Download Movement of subsances in and out of cells Questions and Answers

Q1.
The table shows the percentage of some gases in the air a boy breathed in and out.
Gases
Air
breathed in
Air
breathed out
carbon dioxide
0.04%
4.0%
oxigen
20.0%
16.0%
water vapour
1.0%
6.0%
(a)
What happens in the lungs to change the levels of oxygen and carbon dioxide in
this way?
Oxygen .......................................................................................................................
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Carbon dioxide ...........................................................................................................
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(4)
(b)
Compare the percentage of water vapour in the air breathed out with the
percentage in air breathed in.
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(2)
(Total 6 marks)
Q2.
Plants need chemical energy for respiration and for active transport.
(i)
Write a balanced chemical equation which represents the process of respiration in
plants.
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(2)
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(ii)
Describe the process of active transport in the root hair cells of plants.
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(3)
(Total 5 marks)
Q3.
Plant roots obtain some of their mineral salts from the soil by active transport.
What is involved in active transport?
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(Total 4 marks)
Q4.
(a)
The diagrams show what happens to the shape of a plant cell placed in distilled
water.
(i)
Explain why the cell swells and becomes turgid. Name the process involved.
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(2)
(ii)
Give one feature of the cell wall which allows the cell to become turgid.
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(1)
(b)
Describe the change which will occur if a piece of peeled potato is placed in a
concentrated sugar solution and explain why this change occurs.
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(3)
(Total 6 marks)
Q5.
(a)
The diagram shows a cereal crop.
Complete spaces (i) and (ii).
(2)
(iii)
What sort of weather may cause the cereal crop to wilt?
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(1)
(b)
Describe the process of transpiration in plants.
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(3)
(Total 6 marks)
Q6.
The diagram shows the human breathing system.
(a)
Complete the labels (i) and (ii).
(2)
(b)
Complete the following sentence.
When we breathe out, the mixture of gases which leaves the air sacs contains
more .............................. and less ......................................... than the mixture of
gases which enters the air sacs.
(2)
(Total 4 marks)
Page 5
Q7.
(a)
The graph shows how the mass of oxygen you breathe in changes as you
climb up a mountain.
Describe, in as much detail as you can, how the mass of oxygen in one breath
changes as you climb from sea level to 3000 m.
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(3)
(b)
People who live high up in mountainous areas have more red blood cells than
people who live at sea level. The graph below shows how the number of red blood
cells changes with height above sea level.
(i)
How many more red blood cells does a person living at 3000 m above sea
level have than someone living at sea level? Show clearly how you work out
your answer.
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Increase in number of red blood cells = ....................................millions per m3
(2)
(ii)
What is the advantage of having more red blood cells?
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(1)
(Total 6 marks)
Q8.
The table shows the composition of blood entering and leaving the lungs.
Gas
Concentration in arbitrary units
Blood entering lungs
Blood leaving lungs
Oxygen
40
100
Carbon dioxide
46
40
(a)
Describe, in as much detail as you can, the changes that take place in the
composition of blood as it passes through the lungs.
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(3)
(b)
Which part of the blood:
(i)
transports most carbon dioxide; .....................................................................
(ii)
transports most oxygen? ................................................................................
(2)
(Total 5 marks)
Page 8
Q9.
A potted plant was left in a hot, brightly lit room for ten hours. The plant was not
watered during this period. The drawings show how the mean width of stomata changed
over the ten hour period.
(a)
Why do plants need stomata?
.....................................................................................................................................
(1)
(b)
Name the cells labelled X on the drawing.
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(1)
(c)
The width of the stomata changed over the ten hour period.
Explain the advantage to the plant of this change.
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(2)
(Total 4 marks)
Page 9
Q10.
(a)
The diagram shows four ways in which molecules may move into and out of a
cell. The dots show the concentration of molecules.
The cell is respiring aerobically.
Which arrow, A, B, C or D, represents:
(i)
movement of oxygen molecules;
....................
(ii)
movement of carbon dioxide molecules?
....................
(2)
(b)
Name the process by which these gases move into and out of the cell.
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(1)
(c)
Which arrow, A, B, C or D, represents the active uptake of sugar molecules by the
cell?
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Explain the reason for your answer.
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(2)
(Total 5 marks)
Page 10
Q11.
The diagram shows part of the breathing system in a human.
(a)
Use words from the list to label the parts on the drawing.
alveoli
bronchiole
bronchus
diaphragm
trachea (windpipe)
(4)
(b)
Where in the lungs does oxygen enter the blood?
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(1)
(c)
Which process in cells produces carbon dioxide?
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(1)
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(Total 6 marks)
Q12.A cook prepares a fresh fruit salad by cutting up a variety of fruits and placing them in a
bowl with layers of sugar in between. After two hours the fruit is surrounded by syrup
(concentrated sugar solution).
Explain, as fully as you can, why syrup (concentrated sugar solution) was produced after
two hours.
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(Total 4 marks)
Q13.
(a)
The drawing shows some of the organs in the human thorax.
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On the drawing, use guidelines to label:
(i)
the heart;
(ii)
a rib;
(iii)
the diaphragm;
(iv)
the trachea.
(4)
(b)
The drawing shows a section through an alveolus.
At A, oxygen moves from the air in the alveolus into the blood capillary.
Explain, as fully as you can, how oxygen moves into the blood.
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(2)
(Total 6 marks)
Q14.
Some students set up an experiment using osmosis to find the concentration of
sucrose solution in potato cell sap. They used discs of potato cut to the same size and
weighing approximately 10 gms. The discs were put into each of five beakers.
(a)
(i)
After two hours they reweighed the discs after carefully blotting them first.
Why did the students blot the potato before weighing it?
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(1)
(ii)
Their results are shown in the table below.
Beaker 1 Beaker 2 Beaker 3 Beaker 4 Beaker 5
Final mass in g
13.0
12.2
9.0
7.9
7.3
Initial mass in g
10.0
10.6
10.0
10.1
10.4
The students calculated the % gain or loss in mass of potato. Complete this
table of results for Beakers 2, 4 and 5.
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Beaker 1
Beaker 2
13 – 10.0 = 3.0
Beaker 3
Beaker 4
Beaker 5
9.0 – 10.0 =
–1.0
= –10%
Gain in mass = 30%
Loss in mass =
10%
(3)
(b)
(i)
Draw a graph of % Gain or Loss in mass against sucrose concentration.
(3)
Page 15
(ii)
Use the graph to find the concentration of potato cell sap.
Concentration of cell sap = ........................................... % sucrose solution
(1)
(iii)
Explain in terms of osmosis how you chose this value.
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(2)
(Total 10 marks)
Q15.
The diagram shows an enlargement of structure D.
The arrows show the direction of the gases exchanged in this structure. Name gas X and
gas Y.
X ...........................................................................................................................................
Y ...........................................................................................................................................
(Total 2 marks)
Page 16
Q16.
Some students set up the following apparatus.
The balances show the same mass at the start of the investigation.
After 24 hours the mass of flask B was the same but the mass of flask A had changed.
(i)
Describe and explain the change to the mass of flask A.
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(3)
(ii)
Why did the students need to set up flask B?
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(1)
(Total 4 marks)
Page 17
Q17.
Some students set up this experiment to investigate osmosis. They filled two pieces
of dialysis [visking] tubing with different liquids and left them both in a beaker of 5%
sucrose solution for an hour.
(a)
Describe and explain the likely results after one hour.
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(6)
(b)
Describe two examples where osmosis is used in living things.
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(2)
(Total 8 marks)
Q18.
Some students set up the equipment below to investigate osmosis.
(a)
What is osmosis?
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(3)
(b)
(i)
What will happen to the water level in the capillary tube during the
investigation because of osmosis?
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(1)
(ii)
Use your knowledge of osmosis to explain why this happens.
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(2)
(Total 6 marks)
Q19.
The diagram shows a part of a lung that is involved in gaseous exchange in a
human.
(i)
Draw and label, on the diagram, one arrow to show the direction of movement of
oxygen between the alveolus and capillary.
(1)
(ii)
Draw and label, on the diagram, one arrow to show the direction of movement of
carbon dioxide, between the alveolus and capillary.
(1)
Page 20
(iii)
Give the function of the red blood cell in this process.
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(1)
(Total 3 marks)
Q20.
The table gives information about a geranium plant and a cactus plant.
The geranium grows in gardens in the UK. The cactus grows in hot deserts.
Feature
Geranium
Cactus
5
15
1800
150
Percentage of water storage tissue in stem
50
85
Number of stomata per mm2
59
13
daylight
at night
0.2
5
Thickness of waxy cuticle in micrometres
Total leaf surface area in cm2
Time of day when stomata open
Horizontal spread of roots in metres
Using only information in the table, explain how the cactus is better adapted for living in
hot, dry conditions.
To gain full marks in this question you should write your ideas in good English. Put them
into a sensible order and use the correct scientific words.
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(Total 5 marks)
Q21.Complete the table by writing the correct process next to its description.
Choose your answers from the list in the box
breathing
diffusion
digestion
osmosis
Description
respiration
Process
Moving air in and out of the lungs
The movement of particles of a
substancefrom high to low concentration
The release of energy from glucose
(Total 3 marks)
Q22.
Four leaves were removed from the same plant. Petroleum jelly (a waterproofing
agent) was spread onto some of the leaves, as follows:
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Leaf A: on both surfaces
Leaf B: on the lower surface only
Leaf C: on the upper surface only
Leaf D: none applied
Each leaf was then placed in a separate beaker, as shown in diagram 1.
Diagram 1
Each beaker was weighed at intervals. The results are shown in the graph.
(a)
Give evidence from the graph in answering the following questions.
(i)
Which surface (upper or lower) loses water most rapidly? .............................
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Evidence ...........................................................................................................
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(1)
(ii)
Is water lost from both surfaces of the leaf? ....................................................
Evidence ...........................................................................................................
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(1)
(b)
Diagram 2 shows the appearance of each surface of the leaf as seen through a
microscope.
Upper Surface of Leaf
Lower Surface of Leaf
Diagram 2
(i)
Name space X and cell Y.
X ...................................................................
Y ...................................................................
(2)
(ii)
Use information in diagram 2 to explain why the results are different for leaves
B and C.
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(2)
(Total 6 marks)
Q23.
The diagram shows an alveolus and a blood capillary in the lung.
(i)
During gaseous exchange, oxygen and carbon dioxide are exchanged across the
wall of the alveolus. On the diagram, carefully draw two arrows to show the paths
taken by oxygen and by carbon dioxide during this process. Label each arrow.
(3)
(ii)
Name the process by which oxygen moves across the wall of the alveolus.
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(1)
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(iii)
Each lung contains about 350 million alveoli. How does this help gaseous
exchange?
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(1)
(Total 5 marks)
Q24.
Emphysema is a disease of the lungs. People who smoke cigarettes are more likely
to suffer from emphysema. The diagrams show lung tissue from a healthy person and
lung tissue from a person with emphysema. The diagrams are drawn to the same scale.
Lung tissue from a healthy person
emphysema
Lung tissue from a person with
Explain how emphysema reduces the amount of oxygen which diffuses into the blood
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(Total 2 marks)
Page 26
Q25.
The table shows the concentrations of some mineral ions in the cells of a pond plant
and in the surrounding pond water.
Concentration in mmol per dm3
Potassium
Calcium
Sulphate
Plant cells
49.0
7.0
7.0
Pond water
0.5
0.7
0.4
(i)
The plant cells would not have been able to absorb these mineral ions from the
pond water by diffusion. Explain why not.
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(2)
(ii)
Suggest a process which would allow these ions to be absorbed from the pond
water by the plant cells.
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(1)
(Total 3 marks)
##
A runner might drink a special ‘sports drink’ at intervals during a marathon race. The table
shows the substances present in a sports drink.
Substance
Percentage
Water
Sugar
5.0
Ions
0.2
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(a)
Complete the table to show the percentage of water in the sports drink.
(1)
(b)
The runner sweats and also breathes heavily during the race.
(i)
Why does the runner need to sweat?
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(1)
(ii)
Which two substances in the table are lost from the body in sweat?
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(1)
(iii)
Which substance in the table is lost from the body during breathing?
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(1)
(c)
How does the sugar in the sports drink help the athlete during the marathon?
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(2)
(Total 6 marks)
Q27.
The diagram shows the human breathing system.
Page 28
(a)
Place on the diagram:
(i)
a letter X where oxygen enters the blood;
(1)
(ii)
an arrow showing the direction the diaphragm moves when we breathe in.
(1)
(b)
List the following structures in the order the air passes through them when we
breathe in.
alveoli
bronchi
bronchioles
trachea
1 ..................................................................................
2 ..................................................................................
3 ..................................................................................
4 ..................................................................................
(1)
(c)
By what process does oxygen enter the blood? Draw a ring around your answer.
diffusion
digestion
osmosis
respiration
(1)
(Total 4 marks)
Page 29
Q28.
Long distance runners are advised to take several drinks during a race.
The table gives the composition of two drinks, Isotonic and Cola.
Drink
Sugar
concentration in
grams per litre
Sodium ion
concentration in
mmol per litre
Chloride ion
concentration in
mmol per litre
Isotonic
73
24
12
Cola
105
3
1
Explain why Isotonic would be the best drink for a long distance runner on a hot day.
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(Total 2 marks)
Q29.
A group of students looked at stomata on four different species of plants, A, B, C
and D. They estimated the number of stomata per cm2 on the upper and lower surfaces of
the leaves of the four species.
Their results are shown in the table.
Plant
species
Estimated number of stomata per cm2 of leaf surface
Upper surface of leaf
Lower surface of leaf
A
4000
28 000
B
0
800
Page 30
C
8500
15 000
D
8000
26 000
(a)
Which plant species probably lives in a dry region?
Explain the reason for your answer.
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(3)
(b)
All four species have more stomata on the lower surface of their leaves than on the
upper surface.
Suggest how this could help the plants to survive better.
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(2)
(Total 5 marks)
Q30.
A group of students looked at stomata on four different species of plants, A, B, C
and D. They estimated the number of stomata per cm2 on the upper and lower surfaces of
the leaves of the four species.
Their results are shown in the table.
Page 31
Plant
species
Estimated number of stomata per cm2 of leaf surface
Upper surface of leaf
Lower surface of leaf
A
4000
28 000
B
0
800
C
8500
15 000
D
8000
26 000
(a)
Which plant species probably lives in a dry region?
Explain the reason for your answer.
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(3)
(b)
All four species have more stomata on the lower surface of their leaves than on the
upper surface.
Suggest how this could help the plants to survive better.
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(2)
(Total 5 marks)
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Page 33
##
(a)
oxygen passes from the air/lungs into the body
gains 1 mark
but
oxygen passes from the air/lungs into the blood
gains 2 marks
carbon dioxide passes from the body into the air/lungs
gains 1 mark
but
carbon dioxide passes from the blood into the air/lungs
gains 2 marks
4
(b)
increased/5% more
gains 1 mark
but
6 times more (in air breathed out)
gains 2 marks
2
[6]
M2.
(i)
C6H12O6 + 6O2 → 6CO2 + 6H2O
energy is neutral
1
formulae all correct
with no omissions / deletions
correctly balanced
credit 1 mark if the answer is the exact
reverse of an incorrect answer for (a)
1
(ii)
and three from
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take up of (soluble) substances / ions against the concentration gradient
or when the concentration (of the
substance / ions) is greater inside the
cell / cytoplasm than outside it
through the (semi-permeable) (cell) membrane energy from mitochondria
or energy from respiration
not just energy
3
[5]
M3.
any four from
molecules / ions
do not credit mineral salts
move(d) through / across the cell
wall / membrane
against (a / the) concentration
gradient
by a series of chemical
reactions
(because) diffusion cannot occur
energy (required)
(supplied by) respiration
oxygen required for respiration (to occur)
[4]
M4.
(a)
(i)
water (molecules) enter(s) (the cell)
Page 35
or water (molecules) pass(es) through the (semi-permeable)
cell membrane
1
by osmosis
or because the concentration of water is
greater outside (the cell than inside it
the vacuole)
accept because of the concentration
gradient provided there is no contradiction
1
(ii)
any one from
(it is) elastic
(it is) strong
(it is fully) permeable (to water)
or water can pass through it
do not credit semi-permeable
do not credit cell membrane is semi-permeable
1
(b)
(the piece of) potato shrinks
or loses its turgor
or becomes flabby
or becomes flaccid
or plasmolysis occur
or cytoplasm pulls away from the cell wall
(because) concentration of sugar
or because concentration of water
1
(solution) is greater than concentration inside the cell / vacuole
inside the cell / vacuole is greater than concentration (of
water) outside
1
water is drawn out of the cell
1
[6]
Page 36
M5.
(a)
(i)
photosynthesis
1
(ii)
respiration
do not credit combustion
do not credit decay
1
(iii)
dry
accept hot or windy or drought
1
(b)
any three from
* evaporation (of water)
or loss of water vapour
* (mostly) from the leaf / leaves
do not credit incorrect reference to leaves
* through the stomata
accept through each stoma
accept through the stomas(sic)
* causing a pull
or causing an increase in osmotic potential (at the top of the
plant)
or causing an increase in water potential (at the top of the
plant) or causing a decrease in osmotic pressure (at the top
of the plant)
* (so that) water moves up (through the plant)
do not credit water vapour moves up through the plant
* as the transpiration stream
* water enters through roots (and goes up plants)
3
[6]
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M6.
(a)
(i)
trachea
accept windpipe
1
(ii)
(left) lung or lungs
do not credit right lung
1
(b)
carbon dioxide or water vapour
do not credit just ‘water’
1
oxygen
answers in terms of used air or fresh air or of temperature
differences are not acceptable
1
[4]
M7.
(a)
falls
1
from 0.25
1
to 0.19
but by 0.06 gains two marks
if neither figure given, accept steadily /
at constant rate for one mark
accept mass of oxygen inversely related
/ negative correlation to height above
sea level for 2 marks
1
(b)
(i)
1.8
accept correct readings from graph for (5 and 6.8) if
subtraction incorrect for one mark
allow one mark for correct subtraction from incorrect
readings
2
(ii)
(blood can carry) more oxygen
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1
[6]
M8.
(a)
any three from
(concentration of) oxygen increases
by 60 (units)
allow oxygen more than doubles for 2 marks
(concentration of) carbon dioxide decreases
from 46 to 40 by 6 units
allow ‘by a small amount’
N.B. usually the first 2 marks will be for the change in oxygen
and carbon dioxide.
The third mark will be for a quantitative comment on one of
these changes
3
(b)
plasma
1
red (blood) cell / haemoglobin /
oxyhaemoglobin
1
[5]
M9.
(a)
allow carbon dioxide to enter / gaseous exchange (oxygen neutral)
(transpiration neutral)
for one mark
1
(b)
guard (cells)
for one mark
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1
(c)
stops / reduces the rate of water loss / transpiration (reject if dark initiated)
stops / reduces wilting / description e.g. drooping / maintains turgor
for 1 mark each
2
[4]
M10.
(a)
(i)
(ii)
B
A
for 1 mark each
2
(b)
diffusion
(reject osmosis)
for one mark
1
(c)
C
because uptake against a concentration / diffusion gradient
(reject osmosis)
(if C not given, then idea of movement essential)
for 1 mark each
2
[5]
M11.
(a)
trachea / windpipe
bronchus
alveoli
diaphragm
for 1 mark each
Page 40
4
(b)
alveoli / air sacs
(reject capillaries)
for one mark
1
(c)
respiration
for one mark
1
[6]
M12.
ideas that
sugar has dissolved in moisture (on surface of fruit)
this solution more concentrated than solution inside fruit
osmosis / diffusion movement of water out of fruit
through partially permeable membrane (of fruit cells)
any four for 1 mark each
allow explanations in terms of concentrations of water molecules for full marks
[4]
M13.
(a)
correctly labelled structures (i – iv)
each for 1 mark
(allow labels as words or numbers: allow without guidelines
if unambiguously labelled)
4
(b)
ideas of
diffusion
greater concentration of oxygen in alveolus / high to low oxygen concentration
membrane / alveolus permeability
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any two for 1 mark each
2
[6]
M14.
(a)
(i)
change in weight was due to changes in potato
or osmosis or not due to outside liquid
ignore ‘to make fair test’
1
(ii)
beaker 2 = 15.1(%) gain
allow 15%
1
beaker 4 = 21.8(5) loss
not 21.7
allow –22%
if no minus or no ‘loss’ check graph
1
beaker 5 = 29.8(%) loss
allow –30%
1
(b)
(i)
both axes correct values
and scales > ½ of each axis
ignore lack of minus signs on vertical axis
1
points correct
<
½ square
allow answers in (a)(ii)
1
line correct
allow curve of best fit which can miss 10, 15
or straight lines between points
do not allow one straight line or sketched line
bar graph zero marks
1
Page 42
(ii)
point where line crosses axis (eg 15-16% sucrose)
allow point from candidate’s graph (
0.5%)
1
(iii)
any two from:
looking for understanding that water in equilibrium
no change in mass
not net movement of water
or water entry and exit are equal
because sucrose solution same
concentration as cell sap or sucrose has
same water potential as cell contents
allow because the concentrations are the same (inside and
out)
2
[10]
M15.
X – oxygen
accept O2
Y – carbon dioxide
accept CO2
[2]
M16.
(i)
the mass got less
accept it got lighter
award 1 mark for water was lost from the plant
1
water was taken into the plant or roots
absorbed water
do not accept soaked into plant
1
Page 43
and lost through transpiration or the
leaves or evaporated from the leaves
or stomata
1
(ii)
to check the effect of the plant or to
act as a control or to show that it was
not due to evaporation from water
do not accept to keep it fair or to check that it was fair
do not accept fair test
1
[4]
M17.
(a)
award 3 marks per tube for each key idea
for tube 1:
expands or gets firmer or bigger or inflates
it gains water
because the concentration of water is less than its surroundings
make sure answer is about water movement and not sucrose
solution
3
for tube 2
gets floppy or flaccid or contracts
it loses water
because the concentration of water is greater than its surroundings
3
(b)
any 2 from:
uptake of water by root (hair) or
movement from cell to cell within
plant
do not credit references to diffusion unless it is clear that the
candidate is referring to the diffusion of water
Page 44
guard cell function
maintain turgor
water absorption in the large intestine
reabsorption of water from the
nephron or collecting duct or in
kidney or osmoregulation in kidney
allow osmosis in other animals if some use is shown
2
[8]
M18.
(a)
movement of water [1]
from high concentration (of water) to low concentration (of water)
or
from (an area of) dilute solution to an area of concentrated solution [1]
through a differentially or partially or selectively or semi permeable
membrane [1]
3
(b)
(i)
it will rise
1
(ii)
water enters visking tubing [1]
because the concentration of water outside is greater than the
concentration inside
or
because the concentration of salt or solute is greater inside the tubing than
outside [1]
or
to equalise concentration water has to enter visking tubing [2]
2
[6]
Page 45
M19.
(i)
oxygen into the blood stream
arrow must start inside alveolus and finish outside the
capillary
1
(ii)
carbon dioxide out of the blood stream
arrow must start inside the capillary and finish inside the
alveolus
1
(iii)
carries/takes up/releases oxygen or carbon dioxide
accept forms oxyhaemoglobin
1
[3]
M20.
Quality of written communication
for ideas given in a sensible order;
comparison made for geranium and cactus for each
feature(ie not just list for geranium followed by list for cactus)
+ linking of feature & explanation
1
any four features + explanations from:
cactus has:
accept converse points for geranium plant
Feature
Explanation
thicker cuticle
waterproof / keeps
water in
smaller surface
area
less water loss /
less heat absorbed
Page 46
fewer stomata
less water loss
stomata open at
night / closed in
day
(closed when
warmest) – so less
water loss
more widespread
roots
quickly absorbs
water (after rain) /
access to bigger
area for absorbing
water / absorb
more water
more water
storage tissue
little water
available in
environment / can
survive drought /
avoids
dehydration
4
[5]
M21.
in correct sequence:
breathing
1
diffusion
1
respiration
1
[3]
M22.
(a)
(i)
lower – B loses less (water / mass) than C
or
described in terms of petroleum jelly
accept converse re Leaf C
1
Page 47
(ii)
yes - B and C lose less than D or
B and C lose more than A or
D loses the most or
A loses the least
do not accept just ‘all leaves lose some weight’
1
(b)
(i)
X = stoma
accept stomata / stomatal pore
do not accept air space
1
Y = guard cell
1
(ii)
petroleum jelly blocks stomata / pores
or petroleum jelly prevents water loss
or petroleum jelly waterproofs
allow pores are blocked in B
1
water (mainly) lost via stomata / pores / X
or stomata on lower surface only
1
[6]
M23.
(i)
On diagram:
oxygen arrow to blood from air and CO2 arrow to air from blood
1
oxygen arrow to red blood cell
1
CO2 arrow from plasma
1
(ii)
diffusion
1
(iii)
large surface or large area
do not accept space
1
[5]
Page 48
M24.
thicker surface
1
reduced surface area
accept fewer alveoli
1
[2]
M25.
(i)
in diffusion: material moves high to low concentration
1
here: concentration in cells > concentration in water or
uptake is against the concentration gradient or by diffusion ions would move out
1
(ii)
active transport / active uptake
1
[3]
M26.
(a)
94.8
1
(b)
(i)
to cool (the body) / maintain (body) temperature
do not accept let out heat
1
(ii)
water and ions
1
Page 49
(iii)
water ignore CO2, and vapour
1
(c)
any two from:
used in respiration
provides energy
(energy) needed for movement / running / muscle action
2
[6]
M27.
(a)
correctly labelled on diagram
(i)
‘X’ on an alveolus
centre of X on the alveolus wall or
inside the alveolus
not if the centre is outside
1
(ii)
arrow pointing downwards
accept anywhere but must point down
1
(b)
in sequence
1 trachea
2 bronchi
3 bronchioles
4 alveoli
1
(c)
diffusion
accept positive indicator
1
[4]
Page 50
M28.
any two from:
•
more or most ions / sodium / chloride or replaces ions / sodium / chloride
do not accept more ions / sodium / chloride for energy
•
lost in sweat
•
to keep blood concentration constant
•
less sugar therefore less chance of ‘sugar rush’
[2]
M29.
(a)
B
1
(B has) low(est) number of stomata
or no stomata on upper surface
or only 800 (on lower surface)
1
less transpiration / evaporation / water loss owtte
or water (vapour) is lost via stomata
only allow zero water loss if linked to no stomata on upper
surface / linked to leaf B upper surface
ignore references to leaf surface area
1
(b)
reduce loss / amount of water (vapour)
accept converse
or
reduced transpiration (from upper surface)
do not allow no water is lost
1
warmer above leaf
accept converse
or wilted leaf folds over lower surface
Page 51
or lower leaf in shade
ignore reference to dust
or less light / heat / sun on lower side
1
[5]
M30.
(a)
B
1
(B has) low(est) number of stomata
or no stomata on upper surface
or only 800 (on lower surface)
1
less transpiration / evaporation / water loss owtte
or water (vapour) is lost via stomata
only allow zero water loss if linked to no stomata on upper
surface / linked to leaf B upper surface
ignore references to leaf surface area
1
(b)
reduce loss / amount of water (vapour)
accept converse
or
reduced transpiration (from upper surface)
do not allow no water is lost
1
warmer above leaf
accept converse
or wilted leaf folds over lower surface
or lower leaf in shade
ignore reference to dust
or less light / heat / sun on lower side
1
[5]
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