Human Biology 4HB1 | Paper 2 Mock 01 | Written Paper 2
Question 1 Report
Fig. 1 shows the antibody concentration in the blood of a student who received two injections containing antigens from the same disease-causing bacterium. The student did not have the disease before the first injection.
(a) Complete the statements about the first injection.
The antigens cause lymphocytes to make __________. These substances are carried around the body in the __________. [2]
(b) Describe the change in antibody concentration after the first injection, shown in Fig. 1. [3]
(c) Explain why the second injection gives the student better immunity to this disease. [4]
(d) Use Fig. 1 to state the approximate day on which the second injection was given. [1]
(a) Antigens cause lymphocytes to make antibodies. Antibodies are carried around the body in the blood. [2]
(b) After the first injection, antibody concentration remains low initially, then rises to a small peak, and falls again before the second injection. [3]
(c) The first injection produces memory lymphocytes, which remain in the body. On a second exposure to the same antigen, these cells divide rapidly and make antibodies rapidly. More antibodies are produced, giving a higher concentration and faster protection. [4]
(d) The second injection was given at approximately day 38. [1]
Answer Details
(a) Antigens cause lymphocytes to make antibodies. Antibodies are carried around the body in the blood. [2]
(b) After the first injection, antibody concentration remains low initially, then rises to a small peak, and falls again before the second injection. [3]
(c) The first injection produces memory lymphocytes, which remain in the body. On a second exposure to the same antigen, these cells divide rapidly and make antibodies rapidly. More antibodies are produced, giving a higher concentration and faster protection. [4]
(d) The second injection was given at approximately day 38. [1]
Question 2 Report
[Diagram: B&W exam-style line-art schematic on white: football goalkeeper lower leg nervous pathway, receptor in foot connected by sensory neurone to spinal cord and brain, motor neurone from spinal cord to cal]
Fig. 1 shows a simplified pathway controlling movement of the lower leg when a football goalkeeper kicks a ball. A scan later showed damage to the insulating layer around some neurones in the player's leg. The player could still detect touch but the leg muscle contracted slowly.
(a) Name the type of neurone carrying impulses from the spinal cord to the leg muscle. [1]
(b) Describe the direction of travel of an impulse from the receptor in the foot to the brain. [2]
(c) Complete the sequence for a reflex arc by writing the missing terms.
receptor → sensory neurone → ............ → motor neurone → effector [2]
(d) Explain how damage to the insulating layer can reduce the speed of movement of the leg. [5]
(a) The neurone carrying impulses from the spinal cord to the leg muscle is a motor neurone. [1]
(b) The impulse travels from the foot receptor along a sensory neurone to the spinal cord. It then travels through neurones to the brain. [2]
(c) The completed sequence is:
receptor → sensory neurone → relay neurone → motor neurone → effector
The relay neurone is in the spinal cord, which is part of the central nervous system. [2]
(d) The insulating layer normally prevents electrical signal loss and increases the speed of impulse conduction. Damage to it slows impulses along the neurone. Motor impulses then arrive at the leg muscle later, so the muscle contracts later and movement or reaction when kicking is slower. [5]
Answer Details
(a) The neurone carrying impulses from the spinal cord to the leg muscle is a motor neurone. [1]
(b) The impulse travels from the foot receptor along a sensory neurone to the spinal cord. It then travels through neurones to the brain. [2]
(c) The completed sequence is:
receptor → sensory neurone → relay neurone → motor neurone → effector
The relay neurone is in the spinal cord, which is part of the central nervous system. [2]
(d) The insulating layer normally prevents electrical signal loss and increases the speed of impulse conduction. Damage to it slows impulses along the neurone. Motor impulses then arrive at the leg muscle later, so the muscle contracts later and movement or reaction when kicking is slower. [5]
Question 3 Report
Fig. 1 shows a short section of a DNA molecule built by a student using coloured model pieces. Each rung represents a pair of bases. The student removes the colours before asking another student to identify how the bases are paired. DNA in the nucleus of nearly all human body cells carries the instructions for making proteins.
(a) Name the structure labelled X. [1]
(b) Complete the base-pair rule: adenine pairs with __________ and cytosine pairs with __________. [2]
(c) Use Fig. 1 to write the sequence of bases on the right-hand strand, from top to bottom. [1]
(d) Explain how the base sequence in DNA can affect a feature of a person. [3]
(a) X labels the DNA molecule, which has a double helix structure. [1]
(b) Adenine pairs with thymine, and cytosine pairs with guanine. These are complementary base pairs. [2]
(c) Reading the right-hand strand from top to bottom gives T G C A. [1]
(d) A sequence of DNA bases forms a gene, or contains a genetic code. The code determines the sequence of amino acids used to make a protein. Different proteins affect cell structure or function, so they can affect a person's features. [3]
Answer Details
(a) X labels the DNA molecule, which has a double helix structure. [1]
(b) Adenine pairs with thymine, and cytosine pairs with guanine. These are complementary base pairs. [2]
(c) Reading the right-hand strand from top to bottom gives T G C A. [1]
(d) A sequence of DNA bases forms a gene, or contains a genetic code. The code determines the sequence of amino acids used to make a protein. Different proteins affect cell structure or function, so they can affect a person's features. [3]
Question 4 Report
During a practical lesson, a student compares oxygen uptake by soaked, germinating pea seeds with that of boiled seeds. Each flask is kept in a water bath at 25 degrees C. Fig. 1 shows the sealed apparatus used for the flask containing living seeds. The capillary has an internal cross-sectional area of 0.50 mm2. After 15 minutes, the coloured drop labelled P has moved 24 mm towards the flask.
(a) Name the parts labelled P and Q. [2]
(b) Use the movement of P and the cross-sectional area of the capillary to calculate the volume of oxygen taken up each minute. Give your answer in mm3 per minute. [3]
(c) Explain why the coloured drop moves towards the flask when the pea seed cells respire. [3]
(d) Suggest two features that would make the boiled-seed flask a valid control for this investigation. [2]
(e) State the type of respiration occurring in the germinating pea cells. [1]
(a) P is the coloured liquid drop, or marker drop. Q is soda lime, which absorbs carbon dioxide. [2]
(b) The volume of oxygen taken up in 15 minutes is:
\[\text{volume}=\text{distance moved}\times\text{cross-sectional area}\]
\[24\text{ mm}\times0.50\text{ mm}^2=12\text{ mm}^3\]
\[\text{rate}=\frac{12\text{ mm}^3}{15\text{ min}}=0.80\text{ mm}^3\text{ min}^{-1}\]
The oxygen uptake rate is 0.80 mm3 per minute. [3]
(c) Pea cells use oxygen in aerobic respiration. Carbon dioxide produced by respiration is absorbed by soda lime. Therefore the total gas volume and pressure in the flask fall, drawing the coloured drop towards the flask. [3]
(d) Use the same apparatus and the same mass or number of seeds in the boiled-seed flask. Keeping it at the same temperature, using the same time period and using the same amount of soda lime are also valid controls. Any two are required. [2]
(e) The germinating pea cells carry out aerobic respiration. [1]
Answer Details
(a) P is the coloured liquid drop, or marker drop. Q is soda lime, which absorbs carbon dioxide. [2]
(b) The volume of oxygen taken up in 15 minutes is:
\[\text{volume}=\text{distance moved}\times\text{cross-sectional area}\]
\[24\text{ mm}\times0.50\text{ mm}^2=12\text{ mm}^3\]
\[\text{rate}=\frac{12\text{ mm}^3}{15\text{ min}}=0.80\text{ mm}^3\text{ min}^{-1}\]
The oxygen uptake rate is 0.80 mm3 per minute. [3]
(c) Pea cells use oxygen in aerobic respiration. Carbon dioxide produced by respiration is absorbed by soda lime. Therefore the total gas volume and pressure in the flask fall, drawing the coloured drop towards the flask. [3]
(d) Use the same apparatus and the same mass or number of seeds in the boiled-seed flask. Keeping it at the same temperature, using the same time period and using the same amount of soda lime are also valid controls. Any two are required. [2]
(e) The germinating pea cells carry out aerobic respiration. [1]
Question 5 Report
Table 1 shows measurements made during an altitude-training assessment. Two healthy people cycled for 12 minutes at the same workload. One test was carried out at sea level and the other in a chamber where the air had a lower oxygen concentration. Blood samples were taken immediately after exercise. The body cells of each person need oxygen for aerobic respiration.
| Test condition | Oxygen concentration in inhaled air (%) | Breathing rate (breaths per minute) | Oxygen concentration in blood leaving lungs (%) |
|---|---|---|---|
| Sea level | 21.0 | 22 | 19.2 |
| Low-oxygen chamber | 15.0 | 31 | 16.8 |
(a) Use Table 1 to state the difference in breathing rate between the two tests. [1]
(b) Explain why the person in the low-oxygen chamber breathed faster. [3]
(c) Describe the effect of the lower oxygen concentration in inhaled air on the oxygen concentration in blood leaving the lungs. [2]
(d) Explain how reduced oxygen delivery to body cells could affect the person’s cycling performance. [3]
(e) Suggest one control used in this assessment that makes the comparison between the two people more valid. [2]
(a) \[31 - 22 = 9\] The breathing rate is 9 breaths per minute higher in the low-oxygen chamber. [1]
(b) The inhaled air in the chamber contains less oxygen. Therefore less oxygen diffuses into the blood and blood oxygen concentration falls. Faster breathing increases ventilation, bringing more air, and therefore more oxygen, to the air spaces each minute. [3]
(c) Lower oxygen concentration in inhaled air decreases the oxygen concentration in blood leaving the lungs, from \(19.2\%\) to \(16.8\%\). This is a decrease of \(2.4\) percentage points. [2]
(d) Less oxygen delivered to body cells reduces the rate of aerobic respiration. Less energy is released for muscle contraction, so muscles fatigue sooner or produce less force. Cycling performance is therefore reduced. [3]
(e) Valid controls include making both people cycle at the same workload and for the same time. Using the same equipment and taking blood samples at the same time after exercise are also acceptable. These controls mean oxygen concentration is the main intended difference. [2]
Answer Details
(a) \[31 - 22 = 9\] The breathing rate is 9 breaths per minute higher in the low-oxygen chamber. [1]
(b) The inhaled air in the chamber contains less oxygen. Therefore less oxygen diffuses into the blood and blood oxygen concentration falls. Faster breathing increases ventilation, bringing more air, and therefore more oxygen, to the air spaces each minute. [3]
(c) Lower oxygen concentration in inhaled air decreases the oxygen concentration in blood leaving the lungs, from \(19.2\%\) to \(16.8\%\). This is a decrease of \(2.4\) percentage points. [2]
(d) Less oxygen delivered to body cells reduces the rate of aerobic respiration. Less energy is released for muscle contraction, so muscles fatigue sooner or produce less force. Cycling performance is therefore reduced. [3]
(e) Valid controls include making both people cycle at the same workload and for the same time. Using the same equipment and taking blood samples at the same time after exercise are also acceptable. These controls mean oxygen concentration is the main intended difference. [2]
Question 6 Report
Fig. 1 shows part of the skin of a runner near the end of a 10 km race on a warm day. Her body temperature has increased above 37 degrees C. Blood vessels close to the skin surface are shown, together with a sweat gland. The person drinks water after the race.
(a) Describe two changes in the skin shown in Fig. 1 that help the runner lose heat. [2]
(b) Explain how evaporation of sweat from the skin can lower body temperature. [2]
(c) Use the information about the runner to give two reasons why drinking water is important after the race. [2]
(a) The blood vessels near the skin dilate or become wider. [1] Sweat is released onto the skin surface. [1]
(b) Sweat needs thermal energy to evaporate. [1] It takes this energy from the skin and body, lowering body temperature. [1]
(c) Drinking water replaces water lost in sweat. [1] It also helps prevent dehydration and maintains blood volume for circulation. [1]
Answer Details
(a) The blood vessels near the skin dilate or become wider. [1] Sweat is released onto the skin surface. [1]
(b) Sweat needs thermal energy to evaporate. [1] It takes this energy from the skin and body, lowering body temperature. [1]
(c) Drinking water replaces water lost in sweat. [1] It also helps prevent dehydration and maintains blood volume for circulation. [1]
Question 7 Report
Fig. 1 shows a simplified section through the surface of a human heart. A cardiologist has identified a narrowed coronary artery at P in a person who experiences chest pain when walking uphill.
(a) Name the type of blood vessel at P. [1]
(b) Describe how narrowing at P affects the blood supply to nearby heart muscle cells. [3]
(c) Explain how inserting a stent at P can reduce the person's symptoms. [2]
(d) Use the information in the question to explain why the chest pain is more likely during uphill walking than when resting. [2]
(a) P is an artery, specifically a coronary artery. [1]
(b) Narrowing reduces the amount of blood that can flow through the coronary artery. Consequently, less oxygen reaches nearby heart muscle cells. Less aerobic respiration can occur, so less energy is released for heart muscle contraction. [3]
(c) A stent holds the artery open and widens its lumen. This allows more blood, and therefore more oxygen, to reach the heart muscle, reducing symptoms caused by limited oxygen supply. [2]
(d) Uphill walking makes the heart work harder. Its muscle needs more oxygen, but the narrowed artery cannot provide enough blood and oxygen to meet the increased demand, causing chest pain. [2]
Answer Details
(a) P is an artery, specifically a coronary artery. [1]
(b) Narrowing reduces the amount of blood that can flow through the coronary artery. Consequently, less oxygen reaches nearby heart muscle cells. Less aerobic respiration can occur, so less energy is released for heart muscle contraction. [3]
(c) A stent holds the artery open and widens its lumen. This allows more blood, and therefore more oxygen, to reach the heart muscle, reducing symptoms caused by limited oxygen supply. [2]
(d) Uphill walking makes the heart work harder. Its muscle needs more oxygen, but the narrowed artery cannot provide enough blood and oxygen to meet the increased demand, causing chest pain. [2]
Question 8 Report
A hospital is trialling a treatment in which immune cells are removed from a person with leukaemia. Fig. 1 shows how the cells are genetically changed so that they can recognise cancer cells. The altered immune cells are then grown outside the body before being returned to the person.
(a) Name the substance in a cell that carries the new gene. [2]
(b) Complete the sequence in Fig. 1 by stating what the virus vector delivers and what the altered immune cells can recognise after treatment. [3]
(c) Explain why mitosis is needed while the changed immune cells are grown in the laboratory. [4]
(d) Explain why the treatment should be tested in a small clinical trial before it is used for many people. [5]
(a) A new gene is carried in DNA, and DNA is arranged into chromosomes containing genes. Therefore credit is given for DNA and chromosome or gene. [2]
(b) The virus vector delivers the new gene or DNA into the immune cell. The gene causes production of a receptor or protein on the cell surface. The altered immune cell can therefore recognise cancer cells, including leukaemia cells. [3]
(c) Mitosis produces many cells from one successfully changed immune cell. Each daughter cell receives the same inserted gene. A large number of these immune cells is needed for treatment, and they can recognise and destroy cancer cells. [4]
(d) A small clinical trial checks whether the treatment is safe and identifies side effects. It also tests whether it is effective at reducing cancer cells and helps identify a suitable dose. Unexpected immune reactions can be monitored. Starting with few people limits the number exposed to possible risks. [5]
Answer Details
(a) A new gene is carried in DNA, and DNA is arranged into chromosomes containing genes. Therefore credit is given for DNA and chromosome or gene. [2]
(b) The virus vector delivers the new gene or DNA into the immune cell. The gene causes production of a receptor or protein on the cell surface. The altered immune cell can therefore recognise cancer cells, including leukaemia cells. [3]
(c) Mitosis produces many cells from one successfully changed immune cell. Each daughter cell receives the same inserted gene. A large number of these immune cells is needed for treatment, and they can recognise and destroy cancer cells. [4]
(d) A small clinical trial checks whether the treatment is safe and identifies side effects. It also tests whether it is effective at reducing cancer cells and helps identify a suitable dose. Unexpected immune reactions can be monitored. Starting with few people limits the number exposed to possible risks. [5]
Question 9 Report
Fig. 1 shows a section of human skin from a fingertip. A researcher compared the response of the fingertip with the response of skin on the upper arm when a person was touched with two fine points.
(a) Name the type of cell labelled sensory neurone in Fig. 1. [1]
(b) Describe the route of an impulse from a touch receptor in the skin until the person becomes aware of the touch. [3]
(c) Explain why fingertips can distinguish two close points more easily than skin on the upper arm. [3]
(d) Use Fig. 1 to explain why damage to a sensory neurone may reduce a person's ability to feel touch. [3]
(e) Draw a simple arrow on Fig. 1 to show the direction an impulse travels along the sensory neurone. [3]
(a) A sensory neurone is a nerve cell, or neurone. [1]
(b) A touch receptor is stimulated and produces an electrical impulse. The impulse travels along a sensory neurone or nerve to the brain. The brain interprets the impulse, making the person aware of touch. [3]
(c) Fingertips have more touch receptors per unit area, so their receptors are closer together. Two close points are more likely to stimulate separate receptors. The brain receives separate impulses and can identify two points rather than one. [3]
(d) A sensory neurone carries impulses from a receptor towards the central nervous system. If it is damaged, impulse transmission may be reduced or stopped. The brain receives less information, so touch may be felt less accurately or not at all. [3]
(e) The arrow must begin at a receptor, follow the sensory neurone, and point away from the receptor towards the central nervous system. [3]
Answer Details
(a) A sensory neurone is a nerve cell, or neurone. [1]
(b) A touch receptor is stimulated and produces an electrical impulse. The impulse travels along a sensory neurone or nerve to the brain. The brain interprets the impulse, making the person aware of touch. [3]
(c) Fingertips have more touch receptors per unit area, so their receptors are closer together. Two close points are more likely to stimulate separate receptors. The brain receives separate impulses and can identify two points rather than one. [3]
(d) A sensory neurone carries impulses from a receptor towards the central nervous system. If it is damaged, impulse transmission may be reduced or stopped. The brain receives less information, so touch may be felt less accurately or not at all. [3]
(e) The arrow must begin at a receptor, follow the sensory neurone, and point away from the receptor towards the central nervous system. [3]
