Biology 4BI1 | Paper 1 Mock 01 | Written Paper 1
Question 1 Report
At a coastal animal-feed mill, technicians grow yeast in a tall vessel using diluted molasses. The yeast cells will be dried and mixed into food for farmed fish. Fig. 1 is a diagram of the vessel. Air enters near the base, and warmed water circulates through an outer jacket. Samples are removed at regular intervals to check the number of yeast cells.
(a) Name the process used by yeast cells to release energy when oxygen from the air is available. [1]
(b) Explain why air is bubbled through the liquid culture. [2]
(c) Describe how a student could use samples from the vessel to estimate the number of yeast cells per cm3. [3]
(d) Suggest two reasons why the temperature of the culture is controlled by the water jacket. [2]
(e) Explain why yeast biomass from this process can be a useful source of protein in fish food. [2]
(a) Yeast releases energy with oxygen by aerobic respiration. [1]
(b) Bubbling air supplies oxygen to yeast cells. They need oxygen for aerobic respiration, which releases energy for growth and cell division. [2]
(c) Take a measured volume of culture. Place it on a microscope slide or counting chamber and observe it under a microscope. Count cells in a known volume or area, then calculate the number of cells per cm3. [3]
(d) The water jacket keeps temperature near the optimum for yeast enzymes. It prevents temperatures becoming high enough to denature enzymes or low enough to make respiration and growth slow. Any two. [2]
(e) Yeast cells contain protein. Their biomass can be produced rapidly in large quantities using sugar-containing waste material and requires little land compared with producing animal protein. [2]
Answer Details
(a) Yeast releases energy with oxygen by aerobic respiration. [1]
(b) Bubbling air supplies oxygen to yeast cells. They need oxygen for aerobic respiration, which releases energy for growth and cell division. [2]
(c) Take a measured volume of culture. Place it on a microscope slide or counting chamber and observe it under a microscope. Count cells in a known volume or area, then calculate the number of cells per cm3. [3]
(d) The water jacket keeps temperature near the optimum for yeast enzymes. It prevents temperatures becoming high enough to denature enzymes or low enough to make respiration and growth slow. Any two. [2]
(e) Yeast cells contain protein. Their biomass can be produced rapidly in large quantities using sugar-containing waste material and requires little land compared with producing animal protein. [2]
Question 2 Report
The diagram shows a capillary network in the skin of a human runner. Blood enters the capillaries at A and leaves at B. At A, the pressure of the blood is high. Some plasma is forced out of the capillary and forms tissue fluid around body cells. Most of this fluid returns to the blood near B.
(a) Describe how oxygen moves from blood at A into the body cells. [2]
(b) Explain why plasma proteins usually remain in the capillary rather than entering tissue fluid. [2]
(c) State two substances that may be present in tissue fluid but are used by body cells. [2]
(d) Explain one advantage of forming tissue fluid around cells. [2]
(a) Oxygen moves by diffusion from a higher oxygen concentration in the blood at A to a lower oxygen concentration in the body cells. [2]
(b) Plasma-protein molecules are too large to pass through the pores in the capillary wall, so they normally remain in the capillary rather than entering tissue fluid. [2]
(c) Any two suitable substances are oxygen, glucose, amino acids and fatty acids. They may be present in tissue fluid and are taken up for use by body cells. [2]
(d) Tissue fluid provides a medium for exchange between capillary blood and cells. It gives cells a short diffusion distance, helping maintain delivery of useful substances and removal of wastes. [2]
Answer Details
(a) Oxygen moves by diffusion from a higher oxygen concentration in the blood at A to a lower oxygen concentration in the body cells. [2]
(b) Plasma-protein molecules are too large to pass through the pores in the capillary wall, so they normally remain in the capillary rather than entering tissue fluid. [2]
(c) Any two suitable substances are oxygen, glucose, amino acids and fatty acids. They may be present in tissue fluid and are taken up for use by body cells. [2]
(d) Tissue fluid provides a medium for exchange between capillary blood and cells. It gives cells a short diffusion distance, helping maintain delivery of useful substances and removal of wastes. [2]
Question 3 Report
At a neonatal unit, staff study Fig. 1 to show how an unborn human obtains materials for growth. The diagram shows chorionic villi pressed against maternal blood spaces in the uterus. Table 1 records concentrations measured in blood entering and leaving the placenta on the fetal side. The mother and fetus have separate circulatory systems, even though useful substances move across a very short distance.
(a) State the role of the umbilical cord. [2]
(b) Use Table 1 to identify one substance taken into fetal blood at the placenta and one substance removed from fetal blood. [2]
(c) Explain how three features of the placenta make exchange between maternal blood and fetal blood efficient. [4]
(d) Explain why it is important that maternal and fetal blood do not normally mix directly. [2]
(e) Give one substance, other than carbon dioxide, that can pass from fetal blood to the mother’s blood. [1]
(a) The umbilical cord connects the fetus to the placenta [1] and carries fetal blood to and from the placenta. [1]
(b) Oxygen is taken into fetal blood at the placenta. [1] Carbon dioxide, or urea, is removed from fetal blood. [1]
(c) Exchange is efficient because:
- Many chorionic villi provide a large surface area. [1]
- The villi have thin walls, producing a short diffusion distance. [1]
- A rich maternal blood supply maintains concentration gradients. [1]
- A dense fetal capillary network also maintains concentration gradients. [1]
- The membranes are permeable to suitable small molecules. [1]
Any four points are required. Diffusion is fastest when surface area is large, the path is short, and a steep concentration gradient is maintained.
(d) Maternal and fetal blood may have incompatible blood groups. [1] Direct mixing could trigger an immune response or agglutination, harming blood cells. [1]
(e) Urea can pass from fetal blood to the mother's blood. [1]
Answer Details
(a) The umbilical cord connects the fetus to the placenta [1] and carries fetal blood to and from the placenta. [1]
(b) Oxygen is taken into fetal blood at the placenta. [1] Carbon dioxide, or urea, is removed from fetal blood. [1]
(c) Exchange is efficient because:
- Many chorionic villi provide a large surface area. [1]
- The villi have thin walls, producing a short diffusion distance. [1]
- A rich maternal blood supply maintains concentration gradients. [1]
- A dense fetal capillary network also maintains concentration gradients. [1]
- The membranes are permeable to suitable small molecules. [1]
Any four points are required. Diffusion is fastest when surface area is large, the path is short, and a steep concentration gradient is maintained.
(d) Maternal and fetal blood may have incompatible blood groups. [1] Direct mixing could trigger an immune response or agglutination, harming blood cells. [1]
(e) Urea can pass from fetal blood to the mother's blood. [1]
Question 4 Report
Fig. 1 shows stages used to produce a cloned sheep from a prize ewe. The nucleus is taken from a body cell of the prize ewe and placed into an egg cell from another sheep.
(a) Name the cell structure labelled X that must be transferred from the prize ewe. [1]
(b) Complete the description of stages Y and Z.
Stage Y: the egg cell nucleus is removed so that the egg cell contains no __________.
Stage Z: the reconstructed cell divides by __________ to form an embryo. [2]
(c) Explain why the lamb born from the surrogate has the same inherited characteristics as the prize ewe. [4]
(d) Describe how a farmer could use cloning to increase the number of sheep with a valuable characteristic, such as resistance to a disease. [3]
(e) Explain two reasons why a flock made by cloning many sheep could be at risk if a new disease spreads through it. [4]
(a) Structure X is the nucleus. [1]
(b) At stage Y, the egg cell nucleus is removed so the egg contains no nucleus. At stage Z, the reconstructed cell divides by mitosis to form an embryo. [2]
(c) The donor nucleus contains chromosomes, DNA and genes from the prize ewe. Genes control inherited characteristics. Because the egg nucleus was removed, the embryo receives the prize ewe’s nuclear genetic information. Therefore the lamb has the same inherited characteristics as the prize ewe. [4]
(d) Select a sheep with the valuable characteristic. Transfer nuclei from its body cells into egg cells whose nuclei have been removed. Stimulate the cells to form embryos and implant them into surrogate sheep, producing many cloned lambs. [3]
(e) Cloned sheep have little or no genetic variation. They are likely to have the same susceptibility to a pathogen, so a new disease could infect many or all of them. Few resistant individuals would survive and reproduce, leaving the flock especially vulnerable. [4]
Answer Details
(a) Structure X is the nucleus. [1]
(b) At stage Y, the egg cell nucleus is removed so the egg contains no nucleus. At stage Z, the reconstructed cell divides by mitosis to form an embryo. [2]
(c) The donor nucleus contains chromosomes, DNA and genes from the prize ewe. Genes control inherited characteristics. Because the egg nucleus was removed, the embryo receives the prize ewe’s nuclear genetic information. Therefore the lamb has the same inherited characteristics as the prize ewe. [4]
(d) Select a sheep with the valuable characteristic. Transfer nuclei from its body cells into egg cells whose nuclei have been removed. Stimulate the cells to form embryos and implant them into surrogate sheep, producing many cloned lambs. [3]
(e) Cloned sheep have little or no genetic variation. They are likely to have the same susceptibility to a pathogen, so a new disease could infect many or all of them. Few resistant individuals would survive and reproduce, leaving the flock especially vulnerable. [4]
Question 5 Report
A horticulturist photographs a cactus growing in a dry botanical garden. Fig. 1 shows the structure of the plant. The garden has very little rainfall for several months each year, so conserving water is important for the cactus cells.
(a) Describe two visible features of the cactus in Fig. 1 that help it survive in a dry habitat. [4]
(b) Explain how each of these features reduces water loss or improves water supply to the plant body. [4]
(c) State the name of the process by which water vapour leaves a plant leaf or stem. [1]
(d) Suggest why a cactus with a large leaf surface area would be less likely to survive in this habitat. [2]
(a) Two visible xerophytic features are leaves reduced to spines and a thick, fleshy green stem. Other acceptable visible features are wide-spreading shallow roots and a waxy outer surface. For each feature, description gains credit. [4]
(b) Spines have a small surface area, reducing transpiration or evaporation. The fleshy stem stores water. Wide shallow roots rapidly absorb water from a large area after rain, and a waxy layer reduces evaporation. Explain the two features selected in part (a). [4]
(c) Water vapour leaves a plant by transpiration. [1]
(d) A large leaf surface area exposes more cells to the air. More water would then be lost by transpiration, making survival in a dry habitat less likely. [2]
Answer Details
(a) Two visible xerophytic features are leaves reduced to spines and a thick, fleshy green stem. Other acceptable visible features are wide-spreading shallow roots and a waxy outer surface. For each feature, description gains credit. [4]
(b) Spines have a small surface area, reducing transpiration or evaporation. The fleshy stem stores water. Wide shallow roots rapidly absorb water from a large area after rain, and a waxy layer reduces evaporation. Explain the two features selected in part (a). [4]
(c) Water vapour leaves a plant by transpiration. [1]
(d) A large leaf surface area exposes more cells to the air. More water would then be lost by transpiration, making survival in a dry habitat less likely. [2]
Question 6 Report
A coastal fish farm uses floating sea cages to rear salmon. Fig. 1 shows a cross-section through one cage. Workers add measured amounts of pellet food each day and inspect the fish for disease. Nets stop the fish escaping, but water can move through the cage. The farm manager wants rapid fish growth without causing pollution in the surrounding area.
(a) Describe two features of the cage that allow the salmon to remain alive. [2]
(b) Explain why uneaten food and fish faeces may reduce the number of organisms living on the sea bed below a cage. [2]
(c) State two ways the farm could reduce the spread of disease between fish in the cage. [2]
(a) The net prevents salmon escaping. Water moves through the cage, providing oxygen and removing wastes. [2]
(b) Uneaten food and faeces increase the organic matter or nutrient concentration below the cage. Decomposers break down this waste and use oxygen, leaving less oxygen for sea-bed organisms, so fewer can survive. [2]
(c) Two suitable measures are to isolate infected fish and vaccinate fish. Reducing stocking density, disinfecting equipment, promptly removing dead fish, and preventing sharing of contaminated equipment are also valid. [2]
Answer Details
(a) The net prevents salmon escaping. Water moves through the cage, providing oxygen and removing wastes. [2]
(b) Uneaten food and faeces increase the organic matter or nutrient concentration below the cage. Decomposers break down this waste and use oxygen, leaving less oxygen for sea-bed organisms, so fewer can survive. [2]
(c) Two suitable measures are to isolate infected fish and vaccinate fish. Reducing stocking density, disinfecting equipment, promptly removing dead fish, and preventing sharing of contaminated equipment are also valid. [2]
Question 7 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 8 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 9 Report
The table below shows the diameters of four yeast cells measured by a student during a brewery investigation. Fig. 1 is a diagram showing a yeast cell with a small bud growing from its surface. The brewer wants yeast cells to divide rapidly when sugar solution is added.
| yeast cell | diameter / micrometres |
|---|---|
| 1 | 3.8 |
| 2 | 4.1 |
| 3 | 3.9 |
| 4 | 4.0 |
(a) Calculate the mean diameter of the four yeast cells. Give your answer to two decimal places. [2]
(b) Explain how the diagram shows that yeast can reproduce asexually. [3]
(c) Describe four differences between a yeast cell and a cell from a green plant leaf. [4]
(d) Design a method the student could use to compare the number of living yeast cells in two sugar solutions after 30 minutes. [3]
(a)
\[\text{mean diameter}=\frac{3.8+4.1+3.9+4.0}{4}=\frac{15.8}{4}=3.95\ \mu\text{m}\]
The mean diameter is 3.95 µm. [2]
(b) A small bud grows from the parent yeast cell. The bud receives a nucleus, or genetic material, by mitosis. It separates to form a genetically identical new yeast cell, so only one parent is involved and reproduction is asexual. [3]
(c) Four differences are:
- Yeast has no chloroplasts, whereas a green leaf cell has chloroplasts.
- Yeast has no large permanent vacuole, or has smaller vacuoles.
- Yeast cells are usually oval or round, whereas leaf cells are more regular or box-shaped.
- A green leaf cell contains chlorophyll, whereas yeast does not.
- Leaf cells are arranged in tissues, whereas yeast cells may occur singly.
Any four correct differences gain credit. [4]
(d) Place equal volumes and concentrations of the two sugar solutions into separate containers with equal starting amounts of yeast. Keep temperature, time and other relevant variables the same for 30 minutes. Add a dye such as methylene blue, then use a microscope or haemocytometer to count unstained living cells and compare the counts. [3]
Answer Details
(a)
\[\text{mean diameter}=\frac{3.8+4.1+3.9+4.0}{4}=\frac{15.8}{4}=3.95\ \mu\text{m}\]
The mean diameter is 3.95 µm. [2]
(b) A small bud grows from the parent yeast cell. The bud receives a nucleus, or genetic material, by mitosis. It separates to form a genetically identical new yeast cell, so only one parent is involved and reproduction is asexual. [3]
(c) Four differences are:
- Yeast has no chloroplasts, whereas a green leaf cell has chloroplasts.
- Yeast has no large permanent vacuole, or has smaller vacuoles.
- Yeast cells are usually oval or round, whereas leaf cells are more regular or box-shaped.
- A green leaf cell contains chlorophyll, whereas yeast does not.
- Leaf cells are arranged in tissues, whereas yeast cells may occur singly.
Any four correct differences gain credit. [4]
(d) Place equal volumes and concentrations of the two sugar solutions into separate containers with equal starting amounts of yeast. Keep temperature, time and other relevant variables the same for 30 minutes. Add a dye such as methylene blue, then use a microscope or haemocytometer to count unstained living cells and compare the counts. [3]
Question 10 Report
A camera technician compares the focusing system of a camera with the eye. Fig. 1 shows an eye looking at a nearby mobile phone. The phone is 25 cm from the eye, so the image must be focused sharply on the retina rather than in front of it or behind it.
(a) Complete the table to compare the lens of the eye when viewing a nearby object and a distant object. [2]
| Object viewed | Shape of lens | State of suspensory ligaments |
|---|---|---|
| Nearby mobile phone | ________ | slack |
| Distant road sign | thin / less curved | ________ |
(b) Explain how ciliary muscles and suspensory ligaments make the lens thicker when the mobile phone is viewed. [3]
(c) State two structures, other than the lens, that light passes through before reaching the retina. [2]
(a) For a nearby mobile phone, the lens is thick or more curved. For a distant road sign, the suspensory ligaments are tight or stretched. [2]
(b) To view the nearby phone, the ciliary muscles contract. This makes the suspensory ligaments slack, allowing the lens to become thicker and more curved. The thicker lens refracts light more strongly so that the image is focused on the retina. [3]
(c) Two structures light may pass through before reaching the retina are the cornea and aqueous humour. [2] The pupil and vitreous humour are also acceptable.
Answer Details
(a) For a nearby mobile phone, the lens is thick or more curved. For a distant road sign, the suspensory ligaments are tight or stretched. [2]
(b) To view the nearby phone, the ciliary muscles contract. This makes the suspensory ligaments slack, allowing the lens to become thicker and more curved. The thicker lens refracts light more strongly so that the image is focused on the retina. [3]
(c) Two structures light may pass through before reaching the retina are the cornea and aqueous humour. [2] The pupil and vitreous humour are also acceptable.
Question 11 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]
