Gas exchange

Question 1

Describe, using an example, one way that gas exchange organs are adapted to their function (2)

Answer 1

1. Large surface area
2. Mesophyll cells in plants
3. Thin for short diffusion pathway
4. Walls of tracheoles in insects
5. Steep concentration gradient maintained
6. Counter current system in fish gills

Question 2

Explain why plants that live in the desert often have sunken stomata or stomata surrounded by hairs (2)

Answer 2

1. To trap any moist air near the stomata

2. Reduce the concentration gradient from leaf to air, which reduced water loss

Question 3

(a) How did the scientists ensure they could make a valid comparison between leaves
from different species? (1)

Answer 3

Scientists) used fully grown leaves / used five plants of each (species).
Ignore other references to methodology. Reward only
information provided in the Resource.
Do not accept reference to number of leaves − different
plants were used

Question 4

(b) Describe a method you could use to find the surface area of a leaf. (3)

Answer 4

1. Draw around leaf on graph paper;
   Mark as a trio − MP1, MP2 and MP3 OR MP4, MP5 and
   MP6. Do not mix and match.
   Both aspects needed for mark − drawing and type of paper.
2. Count squares (however described);
   There is no reward for additional detail e.g. dealing with part
   squares.
3. Multiply by 2 (for upper and lower leaf surface);

OR

4. Draw around a leaf on paper of known mass (per unit area);
   Both aspects needed for mark − drawing and mass of paper.
5. Cut out and weigh;
6. Multiply by 2 (for upper and lower leaf surface

Question 5

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

(1).

Answer 5

1. Smaller surface area
   so
   less evaporation / less heat absorbed;
   Correctly selected feature and the explanation required for 1
   mark.
   In all marking points − ‘less water loss’ is insufficient as an
   explanation but accept transpiration for evaporation or
   diffusion.
2. Thicker leaves
   so
   greater diffusion distance (for water);
   Accept ‘thicker leaves so more water storage’.
3. Fewer stomata / lower stomatal density
   so less diffusion / evaporation (of water);
4. Smaller surface area to volume ratio
   so
   less evaporation.

Question 6

(ii) 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.

Answer 6

1. Thick(er) cuticle so increase in diffusion distance / slower (rate of) diffusion; Feature and explanation needed for each mark. Reject other features not related to leaves. Reject features related to water storage. ‘Cuticle’ alone is insufficient (all leaves have a cuticle). Reject suggestion of ‘less’ diffusion, for idea of ‘slower diffusion’, an idea of rate is required.
2. Hairs on leaves so reduction in air movements / increase in humidity / decrease in water potential gradient;
3. Curled leaves so reduction in air movements / increase in humidity / decrease in water potential gradient;
4. Sunken stomata so reduction in air movements / increase in humidity / decrease in water potential gradient.

Question 7

Q2.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. ‘

a) Suggest the purpose of each of the three leaf treatments (3)

Answer 7

1. (No grease)
 means stomata are open
 OR
 allows normal CO2 uptake;
Allow ‘gas exchange’ for CO2 uptake. 

As a control’ is insufficient on its own.
2. (Grease on lower surface)
seals stomata
OR
stops CO2 uptake through
stomata
OR
to find CO2 uptake through
stomata
OR
shows CO2 uptake through cuticle / upper surface;
3. (Grease on both surfaces) shows sealing is effective
OR
stops all CO2 uptake.

Question 8

(b) (i) Describe the results shown for Treatment 1.

2

Answer 8

i) 1. (Mean rate of) carbon dioxide uptake was constant and fell after
the light turned off;
Ignore absence of arbitrary units in both marking points.
Both ideas needed for mark.
Accept ‘stayed at 4.5’ as equivalent to ‘was constant’.

2. Uptake fell from 4.5 to 0 / uptake started to fall at 60 minutes and
   reached lowest at 80 minutes / uptake fell over period of 20
   minutes;
   One correct use of figures required.
   Accept fell to nothing / no uptake for 0.

Question 9

(ii) The stomata close when the light is turned off.

Explain the advantage of this to the plant. (2)

Answer 9

) 1. (Because) water is lost through stomata;
2. (Closure) prevents / reduces water loss;
3. Maintain water content of cells.
This marking point rewards an understanding of reducing
water loss e.g. reduce wilting, maintain turgor, and is not
related to photosynthesis.

Question 10

(c) (i) 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.

Answer 10

) (Carbon dioxide uptake) through the upper surface of the leaf / through
cuticle.

Question 11

i) In both Treatment 1 and Treatment 2, the uptake of carbon dioxide falls to
zero when the light is turned off.
Explain why. (2)

Answer 11

1. No use of carbon dioxide in photosynthesis (in the dark);
2. No diffusion gradient (maintained) for carbon dioxide into leaf /
   there is now a diffusion gradient for carbon dioxide out of leaf (due
   to respiration).

Question 12

Q3.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.

(a) 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. (2)

Answer 12

1. Other gases / nitrogen / water vapour in atmosphere / A;
2. Only oxygen and carbon dioxide in gas mixtures / C and D;
3. Composition of / gases in A not controlled / composition of gas mixtures
   / C and D controlled

Question 13

(b) Use all the data to describe the effect of concentration of carbon dioxide on the
breathing rate of grasshoppers. (3)

Answer 13

1. Breathing rate lowest when no carbon dioxide / in (pure) oxygen /
   B;
   Idea of ‘lowest’ must be stated.
2. (Generally) presence of carbon dioxide increases breathing rate / as
   concentration of carbon dioxide increases breathing rate increases /
   there is a positive correlation;
   A general point incorporating all concentrations.
3. Breathing rate increases when (carbon dioxide) higher than 0.1% /
   concentration in atmosphere / A;
   This MP requires a specific comparison to 0.1% or the
   atmospheric concentration.
   Accept ‘gas mixtures 1 and 2 / C and D’ for ‘higher carbon
   dioxide’.
4. Breathing rate of grasshopper 3 falls in D / 16% / gas mixture 2
   (whereas others increase).
   Restating data alone is insufficient for any mark point.

Question 14

(c) 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.
(i) Use the data to estimate the mean breathing rate of the three grasshoppers in
the meadow. Show your working (2)

Answer 14

(i) 54;
OR
1. Correct data / column A chosen;
A correct answer of 54 gets 2 marks.
MP1 and MP2 allow a possible mark for an incorrect
calculation or choice of wrong data.
2. Correct calculation of mean from data chosen;
Check − the three values must be from same column.

Question 15

(ii) 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. (3)

Answer 15

(ii) 1. Small sample / only 3 (grasshoppers)
so may not be representative (of all grasshoppers / insects);
2. Grasshoppers are not the only insects / species;
so genetic / behavioural / metabolic differences;
3. (Insects) not all mature / are at different stages of development /
different sizes;
so different metabolic rates;
4. Movement not restricted / not at rest in meadow;
so (rate of) respiration higher;
5. (Naturally-occurring) carbon dioxide concentration lower in
meadow;
so breathing rate lower;
Explanations required, therefore both parts of answer
required for credit in each marking point.
Accept appropriate converse answers.
Accept ‘respiration’ for ‘metabolism’ and vice versa.

Question 16

Q4.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.

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

(a) Suggest a difficulty of counting movements of gill covers as a method of measuring
rate of ventilation in fish. (1)

Answer 16

) Fish keep moving / swimming / movement of gill covers too fast to count (at higher
temperatures).
Accept converse.
Reject personal errors e.g. with counting.
Neutral − ‘water not clear’ or ‘difficult to see movement of gill
covers’.

Question 17

(b) The biologist concluded that there was a correlation between rate of ventilation of
the gills and 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.
(1)

Answer 17

1. There is only one dependent variable / there are not two dependent
   variables / water temperature is the independent variable / breathing rate
   is dependent on water temperature;
   Accept either approach for 1 mark.
   For ‘independent’ accept ‘manipulated’.
   Reject −‘need two continuous variables’.
2. Water temperature plus breathing rate are not both properties of
   fish
   or
   water temperature plus breathing rate are not both properties of water.
   Accept reference to the ‘two variables’ (instead of water
   temperature plus breathing rate)

Question 18

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

Answer 18

) As (water) temperature increases, oxygen (concentration / solubility)
falls and ventilation rate increases.
MP requires all 3 aspects before credit is possible. The
correct context is required for each aspect so
e.g. do not reward
‘as oxygen concentration falls, water temperature increases’
or
‘as temperature increases, ventilation rate increases and
oxygen concentration falls’.

Question 19

(ii) Use Figure 1 and Figure 2 to explain the advantage to the fish of the change
in its rate of ventilation. (3)

Answer 19

1. As concentration / solubility of oxygen falls
   less oxygen flows over gills / less oxygen enters gills / less oxygen
   enters fish;
   For MP1 and MP2 accept converse.
   Both aspects needed for mark.
2. (As a result) blood oxygen (concentration) falls / is lower;
3. An increase in ventilation rate increases / maintains the flow of
   oxygen / carbon dioxide across gills / into (or out of) fish;
   Accept idea in relation to either gas or ‘gas exchange’.
4. Maintains diffusion / concentration gradient(s) (in gills);
   Gradient(s) relates to either / both gas(es).
5. To maintain oxygen supply to cells / tissues / organs / to maintain
   respiration.
   Accept a named example of ‘tissues’ e.g. muscle

Question 20

Q5.Breathing out as hard as you can is called forced expiration.
(a) Describe and explain the mechanism that causes forced expiration. (4)

Answer 20

1. Contraction of internal intercostal muscles;
2. Relaxation of diaphragm muscles / of external intercostal muscles;
3. Causes decrease in volume of chest / thoracic cavity;
4. Air pushed down pressure gradient.

Question 21

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.

(c) 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. (4)

Answer 21

) 1. Muscle walls of bronchi / bronchioles contract;

2. Walls of bronchi / bronchioles secrete more mucus;
3. Diameter of airways reduced;
4. (Therefore) flow of air reduced.

Question 22

Q6.(a) Describe how oxygen in the air reaches capillaries surrounding alveoli in the lungs.
Details of breathing are not required. (4)

Answer 22

1. Trachea and bronchi and bronchioles;
2. Down pressure gradient;
3. Down diffusion gradient;
4. Across alveolar epithelium.
   Capillary wall neutral
5. Across capillary endothelium / epithelium.

Question 23

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

(c) 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. (3)

Answer 23

1. (Group B because) breathe out as quickly as healthy / have similar FEV
   to group A;
2. So bronchioles not affected;
3. FVC reduced / total volume breathed out reduced.
   Allow this marking point for group C

Question 24

Q7.(a) Describe and explain how the countercurrent system leads to efficient gas exchange across
the gills of a fish.

Answer 24

1. Water and blood flow in opposite directions;
   Accept: diagram if clearly annotated
2. Maintains concentration / diffusion gradient / equilibrium not reached /
   water always next to blood with a lower concentration of oxygen;
   Must have the idea of ‘maintaining’ or ‘always’ in reference to
   concentration / diffusion gradient
   Accept: constant concentration / diffusion gradient
3. Along whole / length of gill / lamellae;
   Accept: gill plate / gill filament

Question 25

(b) 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 (2)

Answer 25

1. (Thicker lamellae so) greater / longer diffusion distance / pathway;
   Q Neutral: ‘thicker’ diffusion pathway
2. (Lamellae fuse so) reduced surface area;
   Accept: reduced SA:VOL

Question 26

(ii) The volume of water passing over the gills increases if the temperature of the
water increases. Suggest why (1)

Answer 26

1. Increased metabolism / respiration / enzyme activity;
   Accept: enzymes work more efficiently
2. Less oxygen (dissolved in water);
   Neutral: references to increased kinetic energy (of water
   molecules)

Question 27

Q8.(a) (i) Name the process by which oxygen passes from an alveolus in the lungs into
the blood. (1)

Answer 27

(i) (Simple) diffusion;
Reject facilitated diffusion
Accept lipid diffusion

Question 28

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

Answer 28

1. Thin walls / cells;
2. ‘Short diffusion pathway’ alone is an explanation not a
   description
3. Accept squamous epithelia / one cell thick
4. (Total) surface area is large;
5. Ignore references to ‘volume ratio’

Question 29

People who have been fire-breathers for many years often find they cannot
breathe out properly. Explain why. (2)

Answer 29

(b) 1. Loss of elasticity / elastic tissue / increase in scar tissue;
1. Accept elastin
2. Less recoil;