Q1.          Some substances move through membranes.
A student set up an investigation.
The student:
•     tied a thin membrane across the end of a funnel
•     put concentrated sugar solution in the funnel
•     put the funnel in a beaker of water
•     measured the level of the solution in the funnel every 30 minutes.
The diagram shows the apparatus.

The graph shows the results.

(a)     After 3 hours, the level of the solution in the funnel is different from the level at the start.
Explain why, as fully as you can.
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(3)

(b)     The student repeated the investigation using dilute sugar solution instead of concentrated sugar solution.
In what way would you expect the results using dilute sugar solution to be different from the results using concentrated sugar solution?
Give the reason for your answer.
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(2)
(Total 5 marks)
Q2.          The table shows the concentrations of three mineral ions in the roots of a plant and in the water in the surrounding soil.
Mineral ion	Concentration in millimoles per kilogram	
	Plant root	Soil	
Calcium	120	2.0	
Magnesium	80	3.1	
Potassium	250	1.2	

(a)     (i)      The plant roots could not have absorbed these mineral ions by diffusion.
Explain why.
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(2)

(ii)     Name the process by which the plant roots absorb mineral ions.
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(b)     How do the following features of plant roots help the plant to absorb mineral ions from the soil?
(i)     A plant root has thousands of root hairs.
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(ii)     A root hair cell contains many mitochondria.
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(2)
(iii)     Many of the cells in the root store starch.
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(1)
(Total 7 marks)
 Q3.          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.
Beaker 1	Beaker 2	Beaker 3	Beaker 4	Beaker 5	
13 – 10.0 = 3.0
	 	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)

(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)
 Q4.          The drawing shows an animal cell, seen at a very high magnification using an electron microscope.

(a)     (i)      Label a mitochondrion [plural = mitochondria].
(1)
(ii)     What happens in the mitochondria?
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(1)
(b)     (i)      Name and label the structure where you would find chromosomes.
(1)
(ii)     What are chromosomes made of?
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(1)
(c)     What controls the rate of chemical reactions in the cytoplasm?
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(1)
(Total 5 marks)
 
#
          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)
 Q6.          (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)
 Q7.          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)
 Q8.          The diagram shows the same plant cell:
•        after 1 hour in distilled water
•        after 1 hour in strong sugar solution.

(a)     Describe two ways in which the cell in the strong sugar solution is different from the cell in distilled water.
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(2)
(b)     Explain how the differences between the cell in the strong sugar solution and the cell in distilled water were caused.
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(2)
(Total 4 marks)
 Q9.          Plant leaves have many stomata.
          The diagram shows a stoma.
                                     
(a)     Name cell X .................................................................
(1)

(b)     The table shows the mean widths of the stomata at different times of the day for two different species of plant.
Species A normally grows in hot, dry deserts.
Species B grows in the UK.
 
 	Time of day	Mean width of stomata as a percentage of their maximum width	
 	In hours	Species A	Species B	
 	0	95	5	
Dark	2	86	5	
 	4	52	6	
 	6	6	40	
 	8	4	92	
 	10	2	98	
Light	12	1	100	
 	14	0	100	
 	16	1	96	
 	18	5	54	
 	20	86	6	
Dark	22	93	5	
 	24	95	5	
 
The data in the table show that species A is better adapted than species B to living in hot, dry deserts.
Explain how.
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(Total 5 marks)
Q10.          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?
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(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)
 M1.          (a)    water enters (funnel / sugar solution) or water diffuses in (to the funnel)
do not accept if diffusion of sugar
1
membrane partially / selectively / semi permeable or by osmosis
allow description
1
because concentration (of sugar) greater
inside funnel than outside / water / in beaker
assume ‘concentration’ refers to sugar unless candidate indicates otherwise
the position of the solutions may be implied
1
(b)     (level / it) rises more slowly or levels out earlier or does not rise as much
accept inference of less steep gradient (of graph)
allow less / slower osmosis / diffusion / less water passes through or less water enters funnel
allow water enters / passes through slower
1
less difference in concentration (between solution / funnel and water / beaker)
accept due to lower diffusion / concentration gradient / described
1
[5]
 M2.         (a)      (i)     diffusion is down the concentration gradient
for a description of diffusion
ignore along / across gradients
1
to enter must go up / against the concentration gradient
accept by diffusion ions would leave the root
or 
concentration higher in the root / plant
or 
concentration lower in the soil
1
(ii)     active transport
allow active uptake
1
(b)     (i)      (root hairs →) large surface / area
1
(ii)     (aerobic) respiration
do not allow anaerobic
1
releases / supplies / provides / gives energy
accept make ATP (for active transport)
do not allow ‘makes / produces / creates’ energy
1
(iii)    starch is energy source / store (for active transport)
allow starch can be used in respiration
do not allow ‘makes / produces / creates’ energy
1
[7]
 M3.          (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
(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]
 M4.          (a)     (i)      

award 1 mark for any of the mitochondria correctly labelled if a number are labelled and one is incorrect award 0 marks
1
(ii)     respiration or the release or transfer
of energy or it contains the enzymes
for respiration
do not accept energy produced
1
(b)     (i)      nucleus (named and correctly 
labelled)

arrow or line must touch or go inside the nuclear membrane
1
(ii)     DNA or genes or nucleic acids
accept protein or histones or  nucleotides or ATGC
1
(c)     enzymes or nucleus
do not accept factors that affect the rate rather than control it eg pH or temperature
1
[5]
 M5.          (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]
 M6.          (a)     (i)      water (molecules) enter(s) (the cell)
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]
 M7.          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]
 M8.          (a)                          correct names of cell components are required
                    it = cell in sugar solution
any two from:
accept reverse only if clearly stated answer refers to cell in distilled water
•        smaller vacuole
•        smaller / less cytoplasm
allow protoplasm for cytoplasm
•        cell membrane / cytoplasm not (fully) against cell wall
accept plasmolysed / flaccid / less turgid
or
cell membrane / cytoplasm (partly) pulled away from cell wall
ignore reference to nucleus / water
ignore explanations
or
space / liquid / sugar solution between cell
membrane / cytoplasm and cell wall
2
(b)     water passed / moved out (of cell) by osmosis / diffusion
accept reverse answer if clearly refers to cell in distilled water
1
more concentrated (solution) outside
assume reference to
concentration refers to solute
concentration unless answer refers to water concentration
orless concentrated (solution) inside
or
lower water concentration outside
accept references to hypertonic / hypotonic solutions or water potential
or
higher water concentration inside
1
[4]
 M9.          (a)     guard (cell)
ignore stoma / stomata
1
(b)     Species A:
•        stomata open in dark / at night or close in light / in day
1
•        stomata closed during warm(est) period or open when cool(er)
1
•        heat (energy) / warmth increases evaporation / transpiration
must give explicit link between heat and transpiration
1
•        reduces water loss / evaporation / transpiration
ignore photosynthesis
allow converse points for species B
1
[5]
 M10.          (a)     allow carbon dioxide to enter / gaseous exchange (oxygen neutral)
(transpiration neutral)
for one mark
1
(b)     guard (cells)
for one mark
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]