2025 Cambridge IAS-Level Biology (9700) Practice Paper with Answers

At normal mammalian body temperature (37 \(^\circ\text{C}\)), cholesterol interacts with the hydrophobic phospholipid tails, restricting their movement and stabilizing membrane fluidity (preventing it from becoming too fluid). Glycoproteins, which extend from the outer surface of the membrane, act as receptor molecules for cell signalling (e.g., binding hormones and neurotransmitters) as well as playing roles in cell recognition and adhesion. Option B is incorrect because glycoproteins do not act as channel proteins for active transport. Option C is incorrect because cholesterol interacts with the fatty acid tails, not the hydrophilic heads. Option D is incorrect because cholesterol does not create pores to increase fluidity.

1 mark for the correct option (A).
- Reject B, C, D.

Trachea contains cartilage rings (to keep the airway open), ciliated epithelium (to sweep mucus), and smooth muscle (in the posterior wall to allow constriction). Bronchioles lack cartilage entirely but still contain ciliated epithelium (to clear mucus) and smooth muscle (to regulate airflow). Alveoli contain neither cartilage, ciliated epithelium, nor smooth muscle; they consist of simple squamous epithelium supported by elastic fibres to facilitate rapid gas exchange.

1 mark for the correct option (A).
- Reject B, C, D.

1. Calculate base composition:
Total nucleotides = 1200.
Cytosine (C) = 30% of 1200 = 360 nucleotides.
Because DNA is double-stranded, Guanine (G) must also be 30% = 360 nucleotides. This gives 360 C-G base pairs.
The remaining 40% of nucleotides are Adenine (A) and Thymine (T) = 480 nucleotides, which gives 240 A-T base pairs.

2. Calculate hydrogen bonds:
Each C-G base pair has 3 hydrogen bonds: \(360 \times 3 = 1080\).
Each A-T base pair has 2 hydrogen bonds: \(240 \times 2 = 480\).
Total hydrogen bonds = \(1080 + 480 = 1560\).

3. Calculate phosphodiester bonds:
The DNA molecule is double-stranded and linear, meaning it consists of two separate polynucleotide strands.
Each strand contains \(1200 / 2 = 600\) nucleotides.
For a linear strand of \(N\) nucleotides, there are \(N - 1\) phosphodiester bonds.
Number of phosphodiester bonds per strand = \(600 - 1 = 599\).
Total phosphodiester bonds for both strands = \(599 \times 2 = 1198\).

1 mark for the correct option (A).
- Reject B, C, D.

Statement 1 is correct: Haemoglobin is a quaternary protein composed of 4 polypeptide chains (two alpha, two beta), whereas collagen (tropocollagen) is a triple helix of 3 polypeptide chains.
Statement 2 is correct: The globular tertiary structure of haemoglobin is stabilized by ionic bonds, hydrogen bonds, disulfide bridges, and hydrophobic interactions. The three helices of a collagen molecule are held together by hydrogen bonds.
Statement 3 is correct: Haemoglobin contains 4 iron-containing prosthetic haem groups. Collagen has no prosthetic groups.
Statement 4 is incorrect: Haemoglobin is a globular protein (soluble, with hydrophilic amino acids on its outer surface) while collagen is a fibrous protein (insoluble, structural). Thus, statements 1, 2, and 3 are correct.

1 mark for the correct option (A).
- Reject B, C, D.

Inhibitor X increases \(K_m\) without altering \(V_{max}\). This indicates that it reduces the affinity of the enzyme for the substrate, but at high substrate concentrations, the substrate outcompetes the inhibitor to achieve \(V_{max}\). Therefore, Inhibitor X is a competitive inhibitor that binds to the active site.
Inhibitor Y decreases \(V_{max}\) without altering \(K_m\). This indicates that the inhibitor reduces the concentration of active enzyme molecules (usually by binding to an allosteric site), meaning that increasing the substrate concentration cannot overcome its effect. Therefore, Inhibitor Y is a non-competitive inhibitor.

1 mark for the correct option (A).
- Reject B, C, D.

- Ingestion of antibodies in colostrum: Natural passive immunity (natural because it is a maternal-foetal transfer; passive because the baby does not manufacture the antibodies itself).
- Injection of anti-venom containing antibodies: Artificial passive immunity (artificial because it is a medical intervention; passive because pre-made antibodies are introduced).
- Injection of a weakened pathogen (vaccination): Artificial active immunity (artificial because of the injection; active because the body is stimulated to produce its own antibodies and memory cells).
- Infection by a pathogen causing illness: Natural active immunity (natural because of exposure to a wild pathogen; active because the host's immune response produces antibodies and memory cells).

1 mark for the correct option (A).
- Reject B, C, D.

The synthesis and secretion of a protein hormone follow this pathway:
1. Translation begins on ribosomes which bind to the rough endoplasmic reticulum (RER).
2. The polypeptide is synthesized into the lumen of the RER, where it undergoes folding.
3. The protein is packaged into transport vesicles that bud off the RER and fuse with the Golgi body.
4. Within the Golgi body, the protein is modified (e.g., glycosylated).
5. The modified protein is packaged into secretory vesicles that bud off the Golgi body, move towards the cell surface membrane, and release the protein via exocytosis.

1 mark for the correct option (B).
- Reject A, C, D.

- Mature xylem vessel elements are dead cells. They lack cytoplasm, have heavily lignified walls to prevent collapse, and lack a nucleus.
- Mature phloem sieve tube elements are living cells. They contain a thin layer of peripheral cytoplasm (but have no nucleus, ribosomes, or vacuole to reduce resistance to flow) and have unlignified cellulose cell walls.

1 mark for the correct option (A).
- Reject B, C, D.

The secretory pathway of a protein starts on the ribosomes of the rough endoplasmic reticulum (RER), where the polypeptide is synthesized. It is then transported via transport vesicles to the Golgi body, where glycosylation occurs. Modified glycoproteins are packaged into secretory vesicles, which bud off from the Golgi and fuse with the cell surface membrane to release the protein by exocytosis.

Award 1 mark for selecting A. Reject all other sequences as they contain incorrect organelle order or incorrect starting/ending points.

Smooth muscle is present in the walls of the trachea, bronchi, and bronchioles, allowing the regulation of airway diameter. Cartilage is absent in bronchioles, goblet cells are absent in bronchioles (or very rare in larger ones, but absent in smaller bronchioles) and absent in alveoli, and ciliated epithelium is present throughout the trachea, bronchi, and larger bronchioles.

Award 1 mark for B. Reject A because cartilage is not present in bronchioles. Reject C because goblet cells are absent in alveoli. Reject D because ciliated epithelium is present in the bronchi.

A competitive inhibitor competes with the substrate for the active site, decreasing the affinity of the enzyme for its substrate. This increases the Michaelis-Menten constant (\(K_m\)) because a higher substrate concentration is required to reach half the maximum velocity. At infinitely high substrate concentrations, the inhibitor is completely outcompeted, so the maximum velocity (\(V_{max}\)) remains unchanged.

Award 1 mark for B. Reject other options because competitive inhibitors do not change the \(V_{max}\) and always increase the \(K_m\).

Statement 1 is correct because phosphodiester bonds form the sugar-phosphate backbone by linking adjacent nucleotides. Statement 2 is incorrect because hydrogen bonds form between complementary purine and pyrimidine bases (A to T, G to C), not between two purines. Statement 3 is correct because one strand runs in the 5' to 3' direction while the complementary strand runs in the 3' to 5' direction.

Award 1 mark for selecting C (1 and 3 only).

At low temperatures, the kinetic energy of phospholipids decreases, causing them to pack closely together. Cholesterol molecules fit between the fatty acid tails, preventing them from packing too tightly and crystallizing, which maintains membrane fluidity.

Award 1 mark for C. Reject A because cholesterol does not block lateral protein movement in this manner. Reject B because increasing hydrophobic interactions would decrease fluidity. Reject D because active transport of polar molecules does not regulate fluidity.

Hybridoma cells are produced by fusing activated B-lymphocytes (plasma cells producing the desired antibody) with myeloma cells (cancerous B-lymphocytes). This combination allows the resulting hybridoma to produce specific antibodies while dividing indefinitely in culture.

Award 1 mark for B. Reject A, C, and D because T-lymphocytes, phagocytes, and memory cells are not fused with myeloma cells to produce monoclonal antibody-secreting hybridomas.

All three features are characteristic of mature xylem vessels but not phloem sieve tube elements. Phloem sieve tube elements do not have lignified walls, they retain a thin layer of cytoplasm (though they lack a nucleus and most organelles), and they have sieve plates at their end walls rather than completely broken down end walls.

Award 1 mark for selecting A because statements 1, 2, and 3 are all correct differentiators.

The quaternary structure of haemoglobin consists of the association of four separate polypeptide chains (two alpha and two beta chains). This assembly is stabilized by non-covalent interactions between the subunits, which include hydrogen bonds, hydrophobic interactions, and ionic bonds (salt bridges). Peptide bonds are covalent bonds that link amino acids together within individual polypeptide chains (primary structure) and do not stabilize the quaternary structure.

Award 1 mark for B (2, 3 and 4 only). Reject options containing 1 (peptide bonds) as these only stabilize primary structure.

Vibrio cholerae is a bacterium (prokaryote) and therefore contains 70S ribosomes and circular DNA in its cytoplasm. A palisade mesophyll cell is a eukaryotic plant cell; although its cytoplasm contains 80S ribosomes and its nucleus contains linear DNA, its mitochondria and chloroplasts contain both 70S ribosomes and circular DNA. A mature human red blood cell is highly specialized and lacks a nucleus, ribosomes, and other organelles, so it contains neither circular DNA nor 70S ribosomes. A mature xylem vessel element is a dead, hollow structure that has lost all its cytoplasm and organelles. Therefore, only 1 and 2 contain both 70S ribosomes and circular DNA.

1 mark for selecting the correct option (A).
- Accept: A
- Reject: B, C, D

An extreme change in pH alters the charge on the R-groups of amino acids. This disrupts ionic bonds and can also disrupt hydrogen bonds. Mild heating disrupts weaker interactions such as hydrogen bonds and hydrophobic interactions, but does not provide enough energy to break covalent disulfide bonds. Reducing agents specifically break covalent disulfide bonds. High concentrations of urea disrupt hydrogen bonds and hydrophobic interactions, but do not hydrolyze covalent peptide bonds between amino acids in the primary structure.

1 mark for selecting the correct option (A).
- Accept: A
- Reject: B, C, D

Inhibitor X increases the \(K_{\text{m}}\) value from 2.0 to 6.5 \(\text{mmol dm}^{-3}\) while leaving the \(V_{\text{max}}\) unchanged. This is characteristic of a competitive inhibitor, which competes with the substrate for the active site. An increase in \(K_{\text{m}}\) indicates that a higher substrate concentration is required to achieve half the maximum rate of reaction, which reflects a decreased affinity of the enzyme for its substrate. Inhibitor Y decreases \(V_{\text{max}}\) without changing \(K_{\text{m}}\), which indicates non-competitive inhibition; non-competitive inhibitors bind to an allosteric site (not the active site) and their effect cannot be overcome by increasing substrate concentration.

1 mark for selecting the correct option (C).
- Accept: C
- Reject: A, B, D

A bronchus contains irregular plates of cartilage in its wall, smooth muscle, and its epithelial lining contains goblet cells. As the airway narrows to a terminal bronchiole, cartilage is no longer present to support the wall (it is kept open by the surrounding lung tissue and elastic fibres), goblet cells are absent (to prevent mucus blocking the narrow airway), but smooth muscle remains present to control the diameter of the airway.

1 mark for selecting the correct option (A).
- Accept: A
- Reject: B, C, D

In semi-conservative replication, each original strand acts as a template for a new strand.
- Start (Generation 0): 1 molecule of heavy-heavy (\(^{15}\text{N}\)-\(^{15}\text{N}\)) DNA.
- Generation 1: 2 molecules of hybrid (\(^{15}\text{N}\)-\(^{14}\text{N}\)) DNA. Both molecules (100%) contain at least one strand of \(^{15}\text{N}\).
- Generation 2: 4 molecules: 2 hybrid (\(^{15}\text{N}\)-\(^{14}\text{N}\)) and 2 light (\(^{14}\text{N}\)-\(^{14}\text{N}\)) DNA. 2 out of 4 (50%) contain at least one strand of \(^{15}\text{N}\).
- Generation 3: 8 molecules: 2 hybrid (\(^{15}\text{N}\)-\(^{14}\text{N}\)) and 6 light (\(^{14}\text{N}\)-\(^{14}\text{N}\)) DNA. 2 out of 8 (25%) contain at least one strand of \(^{15}\text{N}\).

1 mark for selecting the correct option (B).
- Accept: B
- Reject: A, C, D

A mature xylem vessel element is a dead cell with heavily lignified cell walls. Its cytoplasm has been completely lost to allow water to flow freely, and its end walls are completely broken down to form a continuous tube. In contrast, a mature phloem sieve tube element is a living cell; its cell wall is made of cellulose and is not lignified. It retains a thin, peripheral layer of cytoplasm (although it lacks a nucleus, ribosomes, and vacuole) to keep the cell alive, and its end walls are modified into sieve plates.

1 mark for selecting the correct option (A).
- Accept: A
- Reject: B, C, D

As ventricular systole begins, the muscular wall of the left ventricle contracts, causing the pressure inside the ventricle to rise rapidly. When the pressure in the left ventricle first becomes greater than that in the left atrium, the pressure difference forces the atrioventricular (bicuspid) valve to shut, preventing the backflow of blood into the atrium. The semilunar valve in the aorta only opens later when the ventricular pressure exceeds the pressure inside the aorta.

1 mark for selecting the correct option (A).
- Accept: A
- Reject: B, C, D

Vaccination involves the artificial introduction of antigen to stimulate an immune response without causing disease. Because the child's own immune system produced antibodies and memory cells in response to the vaccine, this is an example of active artificial immunity. When exposed to the live virus years later, memory B-lymphocytes (and memory T-lymphocytes) quickly recognize the antigen, divide rapidly, and differentiate into plasma cells to produce large quantities of antibodies before the virus can cause illness.

1 mark for selecting the correct option (A).
- Accept: A
- Reject: B, C, D

At \( \times 100 \) magnification, 40 eyepiece divisions (epd) align with 10 stage divisions (sd). Since 1 sd = 0.01 mm = 10 \(\mu\text{m}\), 10 sd = 100 \(\mu\text{m}\). This means that 1 epd at \( \times 100 \) magnification is 100 \(\mu\text{m}\) / 40 = 2.5 \(\mu\text{m}\). When the magnification is increased to \( \times 400 \) (a four-fold increase), the actual size represented by each eyepiece division decreases by a factor of 4. Therefore, at \( \times 400 \) magnification, 1 epd represents 2.5 \(\mu\text{m}\) / 4 = 0.625 \(\mu\text{m}\). A cell measuring 12 epd at \( \times 400 \) has a width of 12 \(\times\) 0.625 \(\mu\text{m}\) = 7.5 \(\mu\text{m}\).

1 mark: Correct calculation of cell width (7.5 \(\mu\text{m}\)). Award 1 mark for option A. Reject B, C, and D.

Cysteine contains a sulfhydryl (-SH) group and can form covalent disulfide bonds with another cysteine. Glutamic acid has a negatively charged carboxyl R-group, and lysine has a positively charged amino R-group, allowing them to form an ionic bond. Valine has a non-polar, hydrophobic hydrocarbon R-group and will associate with another valine via hydrophobic interactions in the core of the protein.

1 mark: Correct matching of all three R-group interactions. Award 1 mark for option A. Reject B, C, and D.

A competitive inhibitor competes with the substrate for the active site. It does not alter the maximum velocity (\(V_{max}\)) of the reaction, because at infinitely high substrate concentrations, the substrate molecules will outcompete the inhibitor, allowing \(V_{max}\) to be reached. However, the apparent affinity of the enzyme for its substrate is reduced, which means a higher substrate concentration is required to reach half \(V_{max}\), so the Michaelis-Menten constant (\(K_m\)) increases.

1 mark: Correct identification that competitive inhibition leaves \(V_{max}\) unchanged but increases \(K_m\). Award 1 mark for option A. Reject B, C, and D.

During the S phase (between \(G_1\) and \(G_2\)), DNA replication occurs, so the mass of DNA doubles from \(X\) to \(2X\). However, the replicated chromosomes remain attached at their centromeres as sister chromatids, so they are still counted as single chromosomes. Therefore, the chromosome number remains 12 during \(G_2\) phase.

1 mark: Correct determination of both chromosome number (12) and DNA mass (2X) in G2 phase. Award 1 mark for option A. Reject B, C, and D.

The total water potential of the plant cells (\(\psi_{cell}\)) is calculated as \(\psi_s + \psi_p = -0.8\text{ MPa} + 0.3\text{ MPa} = -0.5\text{ MPa}\). The external sucrose solution has a water potential of \(-0.7\text{ MPa}\). Since water moves from an area of higher water potential (less negative) to an area of lower water potential (more negative), water will move out of the cells by osmosis, which decreases the volume of the protoplast and the vacuole.

1 mark: Correctly identifies that water potential of the cell is higher than the external solution, leading to water loss and vacuole shrinkage. Award 1 mark for option A. Reject B, C, and D.

According to the semi-conservative model of DNA replication: Gen 0 has only heavy DNA (\(^{15}\text{N}/^{15}\text{N}\)). After 1 division in \(^{14}\text{N}\) (Gen 1), all DNA molecules (100%) are hybrid (\(^{15}\text{N}/^{14}\text{N}\)). After 2 divisions in \(^{14}\text{N}\) (Gen 2), the two heavy template strands and the two light template strands are replicated. This results in 50% hybrid DNA (\(^{15}\text{N}/^{14}\text{N}\)) and 50% light DNA (\(^{14}\text{N}/^{14}\text{N}\)). Thus, the proportion of hybrid DNA molecules containing one heavy and one light strand is 50%.

1 mark: Correct identification of the 50% hybrid DNA proportion after two generations of replication. Award 1 mark for option C. Reject A, B, and D.

When ventricular pressure rises above atrial pressure, the atrioventricular (bicuspid/mitral) valve is forced shut to prevent backflow of blood. Since the ventricular pressure has not yet surpassed the high pressure in the aorta, the semi-lunar valve remains closed. This brief period is called isovolumetric ventricular contraction.

1 mark: Correct identification of the state of both valves during early ventricular systole. Award 1 mark for option A. Reject B, C, and D.

The injection of pre-formed antibodies provides immediate but temporary protection because no memory cells are formed by the recipient. Since the antibodies are manufactured outside the body and artificially introduced, this is artificial passive immunity.

1 mark: Correctly identifies anti-venom injection as artificial passive immunity. Award 1 mark for option A. Reject B, C, and D.

To find the actual size, use the formula: Actual Size = Image Size / Magnification. Convert 4.5 cm to micrometres: 4.5 cm = 45 mm = 45,000 \(\mu\)m. Actual Size = 45,000 \(\mu\)m / 15,000 = 3.0 \(\mu\)m.

[1 mark] B is the correct answer. Correctly converts cm to \(\mu\)m and applies the magnification formula.

Both glycogen and amylopectin are polymers of \(\alpha\)-glucose containing 1,4 and 1,6 glycosidic bonds. Glygogen is more highly branched than amylopectin due to a higher frequency of 1,6 glycosidic bonds.

[1 mark] A is the correct answer. Both structures contain the same bond types, but glycogen is more highly branched.

Cholesterol regulates membrane fluidity: at high temperatures, it stabilizes the membrane and decreases fluidity; at low temperatures, it prevents close packing of fatty acid chains, preventing freezing/crystallisation and maintaining fluidity.

[1 mark] A is the correct answer. Dual temperature-dependent role of cholesterol in cell membrane fluidity is correctly described.

The Michaelis-Menten constant (\(K_m\)) is inversely proportional to enzyme affinity. A lower \(K_m\) value (0.2 mmol dm\(^{-3}\) for Substrate X) indicates a higher affinity than for Substrate Y (2.0 mmol dm\(^{-3}\)).

[1 mark] A is the correct answer. A lower \(K_m\) indicates higher substrate affinity.

In double-stranded DNA, adenine (A) equals thymine (T), so A = 28%. Total A + T = 56%. The remaining percentage is G + C = 100% - 56% = 44%. Since cytosine (C) equals guanine (G), C = 44% / 2 = 22%.

[1 mark] A is the correct answer. Calculated correctly based on Chargaff's rules of base pairing.

Xylem vessel elements lose their living contents during development and possess walls thickened with lignin. Sieve tube elements are living cells containing cytoplasm but lack a nucleus, ribosomes, and vacuoles to reduce resistance to flow.

[1 mark] A is the correct answer. Correctly contrasts the cellular and structural properties of xylem and phloem elements.

The left atrioventricular valve closes when the left ventricle contracts (systole) and the intraventricular pressure exceeds the pressure in the left atrium. This prevents the backflow of blood into the atrium.

[1 mark] B is the correct answer. Explains the closing mechanism of the atrioventricular valve.

Immunity is passive because the fetus does not produce its own antibodies or memory cells; it is natural because the antibodies are transferred across the placenta during pregnancy, which is a natural process.

[1 mark] D is the correct answer. Correctly classifies antibody transfer via placenta as natural passive immunity.

(a) The gaseous exchange surface in the lungs consists of alveoli, which are adapted by having:
- A very thin alveolar wall made of a single layer of squamous epithelial cells, minimizing the diffusion distance.
- An extensive network of surrounding blood capillaries that rapidly carries away oxygenated blood and brings deoxygenated blood, maintaining steep concentration gradients.
- A moist layer of surfactant lining the alveoli, which reduces surface tension to prevent collapse and dissolves gases to facilitate diffusion.

(b) Goblet cells and ciliated epithelial cells work together to defend the lungs:
- Goblet cells synthesize and secrete sticky mucus into the airway lumen.
- This mucus traps dust, pollen, and inhaled pathogens (such as bacteria and viruses).
- Ciliated epithelial cells possess hair-like cilia on their apical membranes.
- These cilia beat in a coordinated, rhythmic fashion to move the trapped mucus upward (away from the lungs) to the larynx, where it can be swallowed and destroyed by stomach acid or coughed out.

(c) Cartilage in the trachea and bronchi provides structural support:
- It consists of rigid but flexible rings (or C-shaped plates) that keep the airways open (patent).
- This prevents the trachea and bronchi from collapsing when pressure decreases during inspiration.
- The C-shape of the tracheal cartilage allows flexibility and expansion of the adjacent esophagus during the swallowing of food.

(a) [Max 3 marks]
- One mark for: Alveoli have a single layer of squamous epithelium / very thin walls to minimize diffusion distance.
- One mark for: Dense / extensive network of capillaries to maintain a steep concentration gradient.
- One mark for: Presence of surfactant to reduce surface tension / prevent alveolar collapse / dissolve gases.

(b) [Max 4 marks]
- One mark for: Goblet cells secrete / produce sticky mucus.
- One mark for: Mucus traps inhaled pathogens / dust / particles.
- One mark for: Ciliated cells have cilia that beat in a coordinated / wave-like manner.
- One mark for: Sweeps / moves mucus up and out of the airways (towards throat / trachea) to be swallowed / coughed out.

(c) [Max 3 marks]
- One mark for: Cartilage keeps the trachea / bronchi open / patent.
- One mark for: Prevents collapse of the airways during inspiration / when air pressure falls.
- One mark for: C-shaped rings allow expansion of the esophagus when food is swallowed.

(a) According to the induced-fit hypothesis:
- The active site of the enzyme is not initially a perfect complementary match to the substrate.
- As the substrate collides with and binds to the active site, the active site undergoes a conformational change to fit more tightly around the substrate.
- This induced fit puts strain on the chemical bonds of the substrate, lowering the activation energy required to convert the substrate into products and forming an enzyme-substrate complex.

(b) When substrate concentration increases:
- At low substrate concentrations, substrate concentration is the limiting factor. More substrate molecules increase the frequency of successful collisions with active sites, forming more enzyme-substrate complexes per unit time, so the rate of reaction increases.
- At high substrate concentrations, the rate of reaction levels off and reaches a maximum velocity \(V_{max}\). This is because all active sites become fully saturated with substrate. At this point, enzyme concentration becomes the limiting factor.

(c) Differences between competitive and non-competitive inhibitors at very high substrate concentrations:
- A competitive inhibitor binds reversibly to the active site. Because it competes directly with the substrate, its effect can be completely overcome by adding a very high concentration of substrate. Under these conditions, the substrate outcompetes the inhibitor, and the maximum rate of reaction \(V_{max}\) can still be reached.
- A non-competitive inhibitor binds to an allosteric site on the enzyme, which alters the tertiary structure and shape of the active site. Increasing the substrate concentration does not overcome this inhibition because the substrate cannot bind to the altered active sites. Therefore, the maximum rate of reaction \(V_{max}\) is significantly reduced and cannot be reached.

(a) [Max 3 marks]
- One mark for: Active site is not fully complementary to the shape of the substrate initially.
- One mark for: Binding of the substrate causes a conformational change in the enzyme's active site to fit the substrate closely.
- One mark for: This puts strain on the substrate bonds, lowering activation energy / facilitating the formation of the enzyme-substrate complex (ESC).

(b) [Max 3 marks]
- One mark for: At low substrate concentrations, increasing substrate concentration increases the rate due to more frequent successful collisions.
- One mark for: At high substrate concentrations, the rate plateaus / reaches \(V_{max}\).
- One mark for: All active sites are saturated / occupied, making enzyme concentration the limiting factor.

(c) [Max 4 marks]
- One mark for: Competitive inhibitors bind to the active site, while non-competitive inhibitors bind to an allosteric site.
- One mark for: Competitive inhibition can be overcome by high substrate concentrations; non-competitive inhibition cannot be overcome.
- One mark for: Under competitive inhibition, \(V_{max}\) remains the same / is reached at high substrate concentration.
- One mark for: Under non-competitive inhibition, \(V_{max}\) is lowered / cannot be reached.

(a) Structural differences include:
- Companion cells have a nucleus and many mitochondria/ribosomes, whereas mature sieve tube elements lack a nucleus, have very little cytoplasm, and lack most organelles.
- Companion cells do not have sieve plates, whereas sieve tube elements have perforated sieve plates at their end walls.

(b) Sucrose is loaded into the phloem sieve tubes as follows:
- Hydrogen ions (protons, \(\text{H}^+\)) are actively pumped out of the companion cell cytoplasm into the cell wall space using ATP.
- This active transport is facilitated by proton-pumping proteins in the cell surface membrane of the companion cell, establishing a high concentration of \(\text{H}^+\) in the cell wall.
- This creates an electrochemical gradient for protons.
- Protons then diffuse back down their gradient into the companion cell through co-transporter proteins.
- As the protons move through the co-transporter, they carry sucrose molecules with them against the sucrose concentration gradient (symport).
- Sucrose then diffuses from the companion cell into the sieve tube element through plasmodesmata.

(c) Mass flow occurs as follows:
- The high concentration of sucrose in the sieve tube element at the source decreases its water potential.
- Water moves from the adjacent xylem vessels into the sieve tube by osmosis down a water potential gradient.
- This entry of water increases the hydrostatic pressure inside the sieve tube at the source.
- At the sink, sucrose is actively unloaded, which increases the water potential in the sieve tube and causes water to leave. This lowers the hydrostatic pressure at the sink.
- The difference in hydrostatic pressure between the source and the sink drives the mass flow of sap along the sieve tube.

(a) [Max 2 marks]
- One mark for: Companion cells have a nucleus / dense cytoplasm / more mitochondria / ribosomes AND mature sieve tube elements lack a nucleus / have peripheral / minimal cytoplasm.
- One mark for: Sieve tube elements have sieve plates / sieve pores, whereas companion cells do not.
- Reject: Cell wall differences unless referring to plasmodesmata density.

(b) [Max 5 marks]
- One mark for: Hydrogen ions / protons / \(\text{H}^+\) actively pumped / transported out of the companion cell.
- One mark for: Requires energy from ATP hydrolysis.
- One mark for: Establishes a proton / electrochemical / pH gradient in the cell wall.
- One mark for: Protons diffuse back into the companion cell through a co-transporter protein.
- One mark for: Sucrose is brought into the companion cell against its concentration gradient / co-transported with protons.
- One mark for: Sucrose moves from companion cell into sieve tube element via plasmodesmata.

(c) [Max 3 marks]
- One mark for: Influx of sucrose at the source lowers water potential, causing water to enter from the xylem by osmosis.
- One mark for: This entry of water increases the hydrostatic pressure at the source.
- One mark for: Removal of sucrose at the sink lowers hydrostatic pressure, creating a hydrostatic pressure gradient that drives mass flow from source to sink.

(a) Chromosome behavior:
- During metaphase, chromosomes, each consisting of two identical sister chromatids joined at the centromere, line up individually along the equator (metaphase plate) of the cell.
- Spindle fibers (microtubules) attach to the centromeres (kinetochores) of each chromosome.
- During anaphase, the centromeres split, separating the sister chromatids.
- The spindle fibers shorten, pulling the sister chromatids (now referred to as individual chromosomes) to opposite poles of the spindle, centromeres first.

(b) Importance of mitosis:
- Growth: Allows multicellular organisms to grow by increasing the total number of genetically identical cells.
- Tissue repair and replacement: Replaces damaged, worn-out, or dead cells with new, identical cells to maintain tissue function.
- Asexual reproduction: Enables unicellular eukaryotes or multicellular organisms to reproduce without fertilization, producing genetically identical offspring (clones).

(c) Calculation:
- First, find the total number of cells in the sample:
\(\text{Total cells} = 24 + 12 + 8 + 6 + 350 = 400\) cells.
- Next, find the number of cells undergoing mitosis (prophase + metaphase + anaphase + telophase):
\(\text{Mitotic cells} = 24 + 12 + 8 + 6 = 50\) cells.
- Calculate the mitotic index:
\(\text{Mitotic Index} = \frac{\text{Number of cells in mitosis}}{\text{Total number of cells}} = \frac{50}{400} = 0.125\) (or \(12.5\%\)).

(a) [Max 4 marks]
- One mark for: Chromosomes align individually along the equator / metaphase plate (metaphase).
- One mark for: Spindle fibers attach to the centromeres (metaphase).
- One mark for: Centromeres split / divide (anaphase).
- One mark for: Chromatids / chromosomes are pulled to opposite poles by shortening spindle fibers (anaphase).

(b) [Max 3 marks]
- One mark for: Growth / increasing cell number.
- One mark for: Repair of damaged tissues / replacement of dead or worn-out cells.
- One mark for: Asexual reproduction / cloning to maintain genetic stability / identical genetic information.
- Reject: "Production of gametes" / "meiosis".

(c) [Max 3 marks]
- One mark for correct calculation of total cells: \(400\).
- One mark for correct calculation of mitotic cells: \(50\).
- One mark for correct final Mitotic Index: \(0.125\) or \(12.5\%\) (must show working).

(a) Adaptations of DNA structure:
- The molecule consists of two complementary strands, meaning each individual strand can act as a template for synthesizing a new complementary strand.
- Weak hydrogen bonds hold the complementary bases together, allowing the two strands to be separated ('unzipped') relatively easily by enzymes.
- Specific complementary base pairing (Adenine with Thymine, Cytosine with Guanine) ensures that the exact sequence of bases is conserved with high fidelity.

(b) Roles of enzymes:
- DNA helicase binds to the double helix and travels along the DNA, breaking the hydrogen bonds between the complementary nitrogenous base pairs. This separates ('unzips') the two parental strands to expose the template bases.
- DNA polymerase synthesizes the new complementary strands. It aligns free activated DNA nucleotides alongside the exposed template bases.
- DNA polymerase catalyzes the formation of covalent phosphodiester bonds between the pentose sugar of one nucleotide and the phosphate group of the adjacent nucleotide, moving in a 5' to 3' direction relative to the new strand.

(c) Explanation of Meselson and Stahl's generation 1:
- After one round of replication in the \(^{14}\text{N}\) medium, all the newly formed DNA molecules consist of one 'old' heavy template strand containing \(^{15}\text{N}\) and one 'newly synthesized' light strand containing \(^{14}\text{N}\).
- This results in hybrid-density DNA molecules (\(^{15}\text{N}\)-\(^{14}\text{N}\) hybrids).
- Centrifugation of this DNA produces a single intermediate band located halfway between where pure heavy \(^{15}\text{N}\) and pure light \(^{14}\text{N}\) DNA bands would form, confirming semi-conservative replication.

(a) [Max 3 marks]
- One mark for: Double-stranded nature allows both strands to act as templates.
- One mark for: Hydrogen bonds between bases are weak enough to break easily for unwinding.
- One mark for: Complementary base pairing (A-T and G-C) ensures accurate copying of the sequence.

(b) [Max 4 marks]
- One mark for: DNA helicase breaks hydrogen bonds between base pairs.
- One mark for: DNA helicase separates / unzips the two template strands.
- One mark for: DNA polymerase joins adjacent nucleotides / catalyzes phosphodiester bonds.
- One mark for: DNA polymerase synthesizes DNA in a 5' to 3' direction.

(c) [Max 3 marks]
- One mark for: Every DNA molecule is now a hybrid consisting of one strand with \(^{15}\text{N}\) and one strand with \(^{14}\text{N}\).
- One mark for: This results in DNA of intermediate density.
- One mark for: It forms a single band located halfway between the fully heavy and fully light positions in the centrifuge tube.

(a) Why a hybridoma cell is used:
- Plasma B-lymphocytes produce the specific, required antibody, but they are fully differentiated and cannot divide or survive for long periods in laboratory culture.
- Myeloma (cancer) cells can divide rapidly and indefinitely (they are 'immortal') in culture, but they do not produce the desired antibodies.
- A hybridoma cell is a fusion of a B-lymphocyte and a myeloma cell. It possesses the characteristics of both: it produces the specific target monoclonal antibody and can divide indefinitely in culture, allowing long-term, large-scale production.

(b) Steps of production:
1. An animal (typically a mouse) is injected with the target antigen to stimulate an immune response.
2. B-lymphocytes (plasma cells) producing the specific complementary antibody are extracted from the mouse's spleen.
3. These extracted B-lymphocytes are fused with myeloma cells in vitro using a fusing agent, such as polyethylene glycol (PEG) or an electric shock.
4. The resulting mixture of cells is cultured on a selective HAT medium to allow only successfully fused hybridoma cells to survive.
5. The hybridomas are screened to identify the cells producing the specific desired antibody, which are then cloned to produce large quantities.

(c) Diagnostic or therapeutic uses:
- Diagnostic: Pregnancy testing (detecting human chorionic gonadotropin, hCG); ELISA tests to diagnose infections (e.g., HIV, COVID-19); matching tissue types before transplants.
- Therapeutic: Cancer immunotherapy (delivering drugs/radioisotopes directly to cancer cells or marking cancer cells for destruction by the immune system); treating autoimmune diseases (e.g., neutralizing TNF-alpha in rheumatoid arthritis); neutralising toxins or venom as antivenom.

(a) [Max 3 marks]
- One mark for: B-lymphocytes produce the specific target antibody but cannot divide / survive long-term in culture.
- One mark for: Myeloma cells can divide indefinitely / are immortal but do not make antibodies.
- One mark for: Hybridoma cells combine both traits: they produce the desired specific antibodies and divide indefinitely.

(b) [Max 4 marks]
- One mark for: Immunizing / injecting a mouse / mammal with the target antigen.
- One mark for: Extracting plasma cells / B-lymphocytes from the spleen.
- One mark for: Fusing these B-lymphocytes with myeloma cells using PEG / fusogen / electric shock.
- One mark for: Selection of hybridomas (on selective media) followed by cloning / screening for the desired antibody.

(c) [Max 3 marks]
- Award one mark for each correct diagnostic or therapeutic use, up to 3:
- Pregnancy tests (detecting hCG).
- ELISA / diagnostic tests (e.g. for infectious disease markers).
- Targeting / treating cancer cells (monoclonal antibodies conjugate with drugs / marking cancer cells).
- Treatment of rheumatoid arthritis / Crohn's disease (blocking inflammatory cytokines).
- Blood typing / HLA tissue matching.

**(a) (i)** To prepare a two-fold serial dilution starting with \(2.0\%\) stock solution (G) to make \(10.0\text{ cm}^3\) of each concentration:
- To make \(1.0\%\): Take \(5.0\text{ cm}^3\) of \(2.0\%\) G and add \(5.0\text{ cm}^3\) of distilled water (W).
- To make \(0.5\%\): Take \(5.0\text{ cm}^3\) of \(1.0\%\) solution and add \(5.0\text{ cm}^3\) of distilled water (W).
- To make \(0.25\%\): Take \(5.0\text{ cm}^3\) of \(0.5\%\) solution and add \(5.0\text{ cm}^3\) of distilled water (W).
- To make \(0.125\%\): Take \(5.0\text{ cm}^3\) of \(0.25\%\) solution and add \(5.0\text{ cm}^3\) of distilled water (W).
Note: For the lowest concentration, \(10.0\text{ cm}^3\) is achieved by carrying over \(5.0\text{ cm}^3\) into \(5.0\text{ cm}^3\) water, giving \(10.0\text{ cm}^3\) total volume.

**(a) (ii)**
- Dependent variable: Time taken for the first color change / s.
- Control variables:
1. Volume of Benedict's solution used (e.g., maintained at exactly \(2.0\text{ cm}^3\) using a graduated syringe).
2. Temperature of the water bath (e.g., maintained at \(90\,^{\circ}\text{C}\) using a thermostatically controlled heating element).

**(a) (iii)** Table formatting rules:
- Draw a fully ruled table with borders.
- Column headers must include units: 'Concentration of glucose / \%\)' and 'Time taken for first color change / s'.
- Do not write units in the data cells.
- Sequence the concentrations in ascending or descending order.
- Include sample U as a separate row at the bottom with its recorded time.

**(a) (iv)** Graph plotting guidelines:
- Independent variable (Concentration of glucose / \%\)) on the x-axis, and dependent variable (Time taken for first color change / s) on the y-axis.
- Scales should be linear, simple, and span more than half of the grid.
- Points plotted precisely using a sharp pencil, marked with small crosses (x) or circled dots.
- Points joined using ruled, straight, sharp lines from point to point, or a smooth line of best fit.

**(a) (v)** By plotting the points or interpolating:
- Locate \(145\text{ s}\) on the y-axis (Time).
- Draw a horizontal line to meet the curve/line, then a vertical line down to the x-axis.
- The concentration of reducing sugar in sample U is approximately \(0.40\%\) (acceptable range: \(0.38\%\) to \(0.42\%\)).

**(b)** Sources of error and improvements:
1. *Error:* Subjective determination of the exact start of the 'first color change'. *Improvement:* Use a colorimeter to measure absorbance or transmission quantitatively, or compare against a standard color reference card.
2. *Error:* Temperature fluctuations in the water bath when adding multiple tubes. *Improvement:* Use a larger volume water bath to buffer temperature drops, or utilize a digital thermostatically controlled water bath.
3. *Error:* Evaporation of water from the reaction mixtures during heating, concentrating the reactants. *Improvement:* Cover the tubes with aluminum foil or plastic caps during incubation.

**(a) (i)** [Max 3 marks]
- \(1.0\%\) dilution: \(5.0\text{ cm}^3\) of \(2.0\%\) glucose + \(5.0\text{ cm}^3\) of water (W) [1]
- \(0.5\%\) dilution: \(5.0\text{ cm}^3\) of \(1.0\%\) glucose + \(5.0\text{ cm}^3\) of water (W) [1]
- Remaining dilutions (\(0.25\%\) and \(0.125\%\)) follow the same sequential pattern of transferring \(5.0\text{ cm}^3\) to \(5.0\text{ cm}^3\) of water [1]

**(a) (ii)** [Max 3 marks]
- Identifies dependent variable as 'time taken for first color change' (must include units: seconds/s) [1]
- Identifies first control variable: volume of glucose solution / volume of Benedict's solution (e.g., \(2.0\text{ cm}^3\)) [1]
- Identifies second control variable: temperature of water bath (e.g., \(90\,^{\circ}\text{C}\)) [1]

**(a) (iii)** [Max 5 marks]
- Table fully enclosed with border lines drawn [1]
- Correct column headings with units: 'Concentration of glucose / \%\)' and 'Time taken for first color change / s' (or 'Time / s') [1]
- No units listed in the body of the table [1]
- Concentrations arranged in clear ascending or descending order [1]
- Raw data transcribed accurately, including Sample U with recorded time of \(145\text{ s}\) [1]

**(a) (iv)** [Max 4 marks]
- Axes labeled correctly with units: x-axis: 'Concentration of glucose / \%\)', y-axis: 'Time / s' [1]
- Scale selected so that points occupy more than half of the grid in both directions [1]
- All points plotted accurately within half a small square, using a sharp pencil [1]
- Points connected point-to-point with clean, ruled straight lines OR a smooth curve of best fit drawn with no double-lining [1]

**(a) (v)** [Max 2 marks]
- Shows clear vertical and horizontal construction lines on the graph from \(145\text{ s}\) to the line of best fit [1]
- Correct estimation of U in range \(0.38\%\) to \(0.42\%\) (units must be included) [1]

**(b)** [Max 3 marks]
- Award 1 mark for each matched pair of Error + corresponding Improvement up to 3 marks:
- Error: Subjectivity of color change detection / endpoint is difficult to judge. Improvement: Use a colorimeter / color standards [1]
- Error: Temperature of water bath fluctuates. Improvement: Use a thermostatically controlled water bath [1]
- Error: Evaporation from tubes during heating. Improvement: Cover test tubes with foil/caps [1]

**(a) (i)** Rules for drawing a low-power plan diagram:
1. Clean, continuous, sharp single lines must be drawn using a sharp HB pencil (no sketching or feathering).
2. No individual cells should be drawn; only tissue boundaries are shown.
3. The drawing must occupy at least half of the available space on the page.
4. Proportions must be accurate (e.g., thick cuticle relative to the mesophyll, correct shape of the rolled leaf folds).
5. No shading or coloring is permitted.

**(a) (ii)** Xerophytic modifications of *Ammophilaarenaria*:
1. *Hinged/Bulliform cells:* Lose turgor pressure during dry conditions, causing the leaf to roll tightly. This traps a microclimate of humid air inside.
2. *Stomata located in sunken pits/grooves:* Keeps stomata away from external air currents, trapping moist air and reducing the water potential gradient between the leaf interior and exterior.
3. *Thick outer cuticle:* Provides a highly impermeable waxy barrier to cuticular transpiration, preventing evaporation directly from epidermal cell walls.
4. *Epidermal hairs (trichomes):* Restrict air movement over the stomata, trapping water vapor and maintaining local humidity.

**(b) (i)** Calibration calculation:
- Distance of \(10\) divisions of stage micrometer \(= 10 \times 0.01\text{ mm} = 0.10\text{ mm}\).
- Convert to micrometers: \(0.10\text{ mm} \times 1000 = 100\ \mu\text{m}\).
- \(25\text{ epu} = 100\ \mu\text{m}\).
- \(1\text{ epu} = \frac{100\ \mu\text{m}}{25} = 4.0\ \mu\text{m}\).

**(b) (ii)** Cuticle thickness:
- Thickness in \(\text{epu}\) \(= 3.5\text{ epu}\).
- Actual thickness \(= 3.5 \times 4.0\ \mu\text{m} = 14.0\ \mu\text{m}\).

**(b) (iii)** Magnification calculation:
- Formula: \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\).
- Convert drawing size to micrometers: \(54\text{ mm} = 54,000\ \mu\text{m}\).
- \(\text{Magnification} = \frac{54,000\ \mu\text{m}}{180\ \mu\text{m}} = \times 300\).

**(c)** Comparison Table (Ammophila Leaf vs Mesophytic Stem Slide S1):
| Feature / Organ | *Ammophila* Leaf | Mesophytic Stem (Slide S1) |
| :--- | :--- | :--- |
| **Overall Shape** | Folded/rolled with deep ridges | Circular/oval symmetry |
| **Stomata** | Sunken in pits/grooves | Present on flat surface / superficial |
| **Epidermal Hairs** | Present (inner surface) | Absent (or simple trichomes only) |
| **Cuticle** | Extremely thick | Thin cuticle |
| **Vascular Arrangement** | Bundles arranged in a curve along the leaf folds | Bundles arranged in a neat ring |

**(a) (i)** [Max 4 marks]
- Clean, single, continuous lines drawn with no sketching or overlapping lines [1]
- No individual cells drawn (tissue layers only) [1]
- Drawn to a large size (occupies at least 50% of the space provided) [1]
- Correct proportions of tissues (e.g., width of sclerenchyma relative to mesophyll) [1]
- No shading or stippling [1]

**(a) (ii)** [Max 3 marks]
- Award 1 mark for each valid adaptation + explanation pair (up to 3 marks):
- Thick cuticle / waxy layer on outer epidermis reduces evaporation [1]
- Sunken stomata in grooves trap moist air / reduce the concentration gradient of water vapor [1]
- Hairs / trichomes trap a boundary layer of water vapor / lower rate of diffusion [1]
- Rolled leaf structure / bulliform cells shrink to roll leaf, hiding inner stomata from dry air currents [1]

**(b) (i)** [Max 3 marks]
- Shows calculation of stage micrometer distance: \(10 \times 0.01 = 0.10\text{ mm}\) [1]
- Converts mm to \(\mu\text{m}\) correctly: \(100\ \mu\text{m}\) [1]
- Divides \(100\ \mu\text{m}\) by \(25\) to obtain correct answer of \(4.0\ \mu\text{m}\) per epu [1]

**(b) (ii)** [Max 2 marks]
- Shows multiplication of measured value by calibration factor: \(3.5 \times 4.0\) [1]
- Correct answer: \(14.0\ \mu\text{m}\) (must include units) [1]

**(b) (iii)** [Max 2 marks]
- Converts drawing size correctly: \(54\text{ mm} = 54,000\ \mu\text{m}\) [1]
- Calculates magnification accurately: \(54,000 / 180 = \times 300\) (accept 300 or x300) [1]

**(c)** [Max 6 marks]
- Table constructed with columns labeled properly for both specimens [1]
- At least three comparative structural points recorded (up to 3 marks):
- *Ammophila* has rolled/folded margin VS Stem S1 is round/unrolled [1]
- *Ammophila* has stomata in pits VS Stem S1 has stomata on surface/no pits [1]
- *Ammophila* has internal epidermal hairs VS Stem S1 lacks epidermal hairs [1]
- *Ammophila* vascular bundles are in a curved/wavy line VS Stem S1 bundles are in a ring [1]
- Matching comparative statements written opposite each other in the table [1]
- No cell details included in comparison [1]