Mark Scheme
Question
1
Answer/Indicative content
i
Marks
40 ✓ ✓
2(AO2.8)
(AO2.8)
Guidance
Correct answer = 2 marks even if no
working shown.
IGNORE minus sign
If answer is incorrect, then award 1 mark
for:
dividing by end fig:
66.6 (recurring) / 67
calculating with 0.20 NaCl fig:
81.5 / 82
working: (6.5 – 3.9) ÷ 6.5 x 100
or
2.6 ÷ 6.5 x 100
Examiner’s Comments
Many candidates correctly calculated a
40% decrease. A few candidates did not
notice that the concentration range asked
about was from 0.00 to 0.15 (not the 0.20
figure at the end of the table). Some
candidates found the difference between
the figures for the two concentrations but
divided by the final figure (3.9) not the
initial figure (6.5), gaining one mark only for
error carried forward in their calculation.
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Mark Scheme
Question
Answer/Indicative content
ii
Marks
as NaCl concentration increases:
1 (external) water potential decreases /
solute potential increases ✓
max 2
(AO3.1
x3)
2 water potential gradient decreases ✓
Guidance
1 IGNORE outside vacuole for external
context
2 ALLOW ψ difference decreases / ψ
inside and out becomes more similar
3 less water enters (Paramecium / cell /
cytoplasm) ✓
3 ALLOW water, enters / diffuses, more
slowly
ALLOW takes more time for water to enter
DO NOT ALLOW solution for ‘water’
4 less water needs to be expelled ✓
4 ALLOW removed / got rid of / ejected, for
‘expelled’
DO NOT ALLOW solution for ‘water’ but
ECF from 3
IGNORE water expelled less, often /
frequently or less contractions in a given
time
Examiner’s Comments
This provided a good test of candidates’
ability to describe and explain parallel
trends as two parameters change.
Answers needed to consider the dynamic
trend produced in the dependent variable
as the concentration of sodium chloride
increased (going down the table). There is
a trend of decrease in the external water
potential, leading to less water uptake by
Paramecium and less need to expel water.
Practical work on osmosis should provide
candidates with a background idea that
cytoplasm has a solute concentration
greater than 0.20 mol dm-3, therefore there
will not be water movement from the
cytoplasm into the external solution in this
case.
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Mark Scheme
Question
Answer/Indicative content
iii
Marks
1 making crystals, increases ψ / decreases
max 3
ψs ✓
(AO2.1x4
)
benefit:
2 decreases / less, water entry ✓
1 ALLOW ora dissolving crystals,
decreases ψ / increases ψs
IGNORE removing / releasing, for
‘dissolving’
ALLOW ‘adding’ for ‘making’
ECF from wrong mp1 for an ora of mp 2-4
for 1 mark only
3 (so) less need to expel water ✓
Examiner’s Comments
4 (so) less use of energy ✓
iv
Guidance
This question caused widespread
confusion. Alteration of the water potential
would have to involve either making more
solid crystals or allowing some to dissolve.
Generally candidates thought that more
crystals meant a lower water potential,
whereas more crystallisation would remove
ions from solution and allow water potential
to increase. Those who described crystals
dissolving, reducing the water potential in
the cell, ran into the problem that this
would not be a benefit to the Paramecium,
but would make its osmoregulation
problem worse. Candidates can be
assured that in a novel context like this, an
initial error does not mean they lose all the
marks, so long as they use correct logic to
follow through their argument.
(less) oxygen for aerobic respiration ✓
2(AO2.8)
(less) energy / ATP, for (vacuole)
contraction ✓
(AO2.4)
ALLOW is an active process for ‘energy’
IGNORE active transport
DO NOT ALLOW energy created /
produced
Examiner’s Comments
Some candidates continued to pursue an
osmosis line of thought. Those who
considered respiration often did not qualify
it as aerobic (i.e. less aerobic respiration
can occur if less oxygen is available). Only
a few candidates realised that contraction
of the vacuole is an active, energyrequiring process.
Total
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Mark Scheme
Question
2
Answer/Indicative content
Marks
WP of -100 solution higher than -400 /
ORA;
max 2
Guidance
IGNORE refs to hyper / hypo tonic
solutions
ACCEPT –100 less negative than –400
Note: response must contain clear ref to
both –100 solution and –400 solution
(at -100kPa) water potential gradient
steeper /
described / ORA;
(at -100kPa) water enters Amoeba more
quickly / ORA;
ACCEPT more water enters
Note: ref to osmosis being more rapid only
valid if direction of water movement is clear
Examiner's Comments
This question caused confusion for some
candidates. Many weaker candidates
simply described the process of osmosis to
explain why water entered the Amoeba
without attempting to explain why more
water would enter at the higher water
potential. Some candidates described
water moving into the vacuole rather than
into the cell from its surrounding, while
others had osmosis causing water to leave
the cell despite the contractile vacuole
performing that function several times a
minute. It was pleasing to see that stronger
candidates did have a good grasp of the
concepts and were able to explain why the
contractile vacuole had to empty more
frequently very clearly and succinctly.
Total
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Mark Scheme
Question
3
Answer/Indicative content
i
Marks
(amino acids are) soluble in water / AW ;
increase (in amino acid concentration)
lowers,
water potential / AW ;
2
Guidance
CREDIT correct ref to zwitterions
CREDIT water potential becomes more
negative
CREDIT ORA
Examiner's Comments
This part was generally well-answered.
ii
idea that oxygen is non-polar ;
1
Examiner's Comments
This part proved to be challenging. Good
candidates were able to reason that
oxygen did not affect water potential
because it was not a polar molecule, but
many simply stated that it was because 2%
was a small amount and therefore too
small to affect water potential.
Total
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Mark Scheme
Question
4
Answer/Indicative content
i
1 discs same, size / thickness / surface
area / surface area to volume ratio /
diameter ✔
Marks
Guidance
max 2
Mark first two answers only, ignoring the
numbered sections
IGNORE mass / balance used / soak time /
repeats
IGNORE a list of variables unqualified
1 ACCEPT same cork borer used
ACCEPT ‘pieces of potato’ etc. for ‘discs’
ACCEPT ‘length’ as equivalent to
‘diameter’
IGNORE same shape/similar size etc
2 same (variety / part, of) potato ✔
3 no skin on potato ✔
4 e.g. blotting / shaking
4 ref to removing excess water before
(re)weighing ✔
5 same, number / amount, of discs (in
each solution) ✔
6 same volume (sucrose) solution ✔
7 same temperature ✔
7 ACCEPT in context of room /
environment / solution
8 cover the tubes ✔
Examiner's Comments
This question was relatively well answered
but many candidates stated soak time as a
factor, despite the question specifying four
hours. Some candidates correctly named
the variable but failed to keep it the same.
A significant number of students did not
appreciate that the question referred to the
validity of the results and gave responses
relating to ensuring the accuracy or
reliability of results, e.g. using suitable
measuring equipment for the volumes or to
doing repeats. Candidates did not always
use the term volume rather than ‘amount’,
or refer to the discs rather than just the
potato tuber.
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Mark Scheme
Question
Answer/Indicative content
ii
1 idea that no change of mass occurs
when the water potential of (sucrose)
solution = water potential of potato (tissue)
✔
Marks
Guidance
max 3
ACCEPT Ψ for water potential throughout
IGNORE ref to solute potential / isontonic
2 ref. to no change in mass (of potato)
between 0.2 and 0.3 mol dm–3 ✔
2 correct units must be stated once
ACCEPT ‘between 0.2 and 0.3 mol dm–3
the water potential of the solution and the
potato will be the same’
3 plot graph of concentration of, sucrose /
solution, against (%) change in mass
and
find which (sucrose) concentration gives no
change in mass of potato
3 x and y axes interchangeable
When an axis has been identified it can be
referred to by letter later.
Needs some ref to the mass change being
0.
If the change in mass axis has previously
been identified, then ref to that axis value
being 0 is equivalent to no change in mass
e.g. ‘Should draw a graph of sucrose
concentration on the x axis and change in
mass of potato discs on the y axis. The
point where the line of best fit crosses the
x axis (when the y axis = 0) is the
concentration of sucrose in the potato
discs.’ will get the mark
OR
carry out the experiment again with more
(sucrose) concentration intervals between
0.2 and 0.3 mol dm–3 ✔
‘Draw a graph with change in mass of
potato discs on the y axis and
concentration of sucrose solution on the x
axis and draw a line of best fit. Where the
line intercepts the x axis is where the
change in mass of potato discs is zero.’ will
get the mark
3 correct units must be stated once
Examiner's Comments
4 look up the water potential of the
(sucrose) solution (e.g. on calibration curve
or table), of that concentration / of the
concentration which gives no mass change
✔
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7 of 16
Most candidates did not read the question
carefully enough and just described why
the discs gained or lost mass in the various
sucrose solutions. Typically they gave
statements such as ‘when the increase in
mass is high then the water potential of the
solution is higher than in the potato and
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Mark Scheme
Question
Answer/Indicative content
Marks
Guidance
when mass is lost the water potential of the
solution is lower’. There was no indication
they understood that the water potential of
the potato tissue could be quantified from
the results or the significance of the
sucrose concentration where no mass
change occurred. Several candidates
appreciated that the mass difference
changed from positive to negative between
two stated sucrose solution concentrations,
but did not develop the idea further.
Candidates who, presumably, had done
this as a practical exercise or had analysed
similar data, knew that a graph of the
results would yield an estimate but most of
these said that the water potential could be
obtained directly from the point where the
line of best fit crossed the zero mass value
(rather than the equivalent sucrose
concentration).
Total
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Mark Scheme
Question
5
Answer/Indicative content
a
b
i
Marks
detect the presence of acid / H+ (1)
measure end-point / dependent variable
(1)
1
surface area to volume ratio on x-axis and
time on y-axis (1)
4
Guidance
plotted points occupy at least half of
available area
and
linear scale on both axes
and
line of best fit drawn (1)
DO NOT ALLOW if units given for x-axis
all points plotted correctly (to +/– half a 2
mm grid square) (1)
ALLOW ecf for correctly plotted points on
incorrectly-scaled graph
ii
time taken for diffusion (to centre of cube),
increases as surface area to volume ratio
decreases, ORA
1
Answer must mention surface area to
volume ratio
DO NOT ALLOW if colour change is
discussed in place of diffusion
IGNORE rate
ALLOW a description consistent with the
graph the candidate has drawn
iii
0.44
1
ALLOW answer in the range of 0.40 – 0.48
depending on candidate's plotted graph
Answer must be reported to 2 decimal
places
iv
test cubes of (known) length between 10
and 20 mm
1
0.35 / 0.347 (1) (1)
mm min–1
3
cube A, because…
time for test 2 different from others (1)
use of processed figures to support (1)
2
c
d
axes labelled time (min) and surface area
to volume ratio / AW (1)
i
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ALLOW 0.69 / 0.694 for 1 mark
ALLOW 0.3 or 0.3472 for 1 mark
ALLOW mm/min
ALLOW calculated rates for cube A - E
ALLOW calculated range compared with
that of cubes B - E
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Mark Scheme
Question
Answer/Indicative content
ii
Marks
Limitation
inconsistency in surface area (1)
cube A (1)
3
ALLOW mark only if one of the other two
marks is awarded
Because
It is the smallest cube so small error in
cutting will have proportionately larger
effect in a small cube / idea that error is a
bigger proportion of total time (1)
Limitation
using human eye and judgement to
determine end point (1)
cube E (1)
ALLOW mark only if one of the other two
marks is awarded
Because
largest cube so harder to see through 2cm
of jelly / AW (1)
e
6
idea of involvement of cytoskeleton /
vesicles (1)
1
Total
17
B
1
Total
1
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Guidance
10 of 16
IGNORE reference to different diffusion
resistance
Examiner’s Comments
This was another mathematical question
and only 1 in 5 candidates achieved the
mark.
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Mark Scheme
Question
7
a
Answer/Indicative content
i
Marks
(use a) 100cm3 measuring cylinder ✓
Max 3
Guidance
ACCEPT annotated diagram
mix 80cm3 acid and 20cm3 water ✓
take 50cm3 of the resulting solution and
add 50cm3 water ✓
repeat 50 / 50 dilution for each subsequent
solution required / AW ✓
ii
195 – 200 s ✓
1
iii
range bars ✓
2
the unit must be included
longer range bar indicates more variability /
less repeatable ✓
iv
longer time taken to discolour ✓
2
error becomes smaller proportion of total /
% error reduced ✓
b
cut block in half ✓
3
measure, thickness of colourless region /
distance from edge of block to coloured
region ✓
divide distance (acid diffused) by time ✓
8
Total
11
A✓
1
Total
1
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Mark Scheme
Question
9
Answer/Indicative content
i
Marks
repeat (readings) ✓
Guidance
2 max
calculate mean ✓
this could be mean distance/size of
colourless area, or mean time if cube
allowed to go completely colourless
identifying anomalies ✓
use statistical test to identify difference ✓
ALLOW calculate standard deviation
Examiner’s Comments
The question asks how the student could
ensure confidence in the results.
Confidence is a qualitative judgement
expressing the extent to which a
conclusion is justified by the quality of the
evidence. The majority of candidates
gained one mark here for repeating the
readings. Only the more able candidates
gained a second mark. This second mark
was usually credited for calculating a
mean. Many candidates described how the
student could improve the validity of the
results.
Definitions of the terms associated
with practical work are available in the
practical skills handbook.
Key:
OCR support Identifiable issue or
misconception
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Mark Scheme
Question
Answer/Indicative content
ii
cube A = 0.6 (: 1) ✓
cube B = 1.5 (: 1) ✓
Marks
Guidance
2 max
ALLOW 1 mark for 600 : 1000 and 96 :
64
6 : 10 and 3 : 2
: 5 and 3 : 2
(as correct ratios but not expressed
correctly)
Allow these ratios if written anywhere in the
answer space.
DO NOT ALLOW if units given
Examiner’s Comments
This question asked for the surface area to
volume ratio of two cubes to be calculated.
Less able candidates have always
struggled with this concept and this still
seems to be true. Surface area to volume
ratios should always be calculated as a
surface area to one unit of volume (0.6 :1
rather than 0.6). Less able candidates
often calculated it the other way around – a
volume for one unit of surface area.
Exemplar 1
As seen in Exemplar 1, candidates often
know how to calculate the surface area
and the volume. Less able candidates then
struggle to put these two components
together properly to calculate the surface
area to volume ratio. This exemplar shows
the ratio stated incorrectly as a volume to
one unit of surface area.
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Mark Scheme
Question
Answer/Indicative content
iii
Marks
large(r) organism has small(er) SA : Vol
ratio ✓
(rate of) diffusion (too) slow /
2 max
Guidance
ALLOW ORA for first three mark points
Examiner’s Comments
diffusion distance
(too) long ✓
for (sufficient), delivery / uptake of, oxygen
/ nutrients
OR
for (sufficient) removal of (named) waste
products ✓
for, (aerobic) respiration / metabolic
demands ✓
Many candidates knew that large
organisms have a small surface area to
volume ratio. These candidates
successfully linked the concept of a small
surface area to volume ratio with the need
for a circulatory system. More able
candidates could explain the need in terms
of a slower rate of diffusion which meant
that insufficient oxygen reached the tissues
for respiration or metabolism. Less able
candidates often confused the concept of a
surface area to volume ratio, with surface
area.
Exemplar 2
This exemplar is a good response because
it clearly explains that a small organism
has a large surface area to volume ratio.
This allows rapid diffusion. But in a larger
organism with a small surface area to
volume ratio, the cells will not be supplied
quickly enough.
Total
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Mark Scheme
Question
10
i
Answer/Indicative content
Marks
Substance A
1 for (substance) A the, graph is a straight
line / rate of uptake depends on
concentration ✓
4 max
(AO3.1)
(AO3.2)
2 (so substance) A is (absorbed by simple)
diffusion ✓
ALLOW rate becomes constant
DO NOT ALLOW rate slows
IGNORE stops increasing
4 (so substance) B could be (absorbed by),
facilitated diffusion / active transport ✓
ALLOW channels / carriers working at
maximum capacity
ALLOW transport proteins for either in MP5
DO NOT ALLOW channel proteins for
active transport
5 (because) if facilitated diffusion channels
/ carrier proteins, become saturated
OR
(because) if active transport carrier
proteins /carriers, become saturated ✓
Substance A
effect
(uptake) unaffected / no change ✓
max 4
(AO3.1)
(AO2.5)
explanation
(simple) diffusion, does not require ATP / is
a passive process ✓
OR
effect if facilitated diffusion
(uptake) unaffected / no change ✓
explanation
facilitated diffusion, does not require ATP /
is a passive process ✓
Total
CHECK answer to (b)(i) ALLOW ECF if
answer to part (i) suggests candidate
thinks substance A is taken up by active
transport and Substance B is taken up
entirely by diffusion.
ALLOW does not require energy
Substance B
effect if active transport
slower / little / reduced / no (uptake) ✓
explanation
active transport, requires ATP / is an active
process ✓
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ALLOW rate is (directly) proportional to
concentration
ALLOW as concentration increases rate
increases
DO NOT ALLOW facilitated diffusion
Substance B
3 for (substance) B the curve, reaches a
plateau / levels off ✓
ii
Guidance
ALLOW does not require energy
8
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Mark Scheme
Question
11
Answer/Indicative content
Guidance
i
too large / not fat soluble
1
IGNORE ‘no channels’
ii
water / H2O, and, lactase / enzyme
1
Mark the first two answers. If they are
correct and any other word is written that is
incorrect or contradicts the correct answer
then 0 marks.
DO NOT ALLOW H2O with incorrect case
or subscript
IGNORE refs to pH, buffers,
hydrocarbonate etc.
Total
2
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