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Woodhouse College Mixed Questions Q1. The graph shows the surface area to volume ratio of cubes of different size. (a) Elephant seals are mammals that live in water that has a temperature of between 0 °C and 4 °C. Elephant seals are very large. Use the graph to explain the advantage of a large size to elephant seals. .............................................................. ........................................................ .............................................................. ........................................................ .............................................................. ........................................................ .............................................................. ........................................................ (2) (b) (i) Describe how single-celled organisms exchange respiratory gases. ......................................................... .................................................... Page 1

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Woodhouse College

Mixed Questions

Q1.          The graph shows the surface area to volume ratio of cubes of different size.

(a)     Elephant seals are mammals that live in water that has a temperature of between 0 °C and 4 °C. Elephant seals are very large. Use the graph to explain the advantage of a large size to elephant seals.

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(b)     (i)      Describe how single-celled organisms exchange respiratory gases.

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(ii)     This method of gas exchange is only possible in very small organisms.Explain why.

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Students used the apparatus shown in the diagram to measure the rate at which a leafy shoot took up water.

(c)     The students measure the distance moved by the air bubble every 30 seconds. What other measurement should they take if they wanted to compare water loss from different shoots? Explain your answer.

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(d)     The students plotted the distance moved by the air bubble against time on a graph.Describe how they could use the graph to calculate the mean rate of water uptake.

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(e)     An insect lives in air. Describe how the insect is able to obtain oxygen and limit water loss.

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(Total 15 marks)

 Q2.          The drawing shows part of the lower leaf epidermis of sorghum.

(a)     Calculate the number of stomata per mm2 of the leaf surface. Show your working.

 

 

 

Answer ....................................... stomata per mm2

(2)

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(b)     Sorghum has few stomata per mm2 of leaf surface area. Explain how this is an adaptation to the conditions in which sorghum grows.

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(Total 5 marks)

 Q3.  (a) The table shows the transpiration rate of a group of plants exposed to different humidities at a temperature of 25°C.

 

Humidity / % Transpiration rate /arbitrary units

20 26.0

40 21.0

50 16.5

60 11.0

70 9.5

Describe and explain the relationship between humidity and transpiration rate.

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(b)     The diagrams show a section through a typical leaf and a section through a leaf from a xerophytic plant. The xerophytic leaf has a lower transpiration rate than the typical leaf.

Describe two features shown in the diagram of the xerophytic leaf which reduce transpiration rate. Explain how each of these features contributes to a lower transpiration rate.

Feature 1 ......................................................................................................

Explanation ...................................................................................................

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Feature 2 ......................................................................................................

Explanation ...................................................................................................

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(Total 7 marks)

 

 

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Q4.          (a)     Describe and explain how water moves via the apoplastic and symplastic pathways from the soil to the xylem in a root.

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(b)     The graphs show the daily changes in environmental temperature and light intensity, and changes in the diameter of the trunk of a pine tree.

Use information from the graphs, and your knowledge of the cohesion-tension theory of water movement through a plant, to explain why the diameter of the trunk is smallest at midday.

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(c)     Describe and explain three ways in which the leaves of xerophytic plants may be adapted to reduce water loss.

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(Total 15 marks)

Q5.          A doctor measured the volume of air in the lungs of two people over a period of 7 seconds. Both people were resting. One person was healthy. The other had emphysema. The results are shown in the table.

 

Times / sVolume of air in lungs / dm3

Person A Person B

0 6.5 7.0

1 3.8 6.0

2 3.0 5.6

3 2.3 5.1

4 2.0 4.8

5 1.7 4.5

6 1.6 4.2

7 1.6 3.9

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(a)     The two people were breathing out during the time shown. What evidence in the table supports this statement?

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(b)     Calculate the rate at which person A breathed air out of his lungs between 0 and 3 seconds. Show your working.

 

 

 

Answer ............................................. dm3 s–1

(2)

(c)     Person B has emphysema. Give one piece of evidence from the table that shows this.

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(d)     Emphysema reduces the efficiency of gas exchange in the lungs. Explain why.

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(4)(Total 8 marks)

 Q6.          Emphysema is a disease that affects the alveoli of the lungs and leads to the loss of elastic tissue. The photographs show sections through alveoli of healthy lung tissue and lung tissue from a person with emphysema. Both photographs are at the same magnification.

Source:  Biophoto Associates, Science Photo Library

Using the evidence given above and your own knowledge, explain why a person with emphysema is unable to do vigorous exercise.

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Q7.          (a)     Pulmonary tuberculosis is a disease of the lungs. Describe the transmission and course of infection of pulmonary tuberculosis.

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(b)     Emphysema is another disease of the lungs. People with emphysema may feel weak and tired. Explain why.

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(Total 10 marks)

Q8.          The photographs show sections through alveoli of healthy lung tissue and lung tissue from a person with emphysema. Both photographs are at the same magnification.

Biophoto Associates, Science Photolibrary

(a)     Give two differences that can be seen between the healthy lung tissue and the lung tissue from the person with emphysema.

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(2)

(b)     People with emphysema may find it difficult to climb stairs. Explain why.

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(Total 5 marks)

Q9.          The graph shows changes in the volume of air in a person’s lungs during breathing.

(a)     The person was breathing in between times A and B on the graph.

(i)      Explain how the graph shows that the person was breathing in between times A and B.

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(ii)     Describe and explain what happens to the shape of the diaphragm between times A and B.

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(b)     The person’s pulmonary ventilation changed between times C and D.Describe how the graph shows that the pulmonary ventilation changed.

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(Total 6 marks)

Q10.          (a)     Explain why the rate of diffusion is more rapid at higher temperatures.

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(b)     Fick’s law can be summarised as

Rate of diffusion is proportional to

Complete the table by adding the words maximum or minimum to show the values of the features in Fick’s law which will ensure

(i)      efficient absorption of digested food from the small intestine;

(ii)     reduction of water loss from a leaf. 

Feature Efficient absorption of digested food from the

small intestine

Reduction of water loss from a leaf

Surface area    

Difference in concentration    

Thickness of exchange surface    

(2)

(c)     The graph shows how the concentration of a substance affects its rate of absorption into a cell.

(i)      Substance A enters the cell by simple diffusion. Use Fick’s law to explain the shape of the curve.

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(ii)     Substance B enters the cell by facilitated diffusion. Explain the evidence from the graph which supports this.

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 Q11.          Lung cancer, chronic bronchitis and coronary heart disease (CHD) are associated with smoking. Tables 1 and 2 give the total numbers of deaths from these diseases in the UK in 1974.

Table 1 Men 

Age/years Number of deaths(in thousands)

  lung cancer chronic bronchitis coronary heart disease

35 - 64 11.5 4.2 31.7

65 - 74 12.6 8.5 33.3

75+ 5.8 8.1 29.1

Total (35 - 75+) 29.9 20.8 94.1

Table 2 Women 

Age/years Number of deaths(in thousands)

  lung cancer chronic bronchitis coronary heart disease

35 – 64 3.2 1.3 8.4

65 – 74 2.6 1.9 18.2

75+ 1.8 3.5 42.3

Total (35 – 75+) 7.6 6.7 68.9

(a)     (i)      Using an example from the tables, explain why it is useful to give data for men and women separately.

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(ii)     Data like these are often given as percentages of people dying from each cause.

Explain the advantage of giving these data as percentages.

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(b)     Give two factors, other than smoking, which increase the risk of coronary heart disease.

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(Total 6 marks)

Q12.          (a)     The graph shows the risk of a 50-year-old male having a heart attack during the next ten years, in relation to several risk factors.

Reproduced with permission from New Scientist magazine © RBI Ltd

(i)      Describe what the graph shows about the effect of smoking on the risk of having a heart attack.

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(ii) Explain why an increase in plasma cholesterol concentration increases the risk of a heart attack.

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S       (b)     Cigarette smoke contains nicotine. Nicotine stimulates the sympathetic nervous system and increases the stickiness of blood platelets.

Explain how these effects of nicotine increase the risk of cardiovascular disease.

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(Total 8 marks)

Q13.          (a)     Many different substances enter and leave a cell by crossing its cell surface membrane. Describe how substances can cross a cell surface membrane.

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(b)     Describe and explain how the lungs are adapted to allow rapid exchange of oxygen between air in the alveoli and blood in the capillaries around them.

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(Total 10 marks)

 Q14.          An agar plate was flooded with a culture of a species of bacterium usually found in the mouth. Four sterile paper discs, A, B, C and D, each containing a different brand of mouthwash, were then placed on the agar plate. The drawing shows the appearance of the plate after it had been incubated at 37°C for three days.

(a)     Describe the aseptic techniques that would be used when flooding the agar plate with bacteria.

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(b)     The effectiveness of a mouthwash can be measured by calculating the total area of a paper disc and the clear zone around it. The area of a circle is given by πr2, where r is the radius of the circle. Calculate how many times more effective mouthwash C is than mouthwash B. Show your working.

Mouthwash C is .................................... times more effective than mouthwash B.(2)

S       (c)     Several factors affect the rate at which the antiseptic in the mouthwash from each paper disc diffuses through the agar. Describe the effect of three named factors on this rate.

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(Total 8 marks)

 Q15.          Miner’s lung is a disease caused by breathing in dust in coal mines. The dust causes the alveolar epithelium to become thicker. People with miner’s lung have a lower concentration of oxygen in their blood than healthy people.

(a)     (i)      Describe the path by which oxygen goes from an alveolus to the blood.

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(ii)     Explain why people with miner’s lung have a lower concentration of oxygen in their blood.

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(b)     In healthy lungs, a gradient is maintained between the concentration of oxygen in the alveoli and the concentration of oxygen in the lung capillaries.

(i)      Describe how ventilation helps to maintain this difference in oxygen concentration.

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(ii)     Give one other way that helps to maintain the difference in oxygen concentration.

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(c)     Scientists investigated the number of cases of miner’s lung reported in Britain between 1992 and 2006.

Coal mining in Britain had been dramatically reduced by 1990.

Some scientists concluded that the rise in reported cases of miner’s lung after 1992 shows that the disease takes a long time to develop.

Evaluate this conclusion.

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(2)(Total 8 marks)

Q16.          Gas exchange surfaces allow efficient diffusion of gases. Fick’s law states:

Rate of diffusion is proportional to

(a)     In the gill of a fish, describe how

(i)      a large surface area is provided;

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(ii)     a concentration gradient is maintained.

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(b)     Land-dwelling insects lose water from their gas exchange surface. Use Fick’s law to explain why they lose less water when the air is humid.

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Q17.          Read the following passage.

Several diseases are caused by inhaling asbestos fibres. Most of thesediseases result from the build up of these tiny asbestos fibres in the lungs.

One of these diseases is asbestosis. The asbestos fibres are very small andenter the bronchioles and alveoli. They cause the destruction of phagocytes

5       and the surrounding lung tissue becomes scarred and fibrous. The fibroustissue reduces the elasticity of the lungs and causes the alveolar wallsto thicken. One of the main symptoms of asbestosis is shortness of breathcaused by reduced gas exchange.

People with asbestosis are at a greater risk of developing lung cancer. The time10      between exposure to asbestos and the occurrence of lung cancer is 20–30 years.  

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Use information in the passage and your own knowledge to answer the following questions.

(a)     Destruction of phagocytes (lines 4–5) causes the lungs to be more susceptible to infections. Explain why.

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(b)     (i)      The reduced elasticity of the lungs (lines 6–7) causes breathing difficulty. Explain how.

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(ii)     Apart from reduced elasticity, explain how changes to the lung tissue reduce the efficiency of gas exchange.

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(c)     (i)      Doctors did not make the link between exposure to asbestos and an increased risk of developing lung cancer for many years. Use information in the passage to explain why.

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(ii)     Give one factor, other than asbestos, which increases the risk of developing lung cancer.

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(Total 10 marks)

 Q18.          (a)     The cardiac cycle is controlled by the sinoatrial node (SAN) and the atrioventricular node (AVN). Describe how.

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(b)     What is atheroma and how may it cause myocardial infarction?

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Q19.          (a)     Describe how atheroma may form and lead to a myocardial infarction.

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(b)     The bar chart shows the number of males aged 19-64 admitted to English hospitals with a myocardial infarction within five days of the England football team losing to Argentina by penalty

shoot-out in the 1998 World Cup.

(i)      Suggest how the expected number of admissions might have been calculated.

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(ii)     Describe the difference between the observed and expected numbers of males experiencing a myocardial infarction over the six days.

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(c)     Explain how repeated stress, such as that involved in watching a penalty shoot-out, may lead to a myocardial infarction.

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(d)     A group of male football supporters was shown a video recording of a football match. At the end of the first half, they were each given a beta blocker. The graph shows the heart rate of a typical individual from the investigation.

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Describe and explain the effect of the beta blocker on the heart rate of this person.

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(Total 15 marks)

Q20.          (a)     Gas exchange in fish takes place in gills. Explain how two features of gills allow efficient gas exchange.

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(b)     A zoologist investigated the relationship between body mass and rate of oxygen uptake in four species of mammal. The results are shown in the graph.

(i)      The scale for plotting body mass is a logarithmic scale. Explain why a logarithmic scale was used to plot body mass.

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.............................................................................................................

.............................................................................................................(1)

(ii)     Describe the relationship between body mass and oxygen uptake.

.............................................................................................................

.............................................................................................................(1)

(iii)     The zoologist measured oxygen uptake per gram of body mass. Explain why he measured oxygen uptake per gram of body mass.

.............................................................................................................

.............................................................................................................

.............................................................................................................

.............................................................................................................(2)

(iv)    Heat from respiration helps mammals to maintain a constant body temperature.

Use this information to explain the relationship between body mass and oxygen uptake shown in the graph.

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.............................................................................................................

.............................................................................................................

.............................................................................................................

.............................................................................................................

.............................................................................................................(3)

(Total 9 marks)

 

 

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Mixed Questions ANSWERS

M1.          (a)     Small surface area to volume ratio;Loses less heat (to the water);

2

(b)     (i)      Diffusion through cell/body surface;Q The key term here is diffusion

1

(ii)     Small organisms have large surface area to volume ratio;Rate of diffusion depends on surface area;All parts of cell only a short distance from exchange surface;

2 max

(c)     Surface area of leaves;Different shoots will have leaves with different surface areas;

2

(d)     Draw line/curve of best fit/from line/curve of best fit;Find slope/gradient/divide distance moved by time;

2

(e)     1 Air enters through (open) spiracles;2 Through tracheae;3 Diffusion gradient in trachea4 Tracheae associated with all cells/closely associated with cells;5 Oxygen diffuses into cells;6 Ventilation replacing air in tracheae;7 Body covered with (waterproof) waxy layer/cuticle;8 Spiracles are able to close;

6 max[15]

 M2.          (a)     235–240;;(one mark for an answer between 200-300based on 2 - 3 stomata in 0.01mm2

Alternatively, one mark for calculating the area of therectangle correctly as 0.016 – 0.017mm2)

2

(b)     grows in arid / dry conditions;less surface area;(rate of) transpiration / water loss would be reduced;

3[5]

 

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 M3.          (a)     increased humidity leads to decreased transpiration;high humidity means more water in the air / increased saturation / increased water potential;reduced diffusion gradient / water potential gradient;slower rate of water loss / less evaporation;

3 max

(b)     thick cuticle; impermeable to water / waterproof;

sunken stomata; reduces water diffusion gradient;

shape of leaf / rounded / small surface area;small surface area : volume ratio;

(explanation must be linked to feature)4 max

[7]

M4.          (a)     Apoplastic – Via cell walls / spaces external to cell membrane / external to                   cytoplasm / between cells;                   As far as endodermis / Casparian strip / layer of wax;                   Caused by transpiration pull;                   Cohesion / hydrogen-bonding between water molecules;

          Symplastic – Through cell surface membrane (of epidermis / root hair cell) / ref.                   vacuoles membrane;

                   High to low  / ;                   Diffusion / osmosis;                   Cell-to-cell via plasmodesmata / via strands of cytoplasm;Secretion / active transport of ions into xylem by endodermis;ORActive uptake of ions from soil at epidemis;

Lowers  /  in xylem / increases osmosis into xylem;[If symplast & apoplast are confused – max 5 marks]

max 6

(b)     1. Diameter of trunk minimal at warmest / brightest time of day /    midday = warmest / brightest;2. Stomata open in light → more water loss;3. Water evaporates more when warm / more heat energy for water evaporation;4. Hydrogen-bonding between water molecules;5. Cohesion (/ described) between water molecules;6. Adhesion (described) between water molecules and walls of xylem vessels;7. (Xylem) pulled inwards by faster flow of water / pulled in by tension;8. Reduced pressure at leaves / top of plant / pull from top / from leaves /    tension from leaves / from top of plant due to transpiration / evaporation;9. Water pulled up plant;

max 6

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(c)      Feature Explanation

Think cuticle / wax layer waterproof / impermeable;Sunken stomata saturated layer of still air outside;Hairy saturated layer of still air outside;Leaves small / reduced to spines / needles reduced S.A. for water loss;Leaves roll up in dry weather less S.A. for water loss / stomata covered

/ saturated region of still air;Reduced number of stomata reduced S.A. for water loss;CAM (/ Crassulacean Acid Metabolism) stomata closed in light / in warm / only

open in dark / when cool;

3 features but no explanations – max 1 markmax 3

[15]

M5.          (a)     Volume (of air in lungs) decreases;Accept: Results decrease

1

(b)     Correct answer 1.4;;

Incorrect answer showing (vol. air breathed out =) 6.5 – 2.3 / 4.2 (dm3);2

(c)     Reduced flow rates / less air breathed out / more air left in lungs (afterbreathing out);

Insufficient: More air in lungs / high volume of air in lungs1

(d)     1. Alveoli break down / collapse / rupture / fewer alveoli / larger alveoli or    alveolar wall/epithelium walls thicken;

2. Reduced surface area / increased diffusion pathway;

3. (So) less diffusion;

4. Less elastin / elastic (tissue) / not recoiling / loss of elasticity / elastin permanently stretched;

5. Reduced flow rate / less air expelled;

6. So small / reduced diffusion or concentration gradient;1. Neutral: Damage. Accept alveoli burstLess surface area for diffusion = 2 marks (mark points 2 and 3)3. Accept diffusion less efficient. Reject diffusion of air4. Elastic tissue must be in context of lungs.6. Accept: Not maintaining a steep diffusion/concentration gradient.

4 max

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[8]

 M6.          not enough O2;for increased respiration / for ATP needed for exercise;

2

reference to decreased surface area of alveoli/ longer diffusionpathway; less gas exchange/less diffusion/less oxygen passes into the blood;

OR

reference to decreased elasticity/reduced elastic recoil;meaning breathing becomes more difficult/laboured/shallower/lungsdo not empty;

2[4]

 M7.          (a)     1      (Bacteria transmitted in) droplets/aerosol;

2       (Bacteria) engulfed/ingested by phagocytes/macrophages;

3       (Bacteria) encased in named structuree.g. wall/tubercle/granuloma/nodule;

4       (Bacteria) are dormant/not active/not replicating;

5       If immunosuppressed, bacteria activate/replicate/released;

6       Bacteria destroy alveoli/capillary/epithelial cells;

7       (Leads to) fibrosis/scar tissue/cavities/calcification;

8       (Damage) leads to less diffusion/less surface area/increasesdiffusion distance;

9       (Activation/damage allows bacteria) to enter blood/spreads(to other organs);

1   Accept: TB/‘it’/the disease/air droplets1   Neutral: spread through the air/coughs/sneezes1   Reject: virus2   Neutral: ‘destroyed by’;2   Accept: white blood cells3   Neutral: bacteria contained5   Accept: reference to HIV/old age/stress7   Accept: fibrous tissue8   Neutral: reduced gas exchange8   Accept: reduced SA:VOL

5 max

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(b)     1       Alveoli break down/collapse/rupture/walls thicken;

2       Less surface area/increases diffusion distance/less diffusion;

3       Loss of elastin/elastic tissue/elastase involved;

4       (Alveoli/lungs) cannot recoil/spring back/have reducedelasticity/more difficult to expel air;

5       Reduced diffusion gradient/air not replenished/less air leaves lungs;

6       Less oxygen enters blood/tissues;

7       Less respiration/less energy released/less ATP produced;1   Neutral: alveoli damaged2   Accept: references to a lack of alpha-1-antitrypsin3   This mark is for a structure.Accept: elastin permanently stretched4   This mark is for a mechanism. Do not award reduced     elasticity for 3.4   Neutral: more difficult to inhale air5   This mark is for a consequence     Accept: reduced concentration gradient;     Neutral: less air enters lungs7   Q Reject: ‘less energy produced’/anaerobic respiration7   Accept: ‘less energy produced in the form of ATP’/less     oxygen for respiration

5 max[10]

M8.          (a)     Smaller number of alveoli;Larger air space per alveolus;Thicker walls;

Q Accept converse for normal cells2 max

f alveoli;Diffusion of gases / gas exchange reduced / less oxygen enters blood;Narrower bronchioles reduce gas flow;Loss of elasticity reduces gas flow / unable to ventilate efficiently;Lungs permanently inflated;Less energy available / less respiration possible for muscles;

Q Award maximum of two marks if candidate suggests energy is ‘used’ in respiration.

3 max[5]

 

 

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M9.          (a)     (i)      (Lung volume) increases/reaches a maximum (at B);Do not negate mark for ‘breathing out’ if qualified e.g. when (lung volume) decreases

1

(ii)     Flattens/lowers/moves down;

         (Diaphragm/muscle) contracts;Reject: second mark only if intercostal muscles cause the diaphragm to flatten

2

(b)     Pulmonary ventilation = tidal volume × breathing rate;

Breathing rate increases/more breaths per min(between C and D)/peaks get closer;

Tidal volume/volume of air (inhaled) per breath increases(between C and D)/deeper breaths;

(Tidal volume increase) qualified by data from graphe.g. approximate three-fold increase/appropriate calculation;

Accept: ventilation rate instead of breathing rateNeutral: breathing increasesAccept: breathe quickerNeutral: volume in lungs increasesAccept: distance from bottom to top of peak increases for ‘tidal volume increases’Neutral: higher peaks for ‘tidal volume increases’

3 max[6]

M10.          (a)     More (kinetic) energy;Molecules are moving faster;

Ignore references to collisions2

(b)      

Feature Efficient absorption of digested food from the

small intestine

Reducing water loss from a leaf

Surface area maximum minimum

Difference in concentration maximum minimum

Thickness of exchange surface

minimum maximum

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Mark for each correct column, one mark each. 2

(c)     (i)      Greater the concentration difference/gradient, fasterrate of entry/diffusion;

1

(ii)     Curve flattens out;Channel/carrier proteins / carriers;Become limiting;

max 2[7]

M11.          (a)     (i)      because there are big differences;any correct named example e.g. lung cancer/bronchitis much lowerin women than in men;

2

(ii)     easier to compare if sample size effectively the same;different numbers of people in each group;

2(b)     ANY TWO: more stress / more saturated fats in diet / less time to

                    exercise / reliance on cars;2

[6]

M12.          (a)     (i)      smoking increases risk and the effect increases as plasma cholesterol increases/is higher at high plasma cholesterol;smoking increases risk and the effect is greater at high bloodpressure;

2

(ii)     cholesterol/fatty tissue deposited in lining/wall of arteries;formation of plaques/blood clots;which obstruct blood flow;

2 max

(b)     noradrenaline produced by SNS;stimulates SAN;increase in heart rate/cardiac output;blood pressure increases;increased risk of cerebrovascular accident/stroke;increased risk of blood clot/thrombosis;

4 max[8]

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M13.          (a)     1       (Simple/facilitated) diffusion from high to lowconcentration/down concentration gradient;

Q Do not allow across/along/with concentration gradient

2       Small / non-polar / lipid-soluble molecules pass viaphospholipids / bilayer;

Reject: named molecule passing through membrane by an incorrect routeAccept: diagrams if annotated

OR

Large / polar / water-soluble molecules go through proteins;

3       Water moves by osmosis/from high water potential to low water potential/from less to more negative water potential;

4       Active transport is movement from low to high concentration/against concentration gradient;

Only penalise once if active transport is not namede.g. ‘movement against the concentration gradient involves proteins and requires ATP’ = 2 marks

5       Active transport/facilitated diffusion involves proteins/carriers;Accept: facilitated diffusion involves channelsReject: active transport involves channels

6       Active transport requires energy/ATP;

7       Ref. to Na+/glucose co-transport;Credit ref. to endo/exocytosis as an alternative

5 max

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(b)     1       Many alveoli/alveoli walls folded provide a large surface area;Neutral: alveoli provide a large surface area

2       Many capillaries provide a large surface area;

3       (So) fast diffusion;Neutral: greater/better diffusionNeutral: fast gas exchangeAllow ‘fast diffusion’ only once

4       Alveoli or capillary walls/epithelium/lining are thin/short distance between alveoli and blood;

Reject: thin membranes/cell wallsAccept: one cell thick for ‘thin’

5       Flattened/squamous epithelium;Accept: endothelial

6       (So) short diffusion distance/pathway;

7       (So) fast diffusion;

8       Ventilation/circulation;Accept: descriptions for ventilation/circulation

9       Maintains a diffusion/concentration gradient;

10     (So) fast diffusion;Do not double penalise if description lacks detaile.g. thin membranes so a short diffusion distance = 1 mark

5 max[10]

M14.          (a)     sterilisation of equipment (once);use of pipette/syringe to transfer culture suspension to plate;use of spreader / shake ;detail regarding lid, e.g. keeping over plate during transfer/spreading;

3 max

(b)     2.25 = 2 marks(general principle (1.52 ÷ 12) gains 1 mark)

2

(c)     increased temperature increases rate;increased concentration increases rate;increased molecule size decreases rate;

(allow increased distance decreases rate)3 max

[8]

 

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M15.          (a)     (i)      Through alveolar epithelium;

         Through capillary epithelium/endothelium;Accept: Through lining/wall of alveolus and capillary for 1 markAccept: squamous epithelial cells for ‘epithelium’Neutral: alveolar endotheliumNeutral: references to diffusionQ Correct use of terminology;

2

(ii)     (Thicker alveolar wall) – no markNeutral: less diffusion

         (So) Longer diffusion pathway/slower diffusion;Neutral: references to surface area

1

(b)     (i)      (In alveolus)Need the idea of air moving and oxygen concentration

         Brings in air containing a high(er) oxygen concentration;Neutral: reference to carbon dioxide concentration

         Removes air with a low(er) oxygen concentration;2

(ii)     Circulation of blood/moving blood;Neutral: bloodNeutral: short diffusion pathway

1

(c)     Long time between decrease in mining and increase in cases;

          Graph shows fluctuations;

Correlation does not prove causation/there may be other causes ofminer’s lung;

          Improved diagnosis methods;

          Do not know number of cases/baseline before 1990;

Not all cases reported/not all individuals with miner’s lung visit adoctor;

Accept: correct use of figures from graph for the first marking point:e.g. cases do not increase until after 2000/2001-2004/10 years later.

2 max[8]

 

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M16.          (a)     (i)      Many gill lamellae/gill filaments;(ignore refs to ‘highly divided’)

1

(ii)     Counter-current mechanism/blood and water flow in oppositedirections;Not enough time for equalisation of concentrations/maintainsconcentration gradient over length of gills/never reachesequilibrium;

max 2

(b)     Humidity reduces difference in concentration of water (vapour)between body and air;reduces rate of diffusion (of water vapour)(as are proportional);

2[5]

M17.          (a)     Phagocytes engulf/ingest pathogens/microorganisms/bacteria/viruses;

Phagocytes destroy pathogens/microorganisms/bacteria/viruses;

Lung diseases are caused by pathogens/microorganisms/bacteria/viruses;Q Allow description of process of engulfing

2 max

(b)     (i)      Alveoli/lungs will not inflate/deflate fully/reduced lung capacity;

Breathing out particularly affected/no longer passive;

Concentration/diffusion gradient / rate of diffusion reduced;2 max

(ii)     Alveolar walls thicken;

Longer diffusion pathway;

Scarred/fibrous tissue;

Reduces surface area (for gaseous exchange);Q Diffusion is essential for 2nd point and surface area for 4th point.

4

(c)     (i)      Cancer develops 20 – 30 years after exposure (to asbestos);1

(ii)     Smoking / air pollution / specified industrial source;1

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[10]

 

M18.          (a)     1.      SAN initiates heartbeat/acts as a pacemaker/myogenic;Q Must be in context

2.      (SAN) sends wave of electrical activity/impulses (across atria)causing atrial contraction;

Reject: signals/electronic/messages/nerve impulses once only

3.      AVN delays (electrical activity/impulses);Neutral: reference to non-conducting tissue delaying impulses instead of the AVN

4.      (Allowing) atria to empty before ventricles contract/ventriclesto fill before they contract;

5.      (AVN) sends wave of electrical activity/impulses downBundle of His/Purkyne fibres;

6.      (Causing) ventricles to contract (from base up)/ventricularsystole;

5 max

(b)     1.      Cholesterol/plaque/lipoprotein/LDL/fatty material/cells;Accept: LDL/triglyceride/cell debris;Reject: fatty acids/HDL;

2.      In artery wall/under lining/endothelium of artery/blood vessel;Q Do not accept references to veins or capillaries as equivalent to blood vessels

3.      Atheroma linked to blood clot/thrombosis;Q Must be in the correct context

4.      (Blocks) coronary artery/artery supplying heart muscle/tissue/cells;

If coronary artery is not mentioned or described, reference to heart muscle/tissue/cells is needed for 5.

5.      Reduces oxygen/glucose supply (to heart muscle/tissues/cells);

6.      (Heart muscle/tissue/cells) unable to respire/dies;5 max

[10]

 M19.          (a)     1  fatty substance / foam cells / cholesterol in artery wall /    under endothelium;2  atheroma creates turbulence / damage to lining of artery;3  formation of  plaques / atherosclerosis / narrows lumen of artery;4  (turbulence) increases risk of blood clot / embolus;5  blood clot / thrombus breaks off;

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6  (blood clot) lodges in coronary artery;7  reduced blood supply to heart muscle;8  reduced oxygen supply;9  leads to death of heart muscle; max 6

(b)     (i)      average number of admissions on ordinary day;when no football match being played;similar time of year / conditions;

max 2

(ii)     large / significant difference for three days;then small difference;

2

(c)     increases heart rate;raises blood pressure / causes hypertension;blood supply to heart / oxygen use by heart increased;atheroma restricts blood / oxygen supply to heart muscle;

max 2

(d)     reduces heart rate;beta-blocker fits receptor sites;on walls of heart / blood vessel;(receptor sites for) adrenaline / moradrenaline / stops adrenaline /noradrenaline binding;

max 3[15]

 M20.          (a)     Filaments/lamellae provide large surface area;

Thin/flattened epithelium/one/two cell layers so short diffusionpathway (between water and blood);

Countercurrent/blood flow maintains concentration/diffusion gradient;Q Do not credit thin cell walls/membranes

2 max

(b)     (i)      Large/wide range of values (so can fit on graph);1

(ii)     Decrease in uptake with increase in mass/negative correlation;1

(iii)     Enables comparison;

As animals differ in size/mass;2

(iv)    Smaller animals have larger surface area to volume ratio;Allow converse for larger animals.Allow appropriately named animal as an alternative to smaller or larger animals.

Lose more heat per gram of tissue;

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Respire more/faster (relative to body mass);

Oxygen used in respiration;3 max

[9]

EXAMINERS REPORT:

 E2.          (a)     Very few candidates correctly worked out the area as 0.0167mm2 and many seemed to have a basic problem with calculating areas. It was common to see 0.1 x 0.1 = 0. l mm2. Many candidates failed to gain credit by carrying out calculations that were not clearly identified. A common approach was to estimate the number of stomata in an area 0. 1mm x 0. 1mm, which gained credit if done correctly. Many gave answers that were clearly incorrect, such as 20-25 or even 0.25!

(b)     Most candidates picked up the 2 marking points but some failed to gain the second point by incorrectly stating that ‘fewer stomata prevent water loss’. Few answers referred to the idea of there being a reduced surface area.

 E3.          Again most candidates scored highly on this question, with many achieving maximum marks.

(a)     Most candidates were able to describe the relationship between humidity and transpiration rate, and then to explain this relationship in terms of how high humidity increases the amount of water in the air, and therefore decreases the diffusion gradient, resulting in a slower rate of water loss. Marks were often not awarded for incorrect or vague explanations, such as ‘water is lost to the air by osmosis’ or ‘humidity affects the aperture of the stomata’.

(b)     There were many excellent answers relating features evident in the diagrams with clear explanations of how each feature contributes to a lower rate of transpiration. For example, many candidates correctly recognised the thicker cuticle of the xerophytic leaf and explained how the waterproof property of the cuticle reduces water loss. A lack of precision often resulted in few marks, such as the vague statement that the cuticle reduces water loss with no explanation of how this is achieved. Sunken stomata reducing the diffusion gradient of water and the shape of the leaf resulting in a smaller surface area:volume ratio, were correctly identified as other explanations, but they were less common.

 E4.          (a)     A few were confused between the symplastic and apoplastic pathways and some thought that movement through the cell walls would be by ‘osmosis’. Additionally, those who gave their answers in terms of water potential gradients were generally more successful than those who attempted to do so using the term ‘concentration’ – the latter being best avoided as it is often ambiguous.

(b)     Most candidates used the graphs to link midday with high light intensity and high temperature. Better candidates related higher temperature to a high rate of water evaporation, but only a few commented on the relationship between high light intensity and stomatal opening. Most understood the principles of cohesion and adhesion with a fair proportion mentioning hydrogen bonding as the cause of these phenomena. While most realised that water was pulled up the plant, driven by transpiration from the leaves, explanations of how this would generate an inward pulling force on the xylem vessels were frequently not very clear. In some cases due to the use of the term ‘pressure’, it was difficult to decide what the candidate intended.

(c)     While most candidates were able to give three adaptations of the leaves of xerophytic plants for reduction of water loss, explanations were often not very thorough. Others were excellent, stating that trapped layers of water vapour would

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reduce the water potential gradient and hence slow down diffusion of water molecules. Some, unfortunately, did not confine themselves to the constraints of the question and included inappropriate references to the root system or to water storage tissue in the stem.

 

E5.          (a)     The vast majority of the candidates answered correctly.

(b)     This question differentiated well with many correctly calculating the rate as 1.4 dm3s-

1. However a significant number were only able to calculate the volume of air breathed out during the time period rather than the rate. Weaker candidates often did not attempt the calculation.

(c)     This question was generally well answered although some candidates did not make clear that the difference was that B had more air left in his lungs after exhalation. To state that the volume of air in the B’s lungs was greater was insufficient as B might just have been a much larger individual. The best answers compared volume of air exhaled or rate of exhalation.

(d)     Many candidates gained all 4 marks here. Most candidates knew that surface area was reduced although many were unclear about how this happened. They often wrote about gas exchange in general terms and failed to mention diffusion. There was some confusion between different lung diseases with candidates referring to excess mucus and the narrowing of the airways.

 E6.          There were mixed responses to this question. Many candidates gained three or four marks. The mark not gained by the more able candidates was usually that for making explicit that there would be not enough oxygen for vigorous exercise, or that increased respiration was need for vigorous exercise. Usually one mark was gained for stating that there would be decreased surface area of the alveoli and sometimes a second mark was gained for saying as a result there would be less gaseous exchange or less oxygen going into the blood.

 E7.          (a)     Many candidates were aware of how pulmonary tuberculosis is transmitted. However, the course of infection was poorly understood. There are nine marking points on the mark scheme for this question. Despite this, only 25% of candidates scored at least three or more marks. The most common marks awarded were for the transmission of bacteria in droplets, their containment in a tubercle and the formation of scar tissue. It was usually only better candidates who were aware of engulfment by phagocytes, the dormant stage of disease and the role of immunosuppression in allowing the activation of bacteria at a later date. Weaker candidates often stated that transmission is via ‘coughing’, ‘sneezing’ or ‘through the air’. Some thought that pulmonary tuberculosis was caused by a virus. Similarly, these candidates frequently went into unnecessary detail about the symptoms of pulmonary tuberculosis and the social conditions that would increase the chance of transmission. Some candidates confused pulmonary tuberculosis with emphysema. A minority of weaker candidates wrote about infection in the pulmonary artery or vein. This is in contrast to some truly excellent descriptions of the course of infection by the best candidates.

(b)     Most candidates scored at least three marks and all marking points were regularly seen, with the exception of the consequence of alveoli being unable to recoil. Most candidates were aware that alveoli break down, although some wrote vaguely about ‘damaged alveoli’. The effect of this on surface area or diffusion was also well known. Better candidates usually wrote about the loss of elastin and the inability of the alveoli to recoil or force air out of the lungs. However, some candidates focused

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on increased difficulty to inhale air due to the loss of elastin. This was not credited. Weaker candidates were usually aware that less oxygen enters the blood but they often disqualified the last marking point by stating that ‘less energy is produced’. A minority confused emphysema with asthma or linked a lack of oxygen with ‘the heart having to work harder’.

 

 E9.          (a)     (i)      90% of candidates gained this mark for noting that the volume of air in the lungs increased. Candidates who failed to score usually referred to ‘fluctuations’ in the graph.

(ii)     Just over 40% of candidates scored full marks. However, most candidates gained one mark for stating that the diaphragm contracts or flattens. Some answers went into unnecessary detail about the role of the intercostal muscles and ribcage in inhalation. This was not asked for. Additionally, a minority of weaker candidates thought that the intercostal muscles move the diaphragm.

(b)     Most candidates gained one mark for noting that the breathing rate increased. However, relatively few mentioned that the tidal volume also increased. The usual attempt at this was along the lines that ‘the volume of air in the lungs increases’, a response which did not gain credit. If the term ‘tidal volume’ was not used, it had to be clear that candidates were referring to the volume of air inhaled per breath. A number of candidates gained this mark for noting that the distance between peaks and troughs increased. Better candidates usually gained at least two marks by starting their answers with the equation for pulmonary ventilation. Some weaker candidates confused the terms ‘stroke volume’ and ‘tidal volume’.

E10.          (a)     There was a good understanding of the effect of heat on increasing the kinetic energy of molecules and its subsequent effect on the rate of diffusion. Although many appeared to be of the opinion that diffusion can only occur across a membrane, such responses in themselves were not penalised. Candidates should be cautioned against simply rewriting the question. Thus, increased kinetic energy leading to faster molecular movement gained full credit. Increased kinetic energy resulting in the rate of diffusion being faster at higher temperatures did not.

(b)     This part of the question was also well answered and most candidates were able to apply a knowledge of Fick’s law to the examples provided. Candidates must be careful, however, to follow the instructions provided. Thus, use of the words maximum and minimum was stipulated and candidates clearly risked losing credit when they used ticks and crosses, or referred to “high”, “low” and “medium”.

(c)     There were two frequent errors which supported the view that some candidates struggled with interpreting the graph. The levelling out of the curve for substance B was described as either meaning that the concentration of the substance outside the cell became constant, or that the substance no longer entered the cell. In part (i), while less able candidates confined their comments to describing the curve, there were many excellent answers in which increased rate of entry was clearly related to an increase in concentration difference across the plasma membrane. Those who gave an accurate description of the relevant curve in part (ii) were generally able to relate the plateau to the limiting factor of carrier proteins. A few confused facilitated diffusion and active transport and incorrectly linked the observation with ATP deficiency.

E11.          (a)     In part (i) most candidates were aware that there are differences in the figures for men and women. However, despite being asked to use an example from the tables, many failed to do so. In part (ii) many candidates were aware that

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percentages lead to ease of comparison, but few referred to differing sample sizes.

(b)     Most candidates could give two suitable factors. Marks were not awarded for imprecise answers. For example, some gave exercise, stress or cholesterol concentration without further qualification, such as ‘lack of exercise’, ‘high levels of stress’ or ‘high concentrations of cholesterol’.

E12.          The question was quite well answered, with most candidates scoring about half marks.

(a)     (i)      The majority of candidates did not consider all of the data in the graph, and therefore gave incomplete answers such as, just stating that smoking increases the risk of a heart attack. The effect of smoking when plasma concentration increased or when blood pressure was higher were not considered.

(ii)     Most candidates scored one mark. Many failed to score the second mark, as they stated that fatty deposits occurred on the inside wall of the artery rather than in the lining.

(b)     Most candidates scored two marks for stating that the heart rate would increase and that there would be a greater risk of blood clots forming. Few candidates developed their answer further to discuss the effect on the SAN, blood pressure and the brain.

E13.          This question discriminated well across the ability range and there were many excellent answers to parts (a) and (b). Weaker candidates were often let down by poor expression and this was particularly notable for part (b).

(a)     Approximately half of the candidates gained at least three marks. Most correctly described movement down a concentration gradient by diffusion and against a concentration gradient by active transport. The requirement for energy or ATP in active transport was frequently noted. Similarly, many candidates were aware that membrane proteins are involved in active transport or facilitated diffusion. However, some disqualified this mark for stating that active transport involves channel proteins. Better candidates also referred to the movement of water by osmosis and related the property of a molecule to its route through the plasma membrane. Weaker candidates sometimes confused active transport and facilitated diffusion. Similarly, a minority described the structure of the membrane, without any reference to transport across it.

(b)     Just over half of candidates gained at least four marks. It was pleasing to see better candidates often scoring full marks. References to a flattened epithelium or many capillaries providing a large surface area were rare. However, all other marking points were frequently seen. Many candidates appreciated the role of ventilation or circulation in maintaining a concentration gradient. Unfortunately, weaker candidates often gave answers that lacked detail or were out of context e.g. ‘thin membranes’, ‘better diffusion’ and ‘faster gas exchange’. Similarly, they did not usually relate ‘large surface area’ to the many alveoli present. A minority of candidates started their answer with Fick’s equation but did not relate this to the question in sufficient detail.

E14.          (a)     A majority of candidates failed to refer to ‘…when flooding the agar plate with bacteria’ required by the question. Thus, the examiner was given long accounts of mopping benches with disinfectant, autoclaving of dishes and safe disposal of dishes. Reference to use of a spreader was comparatively rare.

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(b)     Many candidates achieved the correct answer, but few realised that ‘π‘ cancelled out and went on to do unnecessarily long calculations.

 (c)     Many candidates realised that factors such as concentration, temperature and molecule size would affect the rate, but failed to state how the rate would be affected by these. Weaker candidates gave explanations in terms of bacteria, or of the strength of the antiseptic.

 

 E15.          (a)     (i)      Very few candidates gained two marks for this question. Only the most able used the correct scientific terminology to name the layers of the alveolus and capillary through which oxygen passes. A mark was often awarded for ‘alveolar epithelium’ or referring to the wall of both alveolus and capillary. However, a number of candidates who referred to the capillary simply stated that oxygen entered, without any reference to a layer. A number of weaker candidates referred to ‘one cell thick membranes’ or gave answers that focused solely on diffusion. Similarly, a minority of candidates referred to the passage of air through the ‘bronchial tree’, from trachea to alveoli.

(ii)     This proved to be a good discriminator. Nearly 60% of candidates gained this mark for explaining that a thicker alveolar epithelium would increase the diffusion pathway or reduce the rate of diffusion. Candidates who failed to score usually referred to ‘less diffusion’, ‘less surface area’ or ‘a longer pathway for gas exchange’.

(b)     (i)      Approximately one third of candidates gained one mark for the idea that ventilation brings in air with a high concentration of oxygen. However very few then went on to mention the removal of air with a low concentration of oxygen. References to the removal of air almost always referred to carbon dioxide concentration. This was not credited. Similarly, many candidates did not read the stem of the question carefully and described the need for a concentration gradient between the alveolus and blood. This was usually expressed in terms of where the concentration of oxygen would be high and low. A minority of candidates described the roles of the ribcage, intercostals muscles and diaphragm in ventilation.

(ii)     Many candidates gained the mark for the idea that the circulation of blood also helps to maintain the concentration gradient between the alveolus and capillary. However, some candidates did not attempt this question or gave answers that related to the properties of a gas exchange surface.

(c)     This was well answered and most candidates scored at least one mark. This was usually for the idea that miner’s lung may be caused by other factors. Better candidates noticed fluctuations on the graph and made reference to a suitable year when the number of cases had dramatically increased. Weaker candidates often gave vague answers such as ‘the number of cases gradually increased’ without qualification or they suggested how coal dust may have affected the lungs over time.

 E16.          (a)     In part (i), candidates were frequently uncertain as to the names of the various structures that make up a gill. However, most were aware that, whatever their name, there were a lot of them and that this large number of small structures gave the gill its large surface area. In part (ii), nearly all candidates knew of the counter-current system that operates within the gill, although some called it a counter-current multiplier, clearly confusing it with the system operating in the loop of Henle. Fewer made the points that, because of this system, equilibration of concentrations is not reached or that a ventilation system operates to move the water over the gills.

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(b)     This was a question where candidates needed to pay attention to what they were asked. Too many candidates failed to score any marks when, with a little more care, they could probably have scored both. Given the equation for Fick’s law and the fact that a humid atmosphere reduces water loss from the insects, candidates were clearly required to say more than “a reduced concentration gradient would cut down water loss”. Both those pieces of information are effectively given in the question. Better candidates did identify that there was a reduced concentration gradient between insect and atmosphere, which would reduce the diffusion rate.

 

 E17.          (a)     Although it was evident that most candidates had a good idea of the role of phagocytes, poor use of terminology often resulted in marks not being awarded. It was common to see responses such as phagocytes ‘fighting disease’ or destroying ‘foreign bodies’ or ‘infections’. Nevertheless, approximately a third of candidates obtained both marking points.

(b)     (i)      Most candidates obtained one mark for stating that the lungs would not fully inflate or deflate. However, very few candidates obtained a second mark for suggesting that breathing out would particularly be affected or that the rate of diffusion would be reduced. There was some confusion over the meaning of the term elasticity with many references to ‘lungs contracting and relaxing’.

(ii)     This proved to be an effective discriminator. The vast majority of candidates obtained at least one mark often by referring to the presence of scarred or fibrous tissue in the lungs. Most candidates then gained a second mark by linking this to a reduction in the surface area for effective gaseous exchange. Fewer candidates specifically referred to the thickening of the alveolar walls but instead mentioned thickening of lung tissue. However, a significant number of candidates were able to link this thickening to a longer diffusion pathway.

(c)     (i)      Most candidates obtained this mark using the information in the passage to explain that lung cancer develops 20 – 30 years after exposure to asbestosis.Candidates failing to gain this mark often provided incomplete responses such ae ‘it takes a long time for cancer to develop’.

(ii)     The vast majority of candidates obtained this mark by referring to smoking.

 E18.          There were many excellent answers to parts (a) and (b), with many candidates scoring at least three marks. Both questions proved to be good discriminators.

(a)     Most candidates understood the role of the SAN in initiating the heartbeat and generating electrical impulses, which cause contraction of the atria. Similarly, there were numerous references to the passage of impulses along the Bundle of His or Purkyne fibres and the subsequent contraction of the ventricles. However, there were some inventive spellings of ‘Purkyne’. It was encouraging to see only a small number of candidates referring to electrical impulses as ‘signals’, ‘messages’ or ‘electronic pulses’. It was usually only the most able candidates, who correctly referred to the delay at the AVN and described its significance. A number of candidates described the delay in the wrong context. This was usually in terms of a delay in impulses reaching the AVN or the non-conducting tissue of the heart causing the delay. Weaker candidates often gave a muddled sequence of events or gave a description of the cardiac cycle in terms of blood flow, valves and pressure changes.

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(b)     Most candidates were aware that atheroma involved the build up of fatty tissue. However, a number of weaker candidates incorrectly referred to fatty acids. The location of atheroma proved to be a good discriminator. Weaker candidates often placed it in the artery, lumen, endothelium or on the wall of blood vessels. Thrombosis formation was generally well understood, although a minority thought that atheroma and thrombosis were identical. Consequently, there were some references to the transport of atheroma in the blood. Answers that gained full credit usually referred to the blocking of coronary arteries, reducing the supply of oxygen to heart muscle and the effect of this on respiration or ATP production. Weaker candidates often gave answers that lacked detail or context. These answers typically referred to blood vessels being blocked, less blood reaching the heart or the heart dying. Similarly, a minority of candidates referred to ‘cardiac arteries’ instead of ‘coronary arteries’.

 E19.          (a)     Many candidates were able to score five or six marks here by following the logical progression from atheroma to myocardial infarction. Candidates failed to gain marks through vague references to blood vessels and, even when arteries were specified, not making it clear where atheroma or plaques were forming. It was necessary to say the lumen was narrowing, not just the artery, and that heart muscle, deprived of blood supply and oxygen, might die. Some regard atheroma and blood clots as the same thing.

(b)     As in question 2, the principles underlying graphs and bar charts were not understood and very few could suggest that the expected numbers came from normal admissions on normal days. Much discussion of who England might be playing when the expected numbers were calculated featured in many scripts. The answer required an indication of the main trend but some candidates described each reading in detail. Others placed undue emphasis on the slight variation on the fourth day after the match.

(c)     Once again football dominated the answers while junk food, inactivity and personal opinion also featured in many scripts. Some who did appreciate that raised blood pressure and heart rate could also occur scored two marks. Few recognized that the heart would have an increased oxygen demand.

(d)     This question usually yielded one mark for the lowering in heart rate produced by the beta blocker. A pleasing number could explain its role in preventing adrenaline or noradrenaline from binding to receptor sites, thereby gaining all three marks.

 E20.          (a)     Most candidates gained at least one mark often by explaining that filaments and/or lamellae in the gills provide a large surface area allowing efficient gas exchange. The failure of many candidates to gain both marks was often due to poor use of terminology particularly in relation to the short diffusion pathway between the blood and water and the countercurrent flow mechanism.

(b)     (i)      Approximately half the candidates obtained this mark appreciating that a logarithmic scale enabled the plotting of a large range of values.

(ii)     The vast majority of candidates correctly described the relationship between body mass and oxygen uptake.

(iii)     It was surprising that only one in every five candidates obtained both marks for this question. Many candidates obtained a mark for indicating that measuring oxygen uptake per gram of body mass would enable a comparison to be made. However, only better candidates linked this to the difference in body mass or size of the animals.

(iv)    This proved a very effective discriminator. Better candidates had little difficulty

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using the information provided to obtain all three marks. Some of these answers fully appreciated that smaller animals lose more heat per gram of tissue. Others were able to provide parts of the explanation and scored one or two marks. Weaker candidates showed little understanding, often beginning with the incorrect premise that larger organisms have a larger surface area to volume ratio. Better candidates clearly linked oxygen uptake and respiration but did not always refer to the higher rate of respiration in smaller animals.

 

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