2026 Cambridge IAS-Level Biology (9700) Practice Paper with Answers

Active loading of sucrose begins when proton pumps in the cell surface membrane of companion cells actively pump \(\text{H}^+\) ions out of the cytoplasm and into the cell wall apoplast using ATP. This generates a higher concentration of \(\text{H}^+\) in the cell wall than inside the companion cell (a steep electrochemical gradient). \(\text{H}^+\) ions then diffuse back into the companion cell through cotransporter membrane proteins down their concentration gradient, bringing sucrose molecules with them against the concentration gradient of sucrose. Sucrose then moves into the sieve tube element through plasmodesmata via diffusion.

B is correct [1]; A is incorrect because \(\text{H}^+\) is pumped out into the cell wall, not into the sieve tube lumen directly; C is incorrect because \(\text{H}^+\) ions diffuse through cotransporters back into the companion cell, not active transport of sucrose alone without \(\text{H}^+\); D is incorrect because sucrose enters companion cells with \(\text{H}^+\) moving down its electrochemical gradient.

Unsaturated fatty acid hydrocarbon tails contain one or more carbon-carbon double bonds (\(\text{C}=\text{C}\)), which introduce kinks in the hydrocarbon chains. These kinks prevent the phospholipids from packing closely together, increasing intermolecular spacing and thereby increasing membrane fluidity. The looser packing also increases the permeability of the bilayer to small molecules.

A is correct [1]; B, C, and D are incorrect because kinks prevent close packing of tails, resulting in increased fluidity and increased permeability.

A competitive inhibitor has a molecular shape complementary to the active site of the enzyme and competes directly with the substrate for binding. At very high substrate concentrations, the substrate outcompetes the inhibitor, allowing the enzyme to achieve the same maximum rate (\(V_{\max}\) is unchanged). However, because a higher substrate concentration is required to reach half the maximum rate (\(\frac{1}{2}V_{\max}\)), the Michaelis-Menten constant (\(K_m\)) increases, indicating a lower apparent affinity for the substrate.

C is correct [1]; A and B describe non-competitive inhibition where \(V_{\max}\) decreases and \(K_m\) is unchanged; D is incorrect because competitive inhibitors bind reversibly to the active site, not an allosteric site.

Statement 1 is correct: amylose is an unbranched polymer of \(\alpha\)-glucose linked by \(\alpha\)-1,4-glycosidic bonds. Statement 2 is correct: cellulose is composed of \(\beta\)-glucose monomers linked by \(\beta\)-1,4-glycosidic bonds, requiring consecutive residues to be rotated by \(180^\circ\) relative to their neighbours. Statement 3 is incorrect: the triple helix of collagen is held together by hydrogen bonds (and covalent cross-links in fibrils), not disulfide bonds. Statement 4 is correct: a triglyceride consists of one glycerol molecule joined to three fatty acids via three ester bonds formed through condensation.

D is correct (1, 2, and 4 only) [1]; Statements 1, 2, and 4 are biologically accurate; Statement 3 is incorrect because collagen triple helixes are stabilised by hydrogen bonds, not disulfide bonds.

Penicillin acts as an inhibitor of the bacterial enzyme transpeptidase (glycopeptide transpeptidase), which catalyses the formation of peptide cross-links between peptidoglycan chains in growing bacterial cell walls. The weakened cell wall cannot withstand osmotic pressure, causing the bacterium to burst (lysis). Penicillin has no effect on viruses because viruses are non-cellular, lack peptidoglycan cell walls, and do not possess their own metabolic machinery or transpeptidase enzymes.

B is correct [1]; A is incorrect because penicillin targets transpeptidase and peptidoglycan cross-linking, not 70S ribosomes; C is incorrect because penicillin does not break ester bonds in membranes; D is incorrect because penicillin inhibits cell wall synthesis, not DNA replication enzymes.

In \(\text{G}_1\) phase, the cell has 16 chromosomes, each containing a single chromatid (16 DNA molecules). During the S phase of interphase, semi-conservative DNA replication occurs, so each chromosome consists of two identical sister chromatids joined at a single centromere. Therefore, at metaphase of mitosis, there are 16 chromosomes, each with 2 chromatids, giving a total of \(16 \times 2 = 32\) chromatids and 16 centromeres.

A is correct [1]; B is incorrect because there are 32 chromatids, not 16; C is incorrect because sister chromatids share 1 centromere per chromosome before anaphase; D is incorrect for both values.

In red blood cells, carbon dioxide is converted to carbonic acid by the enzyme carbonic anhydrase. Carbonic acid dissociates into hydrogen ions (\(\text{H}^+\)) and hydrogencarbonate ions (\(\text{HCO}_3^-\)). The \(\text{H}^+\) ions bind to oxyhaemoglobin, acting as a buffer to form haemoglobinic acid (\(\text{HHb}\)). This binding of \(\text{H}^+\) causes an allosteric conformational change in haemoglobin that reduces its affinity for oxygen, promoting the release of oxygen to the respiring tissues (the Bohr effect).

C is correct [1]; A is incorrect because carbaminohaemoglobin is formed by binding CO2 to amino groups of globin, not the main driver of the Bohr effect via \(\text{H}^+\); B is incorrect because \(\text{HCO}_3^-\) leaves the red blood cell in exchange for \(\text{Cl}^-\) (chloride shift), not binding to haem; D is incorrect because carbonic anhydrase catalyses the hydration of \(\text{CO}_2\), not the synthesis of lactic acid.

70S ribosomes have a diameter of approximately 20 nm to 30 nm. The maximum resolution of a standard light microscope is limited by the wavelength of visible light to about 200 nm (0.2 \(\mu\text{m}\)), so ribosomes cannot be resolved by a light microscope and require an electron microscope. Ribosomes are non-membrane-bound structures found freely in the cytoplasm of prokaryotes and within the matrix of mitochondria and stroma of chloroplasts in eukaryotes.

D is correct [1]; A is incorrect because a chloroplast has a double membrane envelope and can be seen with a light microscope; B is incorrect because centrioles are 200 nm diameter and require an electron microscope; C is incorrect because lysosomes are single-membrane bound and have a diameter ~0.1-0.5 um, requiring TEM for detailed structure.

During active loading of sucrose in source tissues, ATP is hydrolysed by proton pumps ($$\mathrm{H^+}\text{-ATPases}$$) to actively transport hydrogen ions ($$\mathrm{H^+}$$) out of the companion cell cytoplasm into the cell wall (apoplast). This generates a steep proton electrochemical gradient. Sucrose-$$\mathrm{H^+}$$ cotransporters (symporters) then transport $$\mathrm{H^+}$$ down its concentration gradient back into the companion cell alongside sucrose molecules against the sucrose concentration gradient.

A ; [1 mark]

[Incorrect options:
B – ATP powers the proton pump, not direct sucrose transport;
C – Protons are pumped out of the companion cell into the apoplast, not into the cytoplasm;
D – Sucrose is synthesized in mesophyll cells, not companion cells, and cotransporters transport into companion cells, not passively into sieve tubes.]

At high temperatures, cholesterol molecules restrict the lateral movement of phospholipids by interacting with their fatty acid tails, thereby reducing membrane fluidity and preventing excessive fluidity. Saturated fatty acids have straight tails that pack closely together, decreasing fluidity. Unsaturated fatty acids introduce kinks (due to carbon-carbon double bonds), which prevent close packing and increase membrane fluidity.

C ; [1 mark]

[A is incorrect because unsaturated fatty acid tails increase fluidity due to kinks; B is incorrect because saturated tails decrease fluidity; D is incorrect because cholesterol restricts phospholipid movement at high temperatures, decreasing fluidity.]

A competitive inhibitor has a complementary shape to the active site and competes directly with the substrate for binding. Because the inhibition is reversible, increasing the substrate concentration increases the probability of substrate collisions over inhibitor collisions, eventually achieving the normal \(V_{\max}\). However, a higher substrate concentration is required to reach half \(V_{\max}\) (\(\frac{1}{2}V_{\max}\)), which means the \(K_{\mathrm{m}}\) increases (indicating a lower apparent affinity for the substrate).

B ; [1 mark]

[A is non-competitive inhibition which lowers \(V_{\max}\); C describes irreversible inhibition which lowers \(V_{\max}\); D describes non-competitive inhibition which does not change \(K_{\mathrm{m}}\) but reduces \(V_{\max}\).]

- \(\mathbf{W}\) gives a blue-black colour with iodine, showing it is starch/amylose.
- \(\mathbf{X}\) gives a purple colour with Biuret reagent, showing it is a protein (haemoglobin).
- \(\mathbf{Y}\) is negative in direct Benedict's test (remains blue) but gives a brick-red precipitate after acid hydrolysis, identifying it as a non-reducing disaccharide (sucrose).
- \(\mathbf{Z}\) is positive in the direct Benedict's test (brick-red), identifying it as a reducing sugar (maltose/glucose/fructose).
Therefore, row A is correct.

A ; [1 mark]

[B is incorrect because glycogen does not give blue-black with iodine and glucose is reducing; C is incorrect because triglyceride does not test positive with Biuret and maltose is reducing; D is incorrect because cellulose does not give a blue-black iodine reaction.]

Malaria is caused by the protoctist Plasmodium (such as P. falciparum) and is transmitted by the vector female Anopheles mosquito.
- Cholera is caused by the bacterium Vibrio cholerae via waterborne transmission (contaminated water/food).
- Tuberculosis is caused by the bacteria Mycobacterium tuberculosis and Mycobacterium bovis, primarily via droplet infection (airborne).
- HIV/AIDS is caused by the human immunodeficiency virus (a virus, not a bacterium).

B ; [1 mark]

[A is incorrect because cholera is bacterial and waterborne; C is incorrect because TB is transmitted by aerosol droplets; D is incorrect because HIV is a virus.]

During the S phase of interphase, semi-conservative DNA replication doubles the quantity of nuclear DNA from \(12\text{ pg}\) to \(24\text{ pg}\). Each replicated chromosome consists of two genetically identical sister chromatids joined at the centromere throughout \(\mathrm{G_2}\), prophase, and metaphase. The sister chromatids only separate at anaphase.

D ; [1 mark]

[A, B, C incorrect: DNA replicates in S phase so mass must be 24 pg; each chromosome at metaphase consists of 2 chromatids.]

As the ventricles contract during ventricular systole:
1. Ventricular pressure rises above atrial pressure, forcing the atrioventricular (bicuspid/mitral or tricuspid) valve to close to prevent backflow into the atrium.
2. Ventricular pressure continues to increase until it exceeds aortic (or pulmonary arterial) pressure, forcing open the semilunar valve so blood is ejected into the arterial system.

C ; [1 mark]

[A, B, D are incorrect because valves open and close according to relative hydrostatic pressure gradients on either side of the valve.]

Both mitochondria and prokaryotes contain \(70\mathrm{S}\) ribosomes (smaller than cytoplasmic \(80\mathrm{S}\) eukaryotic ribosomes) and circular DNA that is not bound to histone proteins (naked DNA). Mitochondria do not have a nuclear envelope or linear chromosomes (feature 3 is eukaryotic nuclear only), and mitochondria do not have a peptidoglycan cell wall (feature 4 is found in bacterial prokaryotic cell walls only).

B ; [1 mark]

[1 and 2 are true for both; 3 is incorrect because neither has a nucleus/linear chromosomes; 4 is incorrect because mitochondria lack cell walls.]

Proton pumps in the cell surface membrane of companion cells actively transport \(\text{H}^+\) ions (protons) out of the companion cell into the cell wall space using ATP. This sets up a steep electrochemical proton gradient. \(\text{H}^+\) ions then move back down their concentration gradient through cotransporter proteins, bringing sucrose molecules into the companion cell against their concentration gradient. Sucrose then diffuses into the sieve tube element through plasmodesmata.

A is correct [1]; B is incorrect as sucrose uptake is coupled to proton cotransport rather than a direct sucrose-ATP pump; C is incorrect as water moves by osmosis rather than active transport; D is incorrect as proton gradients rather than potassium ion gradients power sucrose cotransport.

At 0% change in mass (\(0.35\text{ mol dm}^{-3}\)), the water potential of the sucrose solution equals the internal water potential of the potato cells. In a \(0.20\text{ mol dm}^{-3}\) sucrose solution, the solution has a higher (less negative) water potential than the potato cells. Therefore, water moves by osmosis down a water potential gradient from the solution into the cells, resulting in a net gain in mass.

B is correct [1]; A is incorrect because at \(0.35\text{ mol dm}^{-3}\) dynamic equilibrium exists (water enters and leaves at equal rates); C is incorrect because plant cell membranes are not freely permeable to sucrose; D is incorrect because in \(0.10\text{ mol dm}^{-3}\) solution, water moves into the vacuole, not out.

A competitive inhibitor binds reversibly to the active site, competing directly with the substrate. This increases the substrate concentration required to reach half \(V_{\text{max}}\) (raising \(K_{\text{m}}\)), but with sufficiently high substrate concentration, \(V_{\text{max}}\) is unchanged. A non-competitive inhibitor binds to an allosteric site (a site other than the active site), decreasing the catalytic activity and thus lowering \(V_{\text{max}}\) without affecting the affinity of the uninhibited enzyme active site for the substrate (\(K_{\text{m}}\) remains unchanged).

A is correct [1]; B, C, and D confuse the kinetic effects (\(V_{\text{max}}\) and \(K_{\text{m}}\)) and binding sites of competitive vs non-competitive inhibitors.

Solution 1 gave a negative result (blue) initially with Benedict's reagent, but upon acid hydrolysis into monosaccharide subunits and neutralisation, it gave a positive result (brick-red). This is characteristic of a non-reducing disaccharide such as sucrose. Solution 2 contains a reducing sugar (e.g. glucose or maltose). Solution 3 contains no reducing or non-reducing sugar (could be pure water or a non-hydrolysed polysaccharide under these mild conditions). Solution 4 is chemically invalid as reducing sugars do not lose reducing ability upon boiling with acid.

C is correct [1]; A is incorrect because Solution 1 contains a non-reducing sugar like sucrose, not glucose; B is incorrect because Solution 2 contains reducing sugar; D is incorrect because Solution 3 does not contain a non-reducing disaccharide.

Malaria is caused by protoctists of the genus \(Plasmodium\) (such as \(P.\text{ falciparum}\), \(P.\text{ vivax}\), \(P.\text{ ovale}\), \(P.\text{ malariae}\)) and transmitted by female \(Anopheles\) mosquitoes (insect vector). Cholera is caused by the bacterium \(Vibrio\text{ cholerae}\) and is water-borne. Tuberculosis is caused by the bacterium \(Mycobacterium\text{ tuberculosis}\) and is airborne. HIV/AIDS is caused by a virus and is transmitted via infected body fluids.

D is correct [1]; A is incorrect because \(Vibrio\text{ cholerae}\) is a bacterium, not a virus, and is water-borne; B is incorrect because \(Mycobacterium\text{ tuberculosis}\) is a bacterium transmitted by airborne droplets; C is incorrect because HIV is transmitted by body fluid contact, not insect vectors.

Actual size \(=\frac{\text{Image size}}{\text{Magnification}}\).
\(\text{Image size} = 27\text{ mm} = 27\,000\ \mu\text{m}\).
\(\text{Actual length} = \frac{27\,000\ \mu\text{m}}{6000} = 4.5\ \mu\text{m}\).
Chloroplasts originated via endosymbiosis and contain prokaryote-like 70S ribosomes (unlike the 80S ribosomes found free in eukaryotic cytoplasm or on the rough endoplasmic reticulum).

A is correct [1]; B incorrectly states 80S ribosomes; C and D result from calculation errors (incorrect unit conversion) and/or incorrect ribosome classification.

In active tissues, high \(p\text{CO}_2\) causes \(\text{CO}_2\) to diffuse into red blood cells, where carbonic anhydrase catalyses its hydration to form carbonic acid (\(\text{H}_2\text{CO}_3\)). Carbonic acid dissociates into \(\text{H}^+\) and \(\text{HCO}_3^-\). \(\text{H}^+\) ions combine with oxyhaemoglobin, displacing \(\text{O}_2\) (Bohr effect) and forming haemoglobinic acid (\(\text{HHb}\)), which acts as a buffer. \(\text{HCO}_3^-\) ions diffuse out into the blood plasma down their concentration gradient, and chloride ions (\(\text{Cl}^-\)) move in to maintain electrical neutrality (chloride shift).

C is correct [1]; A is incorrect because \(\text{H}^+\) ions bind to haemoglobin (preventing blood from becoming acidic) rather than actively diffusing out; B is incorrect because \(\text{Cl}^-\) enters the cell while \(\text{HCO}_3^-\) leaves; D is incorrect because carbonic anhydrase catalyses the formation of carbonic acid, not its breakdown, in respiring tissues.

Total number of nucleotides (bases) \(= 3200 \times 2 = 6400\).
Guanine (G) \(= 31\%\), so Cytosine (C) \(= 31\%\).
\(G + C = 62\%\).
Therefore, \(A + T = 100\% - 62\% = 38\%\).
Thymine (T) \(= \frac{38\%}{2} = 19\%\).
Number of thymine bases \(= 0.19 \times 6400 = 1216\).
In a linear single strand of \(n\) nucleotides, there are \(n - 1\) phosphodiester bonds. Each strand has \(3200\) nucleotides, so each strand has \(3200 - 1 = 3199\) phosphodiester bonds.
Total phosphodiester bonds \(= 2 \times 3199 = 6398\).

B is correct [1]; A incorrectly calculates 6400 phosphodiester bonds (assuming a circular molecule); C and D use half the total bases (3200) instead of 6400 or miscalculate phosphodiester bonds per linear strand.

1. Calculate the value of 1 eyepiece graticule unit at \(\times 10\) objective:
\(20\text{ stage micrometer divisions} = 20 \times 10\text{ }\mu\text{m} = 200\text{ }\mu\text{m}\).
\(40\text{ graticule units} = 200\text{ }\mu\text{m}\), so \(1\text{ graticule unit} = \frac{200}{40} = 5.0\text{ }\mu\text{m}\).

2. Adjust for switching from \(\times 10\) to \(\times 40\) objective (a 4-fold increase in magnification):
At \(\times 40\) objective, \(1\text{ graticule unit} = \frac{5.0\text{ }\mu\text{m}}{4} = 1.25\text{ }\mu\text{m}\).

3. Calculate the actual length of the chloroplast:
\(\text{Actual length} = 6 \times 1.25\text{ }\mu\text{m} = 7.5\text{ }\mu\text{m}\).

B ; [1 mark]

[A = incorrect calibration ignoring the 6 graticule length; C = failed to adjust for the change in objective lens from \(\times 10\) to \(\times 40\); D = multiplied by 4 instead of dividing when magnification increased]

Both glycogen and amylopectin are branched polysaccharides composed of \(\alpha\)-glucose units joined by \(\alpha\)-1,4-glycosidic bonds in the linear chains and \(\alpha\)-1,6-glycosidic bonds at branch points. Glycogen has a higher frequency of \(\alpha\)-1,6-glycosidic bonds (branching occurs approximately every 8–10 residues) compared to amylopectin (branching occurs approximately every 20–30 residues), giving glycogen a more compact, highly branched structure.

A ; [1 mark]

[B is incorrect as glycogen contains \(\alpha\)-glucose, not \(\beta\)-glucose; C is incorrect as glycogen is highly branched; D is incorrect as glycogen contains only \(\alpha\)-glucose monomers]

A competitive inhibitor has a shape complementary to the active site of the enzyme and competes with the substrate for binding. This increases the apparent \(K_m\) (meaning higher substrate concentration is required to reach \(\frac{1}{2}V_{max}\)), but at sufficiently high substrate concentrations, the substrate outcompetes the inhibitor, allowing \(V_{max}\) to remain unchanged.

A ; [1 mark]

[B and C describe non-competitive inhibitors which reduce \(V_{max}\) and typically leave \(K_m\) unchanged; D describes irreversible denaturation]

Ethanol is an organic solvent that dissolves the non-polar fatty acid tails of the phospholipid bilayer and disrupts hydrophobic interactions in membrane proteins. This increases the permeability of both the tonoplast and cell surface membrane, allowing betalain pigment to leak out into the surrounding solution, increasing its absorbance.

A ; [1 mark]

[B is incorrect as passive leakage occurs, not active transport failure alone; C is incorrect because ethanol does not increase water potential to cause osmotic bursting; D is incorrect because the cell wall is freely permeable to betalain; the permeability barrier is the membrane]

During metaphase of mitosis:
- The cell has \(16\) chromosomes.
- Each chromosome consists of \(2\) sister chromatids following DNA replication in S phase: \(16 \times 2 = 32\text{ chromatids}\).
- Each chromosome has \(1\) centromere holding the two sister chromatids together: \(16\text{ centromeres}\).
- Each chromatid has \(2\) telomeres (one at each end): \(32 \times 2 = 64\text{ telomeres}\).

A ; [1 mark]

[B is incorrect because sister chromatids share 1 centromere prior to anaphase; C and D confuse chromosome number with chromatid count]

1. Non-transcribed (coding) DNA strand: \(\text{5'-ATG GCT TAC CGA-3'}\).
2. The transcribed mRNA strand is identical to the coding strand, except uracil (U) replaces thymine (T): \(\text{5'-AUG GCU UAC CGA-3'}\).
3. The mRNA codons (\(\text{5'} \rightarrow \text{3'}\)) are: \(\text{AUG}\), \(\text{GCU}\), \(\text{UAC}\), \(\text{CGA}\).
4. The complementary tRNA anticodons (pairing antiparallel \(\text{3'} \rightarrow \text{5'}\)) are:
- For \(\text{5'-AUG-3'}\) \(\rightarrow\) \(\text{3'-UAC-5'}\)
- For \(\text{5'-GCU-3'}\) \(\rightarrow\) \(\text{3'-CGA-5'}\)
- For \(\text{5'-UAC-3'}\) \(\rightarrow\) \(\text{3'-AUG-5'}\)
- For \(\text{5'-CGA-3'}\) \(\rightarrow\) \(\text{3'-GCU-5'}\)

Thus, the anticodons read in the \(\text{3'} \rightarrow \text{5'}\) direction are \(\text{UAC CGA AUG GCU}\).

A ; [1 mark]

[B gives mRNA codons; C contains thymine which is not present in tRNA; D gives anticodons read in the \(\text{5'} \rightarrow \text{3'}\) direction]

Phloem loading involves:
1. Proton pumps (\(\text{H}^+\)-ATPases) actively transporting protons out of companion cells into the apoplast (cell wall spaces).
2. This creates a steep proton electrochemical gradient.
3. Protons re-enter companion cells via cotransporter proteins, bringing sucrose molecules with them against the sucrose concentration gradient.
4. Sucrose then diffuses down a concentration gradient into adjacent sieve tube elements via plasmodesmata (symplast pathway).

A ; [1 mark]

[B is incorrect because sucrose enters via cotransport driven by the proton gradient, not direct ATP pumping into sieve tubes; C is incorrect because protons are pumped out, not diffuse out into sieve tubes; D is incorrect because sucrose is translocated intact, not broken down to hexoses in loading]

1. \(\text{CO}_2\) diffuses into red blood cells and is converted to carbonic acid (\(\text{H}_2\text{CO}_3\)) by carbonic anhydrase.
2. Carbonic acid dissociates into \(\text{H}^+\) and \(\text{HCO}_3^-\).
3. The increased concentration of \(\text{H}^+\) ions lowers pH; \(\text{H}^+\) binds to oxyhaemoglobin, forming haemoglobinic acid (\(\text{HHb}\)).
4. This causes haemoglobin to change shape and release \(\text{O}_2\), shifting the oxygen dissociation curve to the right.

A ; [1 mark]

[B is incorrect as \(\text{CO}_2\) binds to terminal amine groups (forming carbaminohaemoglobin), not haem groups, and high \(\text{CO}_2\) decreases oxygen affinity; C is incorrect because carbonic anhydrase is not inhibited and curve shifts right; D is incorrect because \(\text{HCO}_3^-\)\ leaves in exchange for \(\text{Cl}^-\)\ (chloride shift)]

1. Determine the length represented by 1 eyepiece graticule unit:
\(0.15\text{ mm} = 150\ \mu\text{m}\).
\(20\text{ graticule units} = 150\ \mu\text{m}\).
\(1\text{ graticule unit} = \frac{150}{20} = 7.5\ \mu\text{m}\).

2. Calculate the actual length of the cell:
\(\text{Actual length} = 6 \times 7.5\ \mu\text{m} = 45\ \mu\text{m}\).

B [1]
Calculation of 1 graticule unit = 7.5 µm and cell length = 45 µm.

• Sample 1 is a non-reducing disaccharide such as sucrose, which yields reducing sugars following acid hydrolysis.
• Sample 2 contains starch (giving a blue-black colour with iodine).
• Sample 3 contains protein (biuret test turns purple/lilac due to peptide bonds).
• Sample 4 contains lipid/triglyceride (positive ethanol emulsion test).

A [1]
Identification: 1 = sucrose, 2 = starch, 3 = protein, 4 = triglyceride.

• Inhibitor X is a competitive inhibitor because it increases \(K_m\) (decreases the apparent affinity of the enzyme for the substrate) while \(V_{\text{max}}\) remains unchanged; its inhibition can be overcome by high substrate concentrations as it binds reversibly to the active site.
• Inhibitor Y is a non-competitive inhibitor because it reduces \(V_{\text{max}}\) without changing \(K_m\), binding to an allosteric site.

A [1]
Inhibitor X is competitive: binds to active site and can be overcome by increasing substrate concentration.

The potato tissue has a water potential equivalent to \(0.35\text{ mol}\,\text{dm}^{-3}\) sucrose solution. A \(0.10\text{ mol}\,\text{dm}^{-3}\) sucrose solution has a higher (less negative) water potential than the cell sap. Therefore, water moves by osmosis into the potato cells down a water potential gradient, causing the protoplast to swell and turgor pressure inside the cells to increase.

C [1]
Net movement of water into the cells occurs down a water potential gradient, increasing cell turgidity.

Telomeres are non-coding, repetitive sequences of DNA located at the ends of eukaryotic chromosomes. During DNA replication, DNA polymerase cannot replicate the extreme ends of the lagging strand. Telomeres ensure that vital protein-coding genes are not lost during successive rounds of cell division and prevent chromosome ends from degrading or fusing.

B [1]
Telomeres prevent loss of vital coding DNA during successive rounds of DNA replication.

1. Non-transcribed (coding) DNA strand: \(5'\text{–ATG CCG TTA GCA–}3'\).
2. Template (transcribed) DNA strand: \(3'\text{–TAC GGC AAT CGT–}5'\).
3. mRNA transcript: \(5'\text{–AUG CCG UUA GCA–}3'\).
4. The second mRNA codon is \(5'\text{–CCG–}3'\).
5. The complementary and antiparallel tRNA anticodon is \(3'\text{–GGC–}5'\).

D [1]
Correct deduction of second mRNA codon as 5'–CCG–3' and corresponding complementary anticodon as 3'–GGC–5'.

In phloem loading:
1. Proton pumps in the companion cell surface membrane use ATP to actively pump hydrogen ions (\(\text{H}^+\)) out into the cell wall.
2. This creates a high electrochemical concentration gradient of \(\text{H}^+\) outside the cell.
3. \(\text{H}^+\) ions re-enter companion cells through cotransporter proteins down their concentration gradient, bringing sucrose molecules with them against the sucrose concentration gradient.
4. Sucrose then diffuses into sieve tube elements through plasmodesmata.

A [1]
Proton pumps use ATP to move H+ out, and sucrose-proton cotransporters move sucrose in against its gradient.

In actively respiring tissues, high \(p\text{CO}_2\) leads to the diffusion of \(\text{CO}_2\) into red blood cells, where carbonic anhydrase catalyses the reaction \(\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3\). The carbonic acid dissociates into \(\text{H}^+\) and \(\text{HCO}_3^-\). \(\text{H}^+\) ions bind to oxyhaemoglobin, acting as a buffer to form haemoglobinic acid (\(\text{HHb}\)). This binding alters the tertiary and quaternary structure of haemoglobin, reducing its affinity for oxygen and displacing oxygen to respiring tissues (the Bohr shift).

B [1]
CO2 hydration produces H+ ions that combine with haemoglobin to form haemoglobinic acid, reducing O2 affinity.

1. The Casparian strip is an impermeable band containing suberin located in the radial and transverse cell walls of the endodermal cells.
2. This prevents the apoplastic movement (through cell walls) of water and dissolved mineral ions.
3. Water and ions are forced into the symplast pathway (via cytoplasm and plasmodesmata), requiring them to cross the selectively permeable cell surface membrane, which allows the plant to regulate solute uptake into the stele/xylem.

1. (contains / impregnated with) suberin / waterproof / impermeable (band in cell wall) ;
2. blocks / prevents apoplast pathway / apoplastic movement (of water / mineral ions) ;
3. forces water / ions into symplast pathway / through cytoplasm / across selectively permeable (cell surface) membrane ;
[max 2]

1. At higher temperatures, cholesterol interacts with the fatty acid tails of phospholipids, restricting their kinetic movement and preventing the membrane from becoming overly fluid.
2. At lower temperatures, cholesterol disrupts the regular crystalline packing of phospholipid tails, preventing the membrane from freezing or solidifying.

1. at high(er) temperatures, restricts / reduces movement of phospholipid (fatty acid) tails / phospholipids (to decrease fluidity / maintain membrane stability) ;
2. at low(er) temperatures, prevents close packing / crystallization of fatty acid tails / phospholipids (to increase fluidity / prevent membrane solidifying) ;
[2]

1. A competitive inhibitor binds reversibly to the active site, competing with the substrate. At sufficiently high substrate concentrations, the substrate outcompetes the inhibitor, so the maximum velocity (\(V_{\text{max}}\)) remains unchanged.
2. Because more substrate is required to reach half of the maximum velocity, the Michaelis-Menten constant (\(K_{\text{m}}\)) increases (indicating decreased apparent affinity for the substrate).

1. \(V_{\text{max}}\) is unchanged / remains the same (at high substrate concentration) ;
2. \(K_{\text{m}}\) increases / is higher (indicating lower affinity for substrate) ;
[2]

1. Monomer type: Amylose is a polymer of \(\alpha\)-glucose, while cellulose is a polymer of \(\beta\)-glucose.
2. Chain shape and orientation: Amylose chains are curved/coiled into a helix, whereas cellulose forms straight, unbranched chains where successive \(\beta\)-glucose units are inverted (rotated \(180^\circ\)) relative to each other.

Any two from:
1. monomer is \(\alpha\)-glucose in amylose AND \(\beta\)-glucose in cellulose ;
2. amylose is helical / coiled AND cellulose is straight / linear / unbranched ;
3. successive monomers in cellulose are rotated by \(180^\circ\) / alternate orientation AND same orientation in amylose ;
4. cellulose molecules form inter-chain hydrogen bonds / cross-links (forming microfibrils) AND amylose does not ;
[max 2]

1. Penicillin works by inhibiting bacterial transpeptidase enzymes, preventing cross-link formation in peptidoglycan cell walls during cell division, leading to osmotic lysis.
2. Viruses do not have a cellular structure, do not possess peptidoglycan cell walls, and do not carry out their own metabolic wall synthesis, making them unaffected by penicillin.

1. penicillin inhibits / prevents formation of peptidoglycan / cross-links (in cell walls) / inhibits transpeptidase ;
2. viruses do not have a (peptidoglycan) cell wall / are non-cellular / do not possess transpeptidase / lack bacterial targets ;
[2]

1. During semi-conservative DNA replication, the very end of the lagging strand cannot be fully replicated, resulting in chromosome shortening.
2. Telomeres consist of non-coding repetitive DNA sequences at chromosome ends, ensuring that essential genes are not lost with successive divisions and preventing chromosome fusion or unwanted DNA repair responses.

1. prevent loss of genes / vital genetic information / coding DNA (during DNA replication / cell division) ;
2. prevent ends of chromosomes / DNA molecules fusing / sticking together / prevent chromosome degradation / recognition as damaged DNA ;
[2]

1. Carbonic anhydrase inside red blood cells catalyses the conversion of dissolved carbon dioxide and water into carbonic acid (\(\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3\)).
2. The carbonic acid rapidly dissociates into \(\text{H}^+\) and \(\text{HCO}_3^-\); the hydrogencarbonate ions diffuse out into the plasma (via the chloride shift), facilitating the primary transport mechanism of \(\text{CO}_2\).

1. catalyses the (reversible) reaction between carbon dioxide and water to form carbonic acid / \(\text{H}_2\text{CO}_3\) ;
2. (carbonic acid dissociates allowing) \(\text{CO}_2\) to be converted to / transported as hydrogencarbonate ions / \(\text{HCO}_3^-\) (in blood plasma) ;
[2]

1. Nuclear structure/DNA: Prokaryotes have circular naked DNA free in the cytoplasm (nucleoid), whereas animal cells have linear DNA complexed with histones within a double-membrane nucleus.
2. Ribosomes/Organelles: Prokaryotes contain 70S ribosomes and lack membrane-bound organelles (e.g., mitochondria, ER, Golgi body), whereas animal cells have 80S ribosomes in the cytoplasm and possess membrane-bound organelles.

Any two from:
1. prokaryote has circular DNA / naked DNA / no histones AND animal cell has linear DNA / DNA associated with histone proteins ;
2. prokaryote has no nucleus / no nuclear envelope / nucleoid AND animal cell has a nucleus / double nuclear membrane / nucleolus ;
3. prokaryote has 70S ribosomes (only) AND animal cell has 80S (cytoplasmic) ribosomes ;
4. prokaryote has a (peptidoglycan) cell wall AND animal cell lacks a cell wall ;
5. prokaryote has no membrane-bound organelles / no mitochondria / no ER / no Golgi AND animal cell has membrane-bound organelles ;
[max 2]

The Casparian strip is an impermeable band of suberin located within the radial and transverse walls of endodermal cells. It functions by completely blocking the movement of water and dissolved solutes along the apoplast pathway (through the cell walls and intercellular spaces). As a result, all water and mineral ions are forced into the cytoplasm (the symplast pathway) across the selectively permeable cell surface membrane of the endodermis. This allows the plant to regulate and control the entry of specific mineral ions into the vascular cylinder (stele) and xylem vessels.

1. blocks the apoplast pathway / prevents movement through cell walls (due to suberin / waterproof band) ; [1]
2. forces water and mineral ions into the symplast pathway / through the (partially permeable) cell surface membrane / cytoplasm (allowing selective uptake / regulation of ions into xylem) ; [1]
[Total: 2]

Guidance:
- [A] stele / vascular cylinder / vascular bundle for xylem
- [I] prevents backflow of water

Cholesterol molecules insert between the phospholipid molecules in the lipid bilayer. At higher physiological temperatures, cholesterol interacts with the hydrocarbon tails of phospholipids to restrict their lateral movement, thereby preventing the membrane from becoming overly fluid and unstable. At lower physiological temperatures, cholesterol prevents the phospholipid hydrocarbon tails from packing closely together and crystalising (freezing), which maintains fluidity and membrane flexibility.

1. at high(er) temperatures, reduces fluidity / prevents membrane becoming too fluid by restricting the movement of phospholipid (fatty acid / hydrocarbon) tails / stabilizing the bilayer ; [1]
2. at low(er) temperatures, increases fluidity / maintains fluidity by preventing close packing / crystallization of phospholipid (fatty acid / hydrocarbon) tails ; [1]
[Total: 2]

Guidance:
- [R] cholesterol moving around at different temperatures
- [A] decreases fluidity for reduces fluidity

The Michaelis-Menten constant (\(K_{\text{m}}\)) is defined as the substrate concentration at which the enzyme-catalysed reaction proceeds at half its maximum velocity (\(\frac{1}{2}V_{\text{max}}\)). A low \(K_{\text{m}}\) value means that the enzyme requires only a low concentration of substrate to reach \(\frac{1}{2}V_{\text{max}}\), which indicates that the enzyme has a high affinity for that specific substrate.

1. substrate concentration at which the reaction rate / velocity is half of maximum / \(\frac{1}{2}V_{\text{max}}\) ; [1]
2. (a low \(K_{\text{m}}\) indicates) high affinity of the enzyme for its substrate / reaches \(V_{\text{max}}\) (or \(\frac{1}{2}V_{\text{max}}\)) at lower substrate concentrations ; [1]
[Total: 2]

Guidance:
- [A] 'half maximum rate' for \(\frac{1}{2}V_{\text{max}}\)
- [R] \(K_{\text{m}}\) is the rate of reaction

Penicillin acts by inhibiting the bacterial enzyme transpeptidase (glycopeptide transpeptidase), which catalyses the formation of peptide cross-links between strands of peptidoglycan during cell wall synthesis in growing bacteria. This weakens the cell wall, causing the bacterium to burst due to osmotic lysis. In contrast, viruses are non-cellular entities comprising only genetic material (DNA or RNA) surrounded by a protein coat (capsid); they do not possess a peptidoglycan cell wall or any metabolic machinery that penicillin targets.

1. (bacterial action) inhibits transpeptidase / inhibits cross-link formation in peptidoglycan / bacterial cell wall (causing lysis / cell bursting) ; [1]
2. (viral action) viruses do not have a (peptidoglycan) cell wall / viruses have a protein coat / capsid (only) / lack the target enzyme / transpeptidase ; [1]
[Total: 2]

Guidance:
- [A] murein for peptidoglycan
- [A] viruses are non-cellular / have no metabolism (for mark point 2)
- [R] viruses have a cell membrane only (must specify lack of cell wall / presence of protein coat)

1. Under control conditions, $\text{H}^+$ ions (protons) are actively pumped out of the companion cell cytoplasm into the cell wall apoplast using ATP via proton pumps.
2. This sets up an electrochemical / concentration gradient of $\text{H}^+$ ions (higher outside the companion cell in the cell wall).
3. $\text{H}^+$ ions diffuse back down their electrochemical gradient into the companion cell through $\text{H}^+$/sucrose cotransporter proteins.
4. Sucrose is transported simultaneously into the companion cell against its concentration gradient (secondary active transport).
5. Compound P inhibits the $\text{H}^+$-ATPase proton pump, so no proton gradient is established, preventing sucrose cotransport and resulting in companion cell sucrose concentration dropping from $140\;\text{mmol}\;\text{dm}^{-3}$ to near mesophyll levels ($18\;\text{mmol}\;\text{dm}^{-3}$).

1. $\text{H}^+$ / protons pumped out of companion cell into cell wall / apoplast by active transport / using ATP ;
2. creates an electrochemical / concentration / proton gradient (higher $\text{H}^+$ in wall than cytoplasm) ;
3. $\text{H}^+$ and sucrose enter companion cell together via cotransporter / symporter (protein) ;
4. secondary active transport of sucrose against its concentration gradient / driven by proton movement down its gradient ;
5. compound P prevents $\text{H}^+$ pumping so no / reduced proton gradient generated, preventing active sucrose accumulation / sucrose conc drops to $18\;\text{mmol}\;\text{dm}^{-3}$ ; [max 4]

1. As ethanol concentration increases from 0% to 40%, absorbance increases from 0.04 to 0.94 a.u., indicating an increase in membrane permeability.
2. Ethanol is an organic solvent which dissolves / disrupts the non-polar phospholipid bilayer / hydrophobic core of the membrane.
3. Ethanol disrupts hydrogen bonding and hydrophobic interactions within membrane proteins (including transport proteins and tonoplast proteins), causing protein denaturation.
4. This increases membrane fluidity and creates gaps / pores in both the tonoplast and cell surface membrane, allowing betalain pigment to leak out into the surrounding solution by diffusion down its concentration gradient.

1. comparative data quote: absorbance increases from $0.04\;\text{a.u.}$ at $0\%$ to $0.94\;\text{a.u.}$ at $40\%$ ethanol ;
2. ethanol dissolves / disrupts the phospholipid bilayer / hydrophobic core / non-polar fatty acid tails ;
3. ethanol denatures / disrupts tertiary structure of membrane proteins / transport proteins ;
4. disruption of both cell surface membrane and tonoplast ;
5. membrane permeability increases / larger gaps form, allowing more betalain pigment to diffuse out ; [max 4]

1. Inhibitor W is a competitive inhibitor.
2. Inhibitor W has a molecular shape complementary / similar to the substrate and competes with substrate molecules for binding to the active site of the enzyme.
3. Because it occupies active sites temporarily, a higher substrate concentration is required to achieve half the maximum rate of reaction, which causes $K_{\text{m}}$ to increase from $0.35$ to $1.10\;\text{mmol}\;\text{dm}^{-3}$ (indicating lower enzyme affinity for the substrate).
4. $V_{\text{max}}$ remains unchanged at $120\;\mu\text{mol}\;\text{dm}^{-3}\;\text{min}^{-1}$ because at very high substrate concentrations, substrate molecules outcompete the inhibitor for the active sites, forming enzyme-substrate complexes at the maximum possible rate.

1. (type of inhibition:) competitive (inhibition) ;
2. inhibitor has similar / complementary shape to substrate AND binds to / blocks the active site ;
3. $K_{\text{m}}$ increases because higher substrate concentration is needed to reach half-$V_{\text{max}}$ / affinity of enzyme for substrate decreases ;
4. $V_{\text{max}}$ is unchanged because at high substrate concentration, substrate outcompetes the inhibitor / all active sites occupied by substrate ; [max 4]

(a) Amylose consists of $\alpha$-glucose monomers joined by $\alpha$-1,4-glycosidic bonds and forms a coiled/helical chain. Cellulose consists of $\beta$-glucose monomers joined by $\beta$-1,4-glycosidic bonds where alternate monomers are rotated by $180^\circ$, forming a straight, unbranched chain.

(b) Straight, unbranched chains of $\beta$-glucose lie parallel to one another. Many hydrogen bonds form between hydroxyl ($\text{-OH}$) groups of adjacent chains, bundling them into strong microfibrils and macrofibrils. This provides high tensile strength to the plant cell wall, allowing it to withstand internal turgor pressure without bursting and maintain cell rigidity.

(a) any two differences:
- $\alpha$-glucose in amylose vs $\beta$-glucose in cellulose ;
- $\alpha$-1,4-glycosidic bonds in amylose vs $\beta$-1,4-glycosidic bonds in cellulose ;
- coiled / helical chain in amylose vs straight / unbranched chain in cellulose ;
- all monomers in same orientation in amylose vs alternate monomers rotated by $180^\circ$ in cellulose ; [max 2]

(b)
1. parallel chains held together by many hydrogen bonds (between $\text{-OH}$ groups) ;
2. form microfibrils / fibres / macrofibrils ;
3. gives high tensile strength to resist osmotic / turgor pressure / prevents cell bursting ; [max 2]

(a) Penicillin acts as an enzyme inhibitor of transpeptidase (glycopeptide transpeptidase), the enzyme that forms peptide cross-links between peptidoglycan chains in the bacterial cell wall during growth and division. Without cross-links, newly synthesised cell walls are weak and unable to withstand internal osmotic pressure (turgor pressure). Water enters the bacterium by osmosis, causing the cell wall to rupture and the bacterium to lyse.

(b) Human cells do not possess peptidoglycan or a cell wall, so they lack the target enzyme/structure. Viruses are non-cellular (acellular) and lack cell walls, cytoplasm, and transpeptidase enzymes, relying entirely on host cell machinery.

(a)
1. inhibits transpeptidase / prevents formation of peptide cross-links between peptidoglycan chains ;
2. weakens cell wall / cell wall cannot withstand turgor / osmotic pressure ;
3. cell bursts / lyses as water enters by osmosis during division / growth ; [max 2]

(b)
1. human cells do not have a cell wall / peptidoglycan ;
2. viruses are non-cellular / do not have cell walls / lack enzymes / transpeptidase (A: antibiotics only target living cells/bacteria) ; [max 1]

1. Carbon dioxide diffuses into red blood cells and reacts with water to form carbonic acid ($\text{H}_2\text{CO}_3$), catalysed by carbonic anhydrase.
2. Carbonic acid dissociates into hydrogen ions ($\text{H}^+$) and hydrogencarbonate ions ($\text{HCO}_3^-$).
3. $\text{H}^+$ ions bind to oxyhaemoglobin, acting as a buffer to form haemoglobinic acid ($\text{HHb}$).
4. This causes an allosteric change in haemoglobin structure, lowering its affinity for oxygen and releasing oxygen molecules.
5. In actively respiring tissues, high $p\text{CO}_2$ shifts the curve to the right (Bohr effect), allowing more oxygen to be unloaded at a given $p\text{O}_2$ ($32\%$ vs $58\%$ saturation, meaning an extra $26\%$ of oxygen is delivered) to meet the aerobic metabolic demand of tissues.

1. $\text{CO}_2$ reacts with $\text{H}_2\text{O}$ to form $\text{H}_2\text{CO}_3$ catalysed by carbonic anhydrase ;
2. $\text{H}_2\text{CO}_3$ dissociates into $\text{H}^+$ and $\text{HCO}_3^-$ ;
3. $\text{H}^+$ ions bind to haemoglobin to form haemoglobinic acid / $\text{HHb}$ ;
4. lowers affinity of haemoglobin for oxygen / causes haemoglobin to release $\text{O}_2$ (Bohr shift) ;
5. allows more $\text{O}_2$ to be unloaded / released in actively respiring tissue / muscle where $\text{CO}_2$ is high ; [max 4]

(a) Total number of dividing cells undergoing mitosis = $14 + 6 + 3 + 4 = 27$.
Total number of cells = $180$.
$\text{Mitotic index} = \frac{27}{180} = 0.15$ or $15\%$.

(b) During DNA replication, DNA polymerase cannot replicate the extreme ends of the lagging strand. Telomeres consist of non-coding repetitive DNA sequences at chromosome ends. They act as sacrificial buffers, ensuring that with each cycle of DNA replication, non-coding telomeric DNA is shortened rather than vital gene-coding sequences. They also prevent chromosome ends from being recognised as DNA damage / prevent sticky ends from fusing together.

(a)
$\frac{27}{180} \times 100$ / $\frac{27}{180}$ ;
$= 15\%$ or $0.15$ (must be 2 sig figs) ; [2]

(b) any two from:
1. repetitive non-coding DNA sequences at the tips of chromosomes ;
2. prevent loss of genes / vital genetic information as chromosome shortens during replication ;
3. prevent chromosome ends from fusing / prevent ends being detected as DNA damage / double-strand breaks ; [max 2]

(a) 1. An antigen is injected into a mammal (such as a mouse) to trigger an immune response.
2. B-lymphocytes (plasma cells) producing the specific complementary antibody are extracted from the spleen.
3. These B-cells are fused with mutant myeloma (cancer) cells using a fusing agent such as polyethylene glycol (PEG) to form hybridoma cells.
4. Hybridoma cells are cultured on selective medium, screened for production of the desired antibody, cloned, and grown in fermenters to harvest large quantities of identical monoclonal antibodies.

(b) Normal plasma cells have a limited lifespan and cannot divide in culture. Myeloma cells are cancerous and can divide indefinitely (immortal), conferring longevity and rapid cell division to the hybridoma.

(a)
1. antigen injected into animal / mouse to stimulate B-lymphocytes / plasma cells ;
2. B-cells extracted (from spleen) and fused with myeloma / cancer cells (using PEG / electric shock) ;
3. hybridoma cells screened / selected for required antibody AND cloned / cultured ; [max 2]

(b)
1. myeloma cells can divide indefinitely / are immortal / divide rapidly (unlike plasma cells) ; [1]

(a) Percentage decrease calculation:

$$\text{Decrease} = 14.8 - 1.9 = 12.9\text{ }\mu\text{mol g}^{-1}\text{ h}^{-1}$$
$$\text{Percentage decrease} = \left(\frac{12.9}{14.8}\right) \times 100 = 87.162\% \approx 87.2\%$$ (or \(87\%\))

(b) Explanation of sucrose loading:
1. In the control, ATP is used by proton pumps in the companion cell surface membrane to actively transport protons (\(\text{H}^+\)) out into the cell wall / apoplast.
2. This active transport of \(\text{H}^+\) lowers the external pH (to pH \(5.2\)) and creates a steep electrochemical / proton concentration gradient.
3. Protons diffuse back into the companion cells through cotransporter / symport proteins, moving down their concentration gradient.
4. The movement of \(\text{H}^+\) drives the transport of sucrose against its concentration gradient into the companion cell.
5. Sucrose diffuses from the companion cell into the adjacent sieve tube element through plasmodesmata.

(a) [1 mark total]
- \(87.2\%\) or \(87\%\) ;
[A \(87.16\%\) | A correct working shown: \(\frac{14.8 - 1.9}{14.8} \times 100\) if rounding error occurs]

(b) [3 marks max]
1. ATP (from respiration) is used to actively pump / transport \(\text{H}^+\) / protons out of companion cells (into the cell wall / apoplast) ;
2. (causes lower pH / higher \([\text{H}^+]\) outside) creates a proton / \(\text{H}^+\) concentration / electrochemical gradient ;
3. \(\text{H}^+\) / protons diffuse / move back into companion cell via cotransporter / symporter (protein) AND bring / transport sucrose against its concentration gradient (into companion cell) ;
4. sucrose diffuses / moves into sieve tube element (from companion cell) via plasmodesmata ;

**(a) Determination of \(V_{\max}\) and \(K_m\):**
- From Table 2.1, the maximum rate reaches a plateau at \(60.0\text{ }\mu\text{mol dm}^{-3}\text{ s}^{-1}\). Thus, \(V_{\max} = 60.0\text{ }\mu\text{mol dm}^{-3}\text{ s}^{-1}\).
- Half the maximum rate (\(\frac{1}{2}V_{\max}\)) is \(\frac{60.0}{2} = 30.0\text{ }\mu\text{mol dm}^{-3}\text{ s}^{-1}\).
- The substrate concentration required to achieve an initial rate of \(30.0\text{ }\mu\text{mol dm}^{-3}\text{ s}^{-1}\) can be derived or estimated from the table: at \([S] = 2.5\text{ mmol dm}^{-3}\), \(V = 37.5\); at \([S] = 1.0\text{ mmol dm}^{-3}\), \(V = 22.0\); interpolation gives \(K_m \approx 1.8\) to \(2.0\text{ mmol dm}^{-3}\) (or from standard Michaelis-Menten kinetics \(K_m = 1.72\text{ mmol dm}^{-3}\), or read directly from data where \(K_m\) is the substrate concentration at \(\frac{1}{2}V_{\max}\)).

(b) Explanation of lower initial rate for immobilised enzyme:
1. Substrate molecules must diffuse through the alginate gel matrix / pores to reach the active sites of the immobilised enzymes.
2. At low substrate concentrations, the rate of diffusion into the alginate beads is slow / limiting, so substrate concentration inside the bead active sites is lower than in the surrounding bulk solution.
3. This leads to fewer successful collisions and fewer enzyme-substrate complexes (ESCs) formed per unit time compared to the free enzyme.
4. At high substrate concentrations, the diffusion gradient is sufficiently steep to supply substrate rapidly enough to saturate the active sites, allowing \(V_{\max}\) to be reached.

(a) [2 marks total]
1. \(V_{\max} = 60.0\) (\(\mu\text{mol dm}^{-3}\text{ s}^{-1}\)) ;
2. \(K_m =\) value in range \(1.6\) to \(2.0\) (\(\text{mmol dm}^{-3}\)) ;
[A correct method mark if candidate states \(K_m\) is substrate concentration at \(\frac{1}{2}V_{\max}\) / \(30\text{ }\mu\text{mol dm}^{-3}\text{ s}^{-1}\) but derives an incorrect numerical value]

(b) [2 marks max]
1. alginate / matrix acts as a barrier to diffusion / substrate must diffuse through alginate beads to reach active sites ;
2. (at low \([S]\)) diffusion rate of substrate is slow / limiting OR concentration of substrate inside beads is lower than outside ;
3. fewer collisions between substrate and active site / fewer enzyme-substrate complexes / ESCs formed per unit time ;

To calculate the volume of stock solution \(V_1\) needed to make total volume \(V_2 = 20.0\text{ cm}^3\) of target concentration \(C_2\) from stock \(C_1 = 1.00\text{ mol dm}^{-3}\):
\[V_1 = \frac{C_2 \times V_2}{C_1}\]
- For \(0.80\text{ mol dm}^{-3}\): \(V_1 = \frac{0.80 \times 20.0}{1.00} = 16.0\text{ cm}^3\) of S; Volume of W \(= 20.0 - 16.0 = 4.0\text{ cm}^3\).
- For \(0.40\text{ mol dm}^{-3}\): \(V_1 = \frac{0.40 \times 20.0}{1.00} = 8.0\text{ cm}^3\) of S; Volume of W \(= 20.0 - 8.0 = 12.0\text{ cm}^3\).
- For \(0.20\text{ mol dm}^{-3}\): \(V_1 = \frac{0.20 \times 20.0}{1.00} = 4.0\text{ cm}^3\) of S; Volume of W \(= 20.0 - 4.0 = 16.0\text{ cm}^3\).

The independent variable is the factor changed by the experimenter, which is the concentration of the sucrose solution.

(a) Correct volumes of S and W for all three concentrations;
Total volume for each row adds to 20.0 cm^3;

(b) Concentration of sucrose (solution) / sucrose concentration;
[A] molarity of sucrose
[R] volume of sucrose / amount of sucrose

(a) Percentage change in mass is calculated as:
\[\text{Percentage change} = \left( \frac{\text{Change in mass}}{\text{Initial mass}} \right) \times 100\]
\[\text{Percentage change} = \left( \frac{-0.05}{2.52} \right) \times 100 = -1.984\% \approx -2.0\% \text{ (or } -1.98\% \text{ to 3 s.f.)}\]

(b) Potatoes cylinders cut to the same length still differ in starting mass due to variation in diameter or density. Calculating percentage change normalises for these variations in initial mass, allowing fair comparison.

(a) Correct calculation with correct sign / negative sign shown: -2.0(%) or -1.98(%);
[A] answer rounded to 1 decimal place or 3 significant figures
[R] positive value / omitting negative sign

(b) Initial masses of potato cylinders were not identical / different / varied;
Percentage change accounts for / normalises differences in initial mass (allowing valid comparison / AW);

(a) Distance on stage micrometer \(= 10 \times 0.01\text{ mm} = 0.10\text{ mm} = 100\,\mu\text{m}\).
Since \(40\) graticule units \(= 100\,\mu\text{m}\):
\[1\text{ graticule unit} = \frac{100\,\mu\text{m}}{40} = 2.5\,\mu\text{m}\]

(b) Actual diameter of xylem vessel:
\[\text{Diameter} = 18 \times 2.5\,\mu\text{m} = 45\,\mu\text{m}\]

(a) 100 µm / 0.10 mm divided by 40;
2.5 (µm);

(b) 18 × 2.5 (or ecf from part (a));
45 (µm);

(a) Initial rate \(= \frac{\Delta V}{\Delta t} = \frac{14.5 - 0.0}{30 - 0} = \frac{14.5}{30} = 0.4833...\text{ cm}^3\text{ s}^{-1} \approx 0.483\text{ cm}^3\text{ s}^{-1}\).

(b) Between 60 s and 240 s, the concentration of hydrogen peroxide (substrate) decreases because it is being converted into water and oxygen. Therefore, there are fewer enzyme-substrate complexes formed per unit time as collision frequency drops. At 240 s, all substrate has been converted into product, so no more oxygen can be produced.

(a) 14.5 / 30;
0.483 or 0.48 (cm^3 s^-1);
[R] answers without proper rounding or units omitted if requested

(b) Substrate / hydrogen peroxide is used up / concentration decreases (leading to fewer collisions / fewer enzyme-substrate complexes / ESCs formed);
All substrate is converted to product / substrate becomes limiting / no substrate left by 240 s;

(a) For a cube with side length \(l = 10\text{ mm}\):
\[\text{Surface area} = 6 \times l^2 = 6 \times (10\text{ mm})^2 = 600\text{ mm}^2\]
\[\text{Volume} = l^3 = (10\text{ mm})^3 = 1000\text{ mm}^3\]
\[\text{SA}:\text{V ratio} = \frac{600}{1000} = 0.6\text{ mm}^{-1} \quad (\text{or } 6:1)\]

(b) Control variables include: temperature of the acid bath, concentration of hydrochloric acid, volume/depth of hydrochloric acid covering the cubes, and concentration of indicator/alkali inside the agar.

(a) Correct calculation of surface area (600 mm^2) and volume (1000 mm^3);
0.6 (:1) or 6:1 (or 0.6 mm^-1);

(b) Any two from:
- Temperature (of acid / water bath);
- Concentration of hydrochloric acid;
- Volume of hydrochloric acid / cubes completely submerged;
- Composition / batch / concentration of alkali / indicator in agar cubes;
[I] size of cube / time

(a) Radius \(r = \frac{\text{diameter}}{2} = \frac{0.80\text{ mm}}{2} = 0.40\text{ mm}\).
Length \(h = 75\text{ mm}\).
\[V = \pi \times r^2 \times h = 3.142 \times (0.40)^2 \times 75 = 3.142 \times 0.16 \times 75 = 37.704\text{ mm}^3 \approx 37.7\text{ mm}^3\]

(b) The uncertainty in a ruler measurement (involving two readings, start and end, each \(\pm 0.5\text{ mm}\), or total absolute uncertainty \(\pm 1.0\text{ mm}\), or for single reading \(\pm 0.5\text{ mm}\)):
Standard Cambridge Paper 3 calculation:
\[\text{Percentage uncertainty} = \frac{1.0\text{ mm}}{75\text{ mm}} \times 100 = 1.33\%\]
(or \(\frac{0.5}{75} \times 100 = 0.67\%\) depending on method).

(a) Correct substitution into formula: 3.142 × (0.40)^2 × 75;
37.7 (mm^3) / 37.70 (mm^3);
[R] using diameter (0.80) as radius without dividing by 2

(b) Calculation shown: (1 / 75) × 100 or (0.5 / 75) × 100;
1.33(%) or 1.3(%) / 0.67(%) or 0.7(%);

(a) Biological plan drawing rules:
1. Draw clear, continuous, sharp single lines with no sketchy or overlapping strokes.
2. Do not draw any individual cells.
3. Do not include any shading or cross-hatching.
4. Maintain correct relative proportions of tissue layers.

(b) Total diameter \(= 4.8\text{ mm}\).
Stele diameter \(= 1.2\text{ mm}\).
Diameter occupied by outer tissues (cortex and epidermis) \(= 4.8\text{ mm} - 1.2\text{ mm} = 3.6\text{ mm}\).
\[\text{Percentage} = \left( \frac{3.6}{4.8} \right) \times 100 = 75.0\%\]

(a) Any two from:
- Sharp, clear, continuous / unbroken lines (no sketchiness / no feathering);
- No individual cells drawn (tissue plan only);
- No shading / no cross-hatching / stippling;
- Proportions accurate / representative of specimen;

(b) (4.8 - 1.2) / 4.8 × 100 or 3.6 / 4.8 × 100;
75.0(%) / 75(%);

(a) Blue indicates no reducing sugar (sample C). The green colour with a slow change (185 s) indicates a very low concentration (A). Yellow-orange with a faster change (85 s) indicates moderate concentration (D). Red-brown precipitate appearing quickly (35 s) indicates highest concentration (B).
Order from lowest to highest: C, A, D, B.

(b) Quantitative improvements:
1. Make a series of standard solutions of known reducing sugar (e.g. glucose) concentrations and test them with Benedict's reagent under identical conditions.
2. Use a colorimeter with an appropriate filter (e.g. red filter) to measure the absorbance (or transmission) of the supernatant after centrifuging/filtering, then plot a calibration curve and read off the unknown concentration.
Alternatively, filter, dry, and weigh the precipitate formed on a high-precision mass balance.

(a) C, A, D, B / C < A < D < B;

(b) Any two from:
- Prepare known concentrations / standard solutions of glucose / reducing sugar;
- Perform Benedict's test on standards (under identical conditions) and use a colorimeter to measure absorbance / transmission;
- Plot a calibration / standard curve (absorbance against concentration);
- Read unknown concentration of D from calibration curve;
- [Alternative] Filter, dry to constant mass, and weigh precipitate using a balance (for standards and D);

To produce a high-scoring plan drawing: 1. Draw a clear sector representing approximately one-quarter of the circular stem section. 2. Use sharp, unfeathered, continuous pencil lines with no overlapping lines and no shading. 3. Accurately represent the proportional thickness of the outer epidermis, cortex layer, vascular bundle ring, and central pith region. 4. Clearly delineate xylem towards the inside and phloem towards the outside within the vascular bundles without drawing individual cells. 5. Include a ruled horizontal label line terminating on the xylem tissue with the label 'xylem'.

1. Quality of drawing: clean, single, unbroken continuous lines of uniform thickness, no shading, drawing occupies at least 50% of the available space provided; [1]
2. Proportions and accuracy: sector representing one-quarter of the stem TS showing epidermis as a single outer line, distinct cortex, vascular bundle showing xylem and phloem layers, and central pith, with no individual cells drawn; [1.5]
3. Correct label and label line: ruled label line touching the inner vascular tissue, unambiguously labelled 'xylem' (A: 'xylem tissue' / 'xylem vessel region'; R: if arrowheads are used, R: if label line does not touch the tissue). [1]

(a) Length of 22 stage micrometer divisions = \(22 \times 0.01\text{ mm} = 0.22\text{ mm} = 220\text{ \mu m}\).
Actual length of 1 epu = \(\frac{220\text{ \mu m}}{50} = 4.4\text{ \mu m}\).

(b) Actual lumen diameter = \(16\text{ epu} \times 4.4\text{ \mu m/epu} = 70.4\text{ \mu m}\).

(a) 22 divisions = \(0.22\text{ mm}\) or \(220\text{ \mu m}\); [0.5]
\(\frac{220}{50} = 4.4\text{ (\mu m)}\); [1.0]
(b) Shows multiplication of candidate's 1 epu value by 16 (\(16 \times 4.4\)); [1.0]
Correct final value: \(70.4\text{ \mu m}\) (A: 70 / 70.4 / 70.40) (ecf allowed if calculation in (a) is incorrect). [1.0]

1. Draw exactly four cells: two xylem vessels (with thicker walls, generally larger/angular lumens) touching two parenchyma cells (thinner walls, rounded or polyhedral). 2. Double lines representing cell walls with middle lamella / junction between cells shown. 3. No artistic shading or cross-hatching. 4. Proportions must be biologically accurate (xylem cell wall noticeably thicker than parenchyma wall). 5. Straight label lines drawn with a ruler indicating 'cell wall' and 'lumen'.

1. Drawing composition: exactly four touching cells drawn (two xylem elements + two parenchyma cells), occupies >50% of drawing space, clear continuous lines with no gaps, overlaps, or shading; [1]
2. Anatomical features: cell walls drawn with double lines, xylem cell walls noticeably thicker than parenchyma cell walls, intercellular junctions correctly represented; [1.5]
3. Labels: straight ruled label line touching a cell wall labelled 'cell wall' AND ruled label line pointing inside a cell labelled 'lumen' (R: label lines with arrows, R: ambiguous lines). [1]

Construct a comparison table with clear headings ('Feature', 'Vessel A (Artery)', 'Vessel B (Vein)'). Record two clear observable structural differences (such as thickness of wall, thickness of tunica media, regularity/shape of lumen, presence of folded endothelium) and one structural similarity (both have a lumen, both have multi-layered walls consisting of tunica intima, media, and externa).

1. Table format: fully ruled table with clear, appropriate column and row headings separating features, Vessel A, and Vessel B; [0.5]
2. Two valid observable differences (e.g., Vessel A has thicker wall / thicker tunica media compared to Vessel B; Vessel A has round/regular lumen while Vessel B has irregular/folded/collapsed lumen; Vessel A has smaller lumen diameter to wall thickness ratio than Vessel B) (1 mark per valid comparative difference); [2.0]
3. One valid observable similarity (e.g., both possess three tissue layers / tunica intima, media, and externa; both possess an internal lumen; both possess an endothelial lining). [1.0]

(a)(i) Ratio = 12 : 32. Dividing both by 4 gives 3 : 8.
(a)(ii) Percentage = \(\frac{12}{32} \times 100 = 37.5\%\).
(b) Essential plan drawing rules: 1. Clean, continuous single lines without feathering; 2. No shading; 3. No individual cells drawn; 4. Proportions of tissue layers accurate; 5. Large drawing filling at least half the space.

(a)(i) \(3 : 8\) (A: 3 to 8; R: 12 : 32 or decimal ratios); [1.0]
(a)(ii) \(37.5(\%)\); [1.0]
(b) Any three from: [1.5 max, 0.5 each]
- Single, clear, continuous lines / no sketched or feathered lines;
- No shading / cross-hatching / stippling anywhere on the diagram;
- No individual cells drawn / only tissue layer boundaries shown;
- Drawing occupies at least 50% of the provided space / large drawing;
- Correct proportions of tissue layers shown.

(a) State two clear observable anatomical differences: leaf rolling / hinge cells, presence of trichomes (hairs), sunken stomata in crypts, thicker cuticle on outer surface.
(b) State the feature (e.g. rolled leaf / hairs / sunken stomata) and explain the mechanism: traps moist air/water vapour, increases humidity around stomata, reduces the water vapour potential gradient, reducing water loss by transpiration.

(a) Two valid comparative differences between M1 and M2; [2.0, 1 mark per difference]
e.g., M1 leaf is rolled / curled / folded whereas M2 is flat / unrolled;
e.g., M1 has epidermal hairs / trichomes / ridges whereas M2 lacks hairs / has a smooth surface;
e.g., M1 has stomata sunken in pits / crypts whereas M2 has stomata on the lower surface / not in pits;
e.g., M1 has a thicker outer cuticle than M2.

(b) Feature named (e.g. rolled leaf / hairs / sunken stomata) AND explanation: traps moist air / water vapour / creates microclimate of high humidity; [1.0]
Reduces the water vapour potential gradient between inside the substomatal chamber and the external air (A: transpiration rate decreases / less diffusion of water vapour). [0.5]