2025 Edexcel IAL Biology (YBI11) Practice Paper with Answers

High-density lipoproteins (HDLs) have a higher density due to a higher ratio of protein to lipid compared to low-density lipoproteins (LDLs). HDLs function to transport cholesterol from peripheral tissues back to the liver, where it can be metabolized and excreted, helping to lower blood cholesterol levels.

1 mark: Correctly identifies both the higher protein ratio and the transport from tissues to the liver.

Active transport is the movement of substances against their concentration gradient (from a region of lower concentration to a region of higher concentration). This process is active, meaning it requires energy in the form of ATP, and it relies on specific carrier proteins embedded within the cell membrane.

1 mark: Correctly identifies active transport as requiring ATP, moving against a concentration gradient, and requiring a membrane protein.

Glycogen is a highly branched polysaccharide stored in animals, containing both 1,4-glycosidic bonds along its main chains and 1,6-glycosidic bonds at the branch points. In contrast, amylose is one of the components of starch in plants and is a linear, unbranched helical polymer containing only 1,4-glycosidic bonds.

1 mark: Correctly identifies that glycogen is highly branched containing both 1,4 and 1,6 bonds, and amylose is unbranched containing 1,4 bonds only.

During transcription, RNA polymerase matches complementary RNA nucleotides to the template DNA strand. The template strand runs in the 3' to 5' direction, meaning the mRNA will be synthesized in the 5' to 3' direction. The complementary bases are: Thymine (T) pairs with Adenine (A), Adenine (A) pairs with Uracil (U), Cytosine (C) pairs with Guanine (G), and Guanine (G) pairs with Cytosine (C). Thus, 3'-TACGGTCOTACT-5' is transcribed into 5'-AUGCCAGUAUGA-3'.

1 mark: Correctly transcribes the DNA template sequence to its complementary mRNA sequence with appropriate 5' and 3' orientation and Uracil replacing Thymine.

During anaphase, the spindle fibres shorten, causing the centromeres of each chromosome to divide. This splits the sister chromatids, which are then pulled to opposite poles of the cell.

1 mark: Correctly identifies the mitotic stage as anaphase.

During clotting, damaged tissue and platelets release thromboplastin. In the presence of calcium ions, thromboplastin catalyzes the conversion of the inactive plasma protein prothrombin into the active enzyme thrombin. Thrombin then catalyzes the conversion of the soluble plasma protein fibrinogen into insoluble fibrin fibers, which trap red blood cells to form a clot.

1 mark: Correctly identifies the complete sequence of reactions from thromboplastin activation to fibrin formation.

Both xylem vessels and sclerenchyma fibres are dead at maturity. They both have cell walls that are heavily lignified (impregnated with lignin) to provide mechanical strength and support to the plant.

1 mark: Correctly identifies that both xylem vessels and sclerenchyma fibres consist of dead cells and contain lignin.

The F508del mutation results in the deletion of a phenylalanine residue, causing the CFTR channel protein to misfold. Consequently, it is recognized as defective and degraded by the cell's quality control systems before it can be transported to the apical membrane. Because there are no functional CFTR channels in the membrane, chloride ions cannot leave the epithelial cells, which leads to increased sodium and water absorption from the mucus, causing the mucus in the airways to become dehydrated, thick, and sticky.

1 mark: Correctly identifies that the F508del mutation leads to a misfolded, degraded protein and results in thick, sticky mucus due to blocked chloride ion transport.

To calculate the heart rate from the ECG:
1. Determine the duration of one cardiac cycle (R-R interval):
\(4.2 \text{ squares} \times 0.20 \text{ s/square} = 0.84 \text{ seconds}\).
2. Calculate the heart rate in beats per minute (bpm):
\(\text{Heart Rate} = 60 / 0.84 = 71.43 \text{ bpm}\).
3. Rounding to the nearest whole number gives 71 bpm.

- Award 1 mark for calculating the correct duration of one cardiac cycle: \(0.84 \text{ s}\).
- Award 1 mark for the correct final heart rate rounded to the nearest whole number: \(71 \text{ (bpm)}\).
- Allow 1 mark for a correct calculation using an incorrect cycle length (e.g., if cycle length was incorrectly calculated as 0.80 s, leading to 75 bpm).

1. Calculate the number of nucleotides in the mature mRNA:
\(450 \text{ nucleotides} - 75 \text{ nucleotides (intron)} = 375 \text{ nucleotides}\).
2. Calculate the total number of codons:
\(375 / 3 = 125 \text{ codons}\).
3. Subtract the stop codon, which does not code for an amino acid:
\(125 - 1 = 124 \text{ amino acids}\).

- Award 1 mark for calculating the correct remaining nucleotides in the mRNA: \(375\).
- Award 1 mark for dividing the number of nucleotides by 3 to find the number of codons: \(125\).
- Award 1 mark for subtracting the stop codon to get the final answer: \(124\).

Glycogen is ideal for storage because:
1. It is a large, insoluble polysaccharide, meaning it does not dissolve and affect the osmotic balance of the cell (does not cause water to enter by osmosis).
2. It has many 1,4 and 1,6-glycosidic bonds, making it highly branched. This provides many terminal ends for enzymes to rapidly hydrolyze and release glucose when energy demand is high.
3. Its coiled/branched shape makes it highly compact, allowing a high density of glucose storage within the cell.

- Award 1 mark for stating that glycogen is insoluble and therefore has no osmotic effect on the cell.
- Award 1 mark for stating that it is highly branched (due to 1,6-glycosidic bonds) allowing rapid hydrolysis / release of glucose.
- Award 1 mark for stating that it is compact, meaning it stores a large amount of glucose / energy in a small space.

Using Fick's Law:
\(\text{Rate} \propto \frac{\text{Surface Area} \times \text{Concentration Difference}}{\text{Thickness}}\).
Substituting the changes:
\(\text{New Rate} \propto \frac{2 \times 0.5}{4} = \frac{1}{4} = 0.25\).
Therefore, the rate of diffusion is reduced to 0.25 (or \(1/4\)) of its original value.

- Award 1 mark for substituting the values correctly into the ratio: \(\frac{2 \times 0.5}{4}\).
- Award 1 mark for the correct final factor: \(0.25\) (or \(1/4\) or decreased by a factor of 4).

1. Arteries have thick walls containing collagen and elastic fibres to withstand and maintain high blood pressure as blood leaves the heart. Capillaries have walls that are only one cell thick (consisting of a single layer of endothelial cells) to provide a short diffusion distance for the exchange of substances.
2. Arteries contain a layer of smooth muscle that can contract or relax to control blood flow and pressure. Capillaries lack smooth muscle, maximizing their permeability for exchange.
3. Arteries do not allow exchange across their walls due to the thick outer layers, whereas capillary walls have tiny pores/fenestrations to facilitate the rapid movement of tissue fluid, nutrients, and gases.

- Award 1 mark for explaining that arteries have thick walls / elastic fibres to withstand high pressure AND capillaries have thin walls (one cell thick) to provide a short diffusion distance.
- Award 1 mark for explaining that arteries have smooth muscle to regulate blood flow / pressure AND capillaries lack muscle to maximize exchange.
- Award 1 mark for mentioning collagen in arteries for strength vs lack of outer tissue layers in capillaries to permit permeability.

The formula for the temperature coefficient is:
\(\text{Q}_{10} = \frac{\text{Rate of reaction at } (T + 10)^\circ\text{C}}{\text{Rate of reaction at } T^\circ\text{C}}\).
Using the provided rates:
\(\text{Q}_{10} = \frac{11.7}{4.5} = 2.6\).

- Award 1 mark for setting up the correct calculation: \(\frac{11.7}{4.5}\).
- Award 1 mark for the correct answer: \(2.6\).

1. LDLs transport cholesterol from the liver to the bloodstream, where excess LDLs can accumulate in the blood and deposit in the damaged endothelium of artery walls, initiating plaque (atheroma) formation.
2. HDLs transport cholesterol away from body tissues and the bloodstream back to the liver to be broken down and excreted, which reduces blood cholesterol levels and prevents accumulation.
3. Therefore, high levels of LDLs increase the risk of atherosclerosis, whereas high levels of HDLs lower the risk.

- Award 1 mark for stating that LDLs transport cholesterol from the liver to body tissues/blood, leading to deposition in artery walls / atheroma formation.
- Award 1 mark for stating that HDLs transport cholesterol from tissues/blood back to the liver for excretion/metabolism, reducing plaque risk.
- Award 1 mark for stating that a high LDL:HDL ratio / high concentration of LDLs increases the risk of atherosclerosis (or vice versa for HDLs).

Differences:
1. Active transport requires metabolic energy in the form of ATP, whereas facilitated diffusion is a passive process that does not require ATP.
2. Active transport moves substances against their concentration gradient (from low to high concentration), whereas facilitated diffusion moves substances down their concentration gradient (from high to low concentration).

Similarity:
1. Both processes require specific transport proteins (carrier proteins for active transport, carrier/channel proteins for facilitated diffusion) to allow substances to cross the phospholipid bilayer.

- Award 1 mark for each correct difference up to a maximum of 2 marks (e.g., active transport requires ATP vs facilitated diffusion does not; active transport is against gradient vs facilitated diffusion is down gradient).
- Award 1 mark for a correct similarity (e.g., both use transmembrane/carrier proteins; both transport polar/charged/large molecules that cannot pass directly through the bilayer).

Using the formula: Cardiac Output = Heart Rate * Stroke Volume. Rearranging this gives: Stroke Volume = Cardiac Output / Heart Rate. Substituting the values: Stroke Volume = 5.4 dm^3 min^-1 / 72 beats min^-1 = 0.075 dm^3 beat^-1. Converting dm^3 to cm^3 by multiplying by 1000: 0.075 * 1000 = 75 cm^3.

[1 mark] Rearranging formula correctly and substituting values: Stroke volume = 5.4 / 72 = 0.075 dm^3. [1 mark] Conversion of dm^3 to cm^3 by multiplying by 1000. [0.5 mark] Correct final answer of 75.

Amylose and amylopectin are both storage polysaccharides made from alpha-glucose molecules linked by 1,4-glycosidic bonds. However, amylose has an unbranched, helical structure, whereas amylopectin has a branched structure due to the presence of 1,6-glycosidic bonds.

[1 mark] Similarity: both consist of alpha-glucose monomers linked by 1,4-glycosidic bonds. [1 mark] Difference: amylose is unbranched/coiled whereas amylopectin is branched. [0.5 mark] Difference: amylopectin contains 1,6-glycosidic bonds whereas amylose does not.

Increasing ethanol concentration disrupts the cell membrane structure because ethanol is an organic solvent that dissolves the lipid bilayer. It also denatures membrane proteins. This increases membrane permeability, allowing pigment (betalain) molecules to leak out of the vacuole into the surrounding solution, increasing the intensity of the color and therefore the absorbance of light.

[1 mark] Ethanol dissolves/disrupts the phospholipid bilayer. [1 mark] Ethanol denatures membrane proteins. [0.5 mark] Increased membrane permeability allows more pigment to leak out of the cells, increasing absorbance.

The complementary mRNA sequence transcribed from the template DNA sequence 3'- T A C C G T T A C A T T -5' is 5'- A U G G C A A U G U A A -3'. This mRNA contains 4 codons: AUG, GCA, AUG, and UAA. Since UAA is a stop codon, it does not code for an amino acid. Therefore, the maximum number of amino acids coded for is 3.

[1.5 marks] Correct complementary mRNA sequence: 5'-AUGGCAAUGUAA-3' (award 1 mark if sequence is correct but directionality is missing or incorrect). [1 mark] Correct determination of 3 amino acids because the final codon is a stop codon.

First, calculate the energy contribution of each component: Protein: 80 g * 17 kJ/g = 1360 kJ. Carbohydrate: 300 g * 17 kJ/g = 5100 kJ. Lipid: 90 g * 38 kJ/g = 3420 kJ. Total energy = 1360 + 5100 + 3420 = 9880 kJ. Percentage of energy from lipids = (3420 / 9880) * 100 = 34.615%. Rounded to one decimal place, this is 34.6%.

[1 mark] Correct calculation of total energy intake (9880 kJ). [1 mark] Correct calculation of lipid energy (3420 kJ) and division by total energy. [0.5 mark] Correct percentage of 34.6% (accept 34.62% or 35% with working).

The temperature coefficient (Q10) is calculated using the formula: Q10 = (Rate of reaction at T + 10 degrees C) / (Rate of reaction at T degrees C). Substituting the values: Q10 = 9.1 / 4.2 = 2.1666... Rounding to two decimal places gives 2.17.

[1 mark] Correct formula or substitution (9.1 / 4.2). [1 mark] Correct division value of 2.17. [0.5 mark] Correct rounding to two decimal places.

Arteries have a thick layer of elastic fibres (elastin) in their walls which stretch during ventricular systole to accommodate surges of blood, and then recoil during diastole to maintain high blood pressure. They also have a thick layer of smooth muscle which can contract to constrict the lumen and regulate blood pressure. Furthermore, a strong outer layer of collagen prevents the artery from bursting under high pressure.

[1 mark] Elastic fibres stretch to accommodate blood surges and recoil to maintain blood pressure. [1 mark] Smooth muscle contracts to constrict the lumen and maintain high blood pressure. [0.5 mark] Collagen provides strength to prevent bursting under high pressure.

The deletion of three nucleotides removes one codon from the mRNA transcript. This results in the loss of a single amino acid (phenylalanine) from the polypeptide chain, altering the primary structure of the CFTR protein. Because the primary structure is changed, the folding of the polypeptide chain is altered, changing the secondary and tertiary structures. This prevents the protein from forming its correct 3D shape, meaning it cannot function as an effective chloride channel or is targeted for degradation before reaching the cell membrane.

[1 mark] Deletion of three nucleotides removes one codon, leading to the loss of one specific amino acid in the primary structure. [1 mark] This alters the tertiary structure/3D folding of the CFTR protein due to changes in chemical bonds. [0.5 mark] The protein can no longer function effectively as a chloride channel.

1. Calculate heart rate: \( 60 / 0.82 = 73.17 \text{ beats per minute} \). 2. Calculate cardiac output: \( 73.17 \times 74 = 5414.6 \text{ cm}^3\text{ min}^{-1} \). 3. Convert to \( \text{dm}^3\text{ min}^{-1} \) by dividing by 1000: \( 5414.6 / 1000 = 5.4146 \text{ dm}^3\text{ min}^{-1} \). 4. Round to 3 significant figures: \( 5.41 \text{ dm}^3\text{ min}^{-1} \).

1 mark for calculating the heart rate (73.17 bpm). 1 mark for multiplying by stroke volume and converting to dm^3 (dividing by 1000). 0.5 mark for the correct answer rounded to 3 significant figures (5.41).

1. Calculate total number of bases: \( 3400 \times 2 = 6800 \). 2. If cytosine is 18%, then guanine is also 18% (total C + G = 36%). 3. This leaves 64% for adenine and thymine (100% - 36% = 64%). 4. Since A = T, adenine is \( 64\% / 2 = 32\% \). 5. Calculate number of adenine bases: \( 32\% \text{ of } 6800 = 0.32 \times 6800 = 2176 \).

1 mark for calculating total bases (6800) or determining percentage of adenine (32%). 1 mark for calculating the correct number of adenine bases (2176). 0.5 mark for showing correct complementary base pairing logic.

1. Use the BMI formula: \( \text{BMI} = \text{mass in kg} / (\text{height in m})^2 \). 2. Calculate BMI: \( 94.5 / (1.78)^2 = 94.5 / 3.1684 = 29.83 \). 3. The BMI of 29.83 falls into the 'overweight' category (25.0 to 29.9). 4. An overweight status increases the risk of developing cardiovascular disease.

1 mark for correct BMI calculation (29.8 or 29.83). 1 mark for identifying the category as overweight and stating that it increases the risk of cardiovascular disease. 0.5 mark for applying the correct formula.

1. For any linear polypeptide chain, the number of peptide bonds is equal to the number of amino acids minus 1. 2. Peptide bonds in Chain A: \( 21 - 1 = 20 \). 3. Peptide bonds in Chain B: \( 30 - 1 = 29 \). 4. Peptide bonds in Chain C: \( 26 - 1 = 25 \). 5. Total peptide bonds: \( 20 + 29 + 25 = 74 \).

1 mark for using the (N-1) rule for linear chains. 1 mark for adding the individual peptide bond counts together. 0.5 mark for the final correct answer of 74.

1. Convert the time from minutes to seconds: \( 3.0 \text{ minutes} = 3.0 \times 60 = 180 \text{ seconds} \). 2. Calculate the rate: \( \text{Rate} = \text{Volume of product} / \text{Time} = 15.0\text{ cm}^3 / 180\text{ s} = 0.08333... \text{ cm}^3\text{ s}^{-1} \). 3. Convert to standard scientific notation to 2 significant figures: \( 8.3 \times 10^{-2} \text{ cm}^3\text{ s}^{-1} \).

1 mark for converting minutes to seconds (180 s). 1 mark for calculating the rate. 0.5 mark for providing the answer in standard scientific notation to 2 significant figures (8.3 x 10^-2).

1. Find the radius of the cell: \( r = \text{diameter} / 2 = 20 / 2 = 10\text{ }\mu\text{m} \). 2. Substitute into the SA:V ratio formula for a sphere, which simplifies to \( \text{SA:V} = 3/r \). 3. Calculate: \( \text{SA:V} = 3 / 10 = 0.3 \).

1 mark for determining the correct radius (10 micrometers). 1 mark for applying the surface area and volume formulas or using the simplified SA:V ratio formula. 0.5 mark for the correct ratio of 0.3.

1. Calculate energy from fat: \( 12 \text{ g} \times 37 \text{ kJ g}^{-1} = 444 \text{ kJ} \). 2. Calculate energy from carbohydrate: \( 25 \text{ g} \times 17 \text{ kJ g}^{-1} = 425 \text{ kJ} \). 3. Calculate energy from protein: \( 15 \text{ g} \times 17 \text{ kJ g}^{-1} = 255 \text{ kJ} \). 4. Sum the energy values: \( 444 + 425 + 255 = 1124 \text{ kJ} \).

1 mark for calculating individual energy contributions for fat, carbohydrate, and protein. 1 mark for summing the values. 0.5 mark for correct final value of 1124 kJ.

1. Convert the image size from mm to micrometers: \( 45 \text{ mm} = 45 \times 1000 = 45000\text{ }\mu\text{m} \). 2. Use the magnification formula: \( \text{Magnification} = \text{Image size} / \text{Actual size} \). 3. Calculate magnification: \( 45000 / 150 = 300 \).

1 mark for correct unit conversion (45 mm to 45000 micrometers). 1 mark for applying the magnification formula. 0.5 mark for the final answer of 300 (or x300).

The answer should describe the progressive development of atherosclerosis and link it logically to the clotting cascade and subsequent muscle damage. 1. Endothelial damage: Caused by factors such as high blood pressure or toxins. 2. Inflammatory response: Monocytes/macrophages enter the wall and ingest lipids to become foam cells. 3. Plaque formation: Accumulation of lipids, calcium salts, and fibrous tissue forms an atheroma, narrowing the artery lumen and reducing elasticity. 4. Plaque rupture: Tears in the plaque expose collagen fibers. 5. Clotting cascade: Platelets adhere to collagen; clotting factors convert prothrombin to thrombin; thrombin converts soluble fibrinogen to insoluble fibrin, forming a thrombus. 6. Ischemia and infarction: The thrombus blocks coronary blood flow, preventing oxygen delivery to cardiac muscle, stopping aerobic respiration, and causing localized cell death.

Level 1 (1-2 marks): A basic description of plaque buildup and how it blocks blood flow, or a simple outline of the clotting process with some omissions. Level 2 (3-4 marks): Explains how endothelial damage leads to plaque formation and narrowing of the lumen, OR explains how a ruptured plaque triggers the clotting cascade to form a thrombus that blocks blood flow. Explains that this cuts off oxygen to the heart muscle. Level 3 (5-6 marks): Provides a detailed, logically structured explanation covering all key stages: endothelial damage and inflammatory response, plaque buildup, plaque rupture exposing collagen, the full enzymatic clotting cascade (prothrombin to thrombin, fibrinogen to fibrin), and the final physiological impact on cardiac muscle (ischemia and cell death).

The sequence must connect the genetic mutation to the protein structure, then to ion transport, and finally to the osmotic movement of water and its physical effect on mucus. 1. Genetic level: Mutation alters DNA bases and mRNA codons. 2. Protein level: Changes primary structure (amino acid sequence) and tertiary structure of the CFTR protein, making it non-functional or absent. 3. Ion transport: Chloride ions are not transported out of epithelial cells. 4. Osmotic effect: Sodium ions are reabsorbed, and water does not move into the mucus by osmosis. 5. Physical mucus properties: Mucus loses water, becoming highly viscous, thick, and sticky. 6. Cilia function: Cilia cannot beat effectively to clear the thick mucus from the airways.

Level 1 (1-2 marks): Identifies that a mutation leads to a faulty CFTR protein and results in thick mucus, but details of ion transport or osmosis are missing or incorrect. Level 2 (3-4 marks): Explains how the altered CFTR protein prevents chloride ion transport, which stops water from entering the mucus by osmosis, making it sticky. Level 3 (5-6 marks): Gives a fully coherent, step-by-step account from the gene mutation changing the primary and tertiary structure of the protein, to the failure of chloride transport, the subsequent retention/reabsorption of sodium, the lack of osmotic water movement into the mucus, and the resulting physical thickening of mucus that impairs cilia function.