Gas exchange ( Practise paper is in revision wallet )

Question 1

Scientist studied three species of plant.

They selected fully grown leaves from five different plants of each species.

For each leaf they measured:

• leaf surface area
• leaf thickness
• the number of stomata per mm^2

The scientists’ results are shown in the table below.

Plant species

1 ) A

2 ) B

3 ) C

Mean leaf surface area / mm^2

1 ) 218.0

2 ) 17.0

3 ) 2.2

Mean leaf thickness / micrometer

1 ) 191.5

2 ) 296.3

3 ) 354.8

Mean number of stomata per mm^2

1 ) 380.0

2 ) 136.0

3 ) 419.0

How did the scientist ensure they could make a valid comparison between leaves from different species?

Answer 1

• Scientist used fully grown leaves

Question 2

Describe a method you could use to find the surface area of a leaf.

( Scientist studied three species of plant.

They selected fully grown leaves from five different plants of each species.

For each leaf they measured:

leaf surface area
leaf thickness
the number of stomata per mm^2

The scientists’ results are shown in the table below.

Plant species

1 ) A

2 ) B

3 ) C

Mean leaf surface area / mm^2

1 ) 218.0

2 ) 17.0

3 ) 2.2

Mean leaf thickness / micrometer

1 ) 191.5

2 ) 296.3

3 ) 354.8

Mean number of stomata per mm^2

1 ) 380.0

2 ) 136.0

3 ) 419.0 )

Answer 2

• Draw around the leaf on graph paper
• Count the squares
• Multiply it by 2

Question 3

Which species, A or B, would you predict grew in a drier environment?
Explain one feature that caused you to choose this species.

( Scientist studied three species of plant.

They selected fully grown leaves from five different plants of each species.

For each leaf they measured:

leaf surface area
leaf thickness
the number of stomata per mm^2

The scientists’ results are shown in the table below.

Plant species

1 ) A

2 ) B

3 ) C

Mean leaf surface area / mm^2

1 ) 218.0

2 ) 17.0

3 ) 2.2

Mean leaf thickness / micrometer

1 ) 191.5

2 ) 296.3

3 ) 354.8

Mean number of stomata per mm^2

1 ) 380.0

2 ) 136.0

3 ) 419.0 )

Answer 3

• Species B

- Smaller surface area so less evaporation

Question 4

Other than the features of leaves in the table above, give two features of leaves of xerophytes.

For each feature explain how it reduces water loss.

( Scientist studied three species of plant.

They selected fully grown leaves from five different plants of each species.

For each leaf they measured:

leaf surface area
leaf thickness
the number of stomata per mm^2

The scientists’ results are shown in the table below.

Plant species

1 ) A

2 ) B

3 ) C

Mean leaf surface area / mm^2

1 ) 218.0

2 ) 17.0

3 ) 2.2

Mean leaf thickness / micrometer

1 ) 191.5

2 ) 296.3

3 ) 354.8

Mean number of stomata per mm^2

1 ) 380.0

2 ) 136.0

3 ) 419.0 )

Answer 4

• Feature 1 - Hairs on leaves
• Explanations - so an increase in humidity
• Feature 2 - curled leaves
• Explanation - so a decrease in water potential gradient

Question 5

Species C has a high number of stomata per mm^2.
Despite this it loses a small amount of water.

Use the date to explain why.

( Scientist studied three species of plant.

They selected fully grown leaves from five different plants of each species.

For each leaf they measured:

leaf surface area
leaf thickness
the number of stomata per mm^2

The scientists’ results are shown in the table below.

Plant species

1 ) A

2 ) B

3 ) C

Mean leaf surface area / mm^2

1 ) 218.0

2 ) 17.0

3 ) 2.2

Mean leaf thickness / micrometer

1 ) 191.5

2 ) 296.3

3 ) 354.8

Mean number of stomata per mm^2

1 ) 380.0

2 ) 136.0

3 ) 419.0 )

Answer 5

• Small surface area so total number of stomata is low

Question 5

Scientists studied the rate of carbon dioxide uptake by grape plant leaves.
Grape leaves have stomata on the lower surface but no stomata on the upper surface.

The scientists recorded the carbon dioxide uptake by grape leaves with three different treatments:

Treatment 1 - No air-sealing grease was applied to either surface of the leaf.

Treatment 2 - The lower surface of the leaf was covered in air-sealing grease that prevents gas exchange.

Treatment 3 - Both the lower surface and the upper surface of the leaf were covered in air-sealing grease that prevents gas exchange.

The scientists measured the rate of carbon dioxide uptake by each leaf for 60 minutes in light and then for 20 minutes in the dark.

The scientists’ results are shown in the diagram below.

( Diagram shows “ Mean rate of carbon dioxide uptake / arbitrary units “ against “ Time / minutes “ )

( Light is turned off after “ 60 mins “ )

( “ Treatment 1 “ starts from “ 4.5 “ and maintains that value until “ 60 mins “ has been reached )

( “ Treatment 1 “ descends rapidly to “ 0 “ from “ 60 mins “ to “ 80 mins “ )

( “ Treatment 2 “ starts from “ 0.5 “ and maintains that value until “ 60 mins “ has been reached )

( “ Treatment 2 “ descends to “ 0 “ from “ 60 mins “ to “ 80 mins “ )

( “ Treatment 3 “ starts at “ 0 “ and maintains that value over the course of “ 80 mins “ )

Suggest the purpose of each of the three leaf treatments.

Answer 5

• Treatement 1 has no grease, so stomata are open
• Treatment 2 has grease on lower surface, which stops CO2 uptake through stomata
• Treatment 3 has grease on both surfaces, which stops all CO2 uptake

Question 6

Describe the results shown in treatment 1.

( Scientists studied the rate of carbon dioxide uptake by grape plant leaves.
Grape leaves have stomata on the lower surface but no stomata on the upper surface.

The scientists recorded the carbon dioxide uptake by grape leaves with three different treatments:

Treatment 1 - No air-sealing grease was applied to either surface of the leaf.

Treatment 2 - The lower surface of the leaf was covered in air-sealing grease that prevents gas exchange.

Treatment 3 - Both the lower surface and the upper surface of the leaf were covered in air-sealing grease that prevents gas exchange.

The scientists measured the rate of carbon dioxide uptake by each leaf for 60 minutes in light and then for 20 minutes in the dark.

The scientists’ results are shown in the diagram below.

( Diagram shows “ Mean rate of carbon dioxide uptake / arbitrary units “ against “ Time / minutes “ )

( Light is turned off after “ 60 mins “ )

( “ Treatment 1 “ starts from “ 4.5 “ and maintains that value until “ 60 mins “ has been reached )

( “ Treatment 1 “ descends rapidly to “ 0 “ from “ 60 mins “ to “ 80 mins “ )

( “ Treatment 2 “ starts from “ 0.5 “ and maintains that value until “ 60 mins “ has been reached )

( “ Treatment 2 “ descends to “ 0 “ from “ 60 mins “ to “ 80 mins “ )

( “ Treatment 3 “ starts at “ 0 “ and maintains that value over the course of “ 80 mins “ ) )

Answer 6

• The mean rate of CO2 uptake was constant and fell after the light turned off
• Uptake fell from 4.5 to 0

Question 7

The stomata close when the light is turned off.

Explain the advantage of this to the plant.

( Scientists studied the rate of carbon dioxide uptake by grape plant leaves.
Grape leaves have stomata on the lower surface but no stomata on the upper surface.

The scientists recorded the carbon dioxide uptake by grape leaves with three different treatments:

Treatment 1 - No air-sealing grease was applied to either surface of the leaf.

Treatment 2 - The lower surface of the leaf was covered in air-sealing grease that prevents gas exchange.

Treatment 3 - Both the lower surface and the upper surface of the leaf were covered in air-sealing grease that prevents gas exchange.

The scientists measured the rate of carbon dioxide uptake by each leaf for 60 minutes in light and then for 20 minutes in the dark.

The scientists’ results are shown in the diagram below.

( Diagram shows “ Mean rate of carbon dioxide uptake / arbitrary units “ against “ Time / minutes “ )

( Light is turned off after “ 60 mins “ )

( “ Treatment 1 “ starts from “ 4.5 “ and maintains that value until “ 60 mins “ has been reached )

( “ Treatment 1 “ descends rapidly to “ 0 “ from “ 60 mins “ to “ 80 mins “ )

( “ Treatment 2 “ starts from “ 0.5 “ and maintains that value until “ 60 mins “ has been reached )

( “ Treatment 2 “ descends to “ 0 “ from “ 60 mins “ to “ 80 mins “ )

( “ Treatment 3 “ starts at “ 0 “ and maintains that value over the course of “ 80 mins “ ) )

Answer 7

• Water is lost through the stomata

- Closed stomata prevents water loss

Question 7

Treatment 2 shows that even when the lower surface of the leaf is sealed there is still some uptake of carbon dioxide.

Suggest how this uptake of carbon dioxide continues.

( Scientists studied the rate of carbon dioxide uptake by grape plant leaves.
Grape leaves have stomata on the lower surface but no stomata on the upper surface.

The scientists recorded the carbon dioxide uptake by grape leaves with three different treatments:

Treatment 1 - No air-sealing grease was applied to either surface of the leaf.

Treatment 2 - The lower surface of the leaf was covered in air-sealing grease that prevents gas exchange.

Treatment 3 - Both the lower surface and the upper surface of the leaf were covered in air-sealing grease that prevents gas exchange.

The scientists measured the rate of carbon dioxide uptake by each leaf for 60 minutes in light and then for 20 minutes in the dark.

The scientists’ results are shown in the diagram below.

( Diagram shows “ Mean rate of carbon dioxide uptake / arbitrary units “ against “ Time / minutes “ )

( Light is turned off after “ 60 mins “ )

( “ Treatment 1 “ starts from “ 4.5 “ and maintains that value until “ 60 mins “ has been reached )

( “ Treatment 1 “ descends rapidly to “ 0 “ from “ 60 mins “ to “ 80 mins “ )

( “ Treatment 2 “ starts from “ 0.5 “ and maintains that value until “ 60 mins “ has been reached )

( “ Treatment 2 “ descends to “ 0 “ from “ 60 mins “ to “ 80 mins “ )

( “ Treatment 3 “ starts at “ 0 “ and maintains that value over the course of “ 80 mins “ ) )

Answer 7

• There is CO2 uptake through the upper surface of the leaf

Question 8

In both treatment 1 and 2, the uptake of carbon dioxide falls to zero when the light is turned off.

Explain why.

( Scientists studied the rate of carbon dioxide uptake by grape plant leaves.
Grape leaves have stomata on the lower surface but no stomata on the upper surface.

The scientists recorded the carbon dioxide uptake by grape leaves with three different treatments:

Treatment 1 - No air-sealing grease was applied to either surface of the leaf.

Treatment 2 - The lower surface of the leaf was covered in air-sealing grease that prevents gas exchange.

Treatment 3 - Both the lower surface and the upper surface of the leaf were covered in air-sealing grease that prevents gas exchange.

The scientists measured the rate of carbon dioxide uptake by each leaf for 60 minutes in light and then for 20 minutes in the dark.

The scientists’ results are shown in the diagram below.

( Diagram shows “ Mean rate of carbon dioxide uptake / arbitrary units “ against “ Time / minutes “ )

( Light is turned off after “ 60 mins “ )

( “ Treatment 1 “ starts from “ 4.5 “ and maintains that value until “ 60 mins “ has been reached )

( “ Treatment 1 “ descends rapidly to “ 0 “ from “ 60 mins “ to “ 80 mins “ )

( “ Treatment 2 “ starts from “ 0.5 “ and maintains that value until “ 60 mins “ has been reached )

( “ Treatment 2 “ descends to “ 0 “ from “ 60 mins “ to “ 80 mins “ )

( “ Treatment 3 “ starts at “ 0 “ and maintains that value over the course of “ 80 mins “ ) )

Answer 8

• No use of CO2 in photosynthesis in the dark

- So no diffusion gradient for CO2 into the leaf

Question 9

A scientist used grasshoppers to investigate the effect of composition of air on breathing rate in insects.
He changed the composition of air they breathed in by varying the
concentrations of oxygen and carbon dioxide.

The scientist collected 20 mature grasshoppers from a meadow.
He placed the grasshoppers in a small chamber where he could adjust and control the composition of air surrounding them.
The small chamber restricted the movement of the grasshoppers.

His results for three of the grasshoppers are shown in the table below in the form in which he presented them.

Gas

1 ) Oxygen

2 ) Carbon dioxide

Breathing rate of grasshopper in different types of air / breaths per minute

3 ) Grasshopper 1

4 ) Grasshopper 2

5 ) Grasshopper 3

Percentage of oxygen and carbon dioxide in different types of air breathed in by grasshopper

A ) Air from atmosthere

B ) Pure oxygen

C ) Gas mixture 1

D ) Gas mixture 2

1A ) 20.9

2A ) 0.1

3A ) 53

4A ) 48

5A ) 61

1B ) 100.0

2B ) 0.0

3B ) 11

4B ) 25

5B ) 13

1C ) 91.0

2C ) 9.0

3C ) 99

4C ) 88

5C ) 96

1D ) 84.0

2D ) 16.0

3D ) 107

4D ) 99

5D ) 93

The percentages of oxygen and carbon dioxide in column A do not add up to 100% but in columns C and D they do.
Suggest two reasons for this difference.

Answer 9

• Other gases in the atmosphere

- Composition of gas in A are not controlled

Question 10

Use all the data to describe the effect of concentration of carbon dioxide on the breathing rate of grasshoppers.

( A scientist used grasshoppers to investigate the effect of composition of air on breathing rate in insects.
He changed the composition of air they breathed in by varying the
concentrations of oxygen and carbon dioxide.

The scientist collected 20 mature grasshoppers from a meadow.
He placed the grasshoppers in a small chamber where he could adjust and control the composition of air surrounding them.
The small chamber restricted the movement of the grasshoppers.

His results for three of the grasshoppers are shown in the table below in the form in which he presented them.

Gas

1 ) Oxygen

2 ) Carbon dioxide

Breathing rate of grasshopper in different types of air / breaths per minute

3 ) Grasshopper 1

4 ) Grasshopper 2

5 ) Grasshopper 3

Percentage of oxygen and carbon dioxide in different types of air breathed in by grasshopper

A ) Air from atmosthere

B ) Pure oxygen

C ) Gas mixture 1

D ) Gas mixture 2

1A ) 20.9

2A ) 0.1

3A ) 53

4A ) 48

5A ) 61

1B ) 100.0

2B ) 0.0

3B ) 11

4B ) 25

5B ) 13

1C ) 91.0

2C ) 9.0

3C ) 99

4C ) 88

5C ) 96

1D ) 84.0

2D ) 16.0

3D ) 107

4D ) 99

5D ) 93 )

Answer 10

• Breathing rate of the grasshoppers is the lowest when there’s no CO2
• Presence of CO2 increases breathing rate
• Breathing rate increases when CO2 is higher than 0.1%

Question 11

One of the different types of air was similar to the air in the meadow where the grasshoppers were collected.
It provides data that might be used to calculate a mean breathing rate for grasshoppers in the meadow.

Use the data to estimate the mean breathing rate of the three grasshoppers in the meadow.
How your working.

( A scientist used grasshoppers to investigate the effect of composition of air on breathing rate in insects.
He changed the composition of air they breathed in by varying the
concentrations of oxygen and carbon dioxide.

The scientist collected 20 mature grasshoppers from a meadow.
He placed the grasshoppers in a small chamber where he could adjust and control the composition of air surrounding them.
The small chamber restricted the movement of the grasshoppers.

His results for three of the grasshoppers are shown in the table below in the form in which he presented them.

Gas

1 ) Oxygen

2 ) Carbon dioxide

Breathing rate of grasshopper in different types of air / breaths per minute

3 ) Grasshopper 1

4 ) Grasshopper 2

5 ) Grasshopper 3

Percentage of oxygen and carbon dioxide in different types of air breathed in by grasshopper

A ) Air from atmosthere

B ) Pure oxygen

C ) Gas mixture 1

D ) Gas mixture 2

1A ) 20.9

2A ) 0.1

3A ) 53

4A ) 48

5A ) 61

1B ) 100.0

2B ) 0.0

3B ) 11

4B ) 25

5B ) 13

1C ) 91.0

2C ) 9.0

3C ) 99

4C ) 88

5C ) 96

1D ) 84.0

2D ) 16.0

3D ) 107

4D ) 99

5D ) 93 )

Answer 11

• 53 + 48 + 61 /3 = 54

Question 12

The estimate does not provide a reliable value for the mean breathing rate of all insect species in the meadow.
Other than being an estimate, suggest and explain three reasons why this value would not be reliable.

Answer 12

• The sample is small so it may not be representative for all grasshoppers
• Grasshoppers aren’t the only insects in the meadow, so there are behavioural differences
• The insects aren’t all mature, so different metabolic rates

Question 13

A biologist investigated the effect of water temperature on the rate of ventilation of gills in a species of fish.
She kept four fish in a thermostatically controlled aquarium and measured the mean ventilation rate by counting movements of their gill covers.
Her results are shown in Figure 1.

( Figure 1 shows a graph, “ Mean ventilation rate of fish “ against “ water temperature “ )

( “ Water temperature “ goes up to “ 30 degrees “ on the scale )

( The graph is a straight linear line which starts from “ 0.9 degrees “ to “ 25 degrees “ )

In this investigation, the biologist also monitored the concentration of oxygen in the water in the aquarium.
The concentration of oxygen in water changes with temperature of the water.
Figure 2 shows how it changes.

( Figure 2 shows a graph, “ Solubility of oxygen / g oxygen per kg water “ against “ Water temperature “ )

( “ Solubility of oxygen “ goes up to “ 0.08 “ in scale and “ Water temperature “ goes up to “ 40 “ in scale )

( The graph is almost a negative straight linear graph, but bends a little upwards towards the end )

( The graph starts from “ 0.07 “ in “ Solubility of water “ and ends at “ 0.04 “ )

( The graph starts starts at “ 0 degrees “ and ends at “ 25 degrees “ )

Suggest a difficulty of counting movements of gill covers as a method of measuring rate of ventilation in fish.

Answer 13

• Fish keep moving

Question 14

The biologist concluded that there was no correlation between rate of ventilation of the gills and the temperature of the water.
A scatter diagram can be used to look for a correlation but, in this investigation, it was not the appropriate graph for her data.
Explain why.

( A biologist investigated the effect of water temperature on the rate of ventilation of gills in a species of fish.
She kept four fish in a thermostatically controlled aquarium and measured the mean ventilation rate by counting movements of their gill covers.
Her results are shown in Figure 1.

( Figure 1 shows a graph, “ Mean ventilation rate of fish “ against “ water temperature “ )

( “ Water temperature “ goes up to “ 30 degrees “ on the scale )

( The graph is a straight linear line which starts from “ 0.9 degrees “ to “ 25 degrees “ )

In this investigation, the biologist also monitored the concentration of oxygen in the water in the aquarium.
The concentration of oxygen in water changes with temperature of the water.
Figure 2 shows how it changes.

( Figure 2 shows a graph, “ Solubility of oxygen / g oxygen per kg water “ against “ Water temperature “ )

( “ Solubility of oxygen “ goes up to “ 0.08 “ in scale and “ Water temperature “ goes up to “ 40 “ in scale )

( The graph is almost a negative straight linear graph, but bends a little upwards towards the end )

( The graph starts from “ 0.07 “ in “ Solubility of water “ and ends at “ 0.04 “ )

( The graph starts starts at “ 0 degrees “ and ends at “ 25 degrees “ ) )

Answer 14

• There is only one dependent variable

Question 15

Describe the relationship between temperature of water, oxygen in water and rate of ventilation.

( A biologist investigated the effect of water temperature on the rate of ventilation of gills in a species of fish.
She kept four fish in a thermostatically controlled aquarium and measured the mean ventilation rate by counting movements of their gill covers.
Her results are shown in Figure 1.

( Figure 1 shows a graph, “ Mean ventilation rate of fish “ against “ water temperature “ )

( “ Water temperature “ goes up to “ 30 degrees “ on the scale )

( The graph is a straight linear line which starts from “ 0.9 degrees “ to “ 25 degrees “ )

In this investigation, the biologist also monitored the concentration of oxygen in the water in the aquarium.
The concentration of oxygen in water changes with temperature of the water.
Figure 2 shows how it changes.

( Figure 2 shows a graph, “ Solubility of oxygen / g oxygen per kg water “ against “ Water temperature “ )

( “ Solubility of oxygen “ goes up to “ 0.08 “ in scale and “ Water temperature “ goes up to “ 40 “ in scale )

( The graph is almost a negative straight linear graph, but bends a little upwards towards the end )

( The graph starts from “ 0.07 “ in “ Solubility of water “ and ends at “ 0.04 “ )

( The graph starts starts at “ 0 degrees “ and ends at “ 25 degrees “ ) )

Answer 15

• As water temperature increases, oxygen concentration falls and ventilation rate increases

Question 16

Use figure 1 and 2 to explain the advantage to the fish of the change in its rate of ventilation.

( A biologist investigated the effect of water temperature on the rate of ventilation of gills in a species of fish.
She kept four fish in a thermostatically controlled aquarium and measured the mean ventilation rate by counting movements of their gill covers.
Her results are shown in Figure 1.

( Figure 1 shows a graph, “ Mean ventilation rate of fish “ against “ water temperature “ )

( “ Water temperature “ goes up to “ 30 degrees “ on the scale )

( The graph is a straight linear line which starts from “ 0.9 degrees “ to “ 25 degrees “ )

In this investigation, the biologist also monitored the concentration of oxygen in the water in the aquarium.
The concentration of oxygen in water changes with temperature of the water.
Figure 2 shows how it changes.

( Figure 2 shows a graph, “ Solubility of oxygen / g oxygen per kg water “ against “ Water temperature “ )

( “ Solubility of oxygen “ goes up to “ 0.08 “ in scale and “ Water temperature “ goes up to “ 40 “ in scale )

( The graph is almost a negative straight linear graph, but bends a little upwards towards the end )

( The graph starts from “ 0.07 “ in “ Solubility of water “ and ends at “ 0.04 “ )

( The graph starts starts at “ 0 degrees “ and ends at “ 25 degrees “ ) )

Answer 16

• As concentration of oxygen falls, less oxygen flows over gills
• So concentration of oxygen in the blood falls
• An increase in ventilation rate increases the flow of oxygen across gills

Question 17

Breathing out as hard as you can is called forced expiration.

Describe and explain the mechanism that causes forced expiration.

Answer 17

• Contraction of internal intercostal muscles
• Relaxation of diaphragm muscles
• Causes decrease in volume of chest
• Air is pushed down the pressure gradient

Question 18

Two groups of people volunteered to take part in an experiment.

• People in group A were healthy.
• People in group B were recovering from an asthma attack.

Each person breathed in as deeply as they could.
They then breathed out by forced expiration.

A scientist measured the volume of air breathed out during forced expiration by each person.

The graph below shows the results.

( Graph shows, “ Mean volume of air breathed out during forced expiration / dm^3 “ against “ Time breathing out / s “ )

( “ Mean volume of air breathed out during forced expiration / dm^3 “ starts at “ 0.0 “ and ends at “ 6.0 “ )

( “ Time breathing out / s “ starts at “ 0 “ and ends at “ 4 “ )

( The graph has two curves, one labelled “ Group A “ and the other is labelled “ Group B “ )

( The “ Group A “ curve starts at “ 0.0 “ in “ Mean volume of air breathed out during forced expiration / dm^3 “ and ends at “ 5.0 “ )

( “ Group A “ starts at “ 0 “ and ends at “ 4 “ in “ Time breathing out / s “ )

( “ Group B “ curve starts at “ 0.0 “ in “ Mean volume of air breathed out during forced expiration / dm^3 “ and ends at “ 2.0 “ )

( “ Group “ starts at “ 0 “ and ends at “ 4 “ in “ Time breathing out / s “ )

Forced expiration volume ( FEV ) is the volume of air a person can breather out in 1 second.
Using data from the first second of forced expiration, calculate the percentage decrease in the FEV for group B compare with group A.

Answer 18

• 19%

Question 19

The people in group B were recovering from an asthma attack.
Explain how an asthma attack caused the drop in the mean FEV shown in the figure below.

( Two groups of people volunteered to take part in an experiment.

• People in group A were healthy.
• People in group B were recovering from an asthma attack.

Each person breathed in as deeply as they could.
They then breathed out by forced expiration.

A scientist measured the volume of air breathed out during forced expiration by each person.

The graph below shows the results.

( Graph shows, “ Mean volume of air breathed out during forced expiration / dm^3 “ against “ Time breathing out / s “ )

( “ Mean volume of air breathed out during forced expiration / dm^3 “ starts at “ 0.0 “ and ends at “ 6.0 “ )

( “ Time breathing out / s “ starts at “ 0 “ and ends at “ 4 “ )

( The graph has two curves, one labelled “ Group A “ and the other is labelled “ Group B “ )

( The “ Group A “ curve starts at “ 0.0 “ in “ Mean volume of air breathed out during forced expiration / dm^3 “ and ends at “ 5.0 “ )

( “ Group A “ starts at “ 0 “ and ends at “ 4 “ in “ Time breathing out / s “ )

( “ Group B “ curve starts at “ 0.0 “ in “ Mean volume of air breathed out during forced expiration / dm^3 “ and ends at “ 2.0 “ )

( “ Group “ starts at “ 0 “ and ends at “ 4 “ in “ Time breathing out / s “ ) )

Answer 19

• Bronchioles contract
• Bronchioles secrete more mucus
• Diameter of airways is reduced
• Therefore flow of air is reduced

Question 20

Describe how oxygen in the air reaches capillaries surrounding alveoli in the lungs.
Details of breathing are not required.

Answer 20

• Trachea, bronchi and bronchioles
• Oxygen moves down the pressure and diffusion gradient
• Then across alveolar epithelium

Question 21

Forced expiratory volume (FEV) is the greatest volume of air a person can breathe out in 1 second.
Forced vital capacity ( FVC ) is the greatest volume of air a person can breathe out in a single breath.
The figure below shows results for the volume of air breathed out by three groups of people, A, B and C.
Group A had healthy lungs. Groups B and C had different lung conditions that affect breathing.

( Figure shows a graph, “ Volume of air breathed out / dm^3 “ against “ Time breathing out / s “ )

( “ Volume of air breathed out / dm^3 “ starts from “ 0.0 “ and ends at “ 6.0 “ in scale )

( “ Time breathing out / s “ starts from “ 0 “ and ends at “ 10 “ in scale )

( In the figure, there are 3 curves, “ Group A “, “ Group B “ and “ Group C “ )

( “ Group A “ starts at “ 0.0 “ in “ Volume of air breathed out / dm^3 “ and ends at “ 5.0 “, the curve becomes stationary at “ 5.0 “ after “ 4 “ seconds of breathing out )

( “ Group B “ starts at “ 0.0 “ in “ Volume of air breathed out / dm^3 “ and ends at “ 3.2 “, the curve becomes stationary at “ 3.2 “ after “ 3 “ seconds of breathing out )

( “ Group C “ starts at “ 0.0 “ in “ Volume of air breathed out / dm^3 “ and ends at “ 3.4 “, the curve doesn’t become stationary, but gradient becomes fixed after “ 4 “ seconds of breathing out )

Calculate the percentage drop in FEV for group C compared with the healthy people.

Answer 21

• 80%

Question 22

Asthma affects bronchioles and reduces flow of air in and out of the lungs.
Fibrosis does not affect bronchioles; it reduces the volume of the lungs.

Which group, B or C, was the one containing people with fibrosis of their lungs?
Use the information provided and evidence from the figure above to explain your answer.

( Forced expiratory volume (FEV) is the greatest volume of air a person can breathe out in 1 second.
Forced vital capacity ( FVC ) is the greatest volume of air a person can breathe out in a single breath.
The figure below shows results for the volume of air breathed out by three groups of people, A, B and C.
Group A had healthy lungs. Groups B and C had different lung conditions that affect breathing.

( Figure shows a graph, “ Volume of air breathed out / dm^3 “ against “ Time breathing out / s “ )

( “ Volume of air breathed out / dm^3 “ starts from “ 0.0 “ and ends at “ 6.0 “ in scale )

( “ Time breathing out / s “ starts from “ 0 “ and ends at “ 10 “ in scale )

( In the figure, there are 3 curves, “ Group A “, “ Group B “ and “ Group C “ )

( “ Group A “ starts at “ 0.0 “ in “ Volume of air breathed out / dm^3 “ and ends at “ 5.0 “, the curve becomes stationary at “ 5.0 “ after “ 4 “ seconds of breathing out )

( “ Group B “ starts at “ 0.0 “ in “ Volume of air breathed out / dm^3 “ and ends at “ 3.2 “, the curve becomes stationary at “ 3.2 “ after “ 3 “ seconds of breathing out )

( “ Group C “ starts at “ 0.0 “ in “ Volume of air breathed out / dm^3 “ and ends at “ 3.4 “, the curve doesn’t become stationary, but gradient becomes fixed after “ 4 “ seconds of breathing out ) )

Answer 22

• Group B because they breathe out as quickly as a healthy person
• So bronchioles aren’t being affected
• FVC is reduced

Question 23

Describe and explain how the countercurrent system leads to efficient gas exchange across the gills of a fish.

Answer 23

• Water and blood flow in opposite directions
• Which maintains a concentration gradient
• Along the whole lamellae

Question 24

Amoebic gill disease (AGD) is caused by a parasite that lives on the gills of some species of fish.
The disease causes the lamellae to become thicker and to fuse together.

AGD reduces the efficiency of gas exchange in fish.
Give two reasons why.

Answer 24

• Thicker lamellae so greater diffusion distance

- Lamellae fuse, so reduced surface area

Question 25

The table below shows some features of gas exchange of a fish at rest.

Volume of oxygen absorbed by the gills from each dm3 of water / cm^3

7

Mass of fish / kg

0.4

Oxygen required by fish / cm^3 kg^-1 hour^-1

90

Calculate the volume of water that would have to pass over the gills each hour to supply the oxygen required by the fish.
Show your working.

Answer 25

• 90 x 0.4 = 36
• 30 / 7 = 5.1428571429
• 5.1 dm^3

Question 26

The volume of water passing over the gills increases of the temperature of the water increases.
Suggest why.

Answer 26

• Less oxygen dissolved in water

Question 27

Name the process by which oxygen passes from an alveolus in the lungs into the blood.

Answer 27

• Simple diffusion

Question 28

Describe two adaptations of the structure of alveoli for efficient gas exchange.

Answer 28

• Thin walls

- Total surface area is large

Question 29

The photograph shows a fire-breather creating a ball of fire.
Fire-breathers do this by blowing a fine mist of paraffin oil onto a flame.
Some of this mist can be inhaled and may eventually lead to fibrosis.

People who have been fire-breathers for many years often find they cannot breathe out properly.
Explain why.

Answer 29

• Loss of elasticity

- Less recoil