Edexcel IGCSE · Thinka-original Practice Paper

2024 Edexcel IGCSE Human Biology Practice Paper with Answers

Thinka Nov 2024 Cambridge International A Level-Style Mock — Human Biology

180 marks210 mins2024
An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 Cambridge International A Level Human Biology paper. Not affiliated with or reproduced from Cambridge.

Paper 1 (4HB1/01)

Answer all questions. Show all the steps in any calculations and state the units. Candidates may use a calculator.
8 Question · 90 marks
Question 1 · Matching & Fill-in-Table
7 marks
The table below contains information about the components of human blood, their structures, and their functions.

Complete the table by writing the correct term, structure, or function in the spaces labelled **A**, **B**, **C**, **D**, **E**, **F**, and **G**.

Blood ComponentKey Structural FeaturePrimary FunctionRed blood cell**A****B**Phagocyte**C****D****E**Cell fragments, no nucleus, irregular shapeInvolved in blood clotting**F**Liquid (mainly water)Transports dissolved substances such as **G**
Show answer & marking scheme

Worked solution

A: Red blood cells are adapted to carry oxygen. Key structural features include their biconcave disc shape (which increases surface-area-to-volume ratio), the absence of a nucleus (providing more space for haemoglobin), and the presence of haemoglobin.
B: The primary function of red blood cells is to transport oxygen from the lungs to respiring tissues.
C: Phagocytes have a distinctive lobed (or multi-lobed) nucleus and a flexible cell membrane, allowing them to change shape easily to engulf pathogens.
D: Phagocytes engulf and digest/destroy pathogens using digestive enzymes in a process called phagocytosis.
E: Platelets (or thrombocytes) are small cell fragments with no nucleus that play a key role in initiating blood clotting.
F: Plasma is the liquid component of blood consisting mainly of water.
G: Plasma is responsible for transporting various dissolved substances including carbon dioxide, urea, digested food molecules (such as glucose or amino acids), hormones, plasma proteins, and antibodies.

Marking scheme

Award 1 mark for each correct answer (maximum 7 marks total):
- **A**: Biconcave shape / no nucleus / lacks a nucleus / contains haemoglobin [1 mark]
- **B**: Transports / carries oxygen [1 mark]
- **C**: Lobed nucleus / multi-lobed nucleus / flexible cell membrane [1 mark]
- **D**: Engulfs and digests / destroys pathogens (accept phagocytosis) [1 mark]
- **E**: Platelets / thrombocytes [1 mark]
- **F**: Plasma [1 mark]
- **G**: Carbon dioxide / CO2 / glucose / amino acids / urea / lactic acid / hormones / antibodies / plasma proteins / heat (accept any one valid substance transported by plasma, reject oxygen) [1 mark]
Question 2 · practical
12 marks
An investigation was carried out to study the effect of different concentrations of sucrose solution on the mass of potato tissue.

Five potato cylinders were cut to equal lengths. The initial mass of each cylinder was recorded. Each cylinder was placed in a different concentration of sucrose solution.

After two hours, the cylinders were removed, gently dried, and reweighed.

The table shows the results of this investigation.

| Sucrose Concentration (\(\text{mol/dm}^3\)) | Initial Mass (g) | Final Mass (g) | Percentage Change in Mass (%) |
| :---: | :---: | :---: | :---: |
| 0.0 | 4.00 | 4.36 | +9.0 |
| 0.2 | 4.00 | 4.16 | +4.0 |
| 0.4 | 4.00 | 3.92 | -2.0 |
| 0.6 | 4.00 | 3.76 | **X** |
| 0.8 | 4.00 | 3.56 | -11.0 |

(a) Calculate the percentage change in mass for the potato cylinder in \(0.6\text{ mol/dm}^3\) sucrose solution (value **X**). Show your working. (2)

(b) Plot a line graph on a grid to show the relationship between the concentration of sucrose solution and the percentage change in mass of the potato cylinders. Draw a straight line of best fit. (5)

(c) Use your graph to estimate the concentration of sucrose solution that is isotonic to the cytoplasm of the potato cells. Explain how you arrived at this estimate. (3)

(d) Explain why the student dried the potato cylinders before weighing them at the end of the investigation. (2)
Show answer & marking scheme

Worked solution

(a)
Change in mass = final mass - initial mass
Change in mass = \(3.76 - 4.00 = -0.24\text{ g}\)
Percentage change = \(\frac{\text{change}}{\text{initial mass}} \times 100 = \frac{-0.24}{4.00} \times 100 = -6.0\%\) (or a loss of \(6.0\%\)).

(b)
Graph plotting criteria:
- Axis choice: x-axis labeled as 'Concentration of sucrose solution (\(\text{mol/dm}^3\))' and y-axis labeled as 'Percentage change in mass (\(\%\))'.
- Scale: Linear scale covering at least half of the grid in both directions. y-axis must accommodate negative values from -12 to +10.
- Points: All 5 points plotted accurately within \(\pm\) half a small grid square.
- Line: A single, straight line of best fit drawn with a ruler that represents the trend of the data points.

(c)
Identify the x-intercept from the line of best fit. Based on the straight line of best fit, the point where it crosses the x-axis (percentage change in mass is 0) is between \(0.32\) and \(0.36\text{ mol/dm}^3\).
At this concentration, there is no net movement of water into or out of the potato cells because the water potential of the external sucrose solution is equal to the water potential of the cytoplasm inside the potato cells.

(d)
The cylinders are dried to remove any excess/surface liquid clinging to the outside of the potato. If not removed, this external liquid would add to the final measured mass, causing an inaccurate reading and underestimating the true mass loss (or overestimating mass gain).

Marking scheme

(a) [2 marks total]
- Method mark: Correct setup showing change in mass divided by initial mass: \(\frac{3.76 - 4.00}{4.00} \times 100\) or \(\frac{-0.24}{4.00}\) [1 mark]
- Accuracy mark: \(-6.0\%\) (accept \(-6\%\) or a loss of \(6\%\)) [1 mark]

(b) [5 marks total]
- Axes: Correctly oriented with sucrose concentration on the x-axis and percentage change in mass on the y-axis, both fully labeled with correct units [1 mark]
- Scale: Suitable linear scales chosen so that plotted points occupy at least half of the grid on both axes [1 mark]
- Plotting: All 5 data points plotted correctly (allow \(\pm 0.5\) small square tolerance) [2 marks]
- Deduct 1 mark if 1 point is plotted incorrectly.
- Deduct 2 marks if 2 or more points are plotted incorrectly.
- Line: Single straight line of best fit drawn cleanly with a ruler [1 mark] (Reject: point-to-point line or a curved line)

(c) [3 marks total]
- Estimate: Gives a value between \(0.32\text{ mol/dm}^3\) and \(0.36\text{ mol/dm}^3\) (or value exactly corresponding to where their line of best fit crosses the x-axis) [1 mark]
- Explanation 1: Identifies that at this point there is no net movement of water into or out of the cells / no change in mass [1 mark]
- Explanation 2: Explains that the water potential of the sucrose solution is equal to the water potential of the cytoplasm / the solution is isotonic [1 mark]

(d) [2 marks total]
- Point 1: To remove excess / surface liquid (or water) from the outside of the potato cylinders [1 mark]
- Point 2: Liquid on the surface would increase the recorded final mass / cause an inaccurate final mass measurement [1 mark]
- Accept: Ensures that only the change in mass of the tissue itself is measured.
Question 3 · Anatomical Labelling & Reflex Evaluation
12 marks
The diagrammatic sequence below represents the pathway of a spinal reflex arc when a person touches a hot object.

$$\text{Stimulus (Heat)} \longrightarrow \text{Receptor in Skin} \longrightarrow \text{Neurone P} \longrightarrow \text{Spinal Cord (containing Neurone Q)} \longrightarrow \text{Neurone R} \longrightarrow \text{Effector S}$$

(a) (i) Name the type of neurone represented by P, Q, and R. (3 marks)
(ii) State the specific response produced if Effector S is the biceps muscle. (1 mark)

(b) Describe how nerve impulses are transmitted across the synapse between Neurone Q and Neurone R, and explain how the structural features of a synapse ensure that impulses only travel in one direction. (4 marks)

(c) A patient has sustained damage to the ventral root of the spinal nerve leading to their arm, while the dorsal root remains fully intact and functional.

Evaluate the effect of this damage on the patient's ability to:
- feel the painful hot stimulus on their hand
- withdraw their hand from the hot stimulus. (4 marks)
Show answer & marking scheme

Worked solution

### Part (a)
(i)
* **Neurone P:** This neurone carries electrical impulses from the receptor to the central nervous system (spinal cord), so it is a **sensory neurone**.
* **Neurone Q:** This neurone lies entirely within the spinal cord and connects the sensory and motor neurones, so it is a **relay neurone** (or interneurone).
* **Neurone R:** This neurone carries impulses away from the spinal cord to the effector organ, so it is a **motor neurone**.

(ii)
* When the effector (biceps muscle) receives the nerve impulse, it will contract, causing the arm to bend (flexion) and pull the hand away from the hot object.

### Part (b)
* **Transmission across the synapse:** An electrical impulse arriving at the axon terminal of the pre-synaptic neurone (Neurone Q) causes vesicles to release **chemical neurotransmitters**. These molecules **diffuse** across the tiny gap called the **synaptic cleft** and bind to specific **receptors** on the post-synaptic membrane of Neurone R, initiating a new electrical impulse.
* **One-way transmission:** This is guaranteed because neurotransmitter vesicles are **only** found in the pre-synaptic terminal, and receptor proteins are **only** found on the post-synaptic membrane. Thus, the signal cannot travel in reverse.

### Part (c)
* **Feeling the stimulus:** The dorsal root contains the axons of sensory neurones. Because the dorsal root is intact, sensory impulses from the temperature/pain receptors in the skin can travel up the sensory neurones, enter the spinal cord, and ascend to the brain. Therefore, the patient **will feel the pain**.
* **Withdrawing the hand:** The ventral root contains the axons of motor neurones. Because the ventral root is damaged, impulses generated in the spinal cord cannot reach the effector muscle (Effector S). Therefore, the patient **cannot withdraw their hand**.

Marking scheme

### Part (a)(i) [3 marks]
* Award 1 mark for correctly identifying Neurone P as the **sensory neurone**.
* Award 1 mark for correctly identifying Neurone Q as the **relay neurone** (accept: interneurone or connector neurone).
* Award 1 mark for correctly identifying Neurone R as the **motor neurone**.

### Part (a)(ii) [1 mark]
* Award 1 mark for **contraction** (accept: contracts / flexes the arm / pulls the arm away).
* *Reject: 'movement' without indicating contraction or direction of withdrawal.*

### Part (b) [4 marks]
* Award 1 mark for mentioning the release of **neurotransmitters** that **diffuse** across the synaptic cleft/gap.
* Award 1 mark for mentioning that they bind to specific **receptors** on the post-synaptic membrane.
* Award 1 mark for explaining that vesicles/neurotransmitters are only located on the pre-synaptic side.
* Award 1 mark for explaining that receptors are only located on the post-synaptic side.

### Part (c) [4 marks]
* Award 1 mark for stating that the patient **can feel the stimulus**.
* Award 1 mark for explaining that the sensory neurones / pathway / dorsal root is undamaged, allowing impulses to reach the brain.
* Award 1 mark for stating that the patient **cannot withdraw their hand / cannot perform the reflex action**.
* Award 1 mark for explaining that the motor neurones / pathway / ventral root is damaged, preventing motor impulses from reaching the effector muscle.
Question 4 · structured
9 marks
(a) High blood glucose levels are regulated by a specific hormone. Explain how this hormone reduces blood glucose levels and how this process can lead to long-term energy storage. (4) (b) An adult individual has a mass of 94.5 kg and a height of 1.75 m. (i) Calculate the Body Mass Index (BMI) for this individual. Show your working and state the unit. (3) (ii) Identify the weight category this individual falls into based on your calculation. State one potential health risk associated with this category. (2)
Show answer & marking scheme

Worked solution

(a) High blood glucose is detected by the pancreas, which secretes the hormone insulin. Insulin travels in the blood and binds to target cells in the liver and muscles, stimulating them to absorb glucose from the blood. Inside these cells, glucose is converted into glycogen for short-term storage. If blood glucose levels remain high, excess glucose is converted into lipids (fats) and stored in adipose tissue, which acts as long-term energy storage. (b)(i) To calculate BMI, use the formula: \( \text{BMI} = \frac{\text{mass}}{\text{height}^2} \). Substituting the values: \( \text{BMI} = \frac{94.5}{1.75^2} = \frac{94.5}{3.0625} = 30.857 \). Rounding to one decimal place gives 30.9. The correct unit is \( \text{kg/m}^2 \). (b)(ii) A BMI of 30.9 is 30 or greater, which classifies the individual in the 'obese' category. A major health risk associated with obesity is Type 2 diabetes (or coronary heart disease, high blood pressure, osteoarthritis, stroke).

Marking scheme

(a) [4 marks total]: 1 mark for identifying insulin is secreted by the pancreas; 1 mark for stating insulin increases glucose uptake by liver/muscle cells; 1 mark for stating glucose is converted to glycogen; 1 mark for stating excess glucose is converted to fat/lipids and stored in adipose tissue. (b)(i) [3 marks total]: 1 mark for correct working/substitution: \( \frac{94.5}{1.75^2} \); 1 mark for correct calculation: 30.9 (accept 30.86 to 31); 1 mark for correct unit: \( \text{kg/m}^2 \) (or \( \text{kg m}^{-2} \)). (b)(ii) [2 marks total]: 1 mark for correct weight category: Obese (allow ecf from calculated BMI); 1 mark for identifying a valid health risk: Type 2 diabetes / coronary heart disease / high blood pressure / cardiovascular disease / osteoarthritis / stroke (reject general 'heart disease' without specific context).
Question 5 · structured
14 marks

Hemophilia A is an inherited bleeding disorder caused by a deficiency in clotting factor VIII.

A clinician studied a family with a history of Hemophilia A. The family pedigree is described below:

  • Generation I: Individual 1 (unaffected male) and Individual 2 (unaffected female) have three children.
  • Generation II: Individual 3 (affected male), Individual 4 (unaffected female), and Individual 5 (unaffected male).
  • Individual 4 partners with Individual 6 (unaffected male) and they have one child.
  • Generation III: Individual 7 (affected male) is the child of Individual 4 and Individual 6.

Table 1 shows the laboratory blood clotting times (in minutes) for selected family members:

IndividualPhenotypeClotting Time (minutes)Individual 1Unaffected Male4Individual 3Affected Male25Individual 4Unaffected Female5

(a) Describe the normal mechanism of blood clotting in humans, including the roles of platelets, prothrombin, thrombin, fibrinogen, and fibrin. (4 marks)

(b) (i) Using evidence from the pedigree, state and explain whether the allele for Hemophilia A is dominant or recessive, and whether it is sex-linked or autosomal. (3 marks)

(b) (ii) Individual 4 and Individual 6 wish to have another child. Draw a genetic diagram (Punnett square) to determine the probability of them having an affected male child. Use the symbols \(X^H\) for the normal allele and \(X^h\) for the hemophilia allele. (4 marks)

(c) Explain the difference in blood clotting times between Individual 1 and Individual 3, and describe how this condition can be managed medically. (3 marks)

Show answer & marking scheme

Worked solution

(a) Normal blood clotting mechanism:

  • When a blood vessel is damaged, platelets aggregate at the site and release clotting factors (thromboplastin).
  • These factors trigger a cascade that converts the inactive plasma protein prothrombin into the active enzyme thrombin.
  • Thrombin then catalyses the conversion of the soluble plasma protein fibrinogen into insoluble protein fibers called fibrin.
  • Fibrin forms a dense mesh network across the wound that traps red blood cells and platelets, forming a solid clot to prevent further blood loss and pathogen entry.

(b) (i) Mode of inheritance:

  • Recessive: Unaffected parents (such as 1 and 2, or 4 and 6) have an affected child (3 and 7 respectively). This indicates the allele can be masked in heterozygous parents.
  • Sex-linked (X-linked): The condition only affects males in this pedigree, and unaffected father 6 has an affected son 7. This shows that the allele is carried on the X chromosome and the unaffected mother 4 is a carrier. Males only inherit one X chromosome (from their mother), so a single recessive allele on their X chromosome results in the disease.

(b) (ii) Genetic diagram:

  • Individual 4 (Carrier female): \(X^H X^h\)
  • Individual 6 (Unaffected male): \(X^H Y\)
\(X^H\)\(Y\)\(X^H\)\(X^H X^H\)
(Unaffected Female)\(X^H Y\)
(Unaffected Male)\(X^h\)\(X^H X^h\)
(Carrier Female)\(X^h Y\)
(Affected Male)

The offspring genotypes are \(X^H X^H\), \(X^H X^h\), \(X^H Y\), and \(X^h Y\).

The probability of having an affected male child is 1 in 4 (or 25% or 0.25) of all births.

(c) Clotting time difference and management:

  • Individual 3 has Hemophilia A, meaning they lack functional clotting factor VIII. Consequently, the chemical cascade that converts fibrinogen to insoluble fibrin is severely disrupted or absent, leading to a much slower clotting process (25 minutes vs 4 minutes in Individual 1).
  • This presents a severe risk of prolonged internal or external bleeding from minor injuries.
  • The condition is managed medically by regular intravenous infusions of recombinant/donor clotting factor VIII (replacement therapy).

Marking scheme

(a) [Maximum 4 marks]

  • MP1: Platelets release clotting factors / thromboplastin at the wound site;
  • MP2: Prothrombin is converted to active enzyme thrombin;
  • MP3: Thrombin catalyses conversion of soluble fibrinogen to insoluble fibrin;
  • MP4: Fibrin forms a mesh that traps red blood cells (forming a clot);

(b) (i) [Maximum 3 marks]

  • MP1: Recessive because unaffected parents have an affected child (e.g., 4 and 6 have 7) / trait skips a generation;
  • MP2: Sex-linked / X-linked because only males are affected in the pedigree / unaffected father (6) has an affected son (7) meaning mother (4) must be a carrier;
  • MP3: Explains mechanism: gene is on the X chromosome, so males (XY) expressing the single recessive allele are affected, while females (XX) need two copies or remain unaffected carriers;

(b) (ii) [Maximum 4 marks]

  • MP1: Correct parental genotypes identified: \(X^H X^h\) and \(X^H Y\);
  • MP2: Correct gametes shown in Punnett square;
  • MP3: Correct offspring genotypes derived: \(X^H X^H\), \(X^H X^h\), \(X^H Y\), \(X^h Y\);
  • MP4: Correct probability of an affected male child stated as 0.25 / 25% / 1 in 4 / 1/4 (Accept 50% of males if clearly qualified);

(c) [Maximum 3 marks]

  • MP1: Individual 3 lacks factor VIII, meaning they cannot efficiently convert soluble fibrinogen to insoluble fibrin / cannot form a clot quickly;
  • MP2: Leads to risk of internal bleeding (into joints/organs) or excessive blood loss from minor wounds;
  • MP3: Managed by regular intravenous injections / infusions of factor VIII / clotting factor replacement therapy;
Question 6 · structural
17 marks
Skeletal muscles are highly specialized tissues adapted for locomotion. (a) Describe four structural differences between a skeletal muscle fibre and a typical mammalian cell. (4) (b) Explain the roles of calcium ions and ATP in the contraction of a muscle myofibril. (5) (c) The elbow joint operates as a third-class lever during forearm flexion. A person holds a 50 N weight stationary in their hand. The distance from the elbow joint (pivot) to the hand (load) is 35 cm. The distance from the elbow joint to the attachment point of the biceps tendon (effort) is 4.0 cm. Calculate the force exerted by the biceps muscle to keep the arm stationary. Show your working and state the unit. (4) (d) Explain how the biceps and triceps muscles work antagonistically to lower the forearm (extend the arm) at the elbow joint. (4)
Show answer & marking scheme

Worked solution

Part (a): Skeletal muscle fibres are multinucleate, containing many nuclei, whereas typical mammalian cells have only one nucleus. Muscle fibres contain specialized contractile myofibrils containing actin and myosin filaments organized into sarcomeres. They have a specialized sarcoplasmic reticulum for calcium storage and a sarcolemma with T-tubules. They also contain a very high density of mitochondria to supply energy. Part (b): Calcium ions bind to troponin on the actin filaments, causing a conformational change in tropomyosin, which shifts to expose the myosin-binding sites on actin. ATP binds to the myosin head, causing it to detach from the actin filament. The hydrolysis of ATP to ADP and inorganic phosphate provides the energy required to cock the myosin head into its high-energy position. The myosin head then binds to the actin filament, forming a cross-bridge, and undergoes the power stroke to pull the actin filament. Part (c): Using the principle of moments in equilibrium: Clockwise moments = Counter-clockwise moments. (Load Force x Load Distance) = (Effort Force x Effort Distance). 50 N x 35 cm = Effort Force x 4.0 cm. 1750 = Effort Force x 4.0. Effort Force = 1750 / 4.0 = 437.5 N. Part (d): To lower the forearm, the triceps muscle contracts (acts as the agonist) and pulls on the ulna bone. At the same time, the biceps muscle relaxes (acts as the antagonist) and is stretched. This coordinated antagonistic action causes extension at the elbow joint, lowering the forearm.

Marking scheme

Part (a): Award 1 mark for each correct difference up to a maximum of 4 marks: 1. Muscle fibres are multinucleate / have multiple nuclei (accept typical cells have one). 2. Muscle fibres contain myofibrils / actin and myosin arranged in sarcomeres. 3. Muscle fibres have a specialized sarcoplasmic reticulum (for calcium storage). 4. Muscle fibres have sarcolemma with T-tubules (transverse tubules). 5. Muscle fibres contain a much higher density of mitochondria. Part (b): Award up to 5 marks: 1. Calcium ions bind to troponin. 2. Tropomyosin shifts / moves to expose myosin-binding sites on actin. 3. Myosin heads bind to actin to form cross-bridges. 4. ATP binds to the myosin head to cause detachment from actin. 5. Hydrolysis of ATP (to ADP + Pi) provides energy to cock / reset the myosin head. 6. Myosin undergoes a power stroke to pull the actin filament. Part (c): Award up to 4 marks: 1. Correct formula stated: Moment = Force x distance (or equivalent equilibrium expression) (1 mark). 2. Correct substitution: 50 x 35 = Force x 4 (or 50 x 0.35 = Force x 0.04) (1 mark). 3. Correct calculation of 437.5 (1 mark). 4. Correct unit: N / Newtons (1 mark). Part (d): Award up to 4 marks: 1. Triceps contracts ( agonist ). 2. Triceps pulls on the ulna. 3. Biceps relaxes ( antagonist ). 4. Causes extension at the elbow joint / lowers the forearm.
Question 7 · practical
8 marks
A student carries out an experiment to estimate the energy content of different dried food samples by burning them under a boiling tube containing water. (a) State two variables that the student should control to ensure that the results can be compared fairly. (2 marks) (b) In one test, the student burns a food sample of mass 0.50 g. The boiling tube contains 20.0 cm3 of water. The initial temperature of the water is 19.0 degrees C and the final temperature is 34.0 degrees C. Calculate the energy content of the food sample in joules per gram (J/g). Show your working. (Specific heat capacity of water = 4.2 J/g degrees C; density of water = 1.0 g/cm3). (3 marks) (c) After digestion, the products such as glucose are absorbed into the blood. Explain how active transport is used to absorb glucose from the lumen of the small intestine into the epithelial cells, even when the concentration of glucose in the lumen is lower than inside the cells. (3 marks)
Show answer & marking scheme

Worked solution

(a) To ensure a fair comparison, the student should control: 1. The distance between the burning food sample and the bottom of the boiling tube. 2. The volume of water in the boiling tube (which must remain constant across all tests). Other acceptable control variables include using the same degree of shielding to reduce heat loss, and ensuring the food sample is completely burned. (b) Step 1: Calculate the mass of water. Since density is 1.0 g/cm3, 20.0 cm3 of water has a mass of 20.0 g. Step 2: Calculate the temperature increase. Temperature change = 34.0 - 19.0 = 15.0 degrees C. Step 3: Calculate the energy absorbed by the water. Energy = mass of water * specific heat capacity * temperature rise = 20.0 * 4.2 * 15.0 = 1260 J. Step 4: Calculate the energy per gram of food. Energy per gram = 1260 J / 0.50 g = 2520 J/g. (c) Active transport absorbs glucose against its concentration gradient, moving it from an area of lower concentration in the intestinal lumen to an area of higher concentration inside the epithelial cells. This process requires energy, which is supplied in the form of ATP produced during aerobic respiration. Specific carrier proteins located in the cell membrane bind to the glucose molecules and pump them across the membrane into the cell.

Marking scheme

(a) Award 1 mark for each correct control variable up to a maximum of 2 marks. Acceptable answers: Distance between the burning food and the boiling tube; Volume of water used; Complete burning of the food sample; Presence of draft shields/insulation. Reject: Mass of food sample (as this is divided out in the calculation). (b) Award 1 mark for calculating the correct temperature rise (15.0 degrees C) or showing the correct formula for heat energy absorbed (20.0 * 4.2 * 15.0 = 1260 J). Award 1 mark for dividing the energy by the mass of the food sample (1260 / 0.50). Award 1 mark for the correct final answer of 2520 (J/g). (c) Award 1 mark for stating that glucose is moved against/up its concentration gradient (or from low to high concentration). Award 1 mark for stating that active transport requires energy in the form of ATP. Award 1 mark for mentioning the role of specific carrier proteins in the cell membrane.
Question 8 · physiological_explanation
11 marks
Tissue fluid surrounds the cells of the body and facilitates the exchange of substances. (a) Explain how high hydrostatic pressure at the arteriole end of a capillary network leads to the formation of tissue fluid. (3 marks) (b) Describe how excess tissue fluid that does not re-enter the blood capillaries directly is returned to the blood circulatory system. (3 marks) (c) Kwashiorkor is a nutritional deficiency disease caused by a severe lack of protein in the diet, leading to a low concentration of plasma proteins. Explain how a low concentration of plasma proteins results in the swelling of the abdomen due to the accumulation of tissue fluid (edema). (5 marks)
Show answer & marking scheme

Worked solution

(a) High hydrostatic pressure at the arteriole end of the capillary is generated by the contraction of the heart. This pressure is greater than the osmotic pressure drawing water in, forcing water and small dissolved molecules (such as glucose, amino acids, and oxygen) out of the capillary walls through tiny pores. Large molecules, such as plasma proteins and red blood cells, are too large to pass through and remain in the blood. (b) Excess tissue fluid that does not diffuse back into the capillary enters the lymphatic capillaries, where it is called lymph. The lymph is transported through lymphatic vessels, which contain valves to prevent backflow, and is propelled by the contraction of surrounding skeletal muscles. The lymphatic vessels eventually empty the lymph back into the blood system at the subclavian veins near the heart. (c) A lack of dietary protein reduces the liver's production of plasma proteins, such as albumin, lowering their concentration in the blood plasma. This increases the water potential of the blood (making the solute potential less negative), which decreases the osmotic pressure gradient between the tissue fluid and the blood. As a result, less water is reabsorbed by osmosis back into the capillary at the venule end. Because tissue fluid is formed faster than it can be reabsorbed or drained by the lymphatic system, fluid accumulates in the tissue spaces of the abdomen, causing edema.

Marking scheme

Part (a) [Max 3 marks]: 1. Credit for identifying that high hydrostatic pressure is generated by heart contraction / high arterial pressure (1 mark). 2. Credit for explaining that hydrostatic pressure exceeds osmotic pressure / tissue fluid pressure (1 mark). 3. Credit for stating that water and small solutes are forced out of capillary pores/gaps, while large proteins/cells remain (1 mark). Part (b) [Max 3 marks]: 1. Credit for stating that excess tissue fluid drains into lymphatic capillaries / lymph vessels (1 mark). 2. Credit for describing transport through lymph vessels using valves to prevent backflow / skeletal muscle contraction (1 mark). 3. Credit for stating that lymph is returned to blood at the subclavian veins (1 mark). Part (c) [Max 5 marks]: 1. Credit for linking protein deficiency to low concentrations of plasma proteins / albumin synthesized by the liver (1 mark). 2. Credit for stating this raises the water potential of the blood plasma / decreases plasma osmotic pressure (1 mark). 3. Credit for explaining that the water potential gradient between tissue fluid and blood is reduced (1 mark). 4. Credit for explaining that less water is reabsorbed by osmosis back into the capillaries at the venule end (1 mark). 5. Credit for concluding that accumulation of fluid occurs because rate of formation exceeds rate of reabsorption/drainage (1 mark).

Paper 2 (4HB1/02)

Answer all questions. Show all the steps in any calculations. Candidates may use a calculator.
6 Question · 75 marks
Question 1 · structured
10 marks
The table shows the success rates of three different contraceptive methods when used perfectly (perfect use) and when used in everyday life (typical use).

| Contraceptive method | Perfect use success rate (%) | Typical use success rate (%) | Main mode of action |
| :--- | :---: | :---: | :--- |
| Male condom | 98.0 | 82.0 | Physical barrier to sperm |
| Combined oral pill | 99.7 | 91.0 | Prevents ovulation |
| Vasectomy | 99.9 | 99.8 | Surgical prevention of sperm release |

(a) Explain why there is a much larger difference between perfect use and typical use success rates for the male condom compared to a vasectomy. (2)

(b) Describe how the hormones in the combined oral contraceptive pill prevent pregnancy. (3)

(c) Describe the surgical procedure of a vasectomy and explain how this prevents fertilisation. (2)

(d) A clinic advises a young couple who are not in a long-term monogamous relationship on contraception.
Evaluate which method or combination of methods from the table they should use. Give reasons for your answer. (3)
Show answer & marking scheme

Worked solution

(a) The male condom requires correct application and preparation every single time sexual intercourse occurs, making it highly susceptible to human errors such as tearing, incorrect sizing, slipping, or delayed application. Conversely, a vasectomy is a one-time surgical procedure that does not require any compliance, preparation, or memory from the user during intercourse once it is confirmed successful.

(b) The combined oral pill contains synthetic progesterone and estrogen. These hormones inhibit the secretion of Follicle-Stimulating Hormone (FSH) and Luteinising Hormone (LH) from the pituitary gland. Without FSH, follicles do not mature in the ovary; without LH, ovulation (the release of a mature egg) is prevented. Additionally, the progesterone thickens cervical mucus, making it difficult for sperm to enter the uterus.

(c) During a vasectomy, the sperm ducts (vas deferens) are cut and tied or sealed. This physically prevents sperm produced in the testes from traveling to mix with fluid from the seminal vesicles and prostate gland. Consequently, semen ejaculated contains no sperm, making fertilisation of an egg impossible.

(d) The couple should use a combination of the male condom and the combined oral pill. The male condom is the only method listed that acts as a physical barrier to prevent the exchange of bodily fluids, thus protecting against sexually transmitted infections (STIs), which is essential since they are not in a monogamous relationship. Combining it with the oral pill ensures a very high level of contraceptive protection (91% typical success rate for the pill compared to 82% for the condom alone) to prevent unwanted pregnancy if the condom breaks or slips. A vasectomy is not recommended as it is permanent and does not protect against STIs.

Marking scheme

(a)
- MP1: Male condom requires consistent, active user compliance / is subject to human errors (e.g. tearing, slipping, incorrect application) (1)
- MP2: Vasectomy is a permanent, one-time surgical procedure that does not require user action or compliance during intercourse (1)

(b)
- MP1: Contains estrogen and progesterone (1)
- MP2: Prevents/inhibits the release of FSH and LH from the pituitary gland (1)
- MP3: Prevents follicle maturation and ovulation OR thickens cervical mucus to block sperm (1)

(c)
- MP1: The sperm ducts / vas deferens are cut and tied/sealed (1)
- MP2: Sperm cannot leave the testes to enter the urethra/mix with semen, preventing fertilisation (1)

(d)
- MP1: Recommend the use of both the male condom and the combined oral pill (1)
- MP2: Condom protects against STIs / pathogens (essential for non-monogamous relationships) (1)
- MP3: The pill provides significantly higher/redundant contraceptive reliability to prevent pregnancy if condom fails (1)
Question 2 · structured
13 marks
An individual tracks their daily food intake. The table shows the masses of macronutrients in their daily diet.

| Macromolecule | Mass consumed (g) | Energy value (kJ/g) |
| :--- | :--- | :--- |
| Carbohydrate | 320 | 17 |
| Protein | 85 | 17 |
| Lipid | 65 | 39 |

(a) (i) Calculate the total energy intake of this individual in kJ. Show your working. (3)
(ii) Calculate the percentage of the total energy intake that is provided by lipids. Give your answer to 1 decimal place. Show your working. (2)

(b) Describe how the individual can chemically test a sample of their food to confirm the presence of:
(i) reducing sugars, such as glucose. (2)
(ii) proteins. (2)

(c) Explain why a high-carbohydrate diet is beneficial for an athlete preparing for a long-distance run, and describe how excess carbohydrate is stored in the human body. (4)
Show answer & marking scheme

Worked solution

**(a) Calculations**

(i) Calculate individual energy values first:
* Energy from carbohydrates: \(320 \text{ g} \times 17 \text{ kJ/g} = 5440 \text{ kJ}\)
* Energy from proteins: \(85 \text{ g} \times 17 \text{ kJ/g} = 1445 \text{ kJ}\)
* Energy from lipids: \(65 \text{ g} \times 39 \text{ kJ/g} = 2535 \text{ kJ}\)

Total Energy = \(5440 + 1445 + 2535 = 9420 \text{ kJ}\)

(ii) Find the percentage of total energy from lipids:
* \(\text{Percentage} = \left(\frac{2535}{9420}\right) \times 100 = 26.9108...\%\)
* Rounded to 1 decimal place = \(26.9\%\)

---

**(b) Chemical Tests**

(i) Reducing sugars:
* Add Benedict's reagent/solution to the food sample.
* Heat the mixture in a water bath (above 65 °C / boiling) for several minutes.
* A color change from blue to green/yellow/orange/brick-red indicates the presence of reducing sugars.

(ii) Proteins:
* Add Biuret reagent (or sodium hydroxide and copper sulfate solution) to the food sample.
* A color change from light blue to purple/violet/lilac indicates the presence of protein (no heating required).

---

**(c) Athlete Diet and Carbohydrate Storage**

* Carbohydrates are digested and broken down into glucose molecules.
* Glucose is the primary respiratory substrate used in aerobic respiration to release energy/ATP needed for muscle contraction during exercise.
* Excess glucose is converted into glycogen.
* This conversion is stimulated by the hormone insulin and the glycogen is stored in the liver and skeletal muscle tissues.

Marking scheme

**(a) (i) Total Energy Calculation (3 Marks max):**
* **1 Mark** for calculating at least one correct individual nutrient energy contribution (e.g., Carb = 5440 kJ, Protein = 1445 kJ, or Lipid = 2535 kJ).
* **1 Mark** for adding all three energy values together: \(5440 + 1445 + 2535\).
* **1 Mark** for the correct final total: **9420 kJ** (accept correct unit if omitted in question, but reject if incorrect units like J or kcal are written alongside the number).

**(a) (ii) Percentage from Lipids (2 Marks max):**
* **1 Mark** for the correct fraction set up using candidate's answer from (a)(i): \(\frac{\text{lipid energy}}{\text{total energy}} \times 100\) (e.g., \(\frac{2535}{9420} \times 100\)).
* **1 Mark** for the correct final answer rounded to 1 decimal place: **26.9%** (allow error carried forward from incorrect (a)(i)).

**(b) (i) Test for Reducing Sugars (2 Marks max):**
* **1 Mark** for stating the addition of Benedict's reagent and heating (water bath over 65 °C).
* **1 Mark** for stating positive color change: blue to green / yellow / orange / brick-red (reject any other initial color or incorrect final color like purple).

**(b) (ii) Test for Proteins (2 Marks max):**
* **1 Mark** for stating the addition of Biuret reagent (or copper sulfate and sodium hydroxide).
* **1 Mark** for stating positive color change: blue to violet / purple / lilac.

**(c) Physiological Explanation (4 Marks max):**
* **1 Mark** for explaining that carbohydrates are broken down to glucose for aerobic respiration to release energy/ATP.
* **1 Mark** for linking this energy to muscle contractions / physical exertion over long duration.
* **1 Mark** for stating that excess glucose is converted into insoluble glycogen (under insulin control).
* **1 Mark** for stating that glycogen is stored in the liver / muscles.
Question 3 · structured
13 marks
The transmission of nerve impulses across synapses is a highly regulated chemical process. (a) Describe how an electrical impulse is transmitted across a synapse from the presynaptic neurone to the postsynaptic neurone. (5 marks) (b) Compound Z is a drug that blocks the voltage-gated calcium ion channels in the presynaptic membrane. Explain the effect of Compound Z on the transmission of nerve impulses across the synapse. (4 marks) (c) Organophosphates are chemicals used in some pesticides that inhibit the enzyme acetylcholinesterase. (i) State the normal role of acetylcholinesterase in synaptic transmission. (2 marks) (ii) Explain the consequences of organophosphate exposure on the postsynaptic neurone and effector muscles. (2 marks)
Show answer & marking scheme

Worked solution

Part (a): The arriving action potential depolarises the presynaptic membrane, opening voltage-gated calcium channels. Calcium influx causes vesicle translocation and exocytosis of neurotransmitters (such as acetylcholine) into the synaptic cleft. Diffusion across the cleft allows neurotransmitters to bind to ligand-gated sodium channels on the postsynaptic membrane, causing sodium entry and depolarisation. Part (b): Blocking calcium channels prevents calcium influx upon impulse arrival. Without calcium, vesicle fusion cannot occur, meaning no neurotransmitters are released into the cleft, preventing postsynaptic action potential generation. Part (c)(i): Acetylcholinesterase breaks down acetylcholine into choline and acetate to clear the cleft and allow the postsynaptic membrane to repolarise and reset. Part (c)(ii): Inhibition of this enzyme leads to accumulation of acetylcholine in the cleft, causing continuous receptor activation, relentless firing of action potentials, and sustained muscle contraction (tetany) or paralysis.

Marking scheme

Part (a) [Max 5 marks]: 1. Impulse causes calcium channels to open / calcium ions (\( \text{Ca}^{2+} \)) enter presynaptic neurone. 2. Calcium influx causes vesicles to move towards / fuse with the presynaptic membrane. 3. Neurotransmitter released into the synaptic cleft by exocytosis. 4. Neurotransmitter diffuses across the synaptic cleft. 5. Neurotransmitter binds to specific receptors on the postsynaptic membrane. 6. Sodium channels open / depolarisation occurs / new action potential generated. Part (b) [Max 4 marks]: 1. Calcium ions cannot enter presynaptic knob. 2. Vesicles cannot fuse with presynaptic membrane / no exocytosis. 3. No neurotransmitter released into the synaptic cleft / no diffusion. 4. Receptors on postsynaptic membrane are not activated. 5. No depolarisation / no action potential generated in postsynaptic neurone. Part (c)(i) [Max 2 marks]: 1. Breaks down / hydrolyses acetylcholine (into choline and ethanoic acid). 2. Prevents continuous stimulation of receptors / allows postsynaptic membrane to repolarise. Part (c)(ii) [Max 2 marks]: 1. Acetylcholine accumulates in the cleft. 2. Leads to continuous stimulation / repeated depolarisation of postsynaptic neurone or muscle membrane. 3. Causes continuous muscle contraction / rapid muscle spasms / tetany / paralysis.
Question 4 · Structured
12 marks
Hemophilia A is a sex-linked recessive genetic disorder caused by a mutation in the gene for clotting factor VIII, located on the X chromosome.

(a) Explain why males are more likely to inherit and express sex-linked recessive disorders, such as hemophilia A, than females. (3)

(b) A pedigree analysis shows that a female (Individual 4) has normal blood clotting, but her father had hemophilia. She partners with a male (Individual 3) who has hemophilia.
Draw a genetic diagram to show the genotypes of the parents, the gametes they can produce, the possible genotypes of their offspring, and state the probability that their first child will be a female with hemophilia.
Use the symbols:
\(X^H\) = allele for normal blood clotting
\(X^h\) = allele for hemophilia
\(Y\) = Y chromosome (no allele for blood clotting) (5)

(c) The genetic defect in Hemophilia A is often a single nucleotide substitution mutation. Explain how a single nucleotide substitution in the DNA sequence can result in a completely non-functional clotting factor VIII protein. (4)
Show answer & marking scheme

Worked solution

(a) Males only have one X chromosome in their sex chromosome pair (\(XY\)), whereas females have two (\(XX\)). If a male inherits a single recessive disease-causing allele on his X chromosome from his mother, he does not have a second X chromosome with a dominant healthy allele to mask it. Therefore, he will always express the disorder. Females must inherit two copies of the recessive allele (one from each parent) to express the phenotype; if they inherit only one, the dominant allele on their other X chromosome masks the condition, making them unaffected carriers.

(b)
- Parental genotypes: Female (Individual 4) is \(X^H X^h\) (she must be a carrier because her father was hemophilic and could only pass on his \(X^h\) chromosome). Male (Individual 3) is \(X^h Y\).
- Gametes: Female produces \(X^H\) and \(X^h\); Male produces \(X^h\) and \(Y\).
- Punnett square / offspring genotypes:
- \(X^H X^h\) (carrier female)
- \(X^h X^h\) (hemophilic female)
- \(X^H Y\) (normal male)
- \(X^h Y\) (hemophilic male)
- Probability of a hemophilic daughter: 1 in 4, 0.25, or 25%.

(c) A single nucleotide substitution changes one base in the DNA triplet code, which changes the corresponding codon in the transcribed mRNA. During translation, this altered codon may code for a different amino acid (missense mutation) or result in a premature stop codon (nonsense mutation).
If it codes for a different amino acid, the primary structure (amino acid sequence) of the clotting factor VIII polypeptide is changed. This alters the chemical interactions (such as hydrogen, ionic, or disulfide bonds) during protein folding, leading to a changed tertiary structure (3D shape). Consequently, the protein's active site or functional domains are disrupted, rendering the clotting factor completely non-functional.

Marking scheme

Part (a) (Maximum 3 marks):
- MP1: Males are hemizygous / have only one X chromosome / have \(XY\) genotype (1)
- MP2: Therefore, males only require one copy of the recessive allele to express the phenotype (1)
- MP3: Females have two X chromosomes / \(XX\) genotype, so they need two copies of the recessive allele to show the condition / a dominant allele on the second X chromosome masks the recessive allele (1)

Part (b) (Maximum 5 marks):
- MP1: Correct parental genotypes: Female is \(X^H X^h\) and Male is \(X^h Y\) (1)
- MP2: Correct gametes identified for both parents: \(X^H\) and \(X^h\) from mother, \(X^h\) and \(Y\) from father (1)
- MP3: Correct offspring genotypes shown (\(X^H X^h\), \(X^h X^h\), \(X^H Y\), \(X^h Y\)) (1)
- MP4: Phenotypes correctly matched to genotypes (carrier female/normal clotting female, affected female, normal clotting male, affected male) (1)
- MP5: Correct probability of having a hemophilic daughter identified as 0.25 / 25% / 1 in 4 / 1/4 (1)
[Note: Reject 50% unless specifically qualified as '50% of daughters']

Part (c) (Maximum 4 marks):
- MP1: Substitution changes a base in a DNA triplet / changes a codon in mRNA (1)
- MP2: Leads to a different amino acid being incorporated (during translation) OR creates a premature stop codon (1)
- MP3: Alters the primary structure / sequence of amino acids in the polypeptide chain (1)
- MP4: Alters the folding / tertiary structure / 3D shape of the protein (due to different bonding) (1)
- MP5: Changes the shape of the active/functional site, making the factor VIII protein unable to function in the blood clotting cascade (1)
Question 5 · long response
12 marks
The circulatory system consists of three main types of blood vessels: arteries, veins, and capillaries. A student investigates how the structure of these vessels relates to blood pressure and hemodynamics.

(a) Describe three structural differences between an artery and a vein. (3 marks)

(b) The table below shows the average blood pressure and total cross-sectional area of different blood vessels in the human systemic circulation.

| Vessel Type | Average Blood Pressure (mmHg) | Total Cross-Sectional Area (cm\(^2\)) |
|---|---|---|
| Aorta | 100 | 4.5 |
| Arteries | 90 | 20.0 |
| Arterioles | 60 | 400.0 |
| Capillaries | 30 | 4500.0 |
| Venules | 15 | 250.0 |
| Veins | 10 | 80.0 |
| Vena Cava | 2 | 18.0 |

(i) State the overall relationship shown by the data between the sequence of blood vessels from the aorta to the vena cava and the average blood pressure. (2 marks)

(ii) Calculate the percentage decrease in average blood pressure as blood moves from the aorta to the capillaries. Show your working. (3 marks)

(c) Explain how the slow velocity of blood flow in capillaries, combined with their structural adaptations, maximizes the exchange of substances between blood and tissues. (4 marks)
Show answer & marking scheme

Worked solution

Part (a):
Arteries have thicker walls with more smooth muscle and elastic fibres to withstand high pressures. They have a narrower lumen to maintain high blood pressure. Veins have thinner walls, a wider lumen to reduce resistance to low-pressure flow, and contain watch-pocket (semilunar) valves to prevent the backflow of blood.

Part (b)(i):
As blood travels further away from the heart from the aorta to the vena cava, the average blood pressure continuously decreases. The most significant decrease in pressure occurs in the high-resistance arterioles (dropping from 90 mmHg to 60 mmHg, and further to 30 mmHg in capillaries).

Part (b)(ii):
Average blood pressure in the Aorta = 100 mmHg
Average blood pressure in Capillaries = 30 mmHg
Decrease in pressure = \(100 - 30 = 70\) mmHg
Percentage decrease = \(\frac{70}{100} \times 100 = 70\%\)

Part (c):
1. Slow velocity of blood flow allows sufficient time for the diffusion/active transport of nutrients, gases (oxygen, carbon dioxide), and metabolic wastes to occur.
2. Capillary walls are only one-cell thick (composed of a single layer of flattened endothelial cells), which minimizes the diffusion distance.
3. Capillaries form highly branched networks, providing a massive total surface area for exchange.
4. Capillary walls are highly permeable, allowing water and small solutes to pass through readily into tissue fluid.

Marking scheme

Part (a) (Max 3 marks):
- Award 1 mark for: Arteries have a thicker muscular / elastic wall OR veins have a thinner wall.
- Award 1 mark for: Arteries have a narrower lumen OR veins have a wider lumen.
- Award 1 mark for: Veins have valves (to prevent backflow) OR arteries do not have valves (except semilunar at start).

Part (b)(i) (Max 2 marks):
- Award 1 mark for identifying that blood pressure continuously/gradually decreases along the pathway from the aorta to the vena cava.
- Award 1 mark for pointing out a specific detail, e.g., the largest pressure drop occurs in the arterioles, OR pressure is at its lowest (2 mmHg) in the vena cava.

Part (b)(ii) (Max 3 marks):
- Award 1 mark for finding the pressure difference: \(100 - 30 = 70\) (mmHg).
- Award 1 mark for dividing by the initial pressure (100 mmHg) and multiplying by 100: \(\frac{70}{100} \times 100\).
- Award 1 mark for the correct calculation: 70%.

Part (c) (Max 4 marks):
- Award 1 mark for linking slow velocity of blood flow to increased time/duration for diffusion/exchange to take place.
- Award 1 mark for stating that capillary walls are one cell thick.
- Award 1 mark for explaining that thin walls provide a short diffusion pathway.
- Award 1 mark for linking the vast capillary network / large total cross-sectional area to a large total surface area for exchange.
- Award 1 mark for mentioning capillary wall permeability/pores allowing substances to escape easily into tissues.
Question 6 · subjective
15 marks
A student investigates the effect of bile salts on the digestion of lipids in milk.

They set up six test tubes, each containing:
* 5 cm³ of full-fat milk
* 1 cm³ of sodium carbonate solution (an alkali)
* 5 drops of phenolphthalein indicator (which is pink in alkaline conditions and colorless in neutral/acidic conditions)
* 1 cm³ of lipase solution
* 1 cm³ of different concentrations of bile salts solution (0%, 1%, 2%, 3%, 4%, and 5%).

The student recorded the time taken for the mixture in each test tube to turn from pink to colorless. The results are shown in the table below.

| Concentration of bile salts (%) | Time taken for pink color to disappear (s) |
| :--- | :--- |
| 0 | 340 |
| 1 | 180 |
| 2 | 110 |
| 3 | 75 |
| 4 | 50 |
| 5 | 45 |

(a) Explain why sodium carbonate is added to the milk before adding lipase. (2)

(b) State the independent variable and two variables that must be controlled in this investigation. (3)

(c) Calculate the rate of reaction for the 2% bile salts concentration using the formula:

$$\text{Rate} = \frac{1}{\text{time taken}}$$

Give your answer to 3 significant figures. Show your working. (2)

(d) Explain the difference in times taken for the reaction to complete between 0% bile salts and 3% bile salts. Use your knowledge of digestion to support your answer. (4)

(e) Describe how the student could modify this method to measure the rate of lipid digestion more objectively and accurately. (2)

(f) Milk also contains protein. State the chemical test used to show the presence of protein in milk and the color change of a positive result. (2)
Show answer & marking scheme

Worked solution

**(a)**
* Sodium carbonate is an alkali, so it raises the pH and turns the phenolphthalein indicator pink at the start (1).
* As lipase digests lipids into fatty acids, the mixture becomes acidic/neutralizes, allowing the color change to be detected as the pH drops below the threshold of phenolphthalein (1).

**(b)**
* Independent variable: Concentration of bile salts (1).
* Control variables (any two from): Temperature (e.g. using a water bath), volume of milk, concentration of lipase, volume of lipase, volume of sodium carbonate solution, or volume of phenolphthalein indicator (2).

**(c)**
* Working: \(\text{Rate} = \frac{1}{110}\) (1)
* \(\text{Rate} = 0.00909\) \(\text{s}^{-1}\) (or \(9.09 \times 10^{-3}\) \(\text{s}^{-1}\)) (1)

**(d)**
* The time taken is much shorter at 3% bile salts (75 s) than at 0% bile salts (340 s) because bile salts emulsify lipids (1).
* This physical digestion breaks large lipid droplets into many smaller droplets (1).
* This increases the surface area-to-volume ratio of the lipid substrate (1).
* Consequently, there are more frequent successful collisions between lipase active sites and lipid molecules, leading to a faster rate of chemical digestion into fatty acids and glycerol (1).

**(e)**
* Use a pH probe / pH sensor connected to a datalogger (1).
* This allows the continuous, objective measurement of pH changes over time instead of relying on subjective human observation of the color change of the indicator (1).

**(f)**
* Test: Biuret reagent / Biuret test (1).
* Positive result color change: blue to lilac / purple / violet (1).

Marking scheme

**(a)**
* **MP1:** To make the solution alkaline / turn phenolphthalein pink at the start (1)
* **MP2:** So that the production of fatty acids (which lowers pH) can be monitored/detected (1)

**(b)**
* **MP1:** Concentration of bile salts (1)
* **MP2 & MP3:** Any two from: Temperature / volume of milk / volume of lipase / concentration of lipase / volume of sodium carbonate (2) [Reject: amount of enzyme/milk]

**(c)**
* **MP1:** Correct substitution: \(1 / 110\) (1)
* **MP2:** Correct evaluation to 3 significant figures: \(0.00909\) or \(9.09 \times 10^{-3}\) (1) [Allow 1 mark for correct calculation with wrong sig figs, e.g., 0.009]

**(d)**
* **MP1:** Digestion is faster with 3% bile salts / times are shorter (1)
* **MP2:** Bile salts emulsify lipids/fats (1) [Reject: 'emulsifies lipase' / 'breaks down fat chemically']
* **MP3:** Large droplets are broken into smaller droplets, increasing surface area (1) [Reject: 'molecules' instead of 'droplets']
* **MP4:** Increases the rate of lipase action / faster production of fatty acids (1)

**(e)**
* **MP1:** Use a pH meter / pH probe / datalogger (1)
* **MP2:** Removes human error/subjectivity in judging color change / allows continuous quantitative recording (1)

**(f)**
* **MP1:** Biuret test / copper sulfate and sodium hydroxide (1)
* **MP2:** (Turns from blue to) lilac / purple / violet (1) [Reject: pink / red / blue]

Wondering how well you actually know this?

Thinka is an AI practice app for DSE students — unlimited questions, instant auto-marking, and detailed step-by-step solutions. 100,000+ students use it to confirm they actually know it, not just think they do.

Want more questions like this? Practice unlimited on Thinka — instant answers included.

Start Practising Free