IGCSE Biology Paper 4 (Theory / Extended)

Topic 3: Movement Into and Out of Cells -- Challenge Exam 3

Duration1 hour 15 minutes

Total Marks80

Questions7

Time Remaining75:00

Syllabus0610

Instructions

Answer all questions in the spaces provided.
Show all working for calculations.
Use appropriate scientific terminology.
Your answers will be automatically graded when you submit.

Question Navigation

This paper covers the whole of Topic 3. Like a real Cambridge paper, the seven questions range across every sub-topic — diffusion, osmosis and active transport — and they are deliberately mixed rather than grouped. All three Topic 3 papers do; they differ in the angle they come at it from, not in what they cover.

Question 1 – Two Potatoes Compared

Total: 12 marks

A student cuts strips of potato tissue 50.0 mm long and leaves them in sucrose solutions for one hour, then measures each strip again. She repeats the whole experiment with a second potato that has been stored in a warm dry cupboard for three weeks.

sucrose concentration / mol dm⁻³	0.0	0.2	0.4	0.6	0.8
change in length, fresh potato / %	+9.0	+4.0	−1.0	−6.0	−11.0
change in length, stored potato / %	+13.0	+8.0	+3.0	−2.0	−7.0

(a) [3]

Explain why the strips in the 0.0 mol dm⁻³ solution became longer.

Model Answer – 1(a)

distilled water has a higher water potential than the cell sap of the potato cells [1]

so there is a net movement of water into the cells by osmosis, through their partially permeable cell membranes [1]

the cells become turgid and slightly larger, so the whole strip of tissue becomes longer [1]

⚠ If you missed marks here: The last step is the one candidates leave out: the question asks about the length of a strip, so the answer must join the change in individual cells to the change in the tissue. Note also that the cells do not burst, however dilute the solution, because the cell wall resists the swelling.

1 mark -- distilled water has a higher water potential than the cell sap 1 mark -- net movement of water into the cells by osmosis through partially permeable membranes 1 mark -- cells become turgid and slightly larger so the strip lengthens

(b) [3]

Estimate the concentration at which there would be no change in length for each potato, and explain what the difference between the two values shows.

Model Answer – 1(b)

fresh potato: about 0.35 mol dm⁻³, where the values change from +4.0 to −1.0 [1]

stored potato: about 0.7 mol dm⁻³, where the values change from −2.0 at 0.6 towards −7.0 at 0.8 [1]

the stored potato needs a more concentrated solution to balance it, so its cell sap must be more concentrated and its water potential lower, because it has lost water during storage [1]

⚠ If you missed marks here: Interpolate rather than quoting the nearest tested concentration — the balance point is where the line crosses zero, which lies between two of the columns. The interpretation mark is for realising that the isotonic point measures the cell sap concentration, so a shift in that point means the tissue itself has changed.

1 mark -- fresh potato balance point estimated at about 0.35 mol per dm cubed 1 mark -- stored potato balance point estimated at about 0.7 mol per dm cubed 1 mark -- stored potato has more concentrated sap and a lower water potential after losing water

Suggest why measuring the change in length is likely to be less accurate than measuring the change in mass.

Model Answer – 1(c)

the percentage changes in length are small and the strips are soft, so the ends are difficult to define and the strip can bend or be stretched while being measured [1]

a change in mass can be measured to 0.01 g on a balance, which is a far finer resolution relative to the size of the change being measured [1]

⚠ If you missed marks here: Accuracy questions are answered by comparing the size of the change with the smallest division of the instrument. Answers such as “rulers are less accurate” state a conclusion without a reason; explain why the measurement itself is hard to take on soft, bendable tissue.

1 mark -- length changes are small and soft strips can bend or stretch when measured 1 mark -- mass can be measured to a much finer resolution relative to the change

(d) [2]

The student uses strips of the same length from the same potato for every concentration. Explain why each of these precautions is necessary.

Model Answer – 1(d)

using strips of the same starting length means the percentage changes are calculated from comparable measurements and the surface area exposed is similar [1]

using one potato keeps the water potential of the cell sap constant, so external concentration is the only variable that changes [1]

⚠ If you missed marks here: Each precaution needs its own reason. The stronger of the two is the second, since the whole point of the second experiment is that different potatoes really do have different sap concentrations — which is exactly what the stored potato results demonstrate.

1 mark -- same starting length gives comparable measurements and similar surface area 1 mark -- same potato keeps the cell sap water potential constant

(e) [2]

Predict what would happen to the length of a strip left in 0.8 mol dm⁻³ sucrose for 24 hours instead of one hour. Explain your answer.

Model Answer – 1(e)

the strip would shorten further at first but the change would then stop, as the cells reach equilibrium with the solution and there is no further net movement of water [1]

the strip would not go on shrinking indefinitely, because once the water potentials are equal water crosses the membrane equally in both directions [1]

⚠ If you missed marks here: The trap is to assume that a longer time always means a bigger change. Osmosis stops changing the tissue once the water potentials are balanced, and the extra 23 hours simply keep the strip at the value it had already reached. Say “no further net movement”, not “osmosis stops”.

1 mark -- strip shortens further at first then the change stops at equilibrium 1 mark -- no further net movement once water potentials are equal

Question 2 – Life in Fresh Water and Sea Water

Total: 12 marks

Paramecium is a single-celled organism that lives in fresh water. Its cytoplasm is much more concentrated than the pond water around it. It contains a structure called a contractile vacuole, which fills with water and then empties it out of the cell. A researcher measures how often the contractile vacuole empties when Paramecium is kept in pond water diluted or concentrated to different degrees.

relative concentration of the surrounding water	very dilute	dilute	same as cytoplasm	concentrated
number of times the contractile vacuole empties per minute	22	13	1	0

(a) [3]

Explain why water continually enters Paramecium when it is living in pond water.

Model Answer – 2(a)

pond water is very dilute, so it has a higher water potential than the cytoplasm of Paramecium [1]

there is therefore a net movement of water molecules into the cell by osmosis, down the water potential gradient [1]

the water crosses the partially permeable cell surface membrane [1]

⚠ If you missed marks here: All three marks are for precise language rather than new ideas: the comparison of water potentials, the phrase net movement, and the naming of the membrane as partially permeable. This is a good example of how the same three sentences earn marks in unfamiliar contexts.

1 mark -- pond water has a higher water potential than the cytoplasm 1 mark -- net movement of water into the cell by osmosis down the gradient 1 mark -- water crosses the partially permeable cell surface membrane

(b) [3]

Describe and explain the pattern in the results.

Model Answer – 2(b)

the more dilute the surrounding water, the more often the contractile vacuole empties, from 22 times per minute in very dilute water down to none in concentrated water [1]

a more dilute solution gives a steeper water potential gradient into the cell, so water enters faster [1]

the vacuole must therefore empty more often to remove the extra water and stop the cell bursting [1]

⚠ If you missed marks here: Describe with figures, then explain with the gradient, then link back to the organism’s problem. The final mark is the biological purpose — an animal-type cell has no cell wall, so unlimited water entry would burst it. Leaving that out reduces a full answer to a description.

1 mark -- describes more frequent emptying in more dilute water with figures 1 mark -- steeper water potential gradient means water enters faster 1 mark -- vacuole empties more often to remove the extra water and prevent bursting

Explain why the contractile vacuole stops working when Paramecium is placed in a solution with the same concentration as its cytoplasm.

Model Answer – 2(c)

the water potential inside and outside the cell is now the same [1]

so there is no net movement of water into the cell and no excess water to be removed [1]

⚠ If you missed marks here: Write “no net movement”. Water molecules still cross the membrane in both directions in equal numbers, and an answer stating that osmosis has stopped is refused even though the observation it explains is correct.

1 mark -- water potential inside and outside are equal 1 mark -- no net movement of water so there is no excess to remove

(d) [2]

Emptying the contractile vacuole requires energy from respiration. Suggest why, and predict what would happen to Paramecium in very dilute water if its respiration were stopped.

Model Answer – 2(d)

water is being moved out of the cell against the water potential gradient, which cannot happen passively, so energy from respiration is needed [1]

if respiration stopped, the vacuole could not empty, water would continue to enter by osmosis, and the cell would swell and burst [1]

⚠ If you missed marks here: This organism is a rare case where water is moved uphill, and it is moved by pumping the vacuole rather than by osmosis. It is still not correct to call it “active transport of water” on a mark scheme, but the underlying principle — uphill movement costs energy from respiration — is exactly the one the syllabus wants you to apply.

1 mark -- water is removed against the water potential gradient so energy is required 1 mark -- without respiration the cell would swell and burst

(e) [2]

A related single-celled organism lives in sea water and has no contractile vacuole. Suggest why it does not need one.

Model Answer – 2(e)

sea water is concentrated and has a water potential similar to, or lower than, that of the organism’s cytoplasm [1]

so there is little or no net entry of water by osmosis and no excess water to remove [1]

⚠ If you missed marks here: A “suggest” command in an unfamiliar context is answered by applying the same rule you have just used, not by recalling a fact. The rule is that water moves down its water potential gradient; if the gradient is flat, nothing needs removing.

1 mark -- sea water has a water potential similar to or lower than the cytoplasm 1 mark -- little or no net entry of water so no excess to remove

Question 3 – The Root Hair Cell

Total: 12 marks

The drawing shows a root hair cell from a young plant root.

Structure P is the outermost layer of the cell, Q is the large central vacuole, R is one of many small organelles scattered in the cytoplasm, and S is the long narrow extension of the cell that reaches between the soil particles.

(a) [3]

Name structures P, Q and R.

Model Answer – 3(a)

P is the cell wall [1]

Q is the vacuole, containing cell sap [1]

R is a mitochondrion [1]

⚠ If you missed marks here: The outermost layer of a plant cell is the cell wall, with the cell membrane pressed against its inner surface, so answers naming the outer layer as the membrane lose the mark. The small scattered organelles in a root cell cannot be chloroplasts, because roots are not exposed to light.

1 mark -- P identified as the cell wall 1 mark -- Q identified as the vacuole 1 mark -- R identified as a mitochondrion

(b) [2]

Explain how structure S helps the cell to absorb water and mineral ions.

Model Answer – 3(b)

the long narrow extension greatly increases the surface area of the cell in contact with soil water [1]

a larger surface area means faster uptake of water by osmosis and of mineral ions by active transport [1]

⚠ If you missed marks here: Surface area is the factor being tested, and the answer should say what it increases the rate of. Adding that the extension also reaches between soil particles into water films is a useful extra detail, but it is the surface area that the mark scheme requires.

1 mark -- long extension greatly increases the surface area in contact with soil water 1 mark -- larger surface area allows faster uptake of water and ions

Explain why structure R is present in unusually large numbers in this cell.

Model Answer – 3(c)

mitochondria are the site of aerobic respiration [1]

respiration releases energy [1]

this energy is needed for the active transport of mineral ions from the soil water into the cell against a concentration gradient [1]

⚠ If you missed marks here: The frequent wrong answer is that the mitochondria supply energy “for osmosis”. Osmosis is passive and costs the cell nothing, so it cannot be the reason. Each of the three links — respiration, energy, active transport against a gradient — carries its own mark.

1 mark -- mitochondria are the site of aerobic respiration 1 mark -- respiration releases energy 1 mark -- energy is needed for active transport of ions against the gradient

(d) [2]

Water enters this cell by osmosis but mineral ions usually enter by active transport. Explain why two different processes are needed.

Model Answer – 3(d)

water moves down its water potential gradient, from the more dilute soil water into the more concentrated cell sap, which requires no energy [1]

mineral ions are usually far more concentrated inside the cell than in the soil water, so they must be moved against their concentration gradient, which requires energy from respiration [1]

⚠ If you missed marks here: The deciding factor is always the direction of the gradient, never the identity of the substance. Both movements are into the same cell across the same membrane at the same time, and it is only because the two gradients run in opposite senses that two processes are needed.

1 mark -- water moves down its water potential gradient without needing energy 1 mark -- ions move against their concentration gradient so energy from respiration is required

(e) [2]

A student floods a pot plant so that the soil is waterlogged for several days. The plant grows poorly and its leaves turn pale. Explain why.

Model Answer – 3(e)

waterlogged soil has almost no air spaces, so the root cells receive very little oxygen and aerobic respiration is greatly reduced [1]

less energy is available for active transport, so fewer mineral ions are taken up and growth is poor [1]

⚠ If you missed marks here: The chain is oxygen, respiration, energy, active transport, ion uptake, growth. Answers that stop at “the roots drown” describe the situation without explaining it. Note that the plant is short of ions, not of water — which is what makes the observation surprising and worth explaining.

1 mark -- waterlogged soil lacks oxygen so aerobic respiration is reduced 1 mark -- less energy for active transport so fewer mineral ions are absorbed

Question 4 – Oxygen Reaching Muscle Tissue

Total: 12 marks

A researcher models the supply of oxygen to a block of respiring tissue. Sheets of muscle tissue of different thicknesses are kept in a well-oxygenated solution, and the oxygen concentration at the centre of each sheet is measured after 20 minutes as a percentage of the concentration in the surrounding solution.

thickness of sheet / mm	0.2	0.5	1.0	2.0	4.0
oxygen at the centre after 20 min / % of surrounding value	96	81	54	19	2

(a) [3]

Describe the relationship shown by the data and explain it.

Model Answer – 4(a)

as the thickness increases the oxygen concentration at the centre falls, from 96 % at 0.2 mm to only 2 % at 4.0 mm [1]

the thicker the sheet, the greater the distance the oxygen must diffuse to reach the centre, and the rate of diffusion falls as distance increases [1]

the tissue is also respiring and using oxygen on the way, so less reaches the centre in thick sheets [1]

⚠ If you missed marks here: The second reason is the one that separates strong answers. Oxygen is not simply travelling through inert jelly; every layer of living tissue it crosses consumes some of it, which is why the fall is so steep. Quote figures from both ends of the table to secure the description mark.

1 mark -- describes falling oxygen at the centre with thickness using figures 1 mark -- greater distance means a slower rate of diffusion 1 mark -- respiring tissue uses oxygen on the way so less reaches the centre

(b) [3]

Use the data to explain why human muscles are supplied by a dense network of capillaries rather than relying on diffusion from the body surface.

Model Answer – 4(b)

diffusion delivers a useful oxygen concentration only over distances of a fraction of a millimetre, since at 2.0 mm only 19 % reaches the centre [1]

human muscles are many centimetres thick, so diffusion from the body surface could never supply the inner cells [1]

capillaries carry oxygenated blood by mass flow to within a fraction of a millimetre of every muscle cell, so oxygen only has to diffuse a very short distance [1]

⚠ If you missed marks here: Support the argument with a figure from the table rather than asserting that diffusion is “too slow”. The final mark is for the role of the capillary network, and the key phrase is that it reduces the diffusion distance, not that it increases the speed of diffusion itself.

1 mark -- uses a figure to show diffusion is only effective over a fraction of a millimetre 1 mark -- human muscles are far too thick to be supplied from the body surface 1 mark -- capillaries reduce the diffusion distance to a fraction of a millimetre

Explain how a muscle cell maintains a concentration gradient for oxygen between the blood and its cytoplasm.

Model Answer – 4(c)

respiration in the cell continually uses up oxygen [1]

so the oxygen concentration inside the cell stays lower than that in the blood, maintaining the concentration gradient into the cell [1]

⚠ If you missed marks here: This two-mark answer is one of the most reusable in the whole syllabus, and it works for carbon dioxide in reverse. Answers claiming that the cell “pumps oxygen in” describe active transport, which is not needed and would waste energy the cell has no reason to spend.

1 mark -- respiration continually uses up oxygen inside the cell 1 mark -- internal concentration stays lower than in the blood so the gradient is maintained

(d) [2]

During hard exercise the oxygen concentration in a muscle cell falls further than at rest. Explain the effect this has on the rate at which oxygen diffuses into the cell.

Model Answer – 4(d)

a lower oxygen concentration inside the cell makes the concentration gradient between blood and cell steeper [1]

so oxygen diffuses into the cell at a faster rate [1]

⚠ If you missed marks here: Concentration gradient is one of the four named factors, and this is the clearest everyday example of it changing. The common slip is to say the cell “needs more oxygen so it takes more in”, which describes a purpose rather than a mechanism and earns nothing.

1 mark -- lower internal concentration makes the gradient steeper 1 mark -- oxygen diffuses into the cell faster

(e) [2]

Suggest two reasons why the results of this model may not apply exactly to living muscle in the body.

Model Answer – 4(e)

in the body the tissue is supplied from a network of capillaries all through it, not only from its outer surfaces, so no cell is as far from a supply as in the model [1]

in the body blood flow keeps replacing the oxygen and the temperature and rate of respiration differ from those in the model, so rates of use and supply are not the same [1]

⚠ If you missed marks here: Evaluation marks are given for differences that would actually change the result, not for generic complaints. Saying “it was only a model” or “there could be errors” identifies nothing specific, whereas naming the missing capillary network directly explains why the model overstates the problem.

1 mark -- in the body capillaries supply oxygen throughout the tissue not only at the surface 1 mark -- blood flow temperature and rate of respiration differ from the model conditions

Question 5 – Planning an Investigation

Total: 10 marks

A student wants to find out how the surface area of a piece of tissue affects the rate at which a substance moves into it. She has a supply of potato, a cork borer, a scalpel, a ruler, an electronic balance, a stopwatch, distilled water, sucrose solutions of a range of concentrations, paper towels and a set of beakers.

(a) [3]

Describe how the student could change the surface area of the potato while keeping the total mass of tissue the same, and state the independent and dependent variables.

Model Answer – 5(a)

cut equal masses of potato and divide each mass into a different number of pieces, for example one large cylinder, two halves, four quarters and eight smaller pieces [1]

the independent variable is the surface area of the tissue, changed by the number of pieces the same mass is cut into [1]

the dependent variable is the percentage change in mass of the tissue after a fixed time [1]

⚠ If you missed marks here: The design must change surface area without changing the amount of tissue, otherwise two variables move at once and no conclusion is possible. Naming the dependent variable as “the mass” rather than “the percentage change in mass” loses the mark, because the starting masses are only approximately equal.

1 mark -- same mass of potato divided into different numbers of pieces 1 mark -- independent variable identified as surface area 1 mark -- dependent variable identified as percentage change in mass

(b) [3]

Describe three variables she must control, and explain the effect on the results if one of them were not controlled.

Model Answer – 5(b)

the concentration of the sucrose solution, the volume of solution and the temperature must be the same for every sample [1]

the time in solution and the way each sample is blotted before weighing must also be the same [1]

if the concentration were not controlled, differences in mass change could be caused by different water potential gradients rather than by surface area, so no valid conclusion could be drawn [1]

⚠ If you missed marks here: Every controlled variable in this topic maps onto one of the four factors affecting the rate of movement, so name the factor as well as the variable. The final mark is for explaining the consequence, and that explanation must refer to a specific alternative cause of the observed change.

1 mark -- names concentration volume and temperature as controlled variables 1 mark -- names time in solution and consistent blotting as controlled variables 1 mark -- explains that an uncontrolled variable gives an alternative cause so the conclusion is invalid

Describe the results she would expect, and explain them.

Model Answer – 5(c)

the samples cut into more pieces would reach their final percentage change in mass sooner, because a larger surface area allows water to enter or leave the tissue faster [1]

the final percentage change would be about the same for every sample, because the water potential of the tissue and of the solution is unchanged by cutting [1]

⚠ If you missed marks here: The distinction between rate and final value is what this part is really testing. Surface area changes how quickly equilibrium is reached; it does not change where equilibrium lies, because the water potentials on either side are unaffected by how the tissue was cut.

1 mark -- pieces with a larger surface area reach the final mass change sooner 1 mark -- the final percentage change is similar because the water potentials are unchanged

(d) [2]

Explain why the student should repeat each sample at least three times, and state what she should do with the readings.

Model Answer – 5(d)

repeats reduce the effect of anomalous results caused by uneven cutting, blotting or weighing, and show how consistent the measurements are [1]

she should calculate a mean percentage change for each surface area, ignoring any clear anomalies [1]

⚠ If you missed marks here: Repeats improve reliability, not accuracy, and the mark scheme distinguishes the two. State what is done with the repeats as well as why they are taken — an answer that says only “to make it more reliable” is a phrase rather than an explanation.

1 mark -- repeats reduce the effect of anomalies and show consistency 1 mark -- calculate a mean percentage change for each surface area

Question 6 – Leaking Beetroot

Total: 12 marks

Beetroot cells contain a red pigment in their vacuoles. Discs of beetroot are rinsed and placed in test tubes of distilled water at different temperatures for 20 minutes. The amount of red pigment that has escaped into the water is then measured with a colorimeter; a higher reading means more pigment has leaked out.

temperature / °C	10	20	30	40	50	60	70
colorimeter reading / arbitrary units	4	6	9	15	48	81	88

(a) [2]

Explain why very little pigment escapes at 10 °C and 20 °C.

Model Answer – 6(a)

at these temperatures the cell membranes are intact and are partially permeable [1]

the pigment molecules are large and cannot pass through an intact partially permeable membrane, so they stay in the vacuole [1]

⚠ If you missed marks here: The reason the pigment stays in is a property of the membrane, not of the pigment’s solubility. Answers that say the pigment “is not dissolved yet” miss the point — it is already dissolved in the cell sap, and only the membranes are stopping it leaving.

1 mark -- membranes are intact and partially permeable at low temperatures 1 mark -- large pigment molecules cannot pass through an intact membrane

(b) [3]

Describe and explain what the results show happens between 40 °C and 60 °C.

Model Answer – 6(b)

the reading rises sharply, from 15 units at 40 °C to 81 units at 60 °C, a more than fivefold increase [1]

the high temperature has damaged the cell membranes, so they are no longer partially permeable [1]

the pigment can now leak out of the vacuole and through the damaged membranes into the water [1]

⚠ If you missed marks here: Two membranes are involved — the one round the vacuole and the cell surface membrane — and damage to both is needed for pigment to appear in the water. Use figures for the description mark and keep the word damaged for the explanation; saying the cells were “killed” describes the outcome rather than the change in the membrane.

1 mark -- describes a sharp rise using figures from the table 1 mark -- high temperature damages the cell membranes so they are no longer partially permeable 1 mark -- pigment leaks out through the damaged membranes

Explain why the discs were rinsed in distilled water before the experiment began.

Model Answer – 6(c)

cutting the discs damages cells at the cut surfaces, and pigment leaks from these damaged cells [1]

rinsing removes that pigment so it is not measured as part of the leakage caused by the temperature treatment, which would give falsely high readings [1]

⚠ If you missed marks here: The mark is for identifying the source of the contamination and its effect on the measurement. Answers that say the rinsing was “to clean the discs” do not connect the step to the result, which is what a validity question always requires.

1 mark -- cutting damages cells at the surfaces which release pigment 1 mark -- rinsing removes this pigment so readings are not falsely high

(d) [3]

Beetroot discs are also left in 1.0 mol dm⁻³ sucrose solution at 20 °C. After an hour the discs are noticeably smaller and the solution stays almost colourless. Explain both observations.

Model Answer – 6(d)

the sucrose solution has a lower water potential than the cell sap, so there is a net movement of water out of the cells by osmosis [1]

the cells lose water and become flaccid or plasmolysed, so the tissue shrinks [1]

the membranes are undamaged at 20 °C and remain partially permeable, so the large pigment molecules cannot escape and the solution stays colourless [1]

⚠ If you missed marks here: Two independent movements must be kept apart: water leaves by osmosis, but pigment does not leave at all. It is tempting to assume that anything which shrinks the cells must also let their contents out, and the colourless solution is the evidence that this is not so.

1 mark -- lower water potential outside causes net movement of water out by osmosis 1 mark -- cells lose water and shrink becoming flaccid or plasmolysed 1 mark -- undamaged partially permeable membranes keep the large pigment molecules inside

(e) [2]

Suggest one reason why a colorimeter gives a better measurement of leakage than judging the colour by eye.

Model Answer – 6(e)

a colorimeter gives a numerical reading on a fixed scale, so results can be compared and plotted [1]

judging by eye is subjective and different observers, or different lighting conditions, would give different judgements [1]

⚠ If you missed marks here: The two credited ideas are quantitative and objective. An answer stating simply that the colorimeter is “more accurate” names the conclusion without giving the reason, and reasons are what evaluation marks are for.

1 mark -- colorimeter gives a numerical reading that can be compared and plotted 1 mark -- judging by eye is subjective and varies between observers or lighting

Question 7 – Preserving Food and Reviving Vegetables

Total: 10 marks

Salting and sugaring have been used to preserve food for thousands of years. Salted fish keeps for months in a warm climate, and jam does not go mouldy while the jar is sealed. In a kitchen, limp lettuce leaves are often revived by soaking them in cold water for half an hour.

(a) [3]

Explain how salting preserves fish, in terms of water potential and the bacteria that would otherwise spoil it.

Model Answer – 7(a)

a heavy coating of salt gives the liquid around the bacteria a very low water potential [1]

so there is a net movement of water out of the bacterial cells by osmosis, through their partially permeable membranes [1]

the bacteria lose so much water that they cannot grow or reproduce, so the fish does not spoil [1]

⚠ If you missed marks here: The answer must be about the microorganisms, not about the fish. A common incomplete answer says “salt kills bacteria”, which names no mechanism at all. Use the water potential language throughout, and remember the same reasoning explains why jam, honey and dried fruit keep so well.

1 mark -- salt gives the surrounding liquid a very low water potential 1 mark -- net movement of water out of the bacterial cells by osmosis 1 mark -- bacteria lose water so cannot grow or reproduce and spoilage is prevented

(b) [3]

Explain why limp lettuce leaves become crisp again after soaking in cold water.

Model Answer – 7(b)

the water has a higher water potential than the cell sap of the lettuce cells [1]

so there is a net movement of water into the cells by osmosis, and the cells become turgid [1]

the pressure of the water inside pressing outwards on the cell walls makes the tissue firm again, so the leaves feel crisp [1]

⚠ If you missed marks here: The final mark is for naming the mechanism of support — turgor pressure acting on the cell wall. Saying only that “the cells fill up with water” describes the change without explaining why filled cells are stiff, and it is the pressure against the wall that provides the stiffness.

1 mark -- water has a higher water potential than the cell sap 1 mark -- net movement of water into the cells by osmosis makes them turgid 1 mark -- turgor pressure on the cell walls makes the tissue firm again

For each of the following, name the process involved and give one reason for your choice.

Movement 1: sodium ions moving from a dilute solution in the gut into a lining cell that already contains far more sodium.
Movement 2: water moving from a lettuce cell into a strong sugar solution.
Movement 3: carbon dioxide moving from the air into a photosynthesising leaf cell.
Movement 4: oxygen moving from a fish gill into the blood.

Model Answer – 7(c)

movement 1 is active transport, because the ions are moving against the concentration gradient and so require energy from respiration [1]

movement 2 is osmosis, because it is water moving down a water potential gradient through a partially permeable membrane [1]

movement 3 is diffusion, because photosynthesis uses carbon dioxide up inside the cell, so it moves down its concentration gradient [1]

movement 4 is diffusion, because oxygen moves down its concentration gradient from the water into the blood [1]

⚠ If you missed marks here: Each mark needs the name and the reason, and the reason is always about the direction of the gradient. Movement 1 is the one most often misnamed: candidates see ions dissolved in water and reach for osmosis, but osmosis moves water and only water, whatever else happens to be dissolved in it.

1 mark -- movement 1 named as active transport with the against-gradient reason 1 mark -- movement 2 named as osmosis with the water potential gradient reason 1 mark -- movement 3 named as diffusion with the correct gradient reason 1 mark -- movement 4 named as diffusion with the correct gradient reason

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