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Question 1 Report
Fig. 1 shows a simplified dialysis machine used by a patient with kidney failure. Table 1 gives solute concentrations before the blood enters and after it leaves the dialyser. The dialysis solution is continuously replaced.
| substance | blood entering / mmol dm-3 | blood leaving / mmol dm-3 | dialysis solution / mmol dm-3 |
|---|---|---|---|
| urea | 18.0 | 7.0 | 0.0 |
| glucose | 5.0 | 5.0 | 5.0 |
| sodium ions | 145 | 140 | 140 |
(a) Name the process by which urea moves from the blood into the dialysis solution. [1]
(b) Calculate the decrease in urea concentration in the blood. Show your working. [2]
(c) Explain why glucose is included in the dialysis solution at 5.0 mmol dm-3. [2]
(d) Describe why fresh dialysis solution must continually enter the machine. [2]
(e) State one component of blood that cannot pass through the membrane. [1]
(f) Explain why dialysis does not fully replace all functions of a kidney. [2]
(a) Urea moves into dialysis solution by diffusion. [1] It moves from a higher concentration in blood to a lower concentration in the solution.
(b)
\[18.0-7.0=11.0\text{ mmol dm}^{-3}\]
The decrease is 11.0 mmol dm-3. [2]
(c) The dialysis solution contains glucose at 5.0 mmol dm-3, the same concentration as normal blood glucose. Therefore there is no net diffusion of glucose out of the blood, so useful glucose is not lost. [2]
(d) Fresh dialysis solution keeps the urea concentration outside the membrane low. This maintains a steep concentration gradient, allowing urea to continue diffusing out of the blood. [2]
(e) A red blood cell cannot pass through the membrane. White blood cells, platelets, and plasma proteins are also acceptable. [1]
(f) Dialysis cannot selectively reabsorb and regulate substances as precisely as a kidney. It also does not make hormones or carry out all kidney functions continuously. [2]
(a) Urea moves into dialysis solution by diffusion. [1] It moves from a higher concentration in blood to a lower concentration in the solution.
(b)
\[18.0-7.0=11.0\text{ mmol dm}^{-3}\]
The decrease is 11.0 mmol dm-3. [2]
(c) The dialysis solution contains glucose at 5.0 mmol dm-3, the same concentration as normal blood glucose. Therefore there is no net diffusion of glucose out of the blood, so useful glucose is not lost. [2]
(d) Fresh dialysis solution keeps the urea concentration outside the membrane low. This maintains a steep concentration gradient, allowing urea to continue diffusing out of the blood. [2]
(e) A red blood cell cannot pass through the membrane. White blood cells, platelets, and plasma proteins are also acceptable. [1]
(f) Dialysis cannot selectively reabsorb and regulate substances as precisely as a kidney. It also does not make hormones or carry out all kidney functions continuously. [2]
Question 2 Report
Fig. 1 shows a simplified diagram of part of the male reproductive system used by a doctor when explaining sperm production to a person attending a fertility clinic.
The labelled parts are not drawn to scale.
(a) Name the structures labelled X, Y and Z. [3]
(b) State where sperm are produced in this system. [1]
(a) X is the bladder, Y is a testis, and Z is the urethra. [3]
(b) Sperm are produced in the testes. [1]
The testes contain the structures where sperm cells are made; the urethra carries semen out of the body but does not produce sperm.
(a) X is the bladder, Y is a testis, and Z is the urethra. [3]
(b) Sperm are produced in the testes. [1]
The testes contain the structures where sperm cells are made; the urethra carries semen out of the body but does not produce sperm.
Question 3 Report
Fig. 1 shows an X-ray-style diagram of a forearm after a fall from a skateboard. The radius has fractured near the wrist. The arm was placed in a plaster cast, and follow-up images were taken over six weeks. A doctor explains that bone is living tissue and that repair depends on specialised bone cells, blood supply and correct alignment.
(a) Name the bone labelled Y. [1]
(b) State what is shown by X. [1]
(c) Describe the role of a plaster cast during early repair. [2]
(d) Explain why an adequate blood supply is needed at the damaged bone. [3]
(e) Describe two changes that occur as the fracture heals. [2]
(f) Which mineral, obtained partly from food, is needed to make bone hard? [1]
(g) State two risks of returning to weight training before the bone has healed. [2]
(a) Y is the ulna. [1]
(b) X shows a fracture, or break, in the radius. [1]
(c) A plaster cast keeps the broken bone ends aligned and prevents excessive movement while repair takes place. [2]
(d) Blood supplies oxygen and glucose or other nutrients to the damaged tissue. These are needed for respiration, which releases energy for cell division and repair. [3]
(e) During healing, a blood clot or callus forms and new bone cells are produced. New bone is deposited across the gap, then remodelled and strengthened. Any two valid changes gain credit. [2]
(f) The mineral is calcium. [1]
(g) Returning early to weight training could reopen the fracture or misalign the bone ends. It could also delay healing or cause pain and further tissue damage. Any two valid risks gain credit. [2]
(a) Y is the ulna. [1]
(b) X shows a fracture, or break, in the radius. [1]
(c) A plaster cast keeps the broken bone ends aligned and prevents excessive movement while repair takes place. [2]
(d) Blood supplies oxygen and glucose or other nutrients to the damaged tissue. These are needed for respiration, which releases energy for cell division and repair. [3]
(e) During healing, a blood clot or callus forms and new bone cells are produced. New bone is deposited across the gap, then remodelled and strengthened. Any two valid changes gain credit. [2]
(f) The mineral is calcium. [1]
(g) Returning early to weight training could reopen the fracture or misalign the bone ends. It could also delay healing or cause pain and further tissue damage. Any two valid risks gain credit. [2]
Question 4 Report
The diagram shows a school laboratory setup used to investigate the blink reflex. A student stands 2 m from a harmless puff of air delivered beside, but not into, the eye. A high-speed camera records the time from the puff to closure of the eyelid. Table 1 shows mean results for five trials at two room temperatures. The eye and its surrounding structures are protected by this rapid response.
| Room temperature / degrees C | Mean blink time / ms |
|---|---|
| 18 | 152 |
| 24 | 149 |
(a) State the mean blink time at 24 degrees C. [1]
(b) Describe the difference between the two mean blink times. [2]
(c) Explain why five trials were used at each temperature. [2]
(d) Name the structure labelled T. [1]
(e) Explain how blinking helps to protect the eye. [2]
(f) Which part of the nervous system coordinates a simple reflex such as blinking: the brain or the spinal cord? [1]
(g) State one variable, other than temperature, that should be kept the same. [1]
(a) The mean blink time at 24 degrees C is 149 ms. [1]
(b) Blink time is lower at 24 degrees C than at 18 degrees C. The difference is \(152-149=3\) ms. [2]
(c) Repeating the test five times allows a mean to be calculated. A mean reduces the influence of anomalous results and random variation, so it is more reliable than one reading. [2]
(d) T is the eyelid. [1]
(e) Blinking closes the eyelid, helping to prevent dust and other foreign objects entering the eye. It also spreads tears over the eye surface, keeping it moist and helping remove particles. [2]
(f) A simple blink reflex is coordinated by the brain. [1]
(g) One suitable control variable is the distance of the air puffer from the eye. The strength or volume of puff, student tested, or light level would also be acceptable. [1]
(a) The mean blink time at 24 degrees C is 149 ms. [1]
(b) Blink time is lower at 24 degrees C than at 18 degrees C. The difference is \(152-149=3\) ms. [2]
(c) Repeating the test five times allows a mean to be calculated. A mean reduces the influence of anomalous results and random variation, so it is more reliable than one reading. [2]
(d) T is the eyelid. [1]
(e) Blinking closes the eyelid, helping to prevent dust and other foreign objects entering the eye. It also spreads tears over the eye surface, keeping it moist and helping remove particles. [2]
(f) A simple blink reflex is coordinated by the brain. [1]
(g) One suitable control variable is the distance of the air puffer from the eye. The strength or volume of puff, student tested, or light level would also be acceptable. [1]
Question 5 Report
Fig. 1 shows a simplified route taken by blood during one complete circuit in a human body. The boxes represent structures in the circulatory system. Arrows show the direction of blood flow.
(a) Name the type of blood vessel at A, carrying blood from body cells towards the heart. [1]
(b) Name the type of blood vessel at B, carrying blood from the heart towards the lungs. [1]
(c) State which gas is taken into the blood at the lungs. [1]
(d) Describe one change in the oxygen concentration of red blood as it travels from the lungs to body cells. [1]
(a) A vessel carrying blood from body cells towards the heart is a vein. [1]
(b) A vessel carrying blood from the heart towards the lungs is an artery, specifically the pulmonary artery. [1]
(c) The gas taken into blood at the lungs is oxygen. [1]
(d) Oxygen concentration decreases as blood travels from the lungs to body cells because oxygen is delivered to the cells. [1]
(a) A vessel carrying blood from body cells towards the heart is a vein. [1]
(b) A vessel carrying blood from the heart towards the lungs is an artery, specifically the pulmonary artery. [1]
(c) The gas taken into blood at the lungs is oxygen. [1]
(d) Oxygen concentration decreases as blood travels from the lungs to body cells because oxygen is delivered to the cells. [1]
Question 6 Report
The table below shows pulse rate measured from the radial artery of one student before, during and after a steady cycling test. The student counted each pulse for 30 seconds and doubled the result.
| Time relative to cycling | Pulse rate / beats per minute |
|---|---|
| Before cycling | 68 |
| After 4 minutes cycling | 142 |
| 2 minutes after cycling | 104 |
| 6 minutes after cycling | 76 |
(a) State the student’s pulse rate after 4 minutes of cycling. [1]
(b) Calculate the increase in pulse rate between before cycling and after 4 minutes cycling. Show your working. [2]
(c) Explain why the heart rate increased during cycling. [2]
(d) Describe one way the student could improve the reliability of these results. [1]
(a) After 4 minutes of cycling, the pulse rate is 142 beats per minute. [1]
(b)
\[142-68=74\text{ beats per minute}\]
The pulse-rate increase is 74 beats per minute. [2]
(c) During cycling, muscle cells respire faster and need more energy. The heart rate rises so blood can deliver more oxygen and glucose for respiration, as well as remove more carbon dioxide. [2]
(d) Improve reliability by repeating measurements and calculating a mean. Using several students and calculating a mean is also acceptable. [1]
(a) After 4 minutes of cycling, the pulse rate is 142 beats per minute. [1]
(b)
\[142-68=74\text{ beats per minute}\]
The pulse-rate increase is 74 beats per minute. [2]
(c) During cycling, muscle cells respire faster and need more energy. The heart rate rises so blood can deliver more oxygen and glucose for respiration, as well as remove more carbon dioxide. [2]
(d) Improve reliability by repeating measurements and calculating a mean. Using several students and calculating a mean is also acceptable. [1]
Question 7 Report
At a remote mountain rescue post, a medic takes blood samples from a cyclist before and after the blood passes through the cells of the thigh. The cyclist has been pedalling a generator to charge a radio. Fig. 1 is a diagram showing a capillary beside human muscle cells. Oxygen and glucose are carried in the blood to provide energy for respiration.
Table 1 shows the oxygen concentration in equal volumes of blood collected at each end of this capillary network.
| Blood sample | Oxygen concentration / arbitrary units |
|---|---|
| Entering the muscle capillaries | 19.6 |
| Leaving the muscle capillaries | 13.9 |
(a) Name the type of blood vessel shown in Fig. 1. [1]
(b) Describe two structural features of this blood vessel that make exchange with body cells efficient. [2]
(c) Use Fig. 1 to explain how oxygen reaches the muscle cells and how carbon dioxide enters the blood. [4]
(d) Calculate the decrease in oxygen concentration between the two blood samples in Table 1. Show your working. [2]
(e) State one reason why the oxygen concentration falls as blood travels through these capillaries. [1]
(a) The vessel is a capillary. [1]
(b) Capillaries have walls one cell thick, giving a short diffusion distance. They also have a narrow lumen, keeping blood close to body cells. [2] A permeable wall or many capillaries providing large surface area are also acceptable.
(c) Oxygen moves from the blood to the muscle cells by diffusion, because oxygen concentration is higher in blood than in the cells. Carbon dioxide moves from muscle cells into blood by diffusion, because its concentration is higher in the cells than in the blood. [4]
(d)
\[19.6-13.9=5.7\text{ arbitrary units}\]
The oxygen concentration decreases by 5.7 arbitrary units. [2]
(e) Oxygen concentration falls because muscle cells use oxygen in aerobic respiration to release energy. [1]
(a) The vessel is a capillary. [1]
(b) Capillaries have walls one cell thick, giving a short diffusion distance. They also have a narrow lumen, keeping blood close to body cells. [2] A permeable wall or many capillaries providing large surface area are also acceptable.
(c) Oxygen moves from the blood to the muscle cells by diffusion, because oxygen concentration is higher in blood than in the cells. Carbon dioxide moves from muscle cells into blood by diffusion, because its concentration is higher in the cells than in the blood. [4]
(d)
\[19.6-13.9=5.7\text{ arbitrary units}\]
The oxygen concentration decreases by 5.7 arbitrary units. [2]
(e) Oxygen concentration falls because muscle cells use oxygen in aerobic respiration to release energy. [1]
Question 8 Report
The diagram shows a vertical section through a human heart from a patient having a routine scan. Labels X and Y point to chambers. The left side of the diagram represents the right side of the person’s heart.
(a) Name the chamber labelled X. [1]
(b) Name the chamber labelled Y. [1]
(c) Describe the difference in thickness between the muscular walls of the two ventricles. [1]
(d) Explain why this difference is needed. [2]
(e) Complete the sentence: valves in the heart prevent blood from flowing in the ______ direction. [1]
(f) State which side of the heart receives blood from the body. [1]
(a) X is the right atrium. [1]
(b) Y is the left ventricle. [1]
(c) The left ventricle has a thicker muscular wall than the right ventricle. [1]
(d) The left ventricle pumps blood around the whole body. It therefore must generate greater force and higher pressure than the right ventricle, which pumps only to the lungs. [2]
(e) Valves prevent blood flowing in the wrong, or reverse, direction. [1]
(f) The right side of the heart receives blood from the body. [1]
(a) X is the right atrium. [1]
(b) Y is the left ventricle. [1]
(c) The left ventricle has a thicker muscular wall than the right ventricle. [1]
(d) The left ventricle pumps blood around the whole body. It therefore must generate greater force and higher pressure than the right ventricle, which pumps only to the lungs. [2]
(e) Valves prevent blood flowing in the wrong, or reverse, direction. [1]
(f) The right side of the heart receives blood from the body. [1]
Question 9 Report
Fig. 1 shows what can happen during meiosis in a person producing egg cells. Normally, the two copies of chromosome 21 separate, so each gamete has one copy. In the diagram, the two chromosome 21 copies fail to separate. A doctor uses this diagram when explaining why a child may have Down syndrome even when there is no known family history of the condition.
(a) State the usual number of chromosomes in a human body cell. [1]
(b) Describe the difference between the two gametes shown in Fig. 1. [2]
(c) Explain how fertilisation of the gamete with two copies of chromosome 21 can lead to Down syndrome in a child. [4]
(d) Complete the chromosome numbers of the two possible zygotes when each gamete shown is fertilised by a normal sperm cell containing 23 chromosomes. [3]
(e) Explain why this chromosome condition can occur in a child with no previous family history. [4]
(a) A human body cell normally contains 46 chromosomes. [1 mark]
(b) One gamete has two copies of chromosome 21, while the other has no copy of chromosome 21. [2 marks]
(c) A normal sperm contributes one copy of chromosome 21. If it fertilises the egg with two copies, the zygote has three copies of chromosome 21, called trisomy 21. This chromosome imbalance causes Down syndrome. [4 marks]
(d) The egg with two copies has 24 chromosomes, so \(24+23=47\) chromosomes in the zygote. The egg with no chromosome 21 has 22 chromosomes, so \(22+23=45\) chromosomes in the zygote. [3 marks]
(e) The condition can arise through an error during meiosis, when chromosome copies fail to separate. This can happen randomly in an egg or sperm cell, so it is not necessarily caused by an inherited disease allele or a previous family history. [4 marks]
(a) A human body cell normally contains 46 chromosomes. [1 mark]
(b) One gamete has two copies of chromosome 21, while the other has no copy of chromosome 21. [2 marks]
(c) A normal sperm contributes one copy of chromosome 21. If it fertilises the egg with two copies, the zygote has three copies of chromosome 21, called trisomy 21. This chromosome imbalance causes Down syndrome. [4 marks]
(d) The egg with two copies has 24 chromosomes, so \(24+23=47\) chromosomes in the zygote. The egg with no chromosome 21 has 22 chromosomes, so \(22+23=45\) chromosomes in the zygote. [3 marks]
(e) The condition can arise through an error during meiosis, when chromosome copies fail to separate. This can happen randomly in an egg or sperm cell, so it is not necessarily caused by an inherited disease allele or a previous family history. [4 marks]
Question 10 Report
Fig. 1 shows a section through one of a student's molar teeth. The student is comparing foods that need different amounts of chewing before swallowing. Table 1 shows the mean number of chews recorded for 20 g portions of two foods.
Table 1
| food | mean number of chews |
|---|---|
| raw carrot | 34 |
| soft bread | 12 |
(a) Name the structures labelled X and Y in Fig. 1. [2]
(b) Describe the difference shown in Table 1. [1]
(c) Explain why chewing is an important stage of digestion before food reaches the stomach. [1]
(a) X is enamel and Y is dentine. Enamel is the hard outer layer; dentine lies beneath it. [2]
(b) Raw carrot needed more chews than soft bread, \(34\) compared with \(12\). [1]
(c) Chewing breaks food into smaller pieces, increasing surface area for enzyme action. It also makes swallowing easier. [1]
(a) X is enamel and Y is dentine. Enamel is the hard outer layer; dentine lies beneath it. [2]
(b) Raw carrot needed more chews than soft bread, \(34\) compared with \(12\). [1]
(c) Chewing breaks food into smaller pieces, increasing surface area for enzyme action. It also makes swallowing easier. [1]
Question 11 Report
A midwife uses Fig. 1 to explain growth from fertilisation to birth. The diagram shows four stages but the time scale is not uniform.
(a) Name the process that causes the fertilised ovum to become a group of cells. [1]
(b) Describe two changes in the developing human between the embryo stage and birth. [2]
(c) Explain why cells in the developing fetus become specialised. [3]
(d) State two substances transferred from the mother to the fetus across the placenta. [2]
(e) Which structure carries blood from the fetus to the placenta? [1]
(a) The fertilised ovum becomes a group of cells by mitosis, or cell division. [1]
(b) Between the embryo stage and birth, the developing human increases greatly in size and organs develop. Development of limbs or other body features, and cell specialisation, are also acceptable changes. [2]
(c) Different types of cells are required to carry out different functions. During development, different genes are switched on, or expressed, in different cells. This produces specialised cells, which form tissues and organs. [3]
(d) Two substances transferred from mother to fetus across the placenta are oxygen and glucose. Amino acids and antibodies are also acceptable. [2]
(e) Blood travels from the fetus to the placenta in the umbilical artery. [1]
(a) The fertilised ovum becomes a group of cells by mitosis, or cell division. [1]
(b) Between the embryo stage and birth, the developing human increases greatly in size and organs develop. Development of limbs or other body features, and cell specialisation, are also acceptable changes. [2]
(c) Different types of cells are required to carry out different functions. During development, different genes are switched on, or expressed, in different cells. This produces specialised cells, which form tissues and organs. [3]
(d) Two substances transferred from mother to fetus across the placenta are oxygen and glucose. Amino acids and antibodies are also acceptable. [2]
(e) Blood travels from the fetus to the placenta in the umbilical artery. [1]
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