2024 AQA AS-Level Biology 7401 Practice Paper with Answers

Pre-mRNA transcription involves helicase unwinding DNA, RNA polymerase joining RNA nucleotides via phosphodiester bonds based on complementary base pairing. Splicing removes non-coding introns. Alternative splicing joins exons in different combinations, yielding different mRNA sequences and therefore different protein products.

1. DNA helicase breaks hydrogen bonds to expose template strands. 2. RNA polymerase joins complementary free RNA nucleotides. 3. Phosphodiester bonds are formed via condensation reactions. 4. Pre-mRNA contains non-coding introns and coding exons. 5. Splicing removes introns and joins exons. 6. Alternative splicing joins exons in different combinations. 7. This creates different mature mRNA transcripts from the same gene. 8. These code for different primary structures/polypeptides.

tRNA has a cloverleaf shape stabilized by hydrogen bonding. It has an amino acid binding site at one end and a specific anticodon at the other, which complementary base-pairs with the mRNA codon during translation inside the ribosome.

1. tRNA transports a specific amino acid to the ribosome. 2. It has an anticodon that is complementary to an mRNA codon. 3. This ensures amino acids are placed in the correct primary sequence. 4. It is single-stranded RNA folded into a cloverleaf shape. 5. This shape is maintained by hydrogen bonds between complementary base pairs. 6. It has an amino acid binding site at the 3' end. 7. ATP is required to attach the amino acid to the tRNA. 8. Ribosomal binding sites accommodate tRNA to form peptide bonds.

N = 50. N(N-1) = 2450. Sum of n(n-1) = 15*14 + 8*7 + 12*11 + 5*4 + 10*9 = 210 + 56 + 132 + 20 + 90 = 508. d = 2450 / 508 = 4.82. Monocultures decrease plant diversity, reducing food resources and habitats.

1. Correct calculation of N = 50 and N(N-1) = 2450. 2. Correct calculation of sum n(n-1) = 508. 3. Correct final answer of 4.82 (accept 4.8). 4. Monoculture means single plant species dominates, reducing plant diversity. 5. Fewer niches/habitats for insect species. 6. Fewer varied food sources/types of food. 7. Use of pesticides kills insects directly. 8. Use of herbicides kills weed species, further reducing insect habitats/food.

Hedgerows provide microhabitats, nesting sites, and diverse food sources. Removing them reduces species richness. Sustainable practices such as leaving field margins uncultivated, crop rotation, and reducing pesticide use preserve and enhance local biodiversity.

1. Removal of hedgerows reduces plant species diversity. 2. Loss of habitats, nesting sites, and shelter for insects/birds/small mammals. 3. Loss of diverse food sources (berries, seeds, nectar). 4. Conservation measure 1: Replant/maintain hedgerows (provides corridors/habitats). 5. Conservation measure 2: Leave uncultivated field margins/wildflower strips (increases plant/insect diversity). 6. Conservation measure 3: Crop rotation / avoid monoculture (supports varied soil organisms/pests). 7. Conservation measure 4: Reduce pesticide/herbicide use or use biological control. 8. Conservation measure 5: Create ponds/wetlands (provides additional niches).

In indirect ELISA, antigen is bound to the well. Patient sample containing antibody is added and binds to antigen. Unbound antibody is washed away. Enzyme-conjugated secondary antibody is added, binds primary antibody. Unbound secondary antibody is washed away. Substrate is added, causing a color change.

1. HIV antigen is bound to the bottom of the well. 2. Patient's sample (serum) containing antibodies is added. 3. Wash step to remove unbound primary antibodies (essential to prevent false positive). 4. Add secondary antibody with attached enzyme, which binds specifically to the primary antibody. 5. Wash step to remove unbound secondary antibody-enzyme. 6. Add substrate which reacts with the enzyme to produce a color change. 7. Intensity of color is proportional to amount of antibody in serum. 8. Secondary antibody is essential because the primary antibody itself does not have an enzyme attached to trigger a color change.

The specific sequence of amino acids (primary structure) determines which R-groups interact. Hydrogen bonds form alpha helices/beta sheets (secondary), and further folding via ionic, disulfide, and hydrophobic interactions forms the active site (tertiary). Excess heat breaks these tertiary bonds, changing active site shape so substrates cannot bind.

1. Primary structure is the specific sequence of amino acids. 2. Secondary structure involves hydrogen bonding between peptide groups forming alpha helices/beta pleated sheets. 3. Tertiary structure is the 3D folding held by bonds between R-groups (hydrogen, ionic, disulfide bridges). 4. Tertiary structure determines the specific shape of the active site complementary to substrate. 5. Low temperatures result in low kinetic energy, fewer collisions, and fewer enzyme-substrate complexes. 6. Higher temperatures (up to optimum) increase kinetic energy, increasing successful collisions. 7. High temperatures above optimum break hydrogen/ionic bonds in the tertiary structure. 8. Active site changes shape (denatures), substrate no longer fits, no enzyme-substrate complexes form.

Sucrose is actively loaded into sieve tubes at the source, lowering water potential. Water enters from xylem by osmosis, creating high hydrostatic pressure. Solutes flow by mass flow to the sink, where sucrose is unloaded and water leaves, lowering pressure.

1. Sucrose is actively loaded into the companion cells and then sieve tubes from source cells. 2. This lowers the water potential in the sieve tube. 3. Water enters the sieve tube from xylem by osmosis. 4. This increases the hydrostatic pressure inside the sieve tube at the source end. 5. At the sink end, sucrose is unloaded (used or stored). 6. This increases the water potential at the sink end. 7. Water leaves the sieve tube by osmosis (reducing hydrostatic pressure). 8. Solutes flow down the hydrostatic pressure gradient from source to sink by mass flow.

Homogenize liver tissue to break cells, filter out debris. Spin filtrate at low speed to pellet nuclei. Spin supernatant at higher speed to pellet mitochondria. Cold prevents enzyme digestion; isotonic prevents osmotic lysis; buffer maintains pH.

1. Homogenize tissue in a homogenizer to break open cell membranes and release organelles. 2. Filter homogenate to remove intact cells and large debris. 3. Centrifuge filtrate at a low speed to pellet heavy nuclei. 4. Remove supernatant and centrifuge at a higher speed to pellet mitochondria. 5. Keep cold to reduce/prevent lysosomal/hydrolytic enzyme activity that could digest organelles. 6. Keep isotonic to ensure no net movement of water by osmosis, preventing organelle swelling/bursting or shrinking. 7. Keep buffered to maintain constant pH to prevent denaturation of proteins/enzymes within organelles. 8. Mitochondria will be found in the second pellet.

### Part (a)
* The first oxygen molecule binds to haemoglobin with difficulty because the polypeptide chains are closely packed.
* The binding of the first oxygen molecule alters the quaternary structure (conformational change) of the haemoglobin molecule.
* This exposes the other haem binding sites, making it easier for the second and third oxygen molecules to bind (cooperative binding), causing the steep rise in the curve.
* It is harder for the fourth oxygen molecule to bind because most binding sites are already occupied.

### Part (b)
* High carbon dioxide concentration lowers pH due to the formation of carbonic acid / release of hydrogen ions.
* This changes the quaternary shape of haemoglobin, causing the oxygen dissociation curve to shift to the right (the Bohr effect).
* This reduces haemoglobin's affinity for oxygen, meaning oxygen is unloaded/released more readily to the respiring tissues to support aerobic respiration.

### Part (c)
* Change in saturation = \(60\% - 44\% = 16\%\)
* Percentage decrease = \(\frac{16}{60} \times 100 = 26.67\%\) (or \(26.7\%\) or \(27\%\)).

**Part (a): (Maximum 3 marks)**
1. Binding of the first oxygen molecule is difficult OR shape of haemoglobin makes first binding difficult;
2. Binding of first oxygen changes quaternary structure / shape of haemoglobin / causes conformational change;
3. This uncovers/exposes other binding sites / haem groups, making it easier for subsequent oxygen molecules to bind (cooperative binding);
4. It is harder for the fourth oxygen to bind because most sites are already occupied.

**Part (b): (Maximum 3 marks)**
1. High carbon dioxide reduces pH / increases acidity / forms carbonic acid;
2. Reduces haemoglobin's affinity for oxygen OR shifts the curve to the right (Bohr effect);
3. Haemoglobin unloads/dissociates/releases oxygen more readily / at higher \(pO_2\);
4. To respiring cells/tissues for aerobic respiration.

**Part (c): (2 marks)**
1. Correct calculation of decrease: \(60 - 44 = 16\) (seen or implied in working);
2. Correct percentage decrease: \(26.7\%\) (or \(26.67\%\) or \(27\%\));
*Note: Award 2 marks for a correct final answer of 26.7% (or 26.67% / 27%) with or without working.*

(a)
1. mRNA binds to the small subunit of the ribosome.
2. The ribosome moves along mRNA until it reaches the start codon (AUG).
3. tRNA with a complementary anticodon binds to the codon on mRNA via complementary base pairing.
4. Each tRNA molecule carries a specific amino acid.
5. Two tRNAs can bind to the ribosome at any one time, bringing two amino acids close together.
6. A peptide bond is formed between these amino acids, catalysed by peptidyl transferase (using energy from ATP hydrolysis).

(b)
1. Bacteria have 70S ribosomes containing a 50S large subunit, which is the specific target/shape that chloramphenicol binds to.
2. Eukaryotes (human cytoplasm) have 80S ribosomes (containing a 60S large subunit), which have a different shape/structure.
3. Therefore, chloramphenicol cannot bind to human cytoplasmic ribosomes, leaving human protein synthesis unaffected while preventing bacterial protein synthesis.

(a) [Max 5 marks]
- 1 mark for stating mRNA binds to the small subunit of the ribosome.
- 1 mark for stating the ribosome moves to / finds the start codon.
- 1 mark for stating tRNA with a complementary anticodon binds to the codon (by hydrogen bonding / complementary base pairing).
- 1 mark for stating that each tRNA carries a specific amino acid.
- 1 mark for stating that two tRNA molecules bind at a time, bringing amino acids close together.
- 1 mark for stating that a peptide bond forms between amino acids, catalysed by peptidyl transferase.

(b) [Max 3 marks]
- 1 mark for stating that bacteria have 70S ribosomes (or a 50S large subunit) that chloramphenicol binds to.
- 1 mark for stating that human cytoplasm has 80S ribosomes (or a 60S subunit) which do not bind chloramphenicol due to a different shape.
- 1 mark for explaining that bacterial translation/protein synthesis is blocked (preventing growth), but human translation continues normally.

(a)
1. Calculate \(N\) (total number of organisms of all species): \(52 + 12 + 6 + 40 = 110\).
2. Calculate \(N(N - 1) = 110 \times 109 = 11990\).
3. Calculate \(\sum n(n-1)\):
- Species 1: \(52 \times 51 = 2652\)
- Species 2: \(12 \times 11 = 132\)
- Species 3: \(6 \times 5 = 30\)
- Species 4: \(40 \times 39 = 1560\)
Sum = \(2652 + 132 + 30 + 1560 = 4374\).
4. Calculate \(d\): \(d = \frac{11990}{4374} \approx 2.74\).

(b)
1. A low index of diversity (2.74) indicates a less stable ecosystem.
2. In a less diverse system, there are fewer species, meaning a change in one population (e.g., due to disease or climate) has a larger relative impact on the food web / there are fewer alternative food sources.
3. This is because only a few species are adapted to tolerate the harsh or uniform abiotic conditions of a commercial coniferous plantation.

(c)
1. Use a grid and random coordinate generator to select trap locations (prevents bias).
2. Keep the traps open for the exact same amount of time (e.g., 24 hours).
3. Ensure all traps have the same design (e.g., identical depth, volume, and cover protecting them from rain).

(a) [3 marks]
- 1 mark for correct calculation of \(N = 110\) and \(N(N-1) = 11990\).
- 1 mark for correct calculation of \(\sum n(n-1) = 4374\).
- 1 mark for correct final answer of 2.74 (accept 2.7 or 2.74; allow follow-through from incorrect sum calculation).

(b) [3 marks]
- 1 mark for stating that a lower index indicates a less stable ecosystem (or more vulnerable to environmental change).
- 1 mark for explaining that a change to one species (e.g. disease) has a larger impact because there are fewer alternative species / food sources.
- 1 mark for linking the low diversity to fewer species being adapted to the uniform monoculture environment.

(c) [2 marks]
- 1 mark for each valid procedure up to 2 marks (e.g. random placement/coordinates; identical duration of collection; identical trap size/design; same season/weather conditions during collection).

(a)
1. An antigen (from cancer cells) is injected into a mouse (or other mammal).
2. This stimulates the mouse's B-cells/lymphocytes to produce antibodies specific to the antigen.
3. B-cells are extracted from the spleen of the mouse.
4. These B-cells are fused with tumor/cancer/myeloma cells to form hybridoma cells (using a fusion agent such as polyethylene glycol).
5. The hybridoma cells are screened to identify those producing the desired antibody, then cloned to produce large quantities of monoclonal antibodies.

(b)
1. Monoclonal antibodies have a highly specific primary structure, which dictates a specific tertiary structure.
2. The antigen-binding site (variable region) has a shape complementary only to that specific antigen (on the cancer cell), so it will not bind to cells lacking this antigen.

(c)
1. The conjugate (antibody-drug complex) only binds to cancer cells that display the complementary antigen.
2. Consequently, the cytotoxic drug is delivered directly to, and concentrated at, the cancer cells, leaving healthy tissues undamaged.

(a) [Max 4 marks]
- 1 mark for injecting the target antigen into a mouse/animal.
- 1 mark for extracting/harvesting B-cells/plasma cells from the spleen.
- 1 mark for fusing B-cells with myeloma/cancer/tumor cells to form hybridoma cells.
- 1 mark for cloning/culturing the selected hybridoma cells to produce large quantities.

(b) [2 marks]
- 1 mark for stating that they have a specific primary structure/amino acid sequence leading to a specific tertiary structure.
- 1 mark for stating that only the target antigen on cancer cells has a complementary shape to the antibody's variable region / antigen-binding site.

(c) [2 marks]
- 1 mark for explaining that the drug is only targeted/delivered to cancer cells (since antibody binds selectively).
- 1 mark for stating that healthy cells are not damaged/killed, minimizing systemic side effects.

(a)
(i) Ice-cold: To reduce/prevent enzyme activity that could break down or damage organelles.
(ii) Isotonic: To prevent osmosis (net movement of water), which could cause the organelles to swell and burst (lyse) or shrink.
(iii) Buffered: To maintain a constant pH so that proteins/enzymes within the organelles do not denature.

(b)
1. Homogenize the plant tissue (grind/blend spinach leaves in the cold, isotonic, buffered solution) to break open cell walls and cell surface membranes, releasing the organelles.
2. Filter the homogenate to remove large, unbroken cell debris and whole cells.
3. Centrifuge the filtrate at a low speed first.
4. This produces a pellet containing the heaviest organelles, which are the nuclei.
5. Pour off the supernatant (liquid above the pellet) and centrifuge it again at a higher speed.
6. The next pellet formed will contain chloroplasts (which are the second heaviest organelles in plant cells).

(a) [3 marks]
- (i) 1 mark for stating it prevents/reduces enzyme activity (which could digest organelles).
- (ii) 1 mark for stating it maintains water potential / prevents osmotic movement of water so organelles do not burst/shrink.
- (iii) 1 mark for stating it maintains constant pH to prevent denaturation of proteins/enzymes.

(b) [Max 5 marks]
- 1 mark for stating homogenize/grind tissue to release organelles.
- 1 mark for stating filter the homogenate to remove cell debris / whole cells.
- 1 mark for stating centrifuge the filtrate at low speed.
- 1 mark for stating that the pellet formed contains nuclei (heaviest organelle) and is discarded/removed.
- 1 mark for stating that the supernatant is spun again at a higher speed.
- 1 mark for identifying that the second pellet contains chloroplasts.

(a)
1. The primary structure is the specific sequence of amino acids in the polypeptide chain.
2. This sequence is determined by the sequence of bases/codons in DNA.
3. The position and type of amino acid R-groups determine where hydrogen, ionic, and disulfide bonds form during folding.
4. This dictates the specific 3D conformation (tertiary structure) of the protein.

(b)
1. The folding of the polypeptide chain creates a specific 3D shape containing a cleft or pocket known as the active site.
2. The active site has a shape that is complementary to the specific substrate (peptidoglycan), allowing enzyme-substrate complexes to form.

(c)
1. A mutation changes the base sequence of DNA, which alters the amino acid sequence (primary structure) of lysozyme.
2. This changes the positions where ionic/hydrogen/disulfide bonds form, leading to a change in the tertiary structure.
3. Consequently, the shape of the active site changes so it is no longer complementary to the substrate, and enzyme-substrate complexes cannot form.

(a) [Max 4 marks]
- 1 mark for stating primary structure is the sequence of amino acids.
- 1 mark for stating that primary structure is determined by the sequence of DNA bases/codons.
- 1 mark for mentioning specific bonding (hydrogen, ionic, or disulfide bonds) between R-groups.
- 1 mark for explaining that the folding/bonds determine the specific 3D/tertiary shape.

(b) [2 marks]
- 1 mark for stating that folding creates a specific active site.
- 1 mark for explaining that the active site is complementary in shape to the substrate, enabling the formation of enzyme-substrate complexes.

(c) [Max 2.33 marks]
- 1 mark for stating that DNA mutation changes the primary structure (amino acid sequence).
- 1 mark for explaining that this changes the bonds formed, altering the tertiary structure / 3D shape.
- 1 mark for explaining that the active site is no longer complementary to the substrate, so no enzyme-substrate complexes can form.

(a)
1. Fetal blood flows very close to maternal blood in the placenta.
2. At the partial pressure of oxygen in the placenta, maternal hemoglobin dissociates/unloads oxygen.
3. Because fetal hemoglobin has a higher affinity, it readily binds/loads the oxygen released from the maternal blood, ensuring the fetus receives an adequate supply for respiration.

(b)
1. An increase in the partial pressure of carbon dioxide (\(CO_2\)) causes the oxygen dissociation curve to shift to the right.
2. This is because high \(CO_2\) concentration lowers blood pH (by forming carbonic acid), altering the shape of hemoglobin and reducing its affinity for oxygen.
3. During exercise, actively respiring muscles produce more \(CO_2\), prompting hemoglobin to unload/release oxygen more readily to tissues that need it.

(c)
Any one of the following:
- Biconcave shape: increases the surface area to volume ratio, allowing rapid diffusion of oxygen into and out of the cell.
- No nucleus/organelles: provides more intracellular space to pack in hemoglobin molecules, maximizing oxygen-carrying capacity.
- Small size / flexibility: allows them to deform and squeeze through narrow capillaries, bringing them closer to body cells and decreasing diffusion distance.

(a) [3 marks]
- 1 mark for stating that maternal hemoglobin unloads oxygen at the placenta.
- 1 mark for explaining that fetal hemoglobin has a higher affinity, so it binds/loads oxygen at lower partial pressures.
- 1 mark for linking this to oxygen uptake across the placenta to meet the respiratory needs of the fetus.

(b) [3 marks]
- 1 mark for stating that high \(CO_2\) concentrations shift the dissociation curve to the right.
- 1 mark for explaining that increased acidity / lower pH decreases hemoglobin's affinity for oxygen.
- 1 mark for stating that this allows active tissues (during exercise) to receive more oxygen for respiration.

(c) [Max 2.33 marks]
- 1 mark for naming a valid adaptation (e.g., biconcave shape, absence of nucleus, or thin/flexible membrane).
- 1.33 marks for fully explaining the benefit (e.g., biconcave shape increases SA:Vol ratio for faster diffusion; lack of nucleus allows more space for hemoglobin / more oxygen carried; flexibility reduces diffusion distance as cells squeeze through capillaries).

(a)
1. Oxygen enters the insect through spiracles, which are pores on the outer body surface.
2. It diffuses down a concentration gradient through the trachea and into branching tracheoles.
3. The tracheoles branch directly into the tissues/cells of the insect.
4. Oxygen dissolves in the moisture at the ends of the tracheoles and diffuses directly into neighboring respiring cells.
5. During high activity, abdominal pumping / muscle contractions can also force air along the tracheae by mass flow.

(b)
1. During high activity, muscles respire anaerobically and produce lactate.
2. This lowers the water potential of the muscle cells, causing water to move from the ends of the tracheoles into the muscle cells by osmosis.
3. This decreases the volume of liquid in the tracheoles, drawing air/oxygen closer to the respiring cells and increasing the surface area for rapid gas diffusion (since diffusion is faster through air than water).

(c)
1. At higher temperatures, metabolic rates / enzymes work faster, increasing the rate of respiration.
2. This increases the demand for oxygen and produces more carbon dioxide, meaning spiracles must open more frequently to maintain gas exchange despite the increased risk of water loss.

(a) [Max 4 marks]
- 1 mark for stating oxygen enters through spiracles.
- 1 mark for stating oxygen diffuses down a concentration/diffusion gradient.
- 1 mark for describing the pathway through trachea to tracheoles.
- 1 mark for stating that tracheoles branch directly into/near respiring tissues.
- 1 mark for mentioning abdominal pumping/ventilation causing mass flow.

(b) [Max 2.33 marks]
- 1 mark for stating that anaerobic respiration produces lactate, lowering water potential in muscle cells.
- 1 mark for explaining that water moves out of tracheoles into muscles by osmosis.
- 1 mark for explaining that this decreases the liquid barrier / allows gases to diffuse faster through air directly to tissues.

(c) [2 marks]
- 1 mark for explaining that higher temperatures increase metabolic rate / rate of respiration (increasing oxygen demand / carbon dioxide production).
- 1 mark for stating that spiracles must open more frequently to meet this gas exchange demand.

(a)
1. Find total number of cells: \(168 + 14 + 8 + 4 + 6 = 200\).
2. Calculate the proportion of cells in metaphase: \(\frac{8}{200} = 0.04\).
3. Calculate time spent in metaphase: \(0.04 \times 1320 = 52.8\) minutes.

(b)
(i) Prophase:
1. Chromosomes condense, becoming shorter, thicker, and visible.
2. Chromosomes appear as two sister chromatids joined at a centromere.
(ii) Anaphase:
1. The centromere divides, separating sister chromatids.
2. Chromatids are pulled to opposite poles of the cell by the shortening/contraction of spindle fibers.

(c)
1. Mitosis / cell division occurs in the meristematic tissue (growth region), which is located specifically at the very tip of the root.
2. Cells in the middle of the root are in the zone of elongation/differentiation, where cells do not actively divide; thus, mitotic stages would not be observed there.

(a) [2 marks]
- 1 mark for finding total cell count (200) and correct proportion (0.04 or 4%).
- 1 mark for correct final calculation: 52.8 minutes (accept 53 minutes).

(b) [4 marks]
- (i) Prophase [Max 2 marks]:
- 1 mark for mentioning chromosomes condense / become visible / shorten and thicken.
- 1 mark for mentioning chromosomes consist of sister chromatids joined by a centromere.
- (ii) Anaphase [Max 2 marks]:
- 1 mark for stating centromeres divide / chromatids split.
- 1 mark for stating chromatids/chromosomes are pulled to opposite poles of the cell by contracting spindle fibers.

(c) [Max 2.33 marks]
- 1 mark for explaining that mitosis/cell division only occurs in the meristem / region of active cell division.
- 1 mark for stating that the meristem is located at the root tip.
- 1 mark for explaining that cells in the middle of the root are differentiating / elongating and are not dividing.

Part a): Explain that haemoglobin undergoes a conformational change (shape change) upon binding the first oxygen molecule. This cooperative binding increases the affinity for the next oxygen molecules, leading to a steep rise in the curve. Part b): Focus on low-oxygen environments. A curve to the left means high affinity. The lugworm can saturate its haemoglobin with oxygen even when the environmental partial pressure of oxygen (\(pO_2\)) is extremely low. Part c): Explain the Bohr effect. Increased \(CO_2\) leads to a lower pH (more acidic), which changes the shape of haemoglobin slightly, decreasing its affinity for oxygen. Consequently, the curve shifts to the right, meaning haemoglobin unloads oxygen more easily to active tissues where respiratory demands are high.

Part a) [2 marks max]: 1 mark for: Binding of the first oxygen molecule changes the tertiary / quaternary structure of haemoglobin (alters shape). 1 mark for: This uncovers or increases access to other binding sites / makes it easier for subsequent oxygen molecules to bind. (Do not accept: makes it easier for the first oxygen to bind). Part b) [3 marks max]: 1 mark for: Lugworm haemoglobin has a higher affinity for oxygen (at any given partial pressure of oxygen). 1 mark for: Allows haemoglobin to load / associate with oxygen at very low partial pressures / in its burrow. 1 mark for: Ensuring sufficient oxygen is supplied to tissues for aerobic respiration. Part c) [3 marks max]: 1 mark for: Shifts the oxygen dissociation curve to the right / Bohr shift. 1 mark for: Carbon dioxide lowers pH / makes conditions more acidic, changing haemoglobin shape to reduce its affinity for oxygen. 1 mark for: Haemoglobin unloads / dissociates from / releases oxygen more readily to respiring cells / tissues.