity does not involve clonal selection/expansion and thus does not produce memory cells, leaving the individual susceptible to reinfection or subsequent bites.
Question 37 · multiple choice
1 marks
During intensive physical exercise, the partial pressure of carbon dioxide in actively respiring muscle tissue increases. What is the effect of this change on the plasma pH, the affinity of haemoglobin for oxygen, and the position of the oxygen dissociation curve?
  1. A.Plasma pH: Decreases | Affinity of haemoglobin: Decreases | Shift of curve: To the right
  2. B.Plasma pH: Decreases | Affinity of haemoglobin: Increases | Shift of curve: To the left
  3. C.Plasma pH: Increases | Affinity of haemoglobin: Decreases | Shift of curve: To the right
  4. D.Plasma pH: Increases | Affinity of haemoglobin: Increases | Shift of curve: To the left
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Worked solution

Increased carbon dioxide concentration leads to higher production of carbonic acid in plasma, which dissociates into hydrogen ions and hydrogencarbonate ions, causing a decrease in plasma pH. The lower pH decreases the affinity of haemoglobin for oxygen (Bohr effect), shifting the oxygen dissociation curve to the right, which facilitates oxygen unloading in the tissues.

Marking scheme

1 mark for option A. [1] Correlating high partial pressure of carbon dioxide with decreased pH, decreased oxygen affinity of haemoglobin, and a rightward shift of the dissociation curve.
Question 38 · multiple choice
1 marks
Water moves from the soil to the xylem of a plant root. Which sequence of structures shows the correct pathway of water that travels mainly via the apoplast pathway in the cortex?
  1. A.cell walls of cortex \(\rightarrow\) cytoplasm of endodermis \(\rightarrow\) cell walls of pericycle \(\rightarrow\) xylem lumen
  2. B.cell walls of cortex \(\rightarrow\) cell walls of endodermis \(\rightarrow\) cell walls of pericycle \(\rightarrow\) xylem lumen
  3. C.plasmodesmata of cortex \(\rightarrow\) cytoplasm of endodermis \(\rightarrow\) plasmodesmata of pericycle \(\rightarrow\) xylem lumen
  4. D.cytoplasm of cortex \(\rightarrow\) Casparian strip \(\rightarrow\) cytoplasm of endodermis \(\rightarrow\) xylem lumen
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Worked solution

Water traveling via the apoplast pathway moves through the cell walls and intercellular spaces of the cortex. At the endodermis, the impermeable Casparian strip blocks the cell walls, forcing water to pass through the cell surface membrane into the cytoplasm of the endodermal cells. Once past the endodermis, water can re-enter the cell walls of the pericycle (apoplast) and then flow into the xylem lumen.

Marking scheme

1 mark for option A. [1] Correctly identifying that apoplastic water must cross the cell membrane into the cytoplasm of endodermal cells due to the presence of the impermeable Casparian strip in the endodermal cell walls.
Question 39 · multiple choice
1 marks
Which row correctly shows the presence (+) or absence (-) of tissues in the bronchus and alveolus of a healthy human lung? (Row A: Cartilage in bronchus: +, Smooth muscle in bronchus: +, Elastic fibres in alveolus: +, Cilia in alveolus: -) (Row B: Cartilage in bronchus: +, Smooth muscle in bronchus: -, Elastic fibres in alveolus: +, Cilia in alveolus: +) (Row C: Cartilage in bronchus: -, Smooth muscle in bronchus: +, Elastic fibres in alveolus: -, Cilia in alveolus: -) (Row D: Cartilage in bronchus: +, Smooth muscle in bronchus: +, Elastic fibres in alveolus: -, Cilia in alveolus: +)
  1. A.Row A
  2. B.Row B
  3. C.Row C
  4. D.Row D
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Worked solution

A bronchus contains cartilage plates to keep the airway open, and smooth muscle to regulate airway diameter. An alveolus has walls composed of simple squamous epithelium that lacks cilia, but has a rich network of elastic fibres to allow stretch during inhalation and elastic recoil during exhalation. Therefore, Row A is correct.

Marking scheme

1 mark for option A. [1] Correctly identifying that bronchi contain both cartilage and smooth muscle, while alveoli lack cilia but contain elastic fibres.
Question 40 · multiple choice
1 marks
A strand of mRNA has the codon sequence 5'-AUG-CCG-AAA-UGC-3'. Which tRNA anticodons will bind to these codons during translation?
  1. A.UAC, GGC, UUU, ACG
  2. B.TAC, GGC, TTT, ACG
  3. C.AUG, CCG, AAA, UGC
  4. D.ATG, CCG, AAA, TGC
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Worked solution

The anticodons on tRNA are complementary to the mRNA codons and must contain RNA bases (Uracil instead of Thymine). - 5'-AUG-3' pairs with anticodon 3'-UAC-5' (written here as UAC). - 5'-CCG-3' pairs with anticodon 3'-GGC-5' (written here as GGC). - 5'-AAA-3' pairs with anticodon 3'-UUU-5' (written here as UUU). - 5'-UGC-3' pairs with anticodon 3'-ACG-5' (written here as ACG). Thus, the correct sequence of anticodons is UAC, GGC, UUU, ACG.

Marking scheme

1 mark for option A. [1] Correctly matching each mRNA codon with its complementary tRNA anticodon containing RNA bases (U instead of T).

Paper 2 (AS Structured)

Answer all six structured questions. Show all working for calculations and state appropriate units.
6 Question · 60 marks
Question 1 · structured
10 marks
Trypsin is a protease enzyme secreted by the pancreas that hydrolyses peptide bonds in proteins. (a) Describe the mechanism of action of trypsin in terms of the lock-and-key or induced-fit hypothesis, explaining how it lowers the activation energy of the reaction. [4] (b) In an investigation, the rate of trypsin activity was measured at different concentrations of a competitive inhibitor, benzamidine, while keeping substrate concentration constant. (i) Explain how a competitive inhibitor like benzamidine decreases the rate of trypsin activity. [3] (ii) State three variables, other than the concentration of the inhibitor, that must be kept constant in this investigation to obtain valid results. [3]
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Worked solution

Part (a) Trypsin has a specific three-dimensional active site that is complementary in shape to its substrate. According to the induced-fit hypothesis, the active site is flexible and changes shape slightly when the substrate binds. This precise fit puts physical strain on the peptide bonds within the substrate, destabilising them and lowering the activation energy required for the hydrolysis reaction to proceed. Once the reaction is complete, the products no longer fit the active site and are released. Part (b)(i) Benzamidine acts as a competitive inhibitor because its chemical structure and molecular shape are very similar to those of trypsin's natural substrate. It competes directly with substrate molecules to bind to the active site of trypsin. While bound, it blocks the active site, preventing the substrate from entering and reducing the rate of enzyme-substrate complex formation. Part (b)(ii) Control variables that must be kept constant include: 1. Temperature (to keep kinetic energy of molecules constant), 2. pH (using a buffer to prevent denaturing or changes in active site charge), and 3. Enzyme concentration (to ensure the concentration of active sites remains constant).

Marking scheme

(a) [Max 4 marks] 1. Trypsin's active site has a specific/complementary shape to the substrate. 2. Substrate binds to active site to form enzyme-substrate (ES) complex. 3. Induced fit: active site changes shape slightly to fit substrate more closely / put strain on peptide bonds. 4. This destabilises bonds / lowers activation energy. 5. Products (amino acids/peptides) no longer fit active site and are released. (b)(i) [Max 3 marks] 1. Benzamidine has a similar structure/shape to the substrate. 2. Competes with substrate for binding to the active site. 3. Prevents substrate from binding / fewer ES complexes formed. (b)(ii) [Max 3 marks] 1. Temperature. 2. pH (accept: use of a buffer). 3. Enzyme concentration. 4. Substrate concentration.
Question 2 · structured
10 marks
Active loading of sucrose into the phloem sieve tubes is essential for translocation in plants. (a) Explain how companion cells are adapted to actively load sucrose into sieve tube elements. [4] (b) Describe how the accumulation of sucrose in sieve tubes leads to the movement of phloem sap from source to sink. [4] (c) State two differences in structure between a companion cell and a phloem sieve tube element. [2]
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Worked solution

Part (a) Companion cells active load sucrose through several coordinated membrane mechanisms. Firstly, proton pumps (\(H^+\) ATPase) in their cell surface membrane actively transport hydrogen ions (protons) out of the companion cell cytoplasm and into the surrounding cell wall. This establishes a high electrochemical gradient of \(H^+\) ions in the apoplast. Protons then diffuse back down this gradient into the companion cell through co-transporter proteins, which simultaneously transport sucrose molecules against their concentration gradient into the companion cell cytoplasm. From the companion cell, sucrose diffuses into the sieve tube element via plasmodesmata. Part (b) When sucrose accumulates in the sieve tube element, it significantly lowers the solute/water potential inside the cell. Consequently, water moves from adjacent xylem vessels and surrounding tissues into the sieve tube down a water potential gradient by osmosis. The entry of water increases the hydrostatic pressure within the sieve tube at the source. At the sink, sucrose is unloaded, increasing the water potential and causing water to leave the sieve tube, lowering the hydrostatic pressure. This difference in hydrostatic pressure creates a pressure gradient that drives the mass flow of phloem sap from source to sink. Part (c) Companion cells possess a functional nucleus and numerous large mitochondria to generate ATP for active transport, whereas mature sieve tube elements lack a nucleus, have a highly degenerate cytoplasm, and contain very few, smaller mitochondria to allow unobstructed mass flow.

Marking scheme

(a) [Max 4 marks] 1. Active transport of \(H^+\) ions out of companion cell into cell wall / apoplast. 2. Uses proton pumps / \(H^+\) ATPase (requires ATP). 3. Establishes a proton gradient / high concentration of \(H^+\) in cell wall. 4. Protons diffuse back into companion cell through co-transporter proteins. 5. Sucrose is co-transported (symported) with protons against its concentration gradient. 6. Sucrose moves into the sieve tube through plasmodesmata. (b) [Max 4 marks] 1. Loading of sucrose decreases water potential inside sieve tube. 2. Water enters sieve tube from xylem/surrounding cells by osmosis. 3. This increases hydrostatic pressure inside sieve tube (at source). 4. Unloading of sucrose at sink increases water potential, causing water to leave. 5. This decreases hydrostatic pressure at the sink. 6. Sap moves down a hydrostatic pressure gradient from source to sink by mass flow. (c) [Max 2 marks] 1. Companion cells have a nucleus, whereas sieve tube elements do not / have a degenerate nucleus. 2. Companion cells have many mitochondria, whereas sieve tube elements have very few/no mitochondria. 3. Companion cells have no sieve plates, whereas sieve tube elements have sieve plates.
Question 3 · structured
10 marks
Monoclonal antibodies are highly specific antibodies produced from a single clone of hybridoma cells. (a) Describe the steps involved in the production of monoclonal antibodies using the hybridoma method. [5] (b) Explain why monoclonal antibodies are useful in the diagnosis of diseases. [3] (c) Distinguish between active immunity and passive immunity. [2]
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Worked solution

Part (a) To produce monoclonal antibodies, a small mammal (such as a mouse) is injected with the target antigen to stimulate an immune response. After some time, plasma B-cells producing the desired antibodies are harvested from the animal's spleen. These B-cells are then fused with cancerous myeloma (tumour) cells using a chemical fusogen, such as polyethylene glycol (PEG), to form hybridoma cells. The hybridoma cells are cultured on a selective medium (such as HAT medium) so only successful fusions survive. These cells are then screened to identify the specific clone that produces the desired antibody. The chosen hybridoma cell is then cloned and grown in fermenters or in vitro to yield large quantities of identical monoclonal antibodies. Part (b) Monoclonal antibodies are invaluable in diagnostics because they are highly specific to a single epitope of a target antigen, avoiding cross-reactivity with similar molecules (preventing false-positive results). They are also highly sensitive and can be chemically linked (conjugated) to detection molecules, such as fluorescent dyes or enzymes, which produce a visible signal upon binding to the pathogen's antigen. Part (c) Active immunity occurs when an individual's own immune system is stimulated to produce antibodies and memory cells in response to an antigen (either via infection or vaccination), providing long-lasting protection. Passive immunity is the temporary protection acquired by receiving pre-made antibodies from another source (e.g., across the placenta, via colostrum, or via injection of antitoxins), meaning no memory cells are formed.

Marking scheme

(a) [Max 5 marks] 1. Inject an antigen into a small mammal / mouse. 2. Harvest/isolate plasma cells / B-lymphocytes from spleen. 3. Fuse plasma cells with myeloma / tumour cells. 4. Use polyethylene glycol (PEG) / electrofusion to facilitate fusion. 5. To produce hybridoma cells. 6. Screen/select hybridomas for the desired antibody. 7. Clone the selected hybridoma cell and grow in vitro / in fermenters. (b) [Max 3 marks] 1. Monoclonal antibodies bind to only one specific epitope / antigen. 2. No cross-reactivity / high specificity prevents false-positive results. 3. High sensitivity / can detect very low concentrations of antigens. 4. Can be conjugated to enzymes / fluorescent markers for visual detection. (c) [Max 2 marks] 1. Active immunity involves the production of antibodies by the individual's own immune system AND produces memory cells. 2. Passive immunity involves receiving antibodies from an external source AND does not produce memory cells.
Question 4 · structured
10 marks
The investigation of cell structure relies on microscopy. (a) Distinguish between the terms magnification and resolution. [2] (b) An organelle has an actual diameter of 2.5 \(\mu\)m. In a micrograph, its image measures 5.0 cm. Calculate the magnification of this image. Show your working and state your final answer. [3] (c) Explain why internal structures of organelles, such as the grana in chloroplasts, can be seen with an electron microscope but not with a light microscope. [3] (d) State two structures visible in a plant cell under a light microscope that are not present in an animal cell. [2]
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Worked solution

Part (a) Magnification refers to how many times larger an image is compared to the actual size of the object being observed. Resolution, however, is the ability to distinguish between two separate points that are extremely close together, showing fine structural detail. Part (b) To calculate the magnification, we use the formula: Magnification = Image Size / Actual Size. First, convert the image measurement to the same units as the actual size (\(\mu\)m): 5.0 cm = 50 mm = 50,000 \(\mu\)m. Now, divide the image size by the actual size: 50,000 / 2.5 = 20,000. Therefore, the magnification is \(\times\)20,000 (or 2.0 \(\times\) 10^4). Part (c) Light microscopes are limited by the wavelength of visible light, which ranges from 400 nm to 700 nm, giving them a maximum resolution of about 200 nm. Electron microscopes use a beam of electrons, which has a much shorter wavelength (less than 1 nm). Because resolution is inversely proportional to wavelength, the electron microscope has a much higher resolution (around 0.5 nm), allowing structures that are closer together than 200 nm, such as individual grana membranes, to be distinguished. Part (d) Plant structures visible under a light microscope that are absent in animal cells include: the cellulose cell wall, a large central permanent vacuole (surrounded by the tonoplast), and chloroplasts.

Marking scheme

(a) [Max 2 marks] 1. Magnification is the number of times larger an image is than the actual object. 2. Resolution is the ability to distinguish between two separate points / objects that are close together. (b) [Max 3 marks] 1. Conversion of units: 5.0 cm = 50 mm = 50,000 \(\mu\)m (1 mark). 2. Magnification = Image size / Actual size = 50,000 / 2.5 (1 mark). 3. Calculation: \(\times\)20,000 or \(\times\)20 000 (1 mark) (accept 2 \(\times\) 10^4). (c) [Max 3 marks] 1. Electron microscopes use a beam of electrons, which has a shorter wavelength than visible light. 2. Shorter wavelength results in a much higher resolution. 3. Resolution of light microscopes is limited to 200 nm, whereas electron microscopes can resolve down to 0.5 nm (allowing small structures like grana to be distinguished). (d) [Max 2 marks] 1. Cellulose cell wall. 2. Large permanent vacuole / tonoplast. 3. Chloroplasts. (Accept any two)
Question 5 · structured
10 marks
Haemoglobin is a globular protein that transports oxygen in the blood. (a) Describe how the quaternary structure of haemoglobin enables it to perform its function of oxygen transport. [3] (b) At high partial pressures of carbon dioxide (\(p\text{CO}_2\)), the oxygen dissociation curve of haemoglobin shifts to the right (the Bohr shift). Explain the physiological significance of the Bohr shift in respiring tissues. [5] (c) Explain the role of the enzyme carbonic anhydrase in the transport of carbon dioxide. [2]
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Worked solution

Part (a) Haemoglobin has a quaternary structure consisting of four polypeptide chains (two alpha chains and two beta chains) bound together. Each polypeptide chain contains a prosthetic haem group with an iron ion (\(\text{Fe}^{2+}\)) at its centre. Each \(\text{Fe}^{2+}\) ion can bind reversibly to one oxygen molecule (\(\text{O}_2\)), allowing a single haemoglobin molecule to transport up to four oxygen molecules (eight oxygen atoms). The arrangement of these chains allows for cooperative binding; as the first oxygen molecule binds, the quaternary conformation of the protein shifts, increasing the affinity of the remaining haem groups for oxygen. Part (b) In actively respiring tissues, carbon dioxide is released as a waste product, leading to a high local \(p\text{CO}_2\). This carbon dioxide diffuses into red blood cells, where it is converted into carbonic acid, which dissociates into hydrogen ions (\(\text{H}^+\)) and hydrogencarbonate ions. The hydrogen ions bind to oxyhaemoglobin, forming haemoglobinic acid. This binding causes a conformational change that reduces haemoglobin's affinity for oxygen, causing the oxygen dissociation curve to shift to the right (Bohr shift). Consequently, haemoglobin releases (unloads) oxygen more readily to tissues that have high metabolic rates and are in high demand of oxygen for aerobic respiration. Part (c) Carbonic anhydrase is an enzyme located inside red blood cells that catalyses the highly rapid, reversible reaction between carbon dioxide and water to form carbonic acid (\(\text{H}_2\text{CO}_3\)). This carbonic acid then rapidly dissociates into hydrogen ions and hydrogencarbonate (\(\text{HCO}_3^-\)) ions. The hydrogencarbonate ions diffuse out into the blood plasma, allowing approximately 85% of carbon dioxide to be transported safely in the blood in a highly soluble form.

Marking scheme

(a) [Max 3 marks] 1. Quaternary structure consists of four polypeptide chains (two alpha, two beta chains). 2. Each chain contains a prosthetic haem group. 3. Each haem group contains an iron ion (\(\text{Fe}^{2+}\)) that binds reversibly to one oxygen molecule (\(\text{O}_2\)). 4. Cooperativity: binding of first oxygen molecule changes the shape of the protein, facilitating easier binding of subsequent oxygen molecules. (b) [Max 5 marks] 1. Respiring tissues produce high concentrations of carbon dioxide. 2. Carbon dioxide reacts with water inside red blood cells to form carbonic acid, which dissociates into hydrogen ions (\(\text{H}^+\)) and hydrogencarbonate ions. 3. Hydrogen ions bind to haemoglobin to form haemoglobinic acid (acting as a buffer). 4. This binding decreases the affinity of haemoglobin for oxygen. 5. This causes the oxygen dissociation curve to shift to the right (the Bohr shift). 6. Oxygen is unloaded / released more easily to tissues that are respiring rapidly / have a high metabolic demand. (c) [Max 2 marks] 1. Carbonic anhydrase catalyses the reaction: \(\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3\) (carbonic acid). 2. Allows carbon dioxide to be quickly converted into hydrogencarbonate ions (\(\text{HCO}_3^-\)) inside red blood cells. 3. Hydrogencarbonate ions are highly soluble and can be transported in blood plasma.
Question 6 · structured
10 marks
Cholera and tuberculosis (TB) are two major infectious diseases that affect human populations globally. (a) Complete the comparison of these two diseases by identifying: (i) the genus of the pathogen that causes cholera, [1] (ii) the genus of the pathogen that causes tuberculosis, [1] (iii) the primary mode of transmission of cholera, [1] (iv) the primary mode of transmission of tuberculosis. [1] (b) Explain why tuberculosis is more difficult to control and eradicate globally compared to other bacterial diseases. [4] (c) State how cholera is treated to prevent death from dehydration. [2]
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Worked solution

Part (a) (i) Cholera is caused by the bacterium Vibrio cholerae, so the genus is Vibrio. (ii) Tuberculosis is caused by Mycobacterium tuberculosis or Mycobacterium bovis, so the genus is Mycobacterium. (iii) Cholera is transmitted primarily through the ingestion of water or food contaminated with the faeces of infected individuals (faecal-oral route). (iv) Tuberculosis is transmitted via the inhalation of airborne droplets released when an infected person coughs or sneezes. Part (b) Tuberculosis is particularly difficult to control and eradicate because the causative bacterium Mycobacterium tuberculosis grows very slowly, and lives intracellularly inside host macrophages. Tubercles form in the lungs, shielding the bacteria from the host's immune response and preventing antibiotics from effectively reaching them. The bacterial cell wall also contains a thick, waxy layer composed of mycolic acid, making it highly impermeable to many standard antibiotics. This requires a long, complex regimen of multiple antibiotics (lasting 6 to 9 months), which leads to poor patient compliance and the emergence of multi-drug-resistant (MDR-TB) strains. Additionally, there is a massive global reservoir of individuals with latent TB infections who are asymptomatic but can reactivate the disease later. Part (c) Cholera results in severe watery diarrhoea due to the action of the cholera toxin, which drives chloride ions into the intestinal lumen, pulling water out of the body by osmosis. The primary treatment to prevent death is Oral Rehydration Therapy (ORT). This involves administering a solution containing a precise balance of salts (electrolytes) and glucose dissolved in clean water. Glucose is vital because its absorption in the gut facilitates the co-transport of sodium ions, which in turn draws water back into the bloodstream by osmosis, rapidly rehydrating the patient.

Marking scheme

(a) [4 marks] (i) Vibrio (accept: Vibrio cholerae) (1 mark). (ii) Mycobacterium (accept: Mycobacterium tuberculosis / Mycobacterium bovis) (1 mark). (iii) Ingestion of water / food contaminated with faeces (accept: waterborne / faecal-oral route) (1 mark). (iv) Inhalation of airborne droplets (accept: droplets / aerosol from coughing/sneezing) (1 mark). (b) [Max 4 marks] 1. Pathogen lives intracellularly / inside host macrophages (making it hard for antibiotics/antibodies to reach). 2. Thick, waxy cell wall / mycolic acid prevents penetration of drugs / resists phagocytosis. 3. Pathogen grows slowly, requiring long courses of treatment (6-9 months). 4. Poor patient compliance leads to development of drug resistance (MDR-TB / XDR-TB). 5. Large global reservoir of latent / asymptomatic infections that can reactivate. (c) [Max 2 marks] 1. Oral Rehydration Therapy (ORT) / rehydration solution. 2. Solution contains water, salts (electrolytes), and glucose. 3. Re-establishes water balance / replaces fluid and ions lost by diarrhoea / glucose aids in salt and water absorption.

Paper 3 (Advanced Practical Skills)

Complete two practical tasks: one experimental investigation and one microscope-based observation and drawing task.
2 Question · 40 marks
Question 1 · practical investigation
22 marks
Amylase is an enzyme that catalyses the hydrolysis of starch into maltose. You are required to investigate the effect of different concentrations of a plant extract, solution I, on the activity of amylase. Solution I acts as an inhibitor.

You are provided with:
- \(10\text{ cm}^3\) of \(1.0\%\) solution of the inhibitor, I
- \(50\text{ cm}^3\) of distilled water, W
- \(20\text{ cm}^3\) of \(1.0\%\) amylase solution, A
- \(20\text{ cm}^3\) of \(1.0\%\) starch solution, S
- Iodine in potassium iodide solution (in a dropper bottle)

(a) You are required to prepare a range of concentrations of the inhibitor solution, I, using simple dilution of the stock \(1.0\%\) solution I with distilled water, W.
Prepare five concentrations of I: \(1.0\%\), \(0.8\%\), \(0.6\%\), \(0.4\%\), and \(0.2\%\). The total volume of each solution should be \(10.0\text{ cm}^3\).
Complete Table 1.1 to show the volumes of \(1.0\%\) solution I and distilled water, W, required to prepare each concentration. [4]

Table 1.1
| Concentration of inhibitor I / % | Volume of 1.0% inhibitor solution I / \(\text{cm}^3\) | Volume of distilled water, W / \(\text{cm}^3\) |
|---|---|---|
| 1.0 | | |
| 0.8 | | |
| 0.6 | | |
| 0.4 | | |
| 0.2 | | |

(b) Carry out the investigation using the following instructions:
1. Label five test-tubes with the concentrations of I prepared in (a).
2. Pipette \(2.0\text{ cm}^3\) of each concentration of I into its respective test-tube.
3. Add \(2.0\text{ cm}^3\) of amylase solution A to each test-tube and mix gently.
4. Leave the mixtures for 2 minutes to allow interaction.
5. Put one drop of iodine solution into each well of a spotting tile.
6. Add \(2.0\text{ cm}^3\) of starch solution S to the first test-tube (containing \(1.0\%\) I and A), mix quickly, and start a timer.
7. Every 30 seconds, use a clean dropper pipette to remove a drop of the mixture and add it to a well containing iodine solution on the spotting tile.
8. Record the time when the starch is completely hydrolysed (when the blue-black colour no longer appears and the iodine remains yellow-brown). If the starch is not hydrolysed after 300 seconds, stop timing and record 'more than 300'.
9. Repeat steps 6 to 8 for the remaining four concentrations of I.

Record your results in an appropriate table. [6]

(c) In another investigation, a student measured the rate of amylase activity at different pH values, with no inhibitor present. The results are shown in Table 1.2.

Table 1.2
| pH | Rate of reaction / arbitrary units (a.u.) |
|---|---|
| 4.0 | 1.2 |
| 5.0 | 3.5 |
| 6.0 | 6.8 |
| 7.0 | 8.5 |
| 8.0 | 4.2 |
| 9.0 | 0.8 |

(i) Plot a line graph of the data in Table 1.2 on a grid. [4]
(ii) State and explain the effect of pH on the activity of amylase shown in the graph. [2]

(d) (i) Identify two sources of error in the method you used in (b). [2]
(ii) Suggest how you would modify the method in (b) to investigate the effect of temperature on the rate of starch hydrolysis. [4]
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Worked solution

(a) Table of Dilutions
To prepare \(10.0\text{ cm}^3\) of each concentration using \(1.0\%\) inhibitor solution:
- \(1.0\%\): \(10.0\text{ cm}^3\) of I, \(0.0\text{ cm}^3\) of W
- \(0.8\%\): \(8.0\text{ cm}^3\) of I, \(2.0\text{ cm}^3\) of W
- \(0.6\%\): \(6.0\text{ cm}^3\) of I, \(4.0\text{ cm}^3\) of W
- \(0.4\%\): \(4.0\text{ cm}^3\) of I, \(6.0\text{ cm}^3\) of W
- \(0.2\%\): \(2.0\text{ cm}^3\) of I, \(8.0\text{ cm}^3\) of W

(b) Table of Results (Specimen Data)
- Heading of 1st column: 'concentration of inhibitor I / %'
- Heading of 2nd column: 'time taken for starch hydrolysis / s'
- All times must be recorded in whole seconds (multiples of 30 seconds).
- Higher concentration of inhibitor results in a longer time taken for hydrolysis.
- Example data:
- 1.0% -> 240 s (or 'more than 300')
- 0.8% -> 180 s
- 0.6% -> 120 s
- 0.4% -> 90 s
- 0.2% -> 60 s

(c) (i) Graph Plotting
- Axes: x-axis labeled 'pH', y-axis labeled 'Rate of reaction / arbitrary units' or 'Rate of reaction / a.u.'.
- Scale: x-axis 1.0 unit per 2 cm; y-axis 1.0 or 2.0 units per 2 cm. Must be linear and use more than half the grid.
- Plotting: All 6 points plotted precisely to within half a small square with a small, sharp cross or circled dot.
- Line: Points connected with straight, ruled lines from point to point, or a smooth, clean curve passing exactly through all points with no double lines.

(c) (ii) Explain the effect of pH
- As pH increases from 4.0 to 7.0, the rate of reaction increases to the optimum at pH 7.0 (8.5 a.u.), and then decreases at higher pH values.
- Explanation: Extreme pH values (both low and high) alter the ionic charge on the amino acid R-groups in the active site of amylase. This disrupts hydrogen and ionic bonds, changing the tertiary structure and denaturing the enzyme. Consequently, the active site is no longer complementary to starch, and fewer enzyme-substrate complexes form.

(d) (i) Sources of Error
1. End-point color change is subjective / difficult to judge exactly when the blue-black color completely disappears.
2. The sampling interval of 30 seconds is too wide, meaning the exact end-point could occur between sampling times.
3. Lack of temperature control / temperature fluctuated during the experiment.

(d) (ii) Modifications for Temperature Investigation
1. Maintain the concentration of the inhibitor I constant (e.g., at 0.5%) or omit it entirely.
2. Use a range of at least five different temperatures (e.g., \(20^\circ\text{C}\), \(30^\circ\text{C}\), \(40^\circ\text{C}\), \(50^\circ\text{C}\), \(60^\circ\text{C}\)) using thermostatically-controlled water baths.
3. Pre-incubate the enzyme solution A and substrate starch solution S separately in the water bath at the target temperature for 5 minutes before mixing.
4. Ensure all other variables remain controlled (e.g., same volumes and concentrations of amylase and starch, same pH using a buffer).

Marking scheme

(a) Dilutions [4 marks]
- 1 mark for all five concentrations listed in order with correct unit in the header.
- 1 mark for correct volumes of stock 1.0% inhibitor I (10.0, 8.0, 6.0, 4.0, 2.0).
- 1 mark for correct volumes of distilled water W (0.0, 2.0, 4.0, 6.0, 8.0).
- 1 mark for all volumes written consistently to 1 decimal place (e.g., 10.0, 8.0, 0.0) or whole numbers.

(b) Results Table [6 marks]
- 1 mark for table with fully drawn borders containing raw data.
- 1 mark for clear column headings with appropriate units: 'concentration of inhibitor I / %' and 'time / s' (do not accept 'time / sec' or 'time / mins').
- 1 mark for independent variable (concentration) in the first column.
- 1 mark for all raw times recorded as whole numbers (multiples of 30 seconds).
- 1 mark for showing the correct trend: time increases as the concentration of inhibitor increases.
- 1 mark for carrying out the experiment at all 5 concentrations.

(c) (i) Graph [4 marks]
- 1 mark (O): Orientation of axes (pH on x-axis, Rate on y-axis) and complete labels with units: 'Rate of reaction / arbitrary units' or 'Rate of reaction / a.u.'.
- 1 mark (S): Scale on both axes must be linear, easy to read, and allow plotted points to occupy more than 50% of the grid area in both directions.
- 1 mark (P): All points plotted accurately within half a small square using small, thin crosses (x or +).
- 1 mark (L): Sharp, thin line connecting all points cleanly, either by straight ruled lines or a smooth curve.

(c) (ii) Explanation of pH [2 marks]
- 1 mark for describing the trend: optimum activity is at pH 7.0; rate decreases as pH deviates below or above pH 7.0.
- 1 mark for explaining that changes in pH alter the charges on amino acid R-groups, breaking hydrogen/ionic bonds, changing active site shape (denaturation) so substrate no longer binds.

(d) (i) Sources of Error [2 marks]
- 1 mark each for any two of:
- End-point color change is subjective / difficult to determine when yellow-brown is reached.
- 30-second sampling interval is too wide.
- Temperature was not controlled / temperature fluctuated.
- Only one trial was completed / no replicates to identify anomalies.

(d) (ii) Modifications [4 marks]
- 1 mark for using at least 5 different temperatures (e.g., \(20, 30, 40, 50, 60^\circ\text{C}\)).
- 1 mark for using thermostatically controlled water baths to maintain constant temperatures.
- 1 mark for pre-incubating solutions A and S separately in the water baths before mixing.
- 1 mark for controlling other key variables (e.g., pH using buffer, keeping volume/concentration of enzyme and substrate constant).
Question 2 · practical
18 marks

Slide J1 is a prepared slide of a transverse section through a herbaceous plant stem.

(a) Draw a large, low-power plan diagram of a sector of the stem shown on slide J1 to show the distribution of tissues. Your drawing should show one complete vascular bundle and the tissues on either side of it, from the outer epidermis to the central pith. Label the xylem and the sclerenchyma. [6]

(b) Use the high-power objective lens of your microscope to locate the cortex of the stem on slide J1. Select three adjacent parenchyma cells. Make a large drawing of these three cells to show their cell walls and arrangement. [5]

(c) (i) A student calibrated an eyepiece graticule using a stage micrometer. Using a high-power objective lens, 1 eyepiece graticule unit (epu) was found to be equivalent to 2.5 micrometres (\(\mu\)m). On slide J1, the width of a single vascular bundle was measured as 68 epu. Calculate the actual width of this vascular bundle in micrometres (\(\mu\)m). Show your working. [3]

(ii) Prepare a table to compare three observable differences between the xylem vessels and the phloem sieve tube elements visible on slide J1. [4]

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Worked solution

(a) Plan Diagram guidelines:
- Clear, single-line sector drawing representing the epidermis, cortex, vascular bundle (with xylem, phloem, and sclerenchyma cap), and pith boundary.
- No cell details drawn.
- Label lines drawn with a ruler, ending precisely on the target tissues (xylem vessels and sclerenchyma cap).

(b) High-power drawing guidelines:
- Drawing shows exactly three adjacent parenchyma cells.
- Cell walls shown as double lines to represent thickness.
- Small intercellular air spaces shown at the junctions between cells.

(c) (i) Calculation:
\(\text{Actual width} = 68 \text{ epu} \times 2.5 \mu\text{m/epu}\)
\(\text{Actual width} = 170 \mu\text{m}\).

(c) (ii) Comparison Table:
A table comparing xylem vessels and phloem sieve tube elements should look like this:

FeatureXylem vesselsPhloem sieve tube elementsCell wall thicknessThick, lignified wallsThin, non-lignified wallsLumen diameterLarger lumen / wider cellsSmaller lumen / narrower cellsEnd wallsNo end walls (open tubes)Perforated end walls (sieve plates) present

Marking scheme

Part (a) [6 marks]:
1. Sharp, continuous, thin lines used; no shading or sketching.
2. Sector drawn is large (occupies at least 50% of the available area).
3. Correct proportion: cortex width is less than half the radial width of the vascular bundle.
4. Vascular bundle drawn as a wedge/oval shape with distinct layers representing xylem, phloem, and sclerenchyma.
5. Label 'xylem' points precisely to the inner region of the vascular bundle with large open vessels.
6. Label 'sclerenchyma' points precisely to the cap of cells on the outer edge of the phloem.

Part (b) [5 marks]:
1. Only three adjacent cells drawn, each at least 40 mm in size.
2. Cell walls shown as double lines with middle lamella visible or represented.
3. Cells are rounded or polygonal, showing parenchyma characteristics.
4. Intercellular spaces drawn correctly at the cell junctions (corners).
5. No cell contents (like nuclei or cytoplasm) drawn or shaded.

Part (c) (i) [3 marks]:
1. Shows correct working: \(68 \times 2.5\). [1]
2. Correct numerical value: \(170\). [1]
3. Correct unit given as \(\mu\)m (or micrometres) AND expressed to appropriate significant figures (2 or 3 s.f.). [1]

Part (c) (ii) [4 marks]:
1. Complete table format with fully ruled lines, columns, and clear headings: 'Feature', 'Xylem vessel', and 'Phloem sieve tube element'. [1]
2. Thickness difference: Xylem walls are thick/lignified AND phloem walls are thin. [1]
3. Lumen diameter difference: Xylem vessels have a larger lumen/diameter AND phloem sieve tubes have a smaller lumen/diameter. [1]
4. End wall structures: Xylem has no end walls/perforated ends AND phloem sieve tubes have sieve plates/end walls. [1]

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