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Mixed Topic Math Questions
Marks= 150 Time= 157
All of the questions below contain a maths questions within them. To still give the full context of the question, the non-maths components of each question are still included.
Q1.Table 1 shows information about some food components in cow’s milk.
Table 1
Value per 500 cm3
Recommended Daily Allowance (RDA) for a
typical adult
Energy in kJ 1046 8700
Fat in g 8.4 70.0
Salt in g 0.5 6.0
Calcium in mg 605 1000
Vitamin B-12 in µg 4.5 2.4
(a) How much more milk would a typical adult have to drink to get their RDA for calcium compared with the amount of milk needed to get their RDA for vitamin B-12?
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Volume of milk = ______________________ cm3
(3)
(b) Describe how a student could test cow’s milk to show whether it contains protein and different types of carbohydrate.
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A scientist investigated the effect of bile on the breakdown of fat in a sample of milk.
The scientist used an indicator that is colourless in solutions with a pH lower than 10, and pink in solutions with a pH above 10.
This is the method used.
1. Add 1 drop of bile to a test tube and one drop of water to a second test tube.
2. Add the following to each test tube:• 5 cm3 of milk• 7 cm3 of sodium carbonate solution (to make the solution above pH 10)• 5 drops of the indicator• 1 cm3 of lipase.
3. Time how long it takes for the indicator in the solutions to become colourless.
The results are shown in Table 2.
Table 2
Time taken for the indicator to become colourless in seconds
Solution with bile 65
Solution without bile 143
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(c) Explain why the indicator in both tubes became colourless.
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(d) Give the reason why the measurement of the time taken for the indicator to become colourless might be inaccurate.
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(e) Explain the difference in the results for the two test tubes in Table 2.
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(Total 16 marks)
Q2.The parts of the blood can be separated from each other by spinning the blood in a centrifuge.
The image below shows the separated parts of a 10 cm3 blood sample.
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(a) Calculate the percentage of the blood that is made up of plasma.
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Answer = _______________ %(2)
(b) Name three chemical substances transported by the plasma.
1. _________________________________________________________________
2. _________________________________________________________________
3. _________________________________________________________________(3)
(c) In this question you will be assessed on using good English, organising information clearly and using specialist terms where appropriate.
White blood cells are part of the immune system. White blood cells help the body to defend itself against pathogens.
Describe how pathogens cause infections and describe how the immune system defends the body against these pathogens.
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(Total 11 marks)
Q3.Catalase is an enzyme found in many different tissues in plants and animals. It speeds up the rate of the following reaction.
hydrogen peroxide water + oxygen
Figure 1 shows a 25-day-old broad bean seedling.
Some students investigated whether different parts of bean seedlings contained different amounts of catalase.
The students:• put hydrogen peroxide into five test tubes
• added a different part of a bean seedling to each tube
• recorded the results after half a minute.
If there was catalase in part of the seedling, oxygen gas was given off.When oxygen gas is given off, foam is produced in the tubes.
Figure 2 shows the results.
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The students made the following conclusions:• most parts of a bean seedling contain catalase
• the seed contains a lot of catalase
• stems and roots have quite a lot of catalase
• the leaves have a little bit of catalase
• the seed coat has hardly any catalase.
The students’ teacher said that the students needed to improve their investigation in order to make valid conclusions.
(a) In this question you will be assessed on using good English, organising information clearly and using specialist terms where appropriate.
Describe how you would carry out an investigation to compare the amounts of catalase in different parts of bean seedlings.
You should include details of how you would make sure your results give a valid comparison of the amounts of catalase.
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(b) Scientists investigated the effect of pH on the activity of the enzyme catalase in a fungus.
The table below shows the scientists’ results.
pHEnzyme activity in arbitrary units
Test 1 Test 2 Test 3 Test 4 Test 5 Mean
3.0 0 0 0 0 0 0
4.0 6 5 8 4 7 6
5.0 38 65 41 42 39
5.5 80 86 82 84 88 84
6.0 100 99 96 103 102 100
6.5 94 92 90 93 91 92
7.0 61 63 61 62 63 62
8.0 22 22 21 24 21 22
(i) Calculate the mean enzyme activity at pH 5.0.
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Mean = _____________ arbitrary units(2)
(ii) On the graph paper in Figure 3, draw a graph to show the scientists’ results.
Remember to:• add a label to the vertical axis
• plot the mean values of enzyme activity
• draw a line of best fit.
Figure 3
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(4)
(iii) At what pH does the enzyme work best?
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(iv) Predict the activity of the enzyme at pH 9.0.
____________________ arbitrary units(1)
(v) Suggest why the enzyme’s activity at pH 3.0 is zero.
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(Total 15 marks)
Q4.Polydactyly is an inherited condition caused by a dominant allele.
(a) The figure below shows the hand of a man with polydactyly. The man has an extra finger on each hand.
The man’s mother also has polydactyly but his father does not.
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© Ifness/iStock
(i) The man is heterozygous for polydactyly.
Explain how the information given above shows that the man is heterozygous for polydactyly.
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(ii) The man marries a woman who does not have polydactyly.
What is the probability that their first child will have polydactyly?
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(b) The man has red hair. His sister has brown hair.
Both of their parents have brown hair.
Brown hair is caused by the dominant allele, B.
Red hair is caused by a recessive allele, b.
Complete the genetic diagram below to show how the man’s parents were able to have some children with red hair and some with brown hair.
Father Mother
Parental phenotypes _________________ _________________
Parental Genotypes _________________ _________________
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Gametes ________ ________ ________ ________
Offspring genotypes: _____________________________________________
Offspring phenotypes: ____________________________________________(5)
(Total 9 marks)
Q5.An athlete ran as fast as he could until he was exhausted.
(a) Figure 1 shows the concentrations of glucose and of lactic acid in the athlete’s blood at the start and at the end of the run.
(i) Lactic acid is made during anaerobic respiration.
What does anaerobic mean?
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(ii) Give evidence from Figure 1 that the athlete respired anaerobically during the run.
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(b) Figure 2 shows the effect of running on the rate of blood flow through the athlete’s
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muscles.
(i) For how many minutes did the athlete run?
Time = ______________________ minutes(1)
(ii) Describe what happens to the rate of blood flow through the athlete’s muscles during the run.
Use data from Figure 2 in your answer.
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(iii) Explain how the change in blood flow to the athlete’s muscles helps him to run.
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(Total 9 marks)
Q6.A student plans an investigation using mould.
(a) Mould spores are hazardous.
Give one safety precaution the student should take when doing this investigation.
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(b) A student made the following hypothesis about the growth of mould:
‘The higher the temperature, the faster the growth of mould’.
The student planned to measure the amount of mould growing on bread.
The student used the following materials and equipment:
• slices of bread
• sealable plastic bags
• a knife
• a chopping board
• mould spores.
Describe how the materials and equipment could be used to test the hypothesis.
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(c) Give one variable the student should control in the investigation.
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(d) Another student did a similar investigation.
The diagram below shows the results.
Determine the rate of mould growth at 42 °C between day 2 and day 7.
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Rate of mould growth = _______________ units per day(2)
(e) The growth of mould shows decomposition of the bread.
Give a conclusion about decomposition from the results in the diagram above.
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(Total 9 marks)
Q7.This question is about the nervous system.
(a) Describe the difference between the function of a receptor and the function of an effector.
In your answer you should give one example of a receptor and one example of an effector.
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(b) Synapses are important in the nervous system.
(i) What is a synapse?
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(ii) Describe how information passes across a synapse.
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(c) Reflexes may be co-ordinated by the brain or by the spinal cord.
(i) The reflexes from sense organs in the head are co-ordinated by the brain.
Name a sense organ involved in a reflex co-ordinated by the spinal cord.
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(ii) The table shows information about reflexes co-ordinated by the brain and reflexes co-ordinated by the spinal cord.
Organ co-ordinating the reflex
Mean length of neurones involved in
cm
Mean time taken for reflex in
milliseconds
Mean speed of impulse in
cm per millisecond
Brain 12 4 3
Spinal cord 80 50
Calculate the mean speed of the impulse for the reflex co-ordinated by the spinal cord.
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Mean speed = ________ cm per millisecond(1)
(iii) In reflexes co-ordinated by the brain there are no relay neurones.
Suggest why there is a difference in the mean speed of the impulse for the two reflexes.
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(Total 12 marks)
Q8.The heart pumps blood to the lungs and to the cells of the body.
(a) Name the blood vessel that transports blood from the body to the right atrium.
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(b) The aorta transports blood from the heart to the body.
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In a person at rest:• blood travels at a mean speed of 10 cm/s in the aorta• blood travels at a mean speed of 0.5 mm/s in the capillaries• the speed of blood decreases at a rate of 0.4 cm/s2 as blood travels from the
aorta to the capillaries.
Calculate the time it takes for blood to travel from the aorta to the capillaries.
Assume that the speed of blood decreases at a constant rate.
Use the equation:
Give your answer to 2 significant figures.
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Time = ____________________ s(4)
(c) Describe the route taken by oxygenated blood from the lungs to the body cells.
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(d) The digestive system and the breathing system both contain specialised exchange surfaces.• In the digestive system, digested food is absorbed into the
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blood stream in structures called villi.• In the breathing system, gases are absorbed into the blood
stream in the alveoli.
The diagram below shows the structure of villi and alveoli.
Explain how the villi and the alveoli are adapted to absorb molecules into the bloodstream.
(6)(Total 15 marks)
Q9.A gardener wants to add compost to the soil to increase his yield of strawberries.
The gardener wants to make his own compost.
(a) An airtight compost heap causes anaerobic decay.
Explain why the gardener might be against producing compost using this method.
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(b) The gardener finds this research on the Internet:
‘A carbon to nitrogen ratio of 25:1 will produce fertile compost.’
Look at the table below.
Type of material to compost
Mass of carbon in
sample in g
Mass of nitrogen
in sample in gCarbon:nitrogen ratio
Chicken manure 8.75 1.25 7:1
Horse manure 10.00 0.50 20:1
Peat moss 9.80 0.20 X
Determine the ratio X in the table above.
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Ratio ________________(1)
(c) Which type of material in the table above would be best for the gardener to use to make his compost?
Justify your answer.
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(d) Some of the leaves from the gardener’s strawberry plant die.
The dead leaves fall off the strawberry plant onto the ground.
The carbon in the dead leaves is recycled through the carbon cycle.
Explain how the carbon is recycled into the growth of new leaves.
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(e) The diagram below shows two strawberries.
• Both strawberries were picked from the same strawberry plant.
• Both strawberries were picked 3 days ago.
• The strawberries were stored in different conditions.
Strawberry A Strawberry B
A © sarahdoow/iStock/Thinkstock, B © Mariusz Vlack/iStock/Thinkstock
Give three possible reasons that may have caused strawberry A to decay.
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(Total 13 marks)
Q10.Metabolism is the sum of all the chemical reactions in the cells of the body.
One metabolic reaction is the formation of lipids.
(a) Give one other metabolic reaction in cells.
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(1)
Table 1 shows the mean metabolic rate of humans of different ages.
Table 1
Age in years
Mean metabolic rate in kJ/m2/hour
Males Females
5 53 53
15 45 42
25 39 35
35 37 35
45 36 35
(b) What two conclusions can be made from the data in Table 1?
Tick two boxes.
As age increases, mean metabolic rate of males and females increases.
Males have a higher metabolic rate than females after five years of age.
The mean metabolic rate of females decreases faster than males up to 25 years of age.
The mean metabolic rate of males and females decreases more quickly after the age of 35.
There is no relationship between age and mean metabolic rate.
(2)
(c) Calculate the percentage decrease in the mean metabolic rate of males between 5 years and 45 years of age.
Use the equation:
Give your answer to 3 significant figures.
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Percentage decrease = ___________________(3)
Regular exercise can increase metabolic rate.
Two people did five minutes of gentle exercise from rest.
Table 2 shows the effect of the exercise on their heart rates.
Table 2
Time in minutes
Heart rate in beats per minute
Person R Person S
0 (at rest) 60 78
1 76 100
2 85 110
3 91 119
4 99 129
5 99 132
(d) Describe two differences in the response of person R and person S to the exercise.
Use information from Table 2.
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(e) Complete the line graph below for person S.
You should:
• add the scale to the x axis• label the x axis.
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(4)
(f) After five minutes of exercise, the heart rate of person S was 132 beats per minute. When person S rested, his heart rate decreased steadily at a rate of 12 beats every minute.
Calculate how much time it would take the heart rate of person S to return to its resting rate.
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Time = ___________________ minutes(2)
(g) A student made the following hypothesis about the heart rate of smokers and non-smokers during exercise.
“During exercise, the heart rate of smokers increases more thanthe heart rate of non-smokers.”
Design an investigation that would allow you to test this hypothesis.
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(Total 20 marks)
Q11.Many scientists think that global air temperature is related to the concentration of carbon dioxide in the atmosphere.
The graph below shows changes in global air temperature and changes in the concentration of carbon dioxide in the atmosphere.
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(a) Complete the table below.Use information from the graph above.
Choose answers from the box.
You may use each answer once, more than once or not at all.
constant decreasing increasing
1960 − 1977 1977 − 2003 2003 − 2015
Trend in carbon dioxide concentration Increasing
Trend in air temperature
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Many scientists think that an increase in carbon dioxide concentration in the atmosphere causes an increase in air temperature.
(b) How would an increase in the concentration of carbon dioxide in the atmosphere cause an increase in air temperature?
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(c) Evaluate evidence for and against the theory that an increase in the concentration of carbon dioxide in the atmosphere causes an increase in air temperature.
Use data from the graph above and your own knowledge.
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In each year, the concentration of carbon dioxide in the atmosphere is higher in the winter than in the summer.
(d) Give one human activity that could cause the higher concentration of carbon dioxide in the winter.
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(e) Give one biological process that could cause the lower concentration of carbon dioxide in the summer.
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(f) Give two possible effects of an increase in global air temperature on living organisms.
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2. _________________________________________________________________
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(Total 11 marks)
Q12.A student carried out an investigation using leaf epidermis.
This is the method used.
1. Peel the lower epidermis from the underside of a leaf.2. Cut the epidermis into six equal sized pieces.3. Place each piece of lower epidermis into a different Petri dish.4. Add 5 cm3 of salt solution to the six Petri dishes. Each Petri dish should have a
different concentration of salt solution.5. After 1 hour, view each piece of epidermis under a microscope at ×400
magnification.6. Count and record the total number of stomata present and the number of open
stomata that can be seen in one field of view.
The student’s results are shown in the table.
Concentration of salt solution in mol / dm3
Number of stomata in
field of view
Number of open
stomata in field of view
Percentage (%) of open
stomata in field of view
0.0 7 7 100
0.1 8 8 100
0.2 7 6 X
0.3 9 6 67
0.4 10 4 40
0.5 9 2 22
(a) Calculate value X in the table above.
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X = ____________________ %(1)
(b) Give one conclusion from the results in the table above.
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(c) How could the student find out what concentration of salt solution would result in half of the stomata being open?
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(d) The student measured the real diameter of the field of view to be 0.375 mm.
Calculate the number of open stomata per mm2 of leaf for the epidermis placed in 0.4 mol / dm3 salt solution.
Use information from the table above.
Take π to be 3.14
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Number of open stomata = ____________________ per mm2
(3)
(e) The diagram below shows two guard cells surrounding a closed stoma and two guard cells surrounding an open stoma.
When light intensity is high potassium ions are moved into the guard cells.
Describe how the movement of potassium ions into the guard cells causes the stoma to open.
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(Total 10 marks)
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Mark schemes
Q1.(a) (for calcium)
allow any correct rounding to minimum 3 significant figuresallow alternative route with correct rounding
1
(for vitamin B-12)
allow alternative route with correct rounding1
560 / 559.8 / 559.78 / 559 (cm3)allow only correct answer based on values given for vitamin B-12 and calcium
1
an answer of 560 / 559.8 / 559.78 / 559 (cm3) scores 3 marksan incorrect answer for one step does not prevent allocation of marks for subsequent steps
(b) Level 2: Scientifically relevant facts, events or processes are identified and given in detail to form an accurate account.
4−6
Level 1: Facts, events or processes are identified and simply stated but their relevance is not clear.
1−3
No relevant content0
Indicative content
• Biuret reagent (allow CuSO4 and NaOH) tests for protein• add Biuret reagent to milk• solution will turn (from blue) to lilac if positive
• iodine solution tests for starch (ignore iodine unqualified)• add iodine solution to milk• solution will turn (from orange / brown) to blue / black if positive
• Benedict’s reagent tests for sugars• add Benedict’s reagent to milk and boil / heat (allow any temperature above
60 °C)• solution will turn (from blue) to (brick) red / brown / orange / yellow / green if
positive
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for level 2, reference to all three food tests is required
(c) lipase breaks down fat into fatty acids (and glycerol)do not accept if ‘glycerol’ is contradicted
1
(and) fatty acids lower the pH1
(and when) fatty acids cause the pH to be below 10 (the indicator becomes colourless)
1
(d) observation of colour change is subjective / based on opinionignore human error unqualifiedignore experimental error or examples of this
1
(e) bile emulsifies fatsallow a correct description of emulsification (i.e. breaks fat from large droplets into smaller droplets)do not accept a description of chemical breakdown
1
creates a larger surface area (of fat)1
(so) lipase can break down fat (to produce fatty acids) more quickly / effectively
allow fatty acids produced by action of lipase more quickly
1[16]
Q2.(a) 55%
2 marks for correct answer aloneaccept 54 − 565.5 / 10 × 100 alone gains 1 mark
2
(b) any three from:
• amino acids• antibodies• antitoxins• carbon dioxide• cholesterol• enzymes• fatty acid• glucose• glycerol• hormones / named hormones• ions / named ions
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• proteins• urea• vitamins• water.
ignore blood cells and plateletsignore oxygenmax 1 named example of each for ions and hormonesallow minerals
3
(c) Marks awarded for this answer will be determined by the Quality of Communication (QC) as well as the standard of the scientific response. Examiners should also refer to the information in the Marking Guidance and apply a ‘best-fit’ approach to the marking.
0 marksNo relevant content.
Level 1 (1 – 2 marks)There is a description of pathogens with errors or roles confused.orthe immune response with errors or roles confused.
Level 2 (3 – 4 marks)There is a description of pathogens and the immune response with some errors or confusionora clear description of either pathogens or the immune response with few errors or little confusion.
Level 3 (5 – 6 marks)There is a good description of pathogens and the immune response with very few errors or omissions.
Examples of biology points made in the response:
• bacteria and viruses are pathogenscredit any ref to bacteria and viruses
• they reproduce rapidly inside the body• bacteria may produce poisons / toxins (that make us feel ill)• viruses live (and reproduce) inside cells (causing damage).
white blood cells help to defend against pathogens by:
• ingesting pathogens / bacteria / (cells containing) virusescredit engulf / digest / phagocytosis
• to destroy (particular) pathogen / bacteria / viruses• producing antibodies• to destroy particular / specific pathogens• producing antitoxins• to counteract toxins (released by pathogens)
credit memory cells / correct description• this leads to immunity from that pathogen.
6[11]
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Q3.(a) Marks awarded for this answer will be determined by the Quality of Communication
(QC) as well as the standard of the scientific response. Examiners should also refer to the information in the Marking guidance and apply a ‘best-fit’ approach to the marking.
0 marksNo relevant content.
Level 1 (1−2 marks)The method described is weak and could not be used to collect valid results, however does show some understanding of the sequence of an investigation.
Level 2 (3−4 marks)The method described could be followed and would enable some valid results to be collected, but lacks detail.
Level 3 (5−6 marks)The method described could be easily followed and would enable valid results to be collected.
Examples of the points made in the response:• bean seedlings of same age• cut material from same part of each organ (for repeats) e.g. top 1 cm of
stem / a whole cotyledon / seed• equal mass of each organ
accept weight for mass• grind / homogenise• in equal amounts of water / buffer• equal volumes of hydrogen peroxide solution• equal concentrations of hydrogen peroxide solution• same temperature• temperature maintained in water bath• quantitative measure of gas production eg height of foam in mm / collect
gas in graduated syringe in cm3
• for same time period• repetitions (3+ times)• calculate mean for each.
6
(b) (i) correct answer: 401 mark for 45 as the anomalous result has been included in the calculationor
1 mark for
or 2
(ii) vertical axis correctly labelled:‘Enzyme activity in arbitrary units’
allow ecf from (b)(i)1
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points plotted correctly ±1 mmdeduct 1 mark for each incorrect plot
2
suitable line of best fitnot feathery, not point to point
1
(iii) 6.0 / 6allow ± 0.1if 6.0 not given, allow correct for candidate’s graph ± 0.1
1
(iv) in range 0 to 14 unitsallow correct for candidate’s graph
1
(v) enzyme denatured / enzyme (active site) shape changedallow substrate no longer fits (active site)ignore reference to temperaturedo not allow enzyme dies
1[15]
Q4.(a) (i) man has (inherited) polydactyly (PD) allele (from mother)
1
man has (inherited) other / normal / recessive allele from father1
because father does not have PD allele or if father had it father would have had PD or father only has normal allele or father is homozygous recessive
1
allow gene for allele
(ii) 0.5 / ½ / 1 in 2 / 1:1 / 50%do not allow 1:2 or 50/50allow 50:50
1
(b) parental phenotypes: both brown1
parental genotypes: both Bb1
gametes: B b and B b 1
allow only on gametes answer lineallow ecf from genotypes
offspring genotypes: BB (2)Bb bballow ecf from gametes
1
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offspring phenotypes correctly assigned to genotypes:BB & Bb = brown bb = red
do not penalise confusion of ‘phenotypes’ & ‘genotypes’ here1
[9]
Q5.(a) (i) without oxygen
allow not enough oxygenignore airignore production of CO2
ignore energy1
(ii) more / high / increased lactic acid (at end)allow approximate figures (to show increase)ignore reference to glucose
1
(b) (i) 1.5allow only 1.5 / 1½ / one and a half
1
(ii) increases at first and levels offignore subsequent decrease
1
suitable use of numbers egrises to 10 / by 9 (dm3 per min)orincreases up to 1.5 (min) / levels off after 1.5 (min) (of x axis timescale)
allow answer in range 1.4 to 1.5orafter the first minute (of the run)
1
(iii) supplies (more) oxygen1
supplies (more) glucose1
need ‘more/faster’ once only for full marksallow removes (more) CO2 / lactic acid / heat as an alternative for either marking point one or two, once only
for (more) respiration1
releases (more) energy (for muscle contraction)do not allow energy production or for respiration
1[9]
Q6.
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(a) wear a face maskallow wear gloves
1
(b) Level 2 (3–4 marks):A detailed and coherent plan covering all the major steps. It sets out the steps needed ina logical manner that could be followed by another person to produce an outcome whichwill address the hypothesis.
Level 1 (1–2 marks):Simple statements relating to steps are made but they may not be in a logical order. The plan may not allow another person to produce an outcome which will address the hypothesis.
0 marks:No relevant content.
Indicative content
Plan:• cut a specified number of pieces of bread to the same size• place mould spores on the bread• the number of mould spores needs to be the same quantity of mould spores
oneach piece of bread
• place bread in different sealable plastic bags• place in different temperatures (minimum of three) eg fridge, room, incubator• leave each for the same amount of time eg four days• measure the percentage cover of mould on each piece of bread• repeat experiment
additional examiner guidance:• good level 2 answer will describe how the growth of mould can be measured
andwill give a range of different temperatures to be used
• allow equivalent levels of credit for alternative methodologies that would clearly produce a measurable outcome in terms of mould growth at various temperatures
4
(c) any one from:• type of mould• amount of mould (put on each piece of bread)• amount of air in the plastic bags• size of the pieces of bread• type of bread• amount of moisture / water added
1
(d) (56 − 4 = 52) / 51
10.4allow 10.4 with no working shown for 2 marks
1
ecf for incorrectly read figures for 1 mark
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(e) (decomposition occurs at a faster rate when the temperature is higheroramount of decomposition is higher when temperature is higher
1[9]
Q7.(a) receptors detect / sense stimuli / change in surroundings or convert stimulus into an
impulseignore send impulses to brain / spinal cord
1
example of a receptorallow any appropriate organ or part of an organ, eg eye / retina or named type of receptor eg light receptor
1
effectors allow / make response or convert an impulse to an actionignore receive impulses from brain / spinal cord
1
(effector) muscle / glandallow an exampleignore eg arm / leg
1
(b) (i) junctionallow idea of a (small) gap / spacedo not allow if implication is that the neurones move
1
between neuron(e)sallow named types of neurones
1
(ii) chemicalallow answers in terms of specific types of neuroneallow neurotransmitter / named neurotransmitter released
1
any one from:• (chemical released) from one neurone
ignore produced• (chemical) passes (across synapse) to next neurone to stimulate / cause
(electrical) impulseallow diffuses for passes (across)
1
(c) (i) skinignore hand / leg
1
(ii) 1.6 (cm per millisecond)allow 2 if evidence of rounding up of 1.6
1
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(iii) any two from:ignore length of neurones
• synapses slow down transmission / impulseallow idea of movement of chemical being slower than electrical impulse
• fewer synapses (via brain)allow one synapse compared to two or only one synapse
• (therefore) fewer delaysallow impulse travels more slowly in relay neurones
2[12]
Q8.(a) vena cava
1
(b) 0.5 mm = 0.05 cm1
time = allow alternative correct substitution
1
24.8751
25 (s)an answer of 25 (s) scores 4 marksallow 24 for 3 marks (no conversion of mm to cm)allow 23.8 / 23.75 for 2 marks (no conversion of mm to cm and incorrect sf)
1
(c) (blood) travels through (the) pulmonary vein1
(blood) enters left atrium1
(blood) enters (the) left ventricle1
(blood) leaves the heart via / through (the) aortaallow blood travels through arteriolesallow blood (travels round the body and) reaches the cells / tissues via / in capillaries
1
ignore ref to valves / systole / diastole throughout
(d) Level 3 (5-6 marks):Relevant points (reasons / causes) are identified, given in detail and logically linked to form a clear account.
Level 2 (3-4 marks):Relevant points (reasons/causes) are identified, and there are attempts at logical linking. The resulting account is not fully clear.
Level 1 (1-2 marks):
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Points are identified and stated simply, but their relevance is not clear and there is no attempt at logical linking.
No relevant content (0 marks)
Indicative content
S = structural F = functional
• (S) both have a large surface area• (S) villi have many microvilli• (S) alveolar walls are not flat / are folded
• (F) to maximise diffusion (of gases) / absorption of (food) molecules
• (S) both have many capillaries / good blood supply / capillaries near the surface
• (F) to maintain concentration / diffusion gradient
• (S) both have thin walls / walls that are one cell thick / one cell thick surface• (F) to provide a short diffusion distance (for molecules to travel)
• (S) villi have many mitochondria• (F) to provide energy for active transport (of food molecules)
• (S) cells of the villi have microvilli / more projections• (F) to further increase the surface area / increase the number of proteins in
the membrane / to allow more active transport to take place[15]
Q9.(a) methane is produced
ignore bad smell1
which is a greenhouse gas / causes global warming1
(b) (9.80 / 0.20 = 49 therefore) 49:11
(c) horse (manure)allow ecf from 11.2
closest to 25:1 (ratio)1
(d) Level 3 (5–6 marks):A detailed and coherent explanation is given, which logically links how carbon is released from dead leaves and how carbon is taken up by a plant then used in growth.
Level 2 (3–4 marks):A description of how carbon is released from dead leaves and how carbon is taken upby a plant, with attempts at relevant explanation, but linking is not clear.
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Level 1 (1–2 marks):Simple statements are made, but no attempt to link to explanations.
0 marks:No relevant content.
Indicative content
statements:• (carbon compounds in) dead leaves are broken down by microorganisms /
decomposers / bacteria / fungi• photosynthesis uses carbon dioxide
explanations:• (microorganisms) respire• (and) release the carbon from the leaves as carbon dioxide• plants take in the carbon dioxide released to use in photosynthesis to produce
glucose
use of carbon in growth:• glucose produced in photosynthesis is used to make amino acids / proteins /
cellulose• (which are) required for the growth of new leaves
6
(e) any three from:(storage conditions)• (at) higher temperature / hotter• (had) more oxygen• (had) more water / moisture• (contained) more microorganisms (that cause decay)
allow reference to bacteria / fungi / mould3
[13]
Q10.(a) any one from:
• respiration• formation of proteins• formation / breakdown of glycogen• breakdown of (excess) protein or formation of urea• photosynthesis or formation of glucose / starch (in plants)
ignore formation of carbohydrates1
allow other correct reference to metabolic reactions in cellsignore reference to digestion
(b) males have a higher metabolic rate than females after five years of age1
the mean metabolic rate of females decreases faster than males up to 25 years of age
1
each additional tick negates a mark
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(c) 1
32.075472…allow correct rounding of this to at least 4 significant figures
1
32.1allow a correct reduction to 3 significant figures from an incorrect calculation for marking point 2
1
an answer of 32.1 scores 3 marks
(d) any two from:allow converse
• (person) R heart rate rose / increased more slowly than (person) S
• (person) R heart rate levelled off whereas (person) S continued to increase
• (person) R heart rate rose less (overall / after 5 minutes of exercise) than S
allow correct use of figurese.g. R increased (overall) by 39 bpm / 65% and S by 54 bpm / 69%ignore lack of units
2
(e) correct scale and axis labelledallow min(s)do not accept ‘m’the zero is not required on the x-axis
1
all points plotted correctly (to within ± ½ square)allow 4 or 5 correct plots for 1 mark
2
line joined point to point or correct curved line of best fit1
(f)
allow sequential deductions of 12 four or five times
1
4.5 (minutes) / 4½ minutes / 4 minutes 30 seconds / 4:30do not accept 4:50 or 4 minutes 50 seconds
1
an answer of 4.5 minutes scores 2 marks
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(g) Level 3: The method would lead to the production of a valid outcome. All key steps are identified and logically sequenced.
5−6
Level 2: The method would not necessarily lead to a valid outcome. Most steps are identified, but the method is not fully logically sequenced.
3−4
Level 1: The method would not lead to a valid outcome. Some relevant steps are identified, but links are not made clear.
1−2
No relevant content0
Indicative content
• two groups of people − non-smokers and smokers• have at least five people in each group or large groups• get each person to do (named) exercise• controlled variables:
− same number of people in each group or large groups− same gender− same level of activity / exercise− same age− no health issues / illnesses− same type of exercise− same time for exercise
• record heart rate for each person before and after exercise• calculate increase in heart rate for each person after exercise• compare results for each group
for level 3, students should refer to at least 5 smokers and 5 non-smokers, carrying out exercise with control variables and a means of determining an increase in heart rate
for level 2, students should refer to ‘groups’ of smokers and non-smokers exercising
[20]
Q11.(a)
1960 − 1977 1977 − 2003 2003 − 2015
trend in carbon dioxide
concentration increasing increasing 1
trend in air temperature decreasing increasing constant /
decreasing1
allow synonyms e.g. level / goes up / goes down
(b) traps heat / energy or (long-wavelength / IR) radiationdo not accept light / UV
or
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less loss of heatallow stops (some) heat escapingdo not accept stops all heat escaping
orinsulates
ignore greenhouse effectignore reference to ozone layer
1
(c) Level 2: Some logically linked reasons are given. There may also be a simple judgement.
3−4
Level 1: Relevant points are made. They are not logically linked.1−2
No relevant content0
Indicative content
for the theory:• (overall increased CO2 parallels) overall increased temperature
(e.g. by 0.4 (°C))• CO2 traps (long-wave) radiation / IR / heat
against the theory:• in some years (e.g. 1960–1977) temperature falls (while CO2 is rising)• many (large and small) erratic rises and falls in temperature• overall correlation does not necessarily mean a causal link• other (unknown) factors may be involved in temperature change
to access level 2 there must be evidence both for and against the theory and use of data from the graph
(d) burning of (fossil) fuelsallow e.g. coal / oil / gasallow driving carsallow any activity which leads to burning fuels − e.g. using central heatingignore power stations unqualifiedignore burning / fires unqualifiedignore deforestation
1
(e) photosynthesisallow full description or full equationallow a symbol equation which is not balanced
1
(f) any two from:• (some) plants grow faster / higher yield• loss of habitat• migration or change in distribution*• extinction*
*if neither is given allow alters biodiversity for 1
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markallow (in terms of extinction) death due to e.g. lack of water / food or increased diseaseignore death unqualified
2
allow points made using examples[11]
Q12.(a) 86
allow this answer onlydo not accept 85.7if no answer given, check for answer in the table
1
(b) as salt concentration increases, percentage of open stomata (in field of view) decreases (above 0.1 mol / dm3)orallow percentage of open stomata stays the same between 0.0 and 0.1 (mol / dm3 then decreases as salt concentration increases)
ignore references to number of open stomataallow converseallow idea that mean concentration (of salt) in guard cells is between 0.3 and 0.4 mol per dm 3
1
(c) use concentrations between 0.3 (mol / dm 3 ) and 0.4 (mol / dm 3)ordraw a graph of the data and read off the value at 50% (open stomata)
allow a list of appropriate concentrations i.e. 0.32 mol / dm 3), 0.34 (mol / dm 3), 0.36 (mol / dm 3) etc.
1
(d) (π × 0.18752) = 0.11 (mm 2)an answer of 36 scores 3 marks
1
1
36 (per mm 2)allow 36.22 / 36.23 or 36.2if answer is incorrect allow for 2 marks for sight of number of open stomata = 9 per mm 2 (diameter used instead of radius)if no other marks awarded allow for 1 mark any one from:• sight of area = 0.44(mm 2) (diameter used instead of
radius)• sight of number of open stomata = 9.1 / 9.05 / 9.06 per
mm 2 (diameter used instead of radius and no rounding)1
(e) (potassium) ions increase the concentration of the solution (inside guard cells)
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or(potassium) ions make cell more concentrated / less dilute
allow (potassium) ions decrease concentration of water / water potential (of guard cells)
1
water moves into the (guard) cell by osmosis1
cell swells unevenly (so stoma opens)1
as inner wall is less flexible than outer wall or thick part of the wall is less flexible than the thin part (of the wall)
1[10]
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