Biology 4BI1 | Paper 1 Mock 01 | Written Paper 1
Question 1 Report
The diagram shows three neurones from an engineering project making a prosthetic hand. The team wants to copy the speed of a motor neurone that carries impulses from the spinal cord to a hand muscle. Some neurones have a fatty myelin sheath around the axon.
(a) Complete the description of the direction of an impulse in the neurone: cell body, ______, neurone ending. [1]
(b) Complete the statement: myelin increases the ______ of nerve impulse transmission. [2]
(c) Explain how a faster impulse along a motor neurone could improve the response of a prosthetic hand. [3]
(d) Design a fair investigation to compare impulse speed in model axons with and without an insulating sheath. [4]
(a) The route is cell body, axon, neurone ending. [1]
(b) Myelin increases the speed, or rate, of nerve impulse transmission. [2]
(c) Faster impulses reach the hand mechanism sooner. The hand begins moving with less delay, giving a quicker response to a stimulus and improved control. [3]
(d) Use two model axons of equal length and give each the same electrical input. Fit an identical insulating sheath to only one model. Measure the time taken for each signal to reach the end, then calculate speed using \(\text{speed}=\frac{\text{distance}}{\text{time}}\). Repeat trials, calculate means, and keep temperature constant. [4]
Answer Details
(a) The route is cell body, axon, neurone ending. [1]
(b) Myelin increases the speed, or rate, of nerve impulse transmission. [2]
(c) Faster impulses reach the hand mechanism sooner. The hand begins moving with less delay, giving a quicker response to a stimulus and improved control. [3]
(d) Use two model axons of equal length and give each the same electrical input. Fit an identical insulating sheath to only one model. Measure the time taken for each signal to reach the end, then calculate speed using \(\text{speed}=\frac{\text{distance}}{\text{time}}\). Repeat trials, calculate means, and keep temperature constant. [4]
Question 2 Report
The diagram shows an apparatus used to model the movement of substances between blood and body cells. A dialysis tube containing starch solution was tied at both ends and placed in iodine solution. The wall of the tube acts as a partially permeable membrane.
After 20 minutes, the liquid inside the tube was blue-black. The iodine solution outside the tube remained orange-brown.
(a) Name the substance that moved into the tube. [1]
(b) Explain why this substance moved into the tube. [3]
(c) State the colour change that would show starch is present. [1]
(d) Describe why starch did not move out of the tube. [2]
(e) Complete the table by giving one variable that should be kept the same when this apparatus is repeated with different concentrations of iodine solution. [2]
| Variable | How it is controlled |
|---|---|
| ________________ | Use the same value in every tube |
(a) The substance that moved into the tube was iodine. [1]
(b) This tests diffusion. Iodine particles were at a higher concentration outside the dialysis tube than inside it. They therefore moved from high to low concentration, through the partially permeable tube wall. This net movement is diffusion. [3]
(c) Starch is identified by the colour change orange-brown to blue-black when iodine is present. [1]
(d) Starch did not leave because starch molecules are too large to pass through the pores in the partially permeable membrane. [2]
(e) One valid control is the volume of starch solution, controlled by using the same volume in every tube. Other valid paired answers include keeping temperature constant or leaving every tube for the same time. [2]
Exam reminder: Diffusion is movement of particles down a concentration gradient; a partially permeable membrane allows only sufficiently small particles through.
Answer Details
(a) The substance that moved into the tube was iodine. [1]
(b) This tests diffusion. Iodine particles were at a higher concentration outside the dialysis tube than inside it. They therefore moved from high to low concentration, through the partially permeable tube wall. This net movement is diffusion. [3]
(c) Starch is identified by the colour change orange-brown to blue-black when iodine is present. [1]
(d) Starch did not leave because starch molecules are too large to pass through the pores in the partially permeable membrane. [2]
(e) One valid control is the volume of starch solution, controlled by using the same volume in every tube. Other valid paired answers include keeping temperature constant or leaving every tube for the same time. [2]
Exam reminder: Diffusion is movement of particles down a concentration gradient; a partially permeable membrane allows only sufficiently small particles through.
Question 3 Report
The diagram shows the lower epidermis of a leaf from a shore plant collected at midday. Two guard cells surround opening Q. The plant grows where the soil is salty, so it must reduce unnecessary water loss while still allowing carbon dioxide to enter leaf cells for photosynthesis. Scientists compared this leaf with a leaf from a water lily.
(a) Name opening Q. [1]
(b) Describe how guard cells cause Q to open. [3]
(c) Explain why closing Q on a hot day reduces water loss from the plant. [2]
(d) State two factors, other than temperature, that can affect the rate of transpiration. [2]
(e) Suggest why water-lily leaves usually have most stomata on their upper surface. [1]
(a) Opening Q is a stoma, or stomatal pore. [1]
(b) Guard cells take up solutes. Water then enters the guard cells by osmosis. The cells become turgid and curve apart, opening the pore. [3]
(c) When Q closes, less water vapour diffuses out through the stomata. Therefore transpiration and water loss decrease. [2]
(d) Two factors other than temperature are light intensity and humidity. [2] Wind speed and leaf surface area are also valid.
(e) Water-lily leaves usually have most stomata on the upper surface because the lower surface is in contact with water, while the upper surface is exposed to air for gas exchange. [1]
Answer Details
(a) Opening Q is a stoma, or stomatal pore. [1]
(b) Guard cells take up solutes. Water then enters the guard cells by osmosis. The cells become turgid and curve apart, opening the pore. [3]
(c) When Q closes, less water vapour diffuses out through the stomata. Therefore transpiration and water loss decrease. [2]
(d) Two factors other than temperature are light intensity and humidity. [2] Wind speed and leaf surface area are also valid.
(e) Water-lily leaves usually have most stomata on the upper surface because the lower surface is in contact with water, while the upper surface is exposed to air for gas exchange. [1]
Question 4 Report
Fig. 1 shows a stained blood smear from a patient attending a health clinic. The large cell is a white blood cell and the many smaller disc-shaped cells are red blood cells. Table 1 gives counts from a separate measured sample of the patient's blood.
| cell type | number of cells per mm3 of blood |
|---|---|
| red blood cells | 5 100 000 |
| white blood cells | 7200 |
(a) Describe three visible features of a red blood cell in Fig. 1. [3]
(b) Explain how the structure of a red blood cell allows efficient transport of oxygen. [3]
(c) Calculate the number of white blood cells expected in 0.50 mm3 of this blood. [2]
(d) Explain why the body needs mitosis to maintain the number of blood cells. [2]
(e) Design two improvements to the student's method if they counted cells in only one microscope field. [3]
(a) Three visible features of red blood cells are any three of: they are disc-shaped; biconcave or thinner in the centre; have no visible nucleus; and many are similar in size. [3]
(b) The biconcave shape gives a large surface area for diffusion. The cell is thin, giving a short diffusion distance. Red blood cells contain haemoglobin, which binds and carries oxygen. [3]
(c)
\[7200\ \text{cells mm}^{-3}\times0.50\ \text{mm}^3=3600\ \text{cells}\]
The expected number is 3600 white blood cells. [2]
(d) Blood cells are continually lost and have limited lifespans. Mitosis produces genetically identical new cells to replace them and maintain blood-cell numbers. [2]
(e) Count cells in several different microscope fields rather than only one. Calculate a mean count. Use the same sample volume each time, ideally using a counting chamber with a known volume, so counts are comparable. [3]
Answer Details
(a) Three visible features of red blood cells are any three of: they are disc-shaped; biconcave or thinner in the centre; have no visible nucleus; and many are similar in size. [3]
(b) The biconcave shape gives a large surface area for diffusion. The cell is thin, giving a short diffusion distance. Red blood cells contain haemoglobin, which binds and carries oxygen. [3]
(c)
\[7200\ \text{cells mm}^{-3}\times0.50\ \text{mm}^3=3600\ \text{cells}\]
The expected number is 3600 white blood cells. [2]
(d) Blood cells are continually lost and have limited lifespans. Mitosis produces genetically identical new cells to replace them and maintain blood-cell numbers. [2]
(e) Count cells in several different microscope fields rather than only one. Calculate a mean count. Use the same sample volume each time, ideally using a counting chamber with a known volume, so counts are comparable. [3]
Question 5 Report
The diagram shows a four-year crop rotation planned for a mixed farm. Each field has the same area. The farmer wants to maintain soil fertility, avoid a build-up of pests and obtain a reliable production of food for people and animals.
(a) Describe the sequence of crops shown in Fig. 1, starting with wheat. [3]
(b) Explain why growing beans can reduce the amount of nitrogen fertiliser needed for a later crop. [3]
(c) State two reasons why a farmer may rotate crops rather than grow wheat every year. [2]
(d) Complete the calculation. A field produces 6.4 tonnes of wheat from an area of 2.0 hectares. Calculate the crop yield in tonnes per hectare.
crop yield = __________ tonnes per hectare [2]
(a) Starting with wheat, the sequence is wheat followed by beans, then cabbage, then clover, before wheat is grown again. [3]
(b) Bean roots contain bacteria that fix nitrogen, converting nitrogen gas into nitrogen compounds. These compounds become available in the soil. A later crop can use them to make proteins for growth, reducing the nitrogen fertiliser needed. [3]
(c) Crop rotation reduces the build-up of crop-specific pests and diseases. It can also prevent depletion of particular mineral ions, improve soil structure, and provide varied products. Any two are required. [2]
(d)
\[\text{crop yield}=\frac{6.4\text{ tonnes}}{2.0\text{ hectares}}=3.2\text{ tonnes per hectare}\]
[2]
Answer Details
(a) Starting with wheat, the sequence is wheat followed by beans, then cabbage, then clover, before wheat is grown again. [3]
(b) Bean roots contain bacteria that fix nitrogen, converting nitrogen gas into nitrogen compounds. These compounds become available in the soil. A later crop can use them to make proteins for growth, reducing the nitrogen fertiliser needed. [3]
(c) Crop rotation reduces the build-up of crop-specific pests and diseases. It can also prevent depletion of particular mineral ions, improve soil structure, and provide varied products. Any two are required. [2]
(d)
\[\text{crop yield}=\frac{6.4\text{ tonnes}}{2.0\text{ hectares}}=3.2\text{ tonnes per hectare}\]
[2]
Question 6 Report
Fig. 1 shows a bell-jar model used by a class to represent breathing in a human. The two balloons represent lungs. A student pulls the rubber sheet at the bottom of the model downwards.
(a) Name the human structure represented by the rubber sheet. [1]
(b) Describe what happens to the volume of air inside the bell jar when the sheet is pulled down. [1]
(c) Explain why air enters the balloons when the sheet is pulled down. [3]
(d) Design an improvement to this model so that it represents the movement of the ribs during breathing in. [4]
(a) The rubber sheet represents the diaphragm. [1]
(b) Pulling the sheet down increases the volume of air inside the bell jar. [1]
(c) Pulling the sheet down increases the volume of the chest model. [1] The pressure inside becomes lower than atmospheric pressure. [1] Air moves into the balloons down this pressure gradient. [1]
(d) To model rib movement, use a flexible cage or box around the balloons. [1] Add hinged rods to represent ribs. [1] Attach material representing intercostal muscles, or strings that pull the ribs. [1] Pull the ribs upwards and outwards [1], increasing the volume around the balloons so air enters through the tube. [1] Any four valid points gain the marks.
Answer Details
(a) The rubber sheet represents the diaphragm. [1]
(b) Pulling the sheet down increases the volume of air inside the bell jar. [1]
(c) Pulling the sheet down increases the volume of the chest model. [1] The pressure inside becomes lower than atmospheric pressure. [1] Air moves into the balloons down this pressure gradient. [1]
(d) To model rib movement, use a flexible cage or box around the balloons. [1] Add hinged rods to represent ribs. [1] Attach material representing intercostal muscles, or strings that pull the ribs. [1] Pull the ribs upwards and outwards [1], increasing the volume around the balloons so air enters through the tube. [1] Any four valid points gain the marks.
Question 7 Report
The diagram shows a researcher using a spirometer to record changes in the volume of air breathed by a person at rest and then after gentle exercise. The person wears a nose clip and breathes through a tube. Carbon dioxide is absorbed in a chamber containing soda lime, so the trace falls slowly over time.
(a) Name the gas absorbed by the soda lime. [1]
(b) Describe the pattern of breathing shown by a spirometer trace for a resting person. [2]
(c) Explain why the trace has larger waves after exercise. [3]
(d) Explain why the overall trace falls when carbon dioxide is absorbed but oxygen is not replaced. [3]
(e) Design a safe comparison of breathing rate before and after two different activities using this apparatus. [4]
(f) Suggest two reasons why the researcher should not test a person with a known serious lung condition without medical advice. [2]
(a) Soda lime absorbs carbon dioxide. [1]
(b) A resting spirometer trace has regular up-and-down waves. [1] Each wave represents breathing in and out, or one tidal volume. [1]
(c) Exercise increases respiration in muscle cells. [1] More oxygen is needed and more carbon dioxide is produced. [1] Breathing becomes deeper, so tidal volume increases and the trace has larger waves. [1]
(d) Carbon dioxide in exhaled air is removed by soda lime. [1] Oxygen is taken from the spirometer air into the blood. [1] Since carbon dioxide is not returned to replace this oxygen, total gas volume decreases and the trace falls. [1]
(e) A safe comparison should obtain informed consent and check health status [1], measure resting breathing rate for a fixed time [1], perform each activity for a fixed duration and intensity [1], then immediately measure breathing rate for the same time [1]. Allow recovery before the second activity, keep the same person, apparatus and room conditions, and repeat trials to calculate mean rates. Any four points gain the marks.
(f) Exercise or breathing through the apparatus could trigger breathing difficulty in someone with serious lung disease. [1] The test may be unsafe or require interpretation by a health professional. [1]
Answer Details
(a) Soda lime absorbs carbon dioxide. [1]
(b) A resting spirometer trace has regular up-and-down waves. [1] Each wave represents breathing in and out, or one tidal volume. [1]
(c) Exercise increases respiration in muscle cells. [1] More oxygen is needed and more carbon dioxide is produced. [1] Breathing becomes deeper, so tidal volume increases and the trace has larger waves. [1]
(d) Carbon dioxide in exhaled air is removed by soda lime. [1] Oxygen is taken from the spirometer air into the blood. [1] Since carbon dioxide is not returned to replace this oxygen, total gas volume decreases and the trace falls. [1]
(e) A safe comparison should obtain informed consent and check health status [1], measure resting breathing rate for a fixed time [1], perform each activity for a fixed duration and intensity [1], then immediately measure breathing rate for the same time [1]. Allow recovery before the second activity, keep the same person, apparatus and room conditions, and repeat trials to calculate mean rates. Any four points gain the marks.
(f) Exercise or breathing through the apparatus could trigger breathing difficulty in someone with serious lung disease. [1] The test may be unsafe or require interpretation by a health professional. [1]
Question 8 Report
A ranger photographed the organisms shown in Fig. 1 beside a mountain stream. Dippers eat mayfly larvae, and trout eat both mayfly larvae and dippers. Algae grow on submerged stones. During a hot summer, the stream level fell and some stones were no longer covered by water.
(a) State one food chain from Fig. 1 that contains four organisms. [2]
(b) Explain why algae are important to this food web. [2]
(c) Suggest how the lower stream level could affect the number of trout. [2]
(a) A four-organism food chain is:
algae → mayfly larva → dipper → trout [2]
The arrows show energy transfer from food to consumer.
(b) Algae are producers. They capture light energy and convert it into food or biomass, providing energy for consumers in the food web. [2]
(c) Trout numbers may decrease. A lower water level can reduce algae and thus reduce food through the web, or it can reduce available aquatic habitat for trout. [2]
Answer Details
(a) A four-organism food chain is:
algae → mayfly larva → dipper → trout [2]
The arrows show energy transfer from food to consumer.
(b) Algae are producers. They capture light energy and convert it into food or biomass, providing energy for consumers in the food web. [2]
(c) Trout numbers may decrease. A lower water level can reduce algae and thus reduce food through the web, or it can reduce available aquatic habitat for trout. [2]
Question 9 Report
A farmer grows bean plants in nutrient solution. Table 1 shows the mean height after 28 days when one mineral ion is omitted from the solution. All plants had equal light, water and starting height. The farmer needs to decide which mineral deficiency is most likely to affect crop yield and to explain why minerals are needed even though plants make their own food by photosynthesis.
| Solution provided | Mean plant height / cm | Leaf observation |
|---|---|---|
| Complete solution | 31 | dark green |
| Without nitrate ions | 14 | pale green |
| Without magnesium ions | 18 | yellow areas |
| Without phosphate ions | 22 | small leaves |
(a) State the mean height difference between plants given complete solution and plants without nitrate ions. [1]
(b) Name the mineral ion required to make chlorophyll. [1]
(c) Explain why plants without magnesium ions may have a lower photosynthesis rate. [2]
(d) Describe the evidence in the table that nitrate ions are important for growth. [2]
(e) Explain why a nitrate shortage can reduce protein production. [2]
(f) Give one reason why several plants should be used for each solution. [1]
(a) \[31\text{ cm}-14\text{ cm}=17\text{ cm}\] The mean height difference is 17 cm. [1]
(b) Magnesium ions are required to make chlorophyll. [1]
(c) Magnesium is needed to make chlorophyll. Less chlorophyll absorbs less light energy, so the rate of photosynthesis is lower. [2]
(d) Plants without nitrate ions have a mean height of 14 cm, compared with 31 cm for plants in complete solution. Their leaves are also pale green rather than dark green. These observations support the conclusion that nitrates are important for healthy growth. [2]
(e) Nitrates are needed to make amino acids, and amino acids are joined to make proteins. [2]
(f) Use several plants so a mean can be calculated, improving reliability and reducing the effect of natural variation. [1]
Answer Details
(a) \[31\text{ cm}-14\text{ cm}=17\text{ cm}\] The mean height difference is 17 cm. [1]
(b) Magnesium ions are required to make chlorophyll. [1]
(c) Magnesium is needed to make chlorophyll. Less chlorophyll absorbs less light energy, so the rate of photosynthesis is lower. [2]
(d) Plants without nitrate ions have a mean height of 14 cm, compared with 31 cm for plants in complete solution. Their leaves are also pale green rather than dark green. These observations support the conclusion that nitrates are important for healthy growth. [2]
(e) Nitrates are needed to make amino acids, and amino acids are joined to make proteins. [2]
(f) Use several plants so a mean can be calculated, improving reliability and reducing the effect of natural variation. [1]
Question 10 Report
A pharmacist is testing whether a new medicine affects communication between neurones. Fig. 1 shows a synapse, the tiny gap between the ending of one neurone and the next cell. In the test, the medicine prevents some neurotransmitter molecules from binding to receptor molecules.
(a) Name the gap between the two neurones shown in Fig. 1. [1]
(b) Name the chemical released from the end of the first neurone. [1]
(c) Explain how preventing neurotransmitter binding could reduce the number of impulses in the next neurone. [3]
(d) Describe two features of a synapse that allow transmission in only one direction. [2]
(e) State one reason why testing this medicine on isolated cells before testing it in a person is useful. [1]
(a) The gap is the synapse, also called the synaptic cleft. [1]
(b) The chemical released from the first neurone is a neurotransmitter. [1]
(c) If fewer neurotransmitter molecules bind to receptors, fewer ion channels open and the next neurone is less likely to depolarise. Its threshold is reached less often, so fewer nerve impulses are produced. [3]
(d) Neurotransmitter-containing vesicles that release transmitter occur only at the end of the first neurone. Receptor molecules occur only on the membrane of the next neurone. These arrangements make transmission one-way. [2]
(e) Testing isolated cells can identify an effect or toxicity before a person is exposed to unnecessary risk. [1]
Answer Details
(a) The gap is the synapse, also called the synaptic cleft. [1]
(b) The chemical released from the first neurone is a neurotransmitter. [1]
(c) If fewer neurotransmitter molecules bind to receptors, fewer ion channels open and the next neurone is less likely to depolarise. Its threshold is reached less often, so fewer nerve impulses are produced. [3]
(d) Neurotransmitter-containing vesicles that release transmitter occur only at the end of the first neurone. Receptor molecules occur only on the membrane of the next neurone. These arrangements make transmission one-way. [2]
(e) Testing isolated cells can identify an effect or toxicity before a person is exposed to unnecessary risk. [1]
Question 11 Report
Fig. 1 shows a river beside a small town. Treated sewage is released from an outfall pipe. Scientists sampled the water 20 m upstream and at several points downstream.
| Distance from outfall | Dissolved oxygen / mg per dm3 | Number of fish caught in a standard net |
|---|---|---|
| 20 m upstream | 9.1 | 18 |
| 100 m downstream | 5.4 | 7 |
| 500 m downstream | 7.8 | 14 |
(a) State two substances in untreated sewage that can cause water pollution. [2]
(b) Explain why the dissolved oxygen concentration falls soon after sewage enters the river. [4]
(c) Describe the relationship between dissolved oxygen concentration and the number of fish caught. [3]
(d) Suggest three ways the town could reduce pollution of the river. [3]
(a) Untreated sewage may contain faeces or other organic matter, urine or urea, detergents, pathogens, or nitrate-containing waste. Any two. [2]
(b) Organic material in sewage is food for decomposer microorganisms. Their population can increase, and they respire aerobically. Aerobic respiration uses dissolved oxygen from the river water, so oxygen concentration falls after the outfall. [4]
(c) Higher dissolved oxygen is associated with more fish caught. At 100 m downstream, oxygen is lowest at 5.4 mg dm-3 and only 7 fish are caught. By 500 m downstream, both oxygen concentration, 7.8 mg dm-3, and fish number, 14, have recovered. This is an association in these data; it does not alone prove oxygen is the only cause. [3]
(d) The town could improve sewage treatment, remove organic solids by settlement or filtration, use biological treatment, disinfect treated water, repair leaking sewers, and prevent chemicals being poured into drains. Any three. [3]
Answer Details
(a) Untreated sewage may contain faeces or other organic matter, urine or urea, detergents, pathogens, or nitrate-containing waste. Any two. [2]
(b) Organic material in sewage is food for decomposer microorganisms. Their population can increase, and they respire aerobically. Aerobic respiration uses dissolved oxygen from the river water, so oxygen concentration falls after the outfall. [4]
(c) Higher dissolved oxygen is associated with more fish caught. At 100 m downstream, oxygen is lowest at 5.4 mg dm-3 and only 7 fish are caught. By 500 m downstream, both oxygen concentration, 7.8 mg dm-3, and fish number, 14, have recovered. This is an association in these data; it does not alone prove oxygen is the only cause. [3]
(d) The town could improve sewage treatment, remove organic solids by settlement or filtration, use biological treatment, disinfect treated water, repair leaking sewers, and prevent chemicals being poured into drains. Any three. [3]
