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A shift to the left (Curve X) indicates that haemoglobin has a higher affinity for oxygen, which occurs at a lower partial pressure of carbon dioxide (\(pCO_2\)) and higher pH (lower concentration of hydrogen ions), typical of blood in the lungs where carbon dioxide is exhaled. A shift to the right (Curve Z) indicates a lower affinity for oxygen (the Bohr effect), which occurs at higher \(pCO_2\) and lower pH (higher concentration of hydrogen ions), typical of actively respiring tissues to facilitate the unloading of oxygen.

Award 1 mark for selecting the correct option (B) which correctly correlates left-shift (X) with lung conditions and right-shift (Z) with respiring tissues.

Protons (\(H^+\) ions) are actively pumped out of the companion cell into the cell wall (apoplast) using ATP. This establishes a high proton concentration gradient outside the cell. Protons then diffuse back into the companion cell down their electrochemical gradient through a co-transporter protein, bringing sucrose molecules with them against their concentration gradient (cotransport). From the companion cell, sucrose enters the sieve tube element via plasmodesmata.

Award 1 mark for identifying the correct mechanism (A) in which protons are pumped out into the cell wall and then cotransported with sucrose back into the companion cell.

Amylose is an unbranched, helical polymer of glucose linked entirely by \(\alpha\)-1,4-glycosidic bonds. Amylopectin and glycogen are branched polymers containing both \(\alpha\)-1,4-glycosidic bonds in the linear chains and \(\alpha\)-1,6-glycosidic bonds at the branch points. Glycogen is more highly branched than amylopectin.

Award 1 mark for the correct option (B) describing the specific glycosidic bonds found in amylose, amylopectin, and glycogen.

Vaccination introduces a foreign antigen to trigger the body's own immune response to produce antibodies and memory cells, which is active artificial immunity. During the second exposure to the pathogen, memory B-lymphocytes and memory T-lymphocytes quickly recognize the antigen and proliferate/differentiate to eliminate the pathogen before symptoms occur.

Award 1 mark for identifying option (A) as the correct classification of vaccine-induced immunity (active artificial) and the cellular mediators (memory lymphocytes).

Since DNA is double-stranded, Adenine (A) pairs with Thymine (T), meaning T also makes up 28% of the bases. Together, A + T = 56%. The remaining bases (G + C) make up 44% (100% - 56%). Since Cytosine (C) pairs with Guanine (G), Cytosine makes up 22% of the total bases (44% / 2). The total number of Cytosine bases is 22% of 1200, which is \(0.22 \times 1200 = 264\).

Award 1 mark for calculating the correct number of cytosine bases (264), showing understanding of Chargaff's rules of base pairing.

If \(V_{max}\) is unchanged but requires a higher substrate concentration to be reached, the inhibitor is competitive. A competitive inhibitor binds reversibly to the active site, competing with the substrate. Because more substrate is required to achieve half of the \(V_{max}\), the Michaelis-Menten constant (\(K_m\)) of the enzyme is increased.

Award 1 mark for identifying option (B), showing understanding that competitive inhibitors bind to the active site and increase the value of \(K_m\) without changing \(V_{max}\).

Bronchioles contain smooth muscle fibres in their walls to regulate the diameter of the airways, whereas alveoli completely lack smooth muscle. Both bronchioles and alveoli contain elastic fibres and collagen to allow expansion and elastic recoil during breathing.

Award 1 mark for selecting the correct tissue component (B) that is present in bronchioles but absent in alveoli.

Proteins destined for secretion are synthesized by ribosomes on the rough endoplasmic reticulum (RER). They are then transported via vesicles to the Golgi body, where they undergo modification (e.g., adding carbohydrate groups to make glycoproteins) and packaging. Secretory vesicles bud off the Golgi body and transport the hormone to the cell surface membrane, fusing with it to release the content via exocytosis.

Award 1 mark for selecting option (B) which correctly outlines the secretional pathway of a glycoprotein hormone from RER to Golgi to secretory vesicles and finally the cell surface membrane.

In actively respiring tissues, carbon dioxide diffuses into the red blood cells and is converted into carbonic acid by carbonic anhydrase. Carbonic acid dissociates into hydrogencarbonate ions and hydrogen ions. Hydrogencarbonate ions diffuse out of the cell down their concentration gradient, while chloride ions diffuse in (the chloride shift) to maintain electrical neutrality. The hydrogen ions bind to hemoglobin, forming haemoglobinic acid, which acts as a buffer.

1 mark for the correct identification of hydrogencarbonate moving out, chloride moving in, and hydrogen ions binding to hemoglobin to form haemoglobinic acid (Option A).

The Casparian strip is a band of suberin (a waterproof substance) located in the cell walls of endodermal cells. It blocks the apoplast pathway (movement through cell walls). As a result, water and dissolved mineral ions must pass through the selectively permeable cell surface membrane into the cytoplasm (the symplast pathway), allowing the plant to control which ions enter the xylem.

1 mark for selecting B, as the Casparian strip blocks the apoplast pathway, forcing entry into the symplast pathway.

Cellulose is made of \(\beta\)-glucose monomers linked by \(\beta\)-1,4-glycosidic bonds. To form these bonds, each successive glucose molecule must be rotated by \(180^\circ\) relative to its neighbour (Statement 1). This results in a straight, unbranched chain (Statement 3), unlike the coiled structure of amylose (making Statement 4 incorrect). Multiple straight chains lie parallel to each other and form hydrogen bonds between their hydroxyl groups, aggregating to form strong microfibrils (Statement 2). Therefore, 1, 2, and 3 are correct.

1 mark for identifying that statements 1, 2, and 3 are correct while statement 4 is incorrect (Option A).

During the primary response, naive B-lymphocytes undergo clonal selection and clonal expansion to produce plasma cells and memory cells. This takes time, making the response slower. Upon second exposure, the memory cells already present can immediately undergo clonal expansion and differentiate into plasma cells, yielding a faster, higher, and more sustained production of antibodies. Helper T-cells do not differentiate into plasma cells (ruling out B), and both responses represent natural active immunity (ruling out D).

1 mark for selecting C, acknowledging that clonal selection in the primary response produces memory cells which enable a faster secondary response.

1. Since the DNA is double-stranded, the percentage of cytosine (C) equals the percentage of guanine (G). Thus, C = 22% and G = 22%, which together make up 44% of the bases. 2. The remaining 56% of the bases must be adenine (A) and thymine (T). Since A = T, the percentage of adenine is 56% / 2 = 28%. 3. In a segment of 100 base pairs, there are 200 total bases. 4. Since C is 22% of 200 bases, we have 44 C bases and 44 G bases, which form 44 C-G base pairs. Similarly, A is 28% of 200 bases, giving 56 A bases and 56 T bases, forming 56 A-T base pairs. 5. Each C-G pair has 3 hydrogen bonds: 44 * 3 = 132 bonds. Each A-T pair has 2 hydrogen bonds: 56 * 2 = 112 bonds. Total hydrogen bonds = 132 + 112 = 244.

1 mark for calculating adenine as 28% and the total hydrogen bonds in 100 base pairs as 244 (Option B).

Inhibitor X increases \(K_m\) but leaves \(V_{\max}\) unchanged, which is characteristic of a competitive inhibitor. Since it competes with the substrate for the active site, its effect can be overcome by increasing substrate concentration. Inhibitor Y decreases \(V_{\max}\) but leaves \(K_m\) unchanged, which is characteristic of a non-competitive inhibitor. It binds to an allosteric site and reduces the overall rate of reaction regardless of how much substrate is added.

1 mark for identifying X as a competitive inhibitor that can be overcome, and Y as a non-competitive inhibitor that cannot be overcome (Option A).

A bronchus contains cartilage plates to keep the airway open, smooth muscle to control airway diameter, goblet cells to secrete mucus, and ciliated epithelial cells to sweep mucus. A terminal bronchiole, being much narrower, lacks cartilage and goblet cells (to prevent airway blockage by mucus), but retains smooth muscle and ciliated epithelial cells to regulate airflow and remove debris.

1 mark for identifying the correct tissue composition of bronchus and terminal bronchiole (Option A).

In G1, there are 16 chromosomes (each composed of 1 chromatid). After DNA replication in the S phase, each chromosome consists of 2 sister chromatids joined at a single centromere. Thus, at metaphase, there are 16 chromosomes, 32 sister chromatids, and 16 centromeres. In anaphase, sister chromatids separate and are pulled to opposite poles. After telophase, each of the two new nuclei contains 16 single-stranded chromosomes, meaning 0 sister chromatids (they are individual chromosomes now) and 16 centromeres.

1 mark for selecting Option A, which correctly details chromosome, sister chromatid, and centromere counts at metaphase and after telophase.

Curve Y represents an increased affinity of hemoglobin for oxygen (shifted to the left). This occurs under conditions of lower temperature, higher pH (lower concentration of hydrogen ions), lower partial pressure of carbon dioxide, or in fetal hemoglobin. Curve Z represents a decreased affinity of hemoglobin for oxygen (shifted to the right / Bohr shift). This occurs under conditions of higher temperature, lower pH (higher concentration of hydrogen ions), or higher partial pressure of carbon dioxide. Therefore, option A is correct because a decreased temperature shifts the curve to the left (Curve Y) and an increased concentration of hydrogen ions (lower pH) shifts the curve to the right (Curve Z).

Award 1 mark for the correct answer A.
- Reject B because increased partial pressure of carbon dioxide shifts the curve to the right, not the left.
- Reject C because decreased pH shifts the curve to the right, not the left.
- Reject D because increased concentration of hydrogen ions shifts the curve to the right, not the left.

During phloem loading, proton pumps active transport protons (\(H^+\)) out of the companion cell into the cell wall apoplast using ATP. This establishes a high electrochemical gradient of protons outside the cell. Protons then enter the companion cell down this gradient by facilitated diffusion via a co-transporter protein (symport). The movement of protons drives the co-transport of sucrose into the companion cell against its concentration gradient.

Award 1 mark for the correct answer A.
- Reject B, C, and D as they incorrect identify the transport mechanisms (e.g., claiming protons are pumped by facilitated diffusion or that sucrose moves down its concentration gradient during loading).

Statement I is correct because amylose, amylopectin, and glycogen are all polymers of \(\alpha\)-glucose linked by \(\alpha\)-1,4-glycosidic bonds. Statement II is correct because only the branched polymers amylopectin and glycogen contain \(\alpha\)-1,6-glycosidic bonds at their branch points (cellulose has \(\beta\)-1,4-glycosidic bonds, and amylose is unbranched). Statement III is correct because cellulose molecules are held together in parallel bundles by hydrogen bonds to form microfibrils. Statement IV is incorrect because glycogen (4) is more highly branched than amylopectin (2).

Award 1 mark for the correct answer A.
- Reject B because statement II is correct.
- Reject C and D because statement IV is incorrect.

The injection of antigen X on day 0 triggers a primary immune response, leaving behind memory B-lymphocytes. When antigen X is re-encountered on day 20 alongside a new antigen Y, a secondary immune response is triggered for X, where memory cells rapidly clone and differentiate into antibody-secreting plasma cells. For antigen Y, this is a primary response, requiring clonal selection of naive B-lymphocytes, which takes longer and results in fewer antibodies.

Award 1 mark for the correct answer A.
- Reject B because T-lymphocytes do not secrete antibodies.
- Reject C because the primary response to Y is slower, not faster.
- Reject D because phagocytosis is a non-specific process and does not determine the rate of specific clonal selection in secondary vs primary responses.

Initially (Gen 0), all DNA molecules contain heavy nitrogen (\(^{15}\text{N}\)-\(^{15}\text{N}\)). After one division in \(^{14}\text{N}\) (Gen 1), all DNA molecules are hybrid (\(^{15}\text{N}\)-\(^{14}\text{N}\)). After two divisions in \(^{14}\text{N}\) (Gen 2), these hybrid strands serve as templates to synthesize new strands using light nitrogen. This results in 50% hybrid DNA molecules (intermediate density) and 50% entirely light DNA molecules (\(^{14}\text{N}\)-\(^{14}\text{N}\)). There are no heavy DNA molecules left.

Award 1 mark for the correct answer A.
- Reject B, C, and D as they incorrect state the molecular ratios resulting from semi-conservative replication after two generations.

Inhibitor P increases the Michaelis-Menten constant (\(K_{\text{m}}\)) but leaves the maximum rate (\(V_{\text{max}}\)) unchanged. This is characteristic of a competitive inhibitor, which competes with the substrate for the active site. Increasing the substrate concentration can overcome this competition. Inhibitor Q decreases \(V_{\text{max}}\) but does not alter \(K_{\text{m}}\), which indicates non-competitive inhibition where the inhibitor binds to an allosteric site.

Award 1 mark for the correct answer A.
- Reject B because inhibitor Q is a non-competitive inhibitor and binds to an allosteric site, not the active site.
- Reject C because inhibitor P is a competitive inhibitor, not non-competitive.
- Reject D because inhibitor Q is a non-competitive inhibitor.

The bronchus is supported by plates of cartilage to maintain patency (prevent collapse) and contains smooth muscle to regulate airflow. The alveolus lacks cartilage and smooth muscle to allow for thin walls, and contains no ciliated epithelium (which would increase the diffusion distance). However, alveoli are surrounded by elastic fibers which allow the alveoli to stretch during inhalation and recoil to expel air during exhalation.

Award 1 mark for the correct answer A.
- Reject B because bronchi contain smooth muscle, and alveoli lack ciliated epithelium.
- Reject C because bronchi contain cartilage, and alveoli contain elastic fibers.
- Reject D because alveoli lack ciliated epithelium, but contain elastic fibers.

The synthesis (translation) of proteins destined for secretion occurs on ribosomes attached to the rough endoplasmic reticulum (RER). The protein is transported within the lumen of the RER and then packaged into transport vesicles that bud off and fuse with the Golgi body. In the Golgi body, proteins are modified (e.g., glycosylated) and packaged into secretory vesicles, which then travel to and fuse with the cell surface membrane to release the protein by exocytosis.

Award 1 mark for the correct answer A.
- Reject B because free ribosomes synthesize intracellular proteins, and lysosomes store rather than secrete digestive enzymes.
- Reject C because the nucleolus is involved in ribosome synthesis, not secretion.
- Reject D because the smooth endoplasmic reticulum is responsible for lipid, not protein synthesis.

Inside respiring tissues, carbon dioxide is produced and diffuses into red blood cells. Statement 1 is correct: carbonic anhydrase rapidly catalyses the reaction between \( \text{CO}_2 \) and water to form carbonic acid (\( \text{H}_2\text{CO}_3 \)). This acid dissociates into hydrogen ions (\( \text{H}^+ \)) and hydrogencarbonate ions (\( \text{HCO}_3^- \)). Statement 2 is correct: \( \text{HCO}_3^- \) ions diffuse out of the cell via facilitated diffusion through specific anion-exchange transport proteins. Statement 3 is correct: the accumulation of \( \text{H}^+ \) causes them to bind to haemoglobin, forming haemoglobinic acid (\( \text{HHb} \)), which decreases the affinity of haemoglobin for oxygen (the Bohr effect) and promotes oxygen release. Statement 4 is incorrect: to maintain electrical neutrality, chloride ions (\( \text{Cl}^- \)) diffuse into the cell as \( \text{HCO}_3^- \) leaves (the chloride shift).

Award 1 mark for identifying that statements 1, 2, and 3 are correct while statement 4 is incorrect because chloride ions enter rather than leave the cell.

During sucrose loading, protons (\( \text{H}^+ \)) are actively pumped out of the companion cell into the cell wall space using ATP-powered proton pumps. This generates an electrochemical gradient of protons. Sucrose is then loaded into the companion cell against its concentration gradient via co-transporter proteins, which couple the movement of sucrose to the down-gradient movement of \( \text{H}^+ \) ions. The immediate/direct energy source for sucrose entry is the proton gradient (secondary active transport), not direct ATP hydrolysis.

Award 1 mark for the correct combination of active transport using ATP for \( \text{H}^+ \) pumping, co-transport for sucrose entry, and the proton gradient as the direct energy source.

Amylopectin is a branched component of starch composed of \(\alpha\)-glucose monomers linked by \(\alpha\)-1,4 glycosidic bonds with \(\alpha\)-1,6 glycosidic bonds at the branch points. Glycogen is also made of \(\alpha\)-glucose with both \(\alpha\)-1,4 and \(\alpha\)-1,6 bonds, but has a much higher frequency of branching than amylopectin. Cellulose is composed of \(\beta\)-glucose monomers linked by \(\beta\)-1,4 glycosidic bonds, forming straight, unbranched chains that form hydrogen bonds with adjacent chains to make microfibrils.

Award 1 mark for identifying the correct monomer, bond types, and molecular shapes for all three polysaccharides.

B-lymphocytes are responsible for humoral immunity. When activated, they divide and differentiate into antibody-secreting plasma cells. Helper T-lymphocytes release cytokines (such as interleukins) that stimulate the clonal expansion and activation of B-lymphocytes. Cytotoxic T-lymphocytes do not produce antibodies, and B-lymphocytes are not phagocytic. Memory B-lymphocytes do not secrete antibodies during the primary response; they persist and rapidly divide to form plasma cells upon a second exposure. Plasma cells are terminally differentiated effector cells and do not divide.

Award 1 mark for correctly identifying the roles of B-lymphocytes, plasma cells, and helper T-lymphocytes in clonal selection/expansion.

Let's track the DNA strands using semi-conservative replication:
- Generation 0 (original): 1 double-stranded DNA molecule where both strands are heavy (\(^{15}\text{N}\)/\(^{15}\text{N}\)).
- Generation 1 (after 1 replication): The two heavy strands separate and act as templates. New complementary strands are synthesized using \(^{14}\text{N}\). This results in 2 hybrid double-stranded DNA molecules (\(^{15}\text{N}\)/\(^{14}\text{N}\)).
- Generation 2 (after 2 replications): The 4 strands from Generation 1 (2 heavy and 2 light) separate and act as templates. New complementary strands are synthesized using \(^{14}\text{N}\). This results in 4 double-stranded DNA molecules: 2 molecules are hybrid (\(^{15}\text{N}\)/\(^{14}\text{N}\)) and 2 molecules are entirely light (\(^{14}\text{N}\)/\(^{14}\text{N}\)). Therefore, 2 out of 4 (which is 50%) of the DNA molecules in the second generation will consist of one heavy strand and one light strand.

Award 1 mark for calculating the correct percentage of hybrid double-stranded DNA molecules after two rounds of semi-conservative replication.

Statement 1 is incorrect because it describes a non-competitive inhibitor binding to an allosteric site. Statement 2 is correct because increasing the substrate concentration increases the probability of substrate molecules binding to the active site rather than the competitive inhibitor. Statement 3 is correct because a competitive inhibitor increases the apparent Michaelis-Menten constant (\(K_{\text{m}}\)) as a higher concentration of substrate is required to reach half of the maximum velocity (\(\frac{1}{2}V_{\max}\)). Statement 4 is incorrect because a competitive inhibitor does not alter the maximum velocity (\(V_{\max}\)) of the reaction; given high enough substrate concentrations, the same \(V_{\max}\) can still be reached.

Award 1 mark for correctly identifying that statements 2 and 3 are correct, and statements 1 and 4 are incorrect.

The trachea contains all four tissues: cartilage (C-shaped rings to prevent collapse), goblet cells (to secrete mucus), smooth muscle (to adjust airway diameter), and elastic fibres (for recoil). Bronchioles do not have cartilage or goblet cells, but they do have smooth muscle and elastic fibres. Alveoli lack cartilage, goblet cells, and smooth muscle, but contain a high density of elastic fibres to allow stretch and recoil during breathing. Therefore, row A is correct.

Award 1 mark for identifying the correct presence/absence pattern of all four tissues across the trachea, bronchioles, and alveoli.

Use the magnification formula: \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\).
First, convert both values to the same unit (micrometres, \(\mu\text{m}\)):
\(\text{Image size} = 80\text{ mm} = 80 \times 1,000\text{ }\mu\text{m} = 80,000\text{ }\mu\text{m}\).
\(\text{Actual size} = 5\text{ }\mu\text{m}\).
Now calculate the magnification:
\(\text{Magnification} = \frac{80,000\text{ }\mu\text{m}}{5\text{ }\mu\text{m}} = 16,000\).
Thus, the magnification is \(\times 16,000\).

Award 1 mark for converting the image size to micrometres and correctly calculating the magnification as \(\times 16,000\).

(a)(i) Enzyme X is carbonic anhydrase, which catalyzes the rapid, reversible conversion of carbon dioxide and water into carbonic acid.

(a)(ii) The movement of hydrogencarbonate ions out of the cell and chloride ions into the cell is known as the chloride shift.

(b) The formation of haemoglobinic acid (\(\text{HHb}\)) has two major roles:
1. Buffering: Free \(\text{H}^+\) ions would lower the pH of the red blood cell, which could denature hemoglobin and other enzymes. Binding them maintains pH stability.
2. Bohr effect: Binding of \(\text{H}^+\) alters the tertiary structure of hemoglobin, reducing its affinity for oxygen. This shifts the oxygen dissociation curve to the right, facilitating the release of oxygen to actively respiring tissues.

(c) Hydrogencarbonate ions (\(\text{HCO}_3^-\)) are negatively charged and diffuse out of the red blood cell into the plasma down their concentration gradient. This loss of negative charge creates an electrical imbalance (the cell interior becomes relatively positive). The facilitated diffusion of negatively charged chloride ions (\(\text{Cl}^-\)) into the cell restores and maintains electrical neutrality.

(a)(i) carbonic anhydrase [1]
(a)(ii) chloride shift [1]

(b) Max 3 marks:
- \(\text{H}^+\) binding acts as a buffer / prevents increase in free hydrogen ion concentration [1]
- prevents drop in pH / maintains stable pH [1]
- prevents denaturation of hemoglobin / other enzymes inside the cell [1]
- alters tertiary structure of hemoglobin to reduce its affinity for oxygen [1]
- promotes dissociation / release of oxygen to respiring tissues / causes the Bohr effect [1]

(c) Max 3 marks:
- hydrogencarbonate ions / \(\text{HCO}_3^-\) diffuse out of the red blood cell [1]
- this causes a loss of negative charge / creates an electrical imbalance [1]
- chloride ions / \(\text{Cl}^-\) move into the cell [1]
- to maintain / restore electrical neutrality [1]

### Part (a)
* Hemoglobin consists of four polypeptide chains (two alpha-globin and two beta-globin chains).
* Each polypeptide chain contains a prosthetic group known as a heme (haem) group.
* Each heme group contains an iron(II) ion (\(Fe^{2+}\)) which binds reversibly to one molecule of oxygen (\(O_2\)).
* The four subunits are held together by non-covalent interactions (hydrophobic interactions, hydrogen bonds, and ionic bonds) to form a compact, spherical (globular) quaternary structure.

### Part (b)(i)
* **Description:** Decreasing pH (increasing carbon dioxide/hydrogen ion concentration) shifts the oxygen dissociation curve to the right, which decreases the affinity of hemoglobin for oxygen.
* **Explanation:** Carbon dioxide reacts with water to form carbonic acid, which dissociates into hydrogen ions (\(H^+\)) and hydrogencarbonate ions (\(HCO_3^-\)).
* The free hydrogen ions (\(H^+\)) bind to specific amino acid residues on the hemoglobin polypeptide chains.
* This binding alters the tertiary and quaternary structure (conformational change) of hemoglobin, stabilizing the deoxygenated form (T-state) and promoting the release (unloading) of oxygen to the tissues that require it most.

### Part (b)(ii)
1. Calculate the percentage of oxygen unloaded at pH 7.4:
\(\text{Oxygen unloaded} = 98\% - 70\% = 28\%\)
2. Calculate the percentage of oxygen unloaded at pH 7.2:
\(\text{Oxygen unloaded} = 95\% - 50\% = 45\%\)
3. Calculate the absolute difference between these two values:
\(45\% - 28\% = 17\%\) (or 17 percentage points).

### Part (c)(i)
* Carbon dioxide (\(CO_2\)) combines with water (\(H_2O\)) to form carbonic acid (\(H_2CO_3\)).
* This reaction is catalyzed by the enzyme **carbonic anhydrase** inside the red blood cell.
* Carbonic acid rapidly dissociates into hydrogen ions (\(H^+\)) and hydrogencarbonate ions (\(HCO_3^-\)).
* Hydrogencarbonate ions (\(HCO_3^-\)) diffuse out of the red blood cell into the blood plasma down their concentration gradient via a transport protein (anion exchange protein) in exchange for chloride ions (\(Cl^-\)) moving into the cell (known as the chloride shift to maintain electrical neutrality).

### Part (c)(ii)
* The hydrogen ions (\(H^+\)) produced by the dissociation of carbonic acid bind to oxyhemoglobin (\(HbO_8\)).
* This forms haemoglobinic acid (\(HHb\)) and releases oxygen, acting as a buffer to remove free hydrogen ions from solution, thereby preventing a drastic drop in the pH of the red blood cell cytoplasm.

### Part (a) [Max 3 marks]
* **MP1:** Four polypeptide chains / subunits (two \(\alpha\)-globin and two \(\beta\)-globin chains). [1]
* **MP2:** Each polypeptide chain contains one prosthetic heme/haem group. [1]
* **MP3:** Each heme group contains an iron ion / \(Fe^{2+}\) that binds reversibly to one oxygen (\(O_2\)) molecule. [1]
* **MP4:** Subunits held together by hydrophobic interactions / ionic bonds / hydrogen bonds (to maintain specific globular shape). [1]

### Part (b)(i) [Max 4 marks]
* **MP1:** Decreased pH / increased \(CO_2\) shifts the oxygen dissociation curve to the right. [1]
* **MP2:** This decreases the affinity of hemoglobin for oxygen (at any given partial pressure). [1]
* **MP3:** Hydrogen ions (\(H^+\)) bind to hemoglobin / polypeptide chains. [1]
* **MP4:** This causes a conformational / structural change in the hemoglobin molecule. [1]
* **MP5:** Resulting in oxygen being released / unloaded more easily to actively respiring tissues. [1]

### Part (b)(ii) [Max 4 marks]
* **MP1:** Correct calculation of oxygen unloaded at pH 7.4: \(98\% - 70\% = 28\%\). [1]
* **MP2:** Correct calculation of oxygen unloaded at pH 7.2: \(95\% - 50\% = 45\%\). [1]
* **MP3:** Correct calculation of the absolute difference: \(45\% - 28\% = 17\%\) (or 17 percentage points). [1]
* **MP4:** Correct units (\% or percentage points) included in final answer. [1]
* *Accept:* Allow full marks for correct final answer of 17% with clear working.

### Part (c)(i) [Max 4 marks]
* **MP1:** \(CO_2\) reacts with \(H_2O\) to form carbonic acid (\(H_2CO_3\)) / catalyzed by carbonic anhydrase. [1]
* **MP2:** Carbonic acid dissociates into hydrogen ions (\(H^+\)) and hydrogencarbonate ions (\(HCO_3^-\)). [1]
* **MP3:** Hydrogencarbonate ions diffuse out of the red blood cell into plasma (down a concentration gradient). [1]
* **MP4:** In exchange for chloride ions (\(Cl^-\)) entering the cell / reference to chloride shift to maintain electrical neutrality. [1]

### Part (c)(ii) [Max 2 marks]
* **MP1:** Hydrogen ions (\(H^+\)) combine with / bind to hemoglobin / oxyhemoglobin. [1]
* **MP2:** To form haemoglobinic acid (\(HHb\)) which removes free \(H^+\), maintaining/buffering intracellular pH. [1]

1. Carbon dioxide (\(\text{CO}_2\)) reacts with water (\(\text{H}_2\text{O}\)) inside the red blood cell to form carbonic acid (\(\text{H}_2\text{CO}_3\)). 2. This reaction is catalysed by the active site of the enzyme carbonic anhydrase. 3. Carbonic acid dissociates rapidly into hydrogen ions (\(\text{H}^+\)) and hydrogencarbonate ions (\(\text{HCO}_3^-\)). 4. Hydrogencarbonate ions diffuse out of the red blood cell into the plasma down a concentration gradient via a transport protein (anion exchanger). 5. To maintain electrical neutrality (the chloride shift), chloride ions (\(\text{Cl}^-\)) diffuse into the red blood cell from the plasma through the same transport protein.

Award 1 mark for each correct point up to a maximum of 5 marks: 1. Carbon dioxide reacts or combines with water to form carbonic acid (\(\text{H}_2\text{CO}_3\)); 2. Catalysed by the enzyme carbonic anhydrase; 3. Carbonic acid dissociates into hydrogen ions (\(\text{H}^+\)) and hydrogencarbonate ions (\(\text{HCO}_3^-\)); 4. Hydrogencarbonate ions diffuse out of the red blood cell into blood plasma; 5. Chloride ions (\(\text{Cl}^-\)) diffuse or move into the red blood cell from plasma; 6. Diffusion occurs via facilitated diffusion or transport protein or anion exchanger or carrier protein; 7. This chloride shift maintains electrical neutrality or balance of charges.

(a) Companion cell plasma membranes contain active proton pumps that transport hydrogen ions (\(H^+\)) out of the cell's cytoplasm and into the cell wall (apoplast). This active transport process requires ATP. This movement creates a high concentration of \(H^+\) in the apoplast, establishing an electrochemical gradient. \(H^+\) ions then diffuse back down their gradient into the companion cell via specialized co-transporter proteins. This movement of \(H^+\) is coupled with the transport of sucrose molecules against their concentration gradient into the companion cell.

(b) 1. Sieve tube elements have no nucleus, very little cytoplasm, and very few organelles, whereas companion cells have a prominent nucleus, dense cytoplasm, and numerous organelles.
2. Sieve tube elements have sieve plates containing sieve pores at their end walls, whereas companion cells have standard cellulose cell walls with no sieve plates.
3. Companion cells contain many mitochondria to produce the ATP required for active transport, whereas sieve tube elements contain very few or no mitochondria.

(c) As sucrose is actively loaded into the companion cells and diffuses into the sieve tube elements at the source, it decreases the solute potential, which lowers the overall water potential (\(\Psi\)) of the sieve tube sap. Water moves from xylem vessels or surrounding tissues into the sieve tube elements by osmosis, down a water potential gradient. The entry of water increases the volume of liquid, which generates a high hydrostatic pressure inside the sieve tube at the source. At the sink, sucrose is unloaded, increasing the water potential and causing water to leave, which lowers the hydrostatic pressure. The sap moves by mass flow down this hydrostatic pressure gradient from the source to the sink.

(d) At the sink, sucrose can be hydrolyzed to glucose and fructose for use in aerobic respiration, or polymerized into starch for long-term storage, or used to synthesize cellulose for cell walls. Unloading sucrose is essential because it maintains the concentration gradient (and thus the hydrostatic pressure gradient) between the source and the sink, ensuring that mass flow continues.

(a) Max 4 marks:
1. Active transport of hydrogen ions / protons / \(H^+\) out of the companion cell cytoplasm into the cell wall / apoplast;
2. Ref. to use of ATP / energy from ATP hydrolysis;
3. Creation of a proton / \(H^+\) concentration gradient / electrochemical gradient (higher concentration in wall than cytoplasm);
4. \(H^+\) ions diffuse back into the companion cell through co-transporter proteins;
5. Sucrose is transported into the companion cell simultaneously / cotransported, against its concentration gradient;

(b) Max 3 marks (must be direct comparisons for each mark):
1. Sieve tube element has no nucleus / very little cytoplasm / few organelles AND companion cell has a nucleus / dense cytoplasm / many organelles;
2. Sieve tube element has sieve plates / pores AND companion cell has no sieve plates / pores;
3. Companion cell has many mitochondria AND sieve tube element has very few / no mitochondria;
4. Sieve tube elements form continuous tubes / joined end-to-end AND companion cells do not form tubes / are located alongside;

(c) Max 4 marks:
1. High concentration of sucrose lowers the water potential (\(\Psi\)) inside the sieve tube element at the source;
2. Water enters the sieve tube element from the xylem / surrounding tissues by osmosis;
3. This entry of water increases the hydrostatic pressure / turgor pressure at the source;
4. At the sink, unloading of sucrose increases water potential, causing water to leave (which lowers hydrostatic pressure);
5. Liquid moves down a hydrostatic pressure gradient / from high hydrostatic pressure to low hydrostatic pressure by mass flow;

(d) Max 2 marks:
1. Used in respiration / converted to glucose and fructose OR stored as starch OR used to make cellulose/cell walls;
2. Maintains the hydrostatic pressure gradient / concentration gradient between source and sink (otherwise pressure would equalize and flow would stop);

(a) In actively respiring tissues, carbon dioxide diffuses into red blood cells. Here, carbonic anhydrase catalyzes the reaction between carbon dioxide and water to form carbonic acid: \(\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3\). Carbonic acid is highly unstable and rapidly dissociates into hydrogen ions (\(\text{H}^+\)) and hydrogencarbonate ions (\(\text{HCO}_3^-\)). The hydrogencarbonate ions then diffuse out of the red blood cells into the blood plasma down their concentration gradient.

(b) As negatively charged hydrogencarbonate ions leave the red blood cell, chloride ions (\(\text{Cl}^-\)) diffuse into the cell from the plasma. This movement of chloride ions is called the chloride shift and it maintains electrochemical neutrality inside the cell. Simultaneously, the accumulated hydrogen ions (\(\text{H}^+\)) are buffered by binding to oxyhemoglobin, triggering the release of oxygen and forming hemoglobinic acid (\(\text{HHb}\)), which prevents a dangerous drop in intracellular pH.

(c) High partial pressures of carbon dioxide reduce the affinity of hemoglobin for oxygen. This results in the oxygen dissociation curve shifting downwards and to the right (the Bohr effect). Mechanistically, the increased concentration of hydrogen ions from carbon dioxide dissociation binds to the polypeptide chains of hemoglobin, altering its tertiary conformational structure. This structural change facilitates the release (unloading) of oxygen to the tissues that are actively respiring.

Part (a) [Max 3 marks]
1. Carbon dioxide reacts with water to form carbonic acid (\(\text{H}_2\text{CO}_3\)), catalyzed by the enzyme carbonic anhydrase.
2. Carbonic acid rapidly dissociates into hydrogen ions (\(\text{H}^+\)) and hydrogencarbonate ions (\(\text{HCO}_3^-\)).
3. Hydrogencarbonate ions diffuse out of the red blood cell into the plasma.

Part (b) [Max 4 marks]
1. Chloride ions (\(\text{Cl}^-\)) diffuse into the red blood cells from the blood plasma.
2. (This chloride shift occurs) to balance the outflow of negative hydrogencarbonate ions / maintain electrochemical neutrality.
3. Hydrogen ions (\(\text{H}^+\)) bind to hemoglobin / oxyhemoglobin.
4. (This forms) haemoglobinic acid (\(\text{HHb}\)) / acts as a buffer to prevent a decrease in cellular pH / maintain stable pH inside the erythrocyte.

Part (c) [Max 3 marks]
1. High carbon dioxide concentration reduces the affinity of hemoglobin for oxygen.
2. (This causes the) oxygen dissociation curve to shift to the right.
3. (Explanation) Hydrogen ions bind to hemoglobin, causing a change in its tertiary structure / conformation, which forces the release of oxygen to the respiring tissues.

Part (a):
When pCO2 is high in active tissues, CO2 diffuses into red blood cells. Here, carbonic anhydrase catalyzes its reaction with H2O to form H2CO3. This acid dissociates into H+ and HCO3-. The increased concentration of H+ ions lowers intracellular pH. H+ ions then bind to oxyhemoglobin, competing with oxygen and acting as a buffer to form haemoglobinic acid (HHb). This hydrogen binding alters the conformation (tertiary structure) of the globin chains, reducing hemoglobin's affinity for oxygen. Consequently, oxygen is released more readily to the tissues where it is needed for respiration.

Part (b):
Once HCO3- is formed from the dissociation of H2CO3 inside the red blood cell, it is transported out of the cell into the blood plasma down its concentration gradient. This occurs via facilitated diffusion using an anion exchange protein. To maintain electrical neutrality inside the red blood cell, chloride ions (Cl-) enter the cell from the plasma, a process known as the chloride shift. The HCO3- ions then remain dissolved in the plasma and are transported by the transport system (veins, heart, pulmonary arteries) to the capillaries of the lungs.

Part (a) [Max 4]:
1. CO2 diffuses into red blood cell and reacts with water to form carbonic acid (H2CO3) / catalyzed by carbonic anhydrase;
2. H2CO3 dissociates into hydrogen ions (H+) and hydrogencarbonate ions (HCO3-);
3. H+ ions bind to hemoglobin / form haemoglobinic acid / HHb;
4. (binding of H+ ions) changes tertiary structure / conformation of hemoglobin;
5. reducing hemoglobin's affinity for oxygen (causing more oxygen to dissociate);
6. Reference to Bohr shift / curve shifts to the right;

Part (b) [Max 3]:
1. HCO3- ions diffuse out of the red blood cell;
2. down a concentration gradient / via transport (carrier) protein / facilitated diffusion;
3. chloride ions (Cl-) enter the red blood cell to maintain electrical neutrality / chloride shift;
4. HCO3- is transported dissolved in (blood) plasma to the lungs;