Year 11 Paper 5: Disease, Energy and Genetics
Covers cell division and stem cells, pathogens and communicable disease, photosynthesis, respiration and exercise, the nervous system and reflexes, hormones and homeostasis, and genetics and inheritance.
Year 11 here means a typical teaching order, not a syllabus rule. No exam board defines what belongs to Year 11, and schools sequence the course differently. Check it against your own scheme of work before using it to decide what a class has covered.
Questions
Question 1 [4 marks]
Pathogens and Communicable Disease
A city of 250,000 people recorded 500 new cases of a disease in one month.
Calculate the incidence rate of the disease, in cases per 1000 people, for that month.
Question 2 [5 marks]
The Nervous System and Reflexes
Some drugs work by blocking the receptor molecules on the postsynaptic neurone at a synapse, so that a neurotransmitter cannot bind to them.
Explain the effect this type of drug would have on the transmission of nerve impulses across that synapse.
Question 3 [5 marks]
Photosynthesis
A grower measures that one tomato plant produces 40 g of new dry biomass per week under standard greenhouse conditions. Under improved lighting, this rises to 58 g per week.
Calculate the percentage increase in biomass production per plant, and calculate the total extra biomass produced per week across a greenhouse of 250 plants, assuming every plant shows the same increase.
Question 4 [5 marks]
Hormones and Homeostasis
After an insulin injection, a patient's blood glucose concentration fell from 12 mmol/l to 6 mmol/l over 45 minutes.
Calculate the mean rate of decrease in blood glucose concentration in mmol/l per hour.
Question 5 [5 marks]
Respiration and Exercise
Explain, in terms of the breakdown of glucose, why aerobic respiration releases much more energy from each glucose molecule than anaerobic respiration.
Question 6 [6 marks]
Photosynthesis
A commercial grower increases the carbon dioxide concentration in a greenhouse from 0.04% to 0.1% while keeping light intensity and temperature high and non-limiting.
Explain why increasing carbon dioxide concentration in the greenhouse is likely to increase the rate of photosynthesis of the plants, and explain why increasing carbon dioxide concentration further, to 0.5%, might not increase the rate any further.
Question 7 [6 marks]
Cell Division and Stem Cells
A laboratory needs to grow 1,000,000 skin cells from a single starting cell for a skin graft, by mitosis. The cells divide every 24 hours.
Calculate the minimum number of complete divisions needed to produce at least 1,000,000 cells, and calculate how many days this would take.
Question 8 [6 marks]
Photosynthesis
An investigation into the effect of temperature on the rate of photosynthesis in a plant found the rate increased steadily up to 25 deg C, reached a maximum rate of 8 arbitrary units at 25 deg C, then fell sharply to 1 arbitrary unit by 40 deg C.
Calculate the percentage decrease in rate between 25 deg C and 40 deg C, and explain, in terms of enzymes, the shape of this graph above 25 deg C.
Question 9 [6 marks]
The Nervous System and Reflexes
A student measured reaction time 3 times and found a range of 0.15 seconds. A second student measured reaction time 10 times under the same conditions and found a range of 0.04 seconds, with five of the ten results being 0.28, 0.24, 0.31, 0.26 and 0.29 seconds.
Calculate the mean of these five results, and explain why the second student's results are likely to be more reliable than the first student's.
Question 10 [6 marks]
Hormones and Homeostasis
A person with poorly controlled type 1 diabetes has a blood glucose concentration that is often much higher than normal.
Explain two ways in which persistently high blood glucose concentration could damage the body over time.
Question 11 [6 marks]
Genetics and Inheritance
Both parents are heterozygous carriers of a recessive allele for a genetic disorder (Ff).
Using a genetic diagram, state the probability that any one child of these parents has the disorder (ff). The couple then has two children. Calculate the probability that both children have the disorder, assuming each pregnancy is independent.
Model solutions
| Question 1[4 marks] | |
|---|---|
| Answer or working | Marks |
| using rate = cases / population | M1 |
| substituting 500 / 250,000 | M1 |
| multiplying by 1000 to convert to cases per 1000 people | M1 |
| 2 cases per 1000 people | A1 |
| Question 2[5 marks] | |
|---|---|
| Answer or working | Marks |
| normally, a neurotransmitter released from the presynaptic neurone diffuses across the synapse and binds to specific receptor molecules on the postsynaptic neurone | B1 |
| this binding normally triggers a new electrical impulse in the postsynaptic neurone | B1 |
| if the drug blocks (occupies) the receptor molecules, the neurotransmitter can no longer bind to them | B1 |
| without the neurotransmitter binding to a receptor, a new impulse cannot be triggered in the postsynaptic neurone | B1 |
| the drug would therefore prevent (or reduce) transmission of the nerve impulse across that synapse | B1 |
| Final answer: Blocking the postsynaptic receptors stops the neurotransmitter binding to them, so a new impulse cannot be triggered in the postsynaptic neurone, preventing transmission of the nerve impulse across that synapse | |
| Question 3[5 marks] | |
|---|---|
| Answer or working | Marks |
| the increase = 58 - 40 = 18 g | M1 |
| percentage increase = (18 / 40) x 100 | M1 |
| 45% | A1 |
| total extra biomass = 18 x 250 | M1 |
| 4500 g | A1 |
| Final answer: A 45% increase per plant; across 250 plants this is an extra 4500 g of biomass per week | |
| Question 4[5 marks] | |
|---|---|
| Answer or working | Marks |
| finding the fall in concentration = 12 - 6 = 6 mmol/l | M1 |
| converting 45 minutes to hours = 45/60 = 0.75 hours | M1 |
| using rate = change / time | M1 |
| rate = 6 / 0.75 = 8 mmol/l per hour | A1 |
| stating the concentration falls because insulin causes cells to take up glucose from the blood | B1independent |
| Final answer: The blood glucose concentration fell at a mean rate of 8 mmol/l per hour, as insulin caused cells to take up glucose from the blood | |
| Question 5[5 marks] | |
|---|---|
| Answer or working | Marks |
| aerobic respiration uses oxygen to completely break down glucose into carbon dioxide and water | B1 |
| this complete breakdown releases all of the energy stored in the glucose molecule | B1 |
| anaerobic respiration does not use oxygen, so it can only partially break down glucose | B1 |
| glucose is only broken down as far as lactic acid, a molecule that still contains a large amount of stored (chemical) energy | B1 |
| because most of the energy originally in the glucose remains locked in the lactic acid rather than being released, anaerobic respiration releases far less usable energy per glucose molecule | B1 |
| Final answer: Aerobic respiration fully breaks glucose down into carbon dioxide and water using oxygen, releasing all the stored energy; anaerobic respiration only partially breaks glucose down to lactic acid, which still holds most of the energy, so far less is released | |
| Question 6[6 marks] | |
|---|---|
| Answer or working | Marks |
| carbon dioxide is a raw material (reactant) needed for photosynthesis | B1 |
| at low concentrations carbon dioxide is often the limiting factor, so increasing it increases the rate provided light and temperature are not limiting | B1 |
| more carbon dioxide molecules are available to react in the light-independent reactions, increasing the rate at which glucose is produced | B1 |
| beyond a certain concentration, carbon dioxide is no longer the limiting factor | B1 |
| another factor, such as light intensity or temperature, becomes limiting instead | B1 |
| increasing carbon dioxide further having no effect on the rate because that other factor now limits how fast photosynthesis can proceed | B1 |
| Final answer: Increasing carbon dioxide increases the rate while it is the limiting factor, providing more raw material for the reaction; beyond a point, another factor such as light intensity or temperature becomes limiting, so further carbon dioxide has no effect | |
| Question 7[6 marks] | |
|---|---|
| Answer or working | Marks |
| recognising cell number after n divisions = 2^n | M1 |
| testing 2^19 = 524,288 (below 1,000,000) | M1 |
| testing 2^20 = 1,048,576 (exceeds 1,000,000) | M1 |
| 20 divisions | A1 |
| time = 20 x 24 = 480 hours | B1independent |
| converting to 20 days | B1independent |
| Final answer: 20 divisions are needed (2^20 = 1,048,576), taking 480 hours, which is 20 days | |
| Question 8[6 marks] | |
|---|---|
| Answer or working | Marks |
| finding the decrease = 8 - 1 = 7 | M1 |
| using percentage decrease = (decrease / original) x 100 | M1 |
| 87.5% decrease | A1 |
| below 25 deg C, higher temperature gives molecules (including enzyme and substrate) more kinetic energy, increasing the rate of the enzyme-controlled reactions of photosynthesis | B1 |
| above 25 deg C, the enzymes involved in photosynthesis begin to denature, as heat breaks the bonds holding their structure/active site shape | B1 |
| denatured enzymes can no longer catalyse the reactions of photosynthesis effectively, causing the rate to fall sharply as temperature rises further | B1 |
| Final answer: The rate fell by 87.5% between 25 deg C and 40 deg C; above 25 deg C the enzymes controlling photosynthesis begin to denature, losing their functional shape, so the rate falls sharply as temperature rises further | |
| Question 9[6 marks] | |
|---|---|
| Answer or working | Marks |
| summing the five results = 0.28 + 0.24 + 0.31 + 0.26 + 0.29 = 1.38 | M1 |
| dividing by 5 | M1 |
| a mean of 0.276, which rounds to 0.28 seconds (2 significant figures) | A1 |
| the second student taking more repeat readings (10 compared with 3) | B1 |
| the second student's results also having a smaller range, meaning the results are more consistent/closer together | B1 |
| more repeats, especially with a smaller range, generally making a mean value more reliable, as the effect of one anomalous or variable reading has less influence on the mean | B1 |
| Final answer: The mean of the five results is 0.28 seconds (2 s.f.); the second student's results are more reliable because they took more repeats (10, not 3) and had a smaller range, showing more consistent results, so the mean is less affected by any single anomalous reading | |
| Question 10[6 marks] | |
|---|---|
| Answer or working | Marks |
| persistently high blood glucose concentration can damage blood vessels over time | B1 |
| damage to blood vessels can reduce blood supply to organs, and increases the risk of conditions such as heart disease or poor circulation to the limbs | B1 |
| high blood glucose concentration can also damage nerves over time | B1 |
| nerve damage can cause loss of sensation, particularly in the extremities such as the feet, meaning injuries may go unnoticed | B1 |
| high blood glucose can damage the small blood vessels in the eyes (the retina) | B1 |
| this eye damage can gradually impair vision and, if untreated, cause blindness | B1 |
| Final answer: Persistently high blood glucose can damage blood vessels, increasing the risk of heart disease and poor circulation, and can damage nerves and the small vessels in the eyes, causing loss of sensation and, over time, damage to vision | |
| Question 11[6 marks] | |
|---|---|
| Answer or working | Marks |
| a correctly completed genetic diagram (Punnett square) crossing Ff with Ff | M1 |
| identifying the probability of one child having the disorder as 25% (1 in 4) | A1 |
| recognising that each pregnancy is an independent event, so the outcome for one child does not affect the probability for the other | B1 |
| using the multiplication rule for two independent events: probability (both) = probability (one) x probability (other) | M1 |
| substituting 0.25 x 0.25 (or 1/4 x 1/4) | M1 |
| 0.0625, or 6.25% (or 1 in 16) | A1 |
| Final answer: The probability of one child having the disorder is 25% (1 in 4); because each pregnancy is independent, the probability of both children having it is 0.25 x 0.25 = 0.0625, or 6.25% (1 in 16) | |