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9 Cell transport mechanisms
Answers to Exam practice questions
1 B
2 A
3 An explanation that includes any four of the following:
• Transported using pinocytosis/endocytosis;
• Membrane is pulled inwards in one small area/to form a depression;
• Requiring the use of ATP;
• Depression is closed off to form a vesicle;
• Vesicle opens at the inner membrane surface to release contents;
Accept points from a clearly annotated diagram
4 a) D = plasmalemma;
E = Cell wall;
b) (concentrated) glucose solution;
c) An explanation that makes reference to any four of the following:
• When immersed in a concentrated solution water will flow out of the cytoplasm by
• This means the turgor pressure becomes zero;
• Because the cytoplasm has shrunk away from the cell wall;
• Turgor pressure is an important method of support in the plant therefore without it the
plant will wilt;
Accept description of symptoms in place of 'wilt'
5 a) An explanation that makes reference to the following:
• Membrane contains both integral and peripheral proteins;
• Integral proteins are embedded in the main structure of the membrane/peripheral
proteins only on the surface of the membrane;
• Therefore peripheral protein much easier/integral protein much harder to extract;
b) An explanation that makes reference to the following:
• Phospholid is a bilayer/two layers;
• But membrane is made up of more than just phospholipid/contains proteins/contains
• Therefore less phospholipid is needed to complete a double layer;
© Hodder & Stoughton Limited 2015
9 Cell transport mechanisms
Answers to Exam practice questions
6 a) Difference in osmotic potentials: −650 − 1245 = −595;
Correct application of formula: WP = TP + OP;
WP = +250 − 595 = −345 kPa;
e.c.f for 2 and 3
Correct answer gains all 3 marks
3rd m.p. must include correct sign and units
Penalise lack of sign and units only once so correct answer without sign or units gains 2 marks[3]
b) An explanation that makes reference to any three of the following:
• Phospholipid hydrocarbon 'tails' are hydrophobic;
• Therefore will prevent passage of water molecules;
• Presence of unsaturated hydrocarbons allows some water to pass;
• Water passes directly through channel proteins;
7 a) • Without cyanide:
Increase = 41 – 16 = 25
Rate = 25 ÷ 9 = 2.78/2.778;
• With cyanide:
Increase = 5 – 6 = –1
Rate = –1 ÷ 9 = –0.11/–0.111;
• Correct units: µM h ;
Penalise lack of units once only but lack of minus sign for cyanide is a separate error
Two correct answers with units = 3 marks
b) • K ions move by active transport;
• Active transport requires ATP;
• Cyanide prevents the formation of ATP in mitochondria;
c) The K ions may be transported out of the root cells/may leak out of the root section;
And are not being replaced;
Do not accept 'used up'
8 a) An explanation that includes any two of the following:
• Red solution allows red light to pass through;
• Therefore little absorption takes place/different red solutions will only have small
differences of absorption;
• There will much more absorption of blue light;
b) Treat at different temperatures for longer time;
© Hodder & Stoughton Limited 2015
9 Cell transport mechanisms
Answers to Exam practice questions
Different temperatures will take longer to reach the given value/whole of cylinder will not
reach the required temperature in 1 minute;
Accept stated time of at least 2 min for m.p.1
c) Discussion that includes any four of the following points:
• Not really valid because no actual evidence of protein damage in these data;
• Possible that temperature affects the pigment colour to give similar results;
• Not really valid because there is an effect below 50#°C (so may not be denaturation);
• Does however support the model that proteins could be affected by the temperature;
• Because outflow of pigment increases suddenly at the expected temperature;
Accept other sound biological explanation for m.p. 2
Reject any reference to 'prove'
Stretch and challenge questions
8 a) The answer here could well stretch to a number of topics but here it is intended to keep to
this chapter. It is a question that could be attempted again during the second year of an Alevel course, where it would have even wider applications.
The main answers here should concentrate on movement of water and movement of ions.
In freshwater:
• Freshwater has a very low osmotic potential (you should attempt to use this terminology
as much as possible). The gills have partially permeable membranes so there will be a
strong tendency for water to enter the fish by osmosis. This would obviously cause severe
problems if left unchecked. The fish would swell and its internal body fluids, including the
blood, would be diluted. This dilution would mean that all of its body cells would quickly
become afflicted by the same problem.
• The second major problem is that of movement of ions. The lack of ions in fresh water
would create a concentration gradient across the gills, which would tend to cause most of
the ions to diffuse out of the fish. Even just considering the two ions mentioned in the
question (Na and Cl ) these are vital to many physiological processes, especially, as we
shall learn later, the transmission of nerve impulses.
• So the fish faces a 'double whammy' of dilution and loss of ions.
In salt water:
• The problems here are the same but reversed.
• The high osmotic potential of sea water will tend to draw water out of the gills by osmosis
and the concentration gradient will tend to cause ions to enter the fish by diffusion. So
© Hodder & Stoughton Limited 2015
9 Cell transport mechanisms
Answers to Exam practice questions
again there is a double problem, increasing concentration of the body fluids of the fish by
water loss and by ions entering.
b) It is a good idea to start thinking of all possible solutions to begin with, some that are using
your knowledge of membrane transport and some that are simply very basic principles.
An example of the latter might be to suggest that fish would be better off with an
impermeable cover. Obviously this would mean that they could not take in oxygen nor lose
carbon dioxide so it's impractical but it is a start. Restricting permeable surfaces to only those
absolutely essential at least begins to control the problems and most animals do this to some
Some answers you might suggest using your biological knowledge:
In fresh water:
• The simple answer to water entry is to get rid of it quickly, just as you would do if you
drank a lot of water. So fresh water fish have simple kidneys to excrete a lot of water.
• However, membranes allow water entry via protein channels as phospholipids limit its
entry, so it may be possible to control these protein channels.
• The ion problem in fresh water should give you more ideas. Cell surface membranes are
selectively permeable so that movement of ions is already restricted and generally
confined to specific carrier proteins. There is no reason why these should not operate to
take up ions selectively, even from very low concentrations. However, this is very energy
demanding, so there is a price to pay!
In salt water:
• Losing water by osmosis also has a simple answer – drink more! However, this does have
serious drawbacks. You might like to research why, if you were adrift in the ocean,
drinking sea water would not be a good idea.
• However, many fish do just that. The problem is obviously that in doing so you ingest
large quantities of ions, especially Na+ and Cl–. Once again, a knowledge of membrane
transport might help. Excreting excess ions is just where active transport might be useful
again and the obvious place for this to happen is across the gills. So you might like to
research this aspect in marine fish.
• Another possibility to control osmotic loss is to go back to basic principles. Osmosis will
not take place unless there is a difference in concentrations across the membrane, so one
solution might be to match the concentration of the body fluids to that of the sea water.
This will bring its own problems but it is another interesting idea to research.
Finally, the remarkable adaptations that must be shown by migratory fish switching from one
environment to another and hence facing opposite problems within a few days is also a very
interesting idea to research.
© Hodder & Stoughton Limited 2015