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Question 1 Report
Fig. 1 shows a worker fitting ear defenders before using a metal cutting machine. The machine produces sound waves that can make the eardrum vibrate. The worker's employer measures the sound level and requires ear defenders when the level is high. The diagram shows how the defenders cover the outer ear, reducing the energy of sound that reaches the ear canal.
(a) Name the structure labelled U. [1]
(b) Describe what happens to the eardrum when sound waves reach it. [2]
(c) Explain how ear defenders reduce the risk of hearing damage. [3]
(d) State the type of energy carried by sound waves. [1]
(e) Which cells in the ear are damaged by very loud sound: receptor cells or red blood cells? [1]
(f) Explain why damage to these cells can cause permanent hearing loss. [2]
(a) U is an ear defender or ear muff. [1]
(b) When sound waves reach the eardrum, it vibrates. Its vibrations have the frequency of the sound waves and are passed onwards through the ear. [2]
(c) Ear defenders absorb or block some sound energy. Consequently, smaller vibrations reach the eardrum and inner ear. This reduces the chance that the inner-ear receptor cells will be damaged. [3]
(d) Sound waves carry sound energy. [1]
(e) Very loud sound can damage receptor cells, not red blood cells. [1]
(f) Hearing receptor cells convert vibrations into nerve impulses. If these cells are damaged, they do not repair or regenerate, so the loss of hearing can be permanent. [2]
(a) U is an ear defender or ear muff. [1]
(b) When sound waves reach the eardrum, it vibrates. Its vibrations have the frequency of the sound waves and are passed onwards through the ear. [2]
(c) Ear defenders absorb or block some sound energy. Consequently, smaller vibrations reach the eardrum and inner ear. This reduces the chance that the inner-ear receptor cells will be damaged. [3]
(d) Sound waves carry sound energy. [1]
(e) Very loud sound can damage receptor cells, not red blood cells. [1]
(f) Hearing receptor cells convert vibrations into nerve impulses. If these cells are damaged, they do not repair or regenerate, so the loss of hearing can be permanent. [2]
Question 2 Report
A diagram shows a section through the female reproductive system. It was prepared for a health lesson about the route followed by an ovum after it is released from an ovary.
(a) Name the parts labelled X, Y and Z. [3]
(b) Describe the route of an ovum from the ovary to the outside of the female body if fertilisation does not occur. [3]
(c) Explain why fertilisation is more likely if sperm are present in the oviduct at the time an ovum is released. [1]
(a) X is an ovary, Y is the uterus, and Z is the vagina. [3]
(b) After release from an ovary, the ovum enters an oviduct, also called a fallopian tube. It moves to the uterus. If fertilisation does not occur, it leaves the body through the vagina with the menstrual flow. [3]
(c) Fertilisation is more likely when sperm are already in the oviduct because the sperm and ovum can meet there, allowing a sperm nucleus to fuse with the ovum nucleus. [1]
(a) X is an ovary, Y is the uterus, and Z is the vagina. [3]
(b) After release from an ovary, the ovum enters an oviduct, also called a fallopian tube. It moves to the uterus. If fertilisation does not occur, it leaves the body through the vagina with the menstrual flow. [3]
(c) Fertilisation is more likely when sperm are already in the oviduct because the sperm and ovum can meet there, allowing a sperm nucleus to fuse with the ovum nucleus. [1]
Question 3 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 4 Report
A student prepared a slide from the growing tip of an onion root. Fig. 1 shows four cells from the same root tip at different stages of mitosis. The chromosomes have been stained so that they can be seen clearly. Each new body cell produced by this process should receive the same genetic information.
(a) Name the stage of mitosis shown by cell B. [1]
(b) Complete this sequence of stages: C, ___, ___, D. [2]
(c) Describe two changes shown in cell D. [2]
(d) Explain why mitosis is important for growth of the onion root. [3]
(a) Cell B is in metaphase. [1 mark] Its chromosomes are arranged across the middle of the cell.
(b) The sequence is C, A, B, D. The missing stages are A, prophase, followed by B, metaphase. [2 marks]
(c) In cell D, the chromosomes or chromatids are at opposite ends of the cell. Also, two nuclei are forming, or the cell is dividing into two cells. [2 marks]
(d) Mitosis produces more cells. The extra cells increase the length and size of the growing onion root. Each daughter cell receives the same chromosomes, or DNA, as the original cell, so the new cells can carry out the same functions. [3 marks]
(a) Cell B is in metaphase. [1 mark] Its chromosomes are arranged across the middle of the cell.
(b) The sequence is C, A, B, D. The missing stages are A, prophase, followed by B, metaphase. [2 marks]
(c) In cell D, the chromosomes or chromatids are at opposite ends of the cell. Also, two nuclei are forming, or the cell is dividing into two cells. [2 marks]
(d) Mitosis produces more cells. The extra cells increase the length and size of the growing onion root. Each daughter cell receives the same chromosomes, or DNA, as the original cell, so the new cells can carry out the same functions. [3 marks]
Question 5 Report
A food technologist is checking whether a low-fat salad dressing mixes oil effectively with water. Fig. 1 shows two test tubes after standing for five minutes. Tube A contains oil and water only. Tube B also contains a small amount of bile salt solution. Table 1 shows the mean diameter of oil droplets seen using a microscope.
Table 1
| tube | mean oil droplet diameter / mm |
|---|---|
| A | 2.4 |
| B | 0.3 |
(a) Describe the difference between the oil droplets in tubes A and B. [2]
(b) Name the substance in the human digestive system that has the same role as the bile salt solution. [1]
(c) Explain how making fat droplets smaller helps fat digestion. [2]
(d) State the organ that produces bile. [1]
(e) Complete the sentence: bile is stored in the ________ before it is released into the small intestine. [2]
(a) Oil droplets in tube B are smaller. Their mean diameter is \(0.3\) mm rather than \(2.4\) mm in tube A, so B has more small droplets. [2]
(b) The equivalent digestive substance is bile. [1]
(c) Smaller fat droplets have a larger total surface area. This gives lipase more surface on which to act, so fat is digested more rapidly. [2]
(d) Bile is produced by the liver. [1]
(e) Bile is stored in the gall bladder before release into the small intestine. [2]
(a) Oil droplets in tube B are smaller. Their mean diameter is \(0.3\) mm rather than \(2.4\) mm in tube A, so B has more small droplets. [2]
(b) The equivalent digestive substance is bile. [1]
(c) Smaller fat droplets have a larger total surface area. This gives lipase more surface on which to act, so fat is digested more rapidly. [2]
(d) Bile is produced by the liver. [1]
(e) Bile is stored in the gall bladder before release into the small intestine. [2]
Question 6 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 7 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 8 Report
Fig. 1 shows a diagram of two vertebrae from the human skeleton. A warehouse worker lifts boxes for several hours each day. The worker reports pain after twisting the body while holding a box. The doctor explains that the structure labelled X between adjacent bones is important when the body bends and absorbs impacts.
(a) Name the structure labelled X. [1]
(b) State the main material found in a healthy structure X. [1]
(c) Explain how X protects the vertebrae during lifting. [2]
(d) Describe two ways the skeleton helps the worker carry out everyday movement. [2]
(e) Which type of joint allows only limited movement between vertebrae? [2]
(a) X is an intervertebral disc. [1]
(b) A healthy disc contains cartilage. [1]
(c) The disc acts as a cushion and absorbs shock during lifting. It prevents neighbouring vertebrae from rubbing or colliding. [2]
(d) Bones provide attachment points for muscles. Joints allow bones to move relative to each other, while the skeleton provides a rigid framework for movement. [2]
(e) Vertebrae are connected by a slightly movable joint. It allows only a small amount of movement. [2]
(a) X is an intervertebral disc. [1]
(b) A healthy disc contains cartilage. [1]
(c) The disc acts as a cushion and absorbs shock during lifting. It prevents neighbouring vertebrae from rubbing or colliding. [2]
(d) Bones provide attachment points for muscles. Joints allow bones to move relative to each other, while the skeleton provides a rigid framework for movement. [2]
(e) Vertebrae are connected by a slightly movable joint. It allows only a small amount of movement. [2]
Question 9 Report
The diagram shows colour boxes from a urine test strip used at a school health screening. The strip was compared with the manufacturer chart after 60 s. A person had eaten breakfast containing carbohydrate two hours earlier.
(a) Which substance is indicated by the grey glucose box on the test strip? [1]
(b) State why glucose in urine may suggest that the person has diabetes. [1]
(c) Explain why glucose is normally absent from the urine of a healthy person. [2]
(d) Describe one change to the test procedure that would make the comparison with the chart more reliable. [2]
(e) Give one reason why a doctor would not diagnose diabetes using this test alone. [1]
(a) The grey glucose box indicates glucose. [1]
(b) Glucose in urine may suggest diabetes because blood glucose may be above the level at which the kidneys can reabsorb all of the glucose. [1]
(c) In a healthy kidney, glucose is first filtered into the nephron. It is then selectively reabsorbed into the blood from the kidney tubule, so it is normally absent from urine. [2]
(d) To improve reliability, use two suitable procedural controls, such as using the stated timing exactly and repeating the test with another sample. Using a fresh, in-date strip or comparing under the same light conditions are also acceptable. [2]
(e) A doctor needs a confirmatory blood-glucose test; a single urine result can be affected by other factors and is not sufficient alone for diagnosis. [1]
(a) The grey glucose box indicates glucose. [1]
(b) Glucose in urine may suggest diabetes because blood glucose may be above the level at which the kidneys can reabsorb all of the glucose. [1]
(c) In a healthy kidney, glucose is first filtered into the nephron. It is then selectively reabsorbed into the blood from the kidney tubule, so it is normally absent from urine. [2]
(d) To improve reliability, use two suitable procedural controls, such as using the stated timing exactly and repeating the test with another sample. Using a fresh, in-date strip or comparing under the same light conditions are also acceptable. [2]
(e) A doctor needs a confirmatory blood-glucose test; a single urine result can be affected by other factors and is not sufficient alone for diagnosis. [1]
Question 10 Report
Fig. 1 shows blood from a small sample after it has been spun in a centrifuge. A nurse then counted red blood cells in samples from two groups. Table 1 shows the mean counts.
| Group | Mean number of red blood cells / million per mm3 |
|---|---|
| Adults living at sea level | 4.8 |
| Adults living at 3500 m altitude | 5.6 |
(a) Name the fluid at the top of the tube in Fig. 1. [1]
(b) Describe the main function of red blood cells. [1]
(c) State the difference between the mean red blood cell counts of the two groups. [1]
(d) Explain why a person living at high altitude may benefit from having more red blood cells. [2]
(e) Name the substance in red blood cells that combines with oxygen. [1]
(f) Describe one adaptation of a red blood cell for carrying oxygen. [2]
(a) The fluid at the top of the centrifuged tube is plasma. [1]
(b) The main function of red blood cells is to transport oxygen. [1]
(c)
\[5.6-4.8=0.8\text{ million per mm}^3\]
The difference is 0.8 million red blood cells per mm3. [1]
(d) At high altitude there is less available oxygen, or lower oxygen pressure, in the air. More red blood cells increase the amount of oxygen that can be transported to body cells. [2]
(e) The oxygen-carrying substance is haemoglobin. [1]
(f) A red blood cell has a biconcave shape, giving it a large surface area for oxygen exchange. It has no nucleus, leaving more space for haemoglobin. [2]
(a) The fluid at the top of the centrifuged tube is plasma. [1]
(b) The main function of red blood cells is to transport oxygen. [1]
(c)
\[5.6-4.8=0.8\text{ million per mm}^3\]
The difference is 0.8 million red blood cells per mm3. [1]
(d) At high altitude there is less available oxygen, or lower oxygen pressure, in the air. More red blood cells increase the amount of oxygen that can be transported to body cells. [2]
(e) The oxygen-carrying substance is haemoglobin. [1]
(f) A red blood cell has a biconcave shape, giving it a large surface area for oxygen exchange. It has no nucleus, leaving more space for haemoglobin. [2]
Question 11 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]
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