A level Biology June 2023 P1 Predicted topics.
Name: _____________ Mark: ____________/85
No
Topic
Mark
1
DNA and protein synthesis
/8
2
Enzyme action
/8
3
Gas exchange
/11
4
Carb and immunology
/10
5
Cell, DNA and classification
/8
6
Breathing and heartbeat
/5
7
Water transport in plants
/10
8
Oxygen affinity
/10
9
Biological molecules
/15
Page 1 of 32
Q1. The diagram below shows part of a DNA molecule.
(a)
Name the type of bond between:
complementary base pairs _______________________________________
adjacent nucleotides in a DNA strand _______________________________
(2)
(b)
The length of a gene is described as the number of nucleotide base pairs it contains.
Use information in above diagram to calculate the length of a gene containing 4.38 ×
103 base pairs.
Answer _______________ nm
(2)
(c)
Describe two differences between the structure of a tRNA molecule and the
structure of an mRNA molecule.
1 _________________________________________________________________
___________________________________________________________________
2 _________________________________________________________________
___________________________________________________________________
(2)
(d)
In a eukaryotic cell, the structure of the mRNA used in translation is different from
the structure of the pre-mRNA produced by transcription. Describe and explain a
difference in the structure of these mRNA molecules.
___________________________________________________________________
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Page 2 of 32
___________________________________________________________________
___________________________________________________________________
(2)
(Total 8 marks)
Q2. (a)
Describe the induced-fit model of enzyme action and how an enzyme acts as a catalyst.
___________________________________________________________________
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(3)
(b)
Scientists investigated the action of the enzyme ATP synthase. They made reaction
mixtures each containing:
•
•
•
ATP synthase
buffer (to control pH)
substrates.
One of the substrates required in these reaction mixtures is inorganic phosphate
(Pi).
Tick (✓) one box to show which other substrate the scientists must add to the
reaction mixtures to produce ATP.
Adenine
Adenosine diphosphate
Glucose
Ribose
(1)
(c)
The scientists investigated the effect of concentration of inorganic phosphate (Pi) on
ATP synthase activity.
Page 3 of 32
After 2 minutes, they stopped each reaction and then measured the concentration of
ATP.
The figure below shows the scientists’ results.
Suggest and explain a procedure the scientists could have used to stop each
reaction.
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
(2)
(d)
Explain the change in ATP concentration with increasing inorganic phosphate
concentration.
___________________________________________________________________
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(2)
(Total 8 marks)
Page 4 of 32
Q3. (a)
Explain the advantage for larger animals of having a specialised system that facilitates
oxygen uptake.
___________________________________________________________________
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(2)
Figure 1 shows two models of oxygen uptake found in animals.
Figure 1
(b)
Suggest how the environmental conditions have resulted in adaptations of systems
using Model A rather than Model B.
___________________________________________________________________
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(2)
Page 5 of 32
(c)
Figure 2 shows changes in concentration of oxygen in two gas exchange systems.
Figure 2
A student studied Figure 2 and concluded that the fish gas exchange system is
more efficient than the human gas exchange system.
Use Figure 2 to justify this conclusion.
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
(2)
(d)
Explain how the counter-current principle allows efficient oxygen uptake in the fish
gas exchange system.
___________________________________________________________________
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(2)
Page 6 of 32
(e)
The table below shows features of two mammals.
Bats are flying mammals; shrews are ground-living mammals.
Mean body mass /
kg
Mean lung volume
/ cm3
Bat
0.096
12.48
Shrew
0.024
0.72
Mammal
Calculate how many times the lung volume per unit of body mass of the bat is
greater than that of the shrew.
Give your answer to an appropriate number of significant figures.
Give one suggestion to explain this difference.
Answer _______________
Explanation ________________________________________________________
___________________________________________________________________
(3)
(Total 11 marks)
Q4. (a)
Alpha-gal is a disaccharide found in red meat.
Alpha-gal is made of two galactose molecules. Galactose has the chemical formula
C6H12O6
Give the chemical formula for the disaccharide, alpha-gal, and describe how it is
formed from two galactose molecules.
Formula ___________________________________________________________
Description _________________________________________________________
___________________________________________________________________
___________________________________________________________________
(2)
Page 7 of 32
(b)
Some people eat red meat for many years without having any reaction, then have
an allergic reaction to the alpha-gal in red meat.
An allergic reaction is caused by an immune response.
Draw a labelled diagram of an antibody and identify the specific alpha-gal binding
site.
(3)
(c)
A tick is a small animal that bites humans and feeds on their blood. This results in
proteins from the tick saliva entering the human body.
Scientists have suggested one hypothesis for the allergic reaction to alpha-gal in red
meat. They think that an earlier immune response to a tick bite can cause a person
to have an allergic reaction to alpha-gal in red meat.
Suggest how one antibody can be specific to tick protein and to alpha-gal.
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
(2)
Page 8 of 32
(d)
Scientists took blood samples from one man over several weeks and measured the
concentration of antibody in the man’s blood. During this time, the man had two tick
bites and had an allergic reaction to alpha-gal in red meat.
The scientists’ results are shown in the graph below.
The scientists’ hypothesis was that an earlier immune response to tick protein
causes the allergic reaction.
Consider whether the graph supports this hypothesis.
___________________________________________________________________
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(3)
(Total 10 marks)
Page 9 of 32
Q5. (a)
Complete Table 1 to show three differences between DNA in the nucleus of a plant cell
and DNA in a prokaryotic cell.
Table 1
DNA in the nucleus of DNA in a prokaryotic
a plant cell
cell
1
2
3
(3)
(b)
Scientists investigated the genetic diversity between several species of sweet
potato. They studied non-coding multiple repeats of base sequences.
Define ‘non-coding base sequences’ and describe where the non-coding multiple
repeats are positioned in the genome.
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___________________________________________________________________
___________________________________________________________________
(2)
Page 10 of 32
The percentage similarities in the non-coding multiple repeats of base sequences of four
species of carrot are shown in Table 2.
Table 2
Species of
carrot
Percentage similarity between non-coding multiple
repeat base sequences
C
C
D
I
N
51.3
23.1
61.2
32.7
51.5
D
51.3
I
23.1
32.7
N
61.2
51.5
(c)
37.4
37.4
Use the information in Table 2 to complete the phylogenetic tree shown in the
diagram below. Write the letter that represents the correct species into each box.
(1)
(d)
The scientists studied five individuals from each species. Within the five individuals
of species N they found a percentage similarity of 66%.
Use Table 2 to evaluate how this information affects the validity of the phylogenetic
tree.
___________________________________________________________________
___________________________________________________________________
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Page 11 of 32
___________________________________________________________________
___________________________________________________________________
(2)
(Total 8 marks)
Q6. (a)
Particulate matter is solid particles and liquid particles suspended in air. Polluted air
contains more particulate matter than clean air.
A high concentration of particulate matter results in the death of some alveolar
epithelium cells. If alveolar epithelium cells die inside the human body they are
replaced by non-specialised, thickened tissue.
Explain why death of alveolar epithelium cells reduces gas exchange in human
lungs.
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(3)
Page 12 of 32
Scientists grew alveolar epithelium cells and exposed the epithelium cells to different
concentrations of particulate matter. They calculated the percentage of these alveolar
epithelium cells that died after 24 hours of exposure to particulate matter. Their results are
shown in the graph below.
(b)
Do the data in the graph above show a linear relationship between concentration of
particulate matter and percentage of dead cells?
Use suitable calculations to justify your answer.
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
Space for your calculations:
(2)
(Total 5 marks)
Page 13 of 32
Q7. The water potential of leaf cells is affected by the water content of the soil.
Scientists grew sunflower plants. They supplied different plants with different volumes of
water.
After two days, they determined the water potential in the leaf cells by using an instrument
that gave a voltage reading.
The scientists generated a calibration curve to convert the voltage readings to water
potential.
Figure 1 shows their calibration curve.
Figure 1
(a)
The scientists needed solutions of known water potential to generate their calibration
curve.
Table 1 shows how to make a sodium chloride solution with a water potential of
−1.95 MPa
Complete Table 1 by giving all headings, units and volumes required to make
20 cm3 of this sodium chloride solution.
Table 1
Water
potential /
MPa
Concentration of
sodium chloride
solution / mol dm−3
Volume of
1 mol dm−3 sodium
chloride solution /
________________
_________________ ___________/ ____
−1.95
0.04
_________________ ________________
(2)
Page 14 of 32
Table 2 shows some of the concentrations of sodium chloride solution the scientists used
and the water potential of each solution.
Table 2
Concentration of Water potential
sodium chloride
/ MPa
solution / mol dm−3
0.04
−1.95
0.10
−4.87
0.12
−5.84
(b)
There is a linear relationship between the water potential and the concentration of
sodium chloride solution.
Use the data in Table 2 to calculate the concentration of sodium chloride solution
with a water potential of −3.41 MPa
Answer = ___________________________ mol dm−3
(2)
In addition to determining the water potential in the leaf cells, the scientists measured the
growth of the leaves.
They recorded leaf growth as a percentage increase of the original leaf area.
Their results are shown in Figure 2.
Figure 2
Page 15 of 32
(c)
One leaf with an original area of 60 cm2 gave a voltage reading of −7 µV
Use Figure 1 and Figure 2 to calculate by how much this leaf increased in area.
Give your answer in cm2
Answer = _________________________ cm2
(2)
(d)
Sunflowers are not xerophytic plants. The scientists repeated the experiment with
xerophytic plants.
Suggest and explain one way the leaf growth of xerophytic plants would be different
from the leaf growth of sunflowers in Figure 2.
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
(2)
(e)
Use your knowledge of gas exchange in leaves to explain why plants grown in soil
with very little water grow only slowly.
___________________________________________________________________
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___________________________________________________________________
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(2)
(Total 10 marks)
Page 16 of 32
Q8. A scientist investigated the affinity for oxygen of horse haemoglobin and mouse haemoglobin.
Some of their results are shown in the table.
Animal
(a)
Partial pressure
of oxygen when
haemoglobin is
50% saturated /
kPa
Partial pressure
of oxygen when
haemoglobin is
25% saturated /
kPa
Body mass of
one animal / g
Horse
3.2
1.9
550 000
Mouse
6.5
3.3
23
Plot the haemoglobin saturation data from the graph and use these points to sketch
the full oxyhaemoglobin dissociation curves for a horse and a mouse.
(3)
Page 17 of 32
(b)
The following equation can be used to estimate the metabolic rate of an animal.
Metabolic rate = 63 × BM−0.27
BM = body mass in grams
Use this equation to calculate how many times faster the metabolic rate of a mouse
is than the metabolic rate of a horse.
Answer = ________________________ times faster
(2)
(c)
The data in the table above show differences between the oxyhaemoglobin
dissociation curve for a mouse and the oxyhaemoglobin dissociation curve for a
horse.
Suggest how these differences allow the mouse to have a higher metabolic rate
than the horse.
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(2)
Page 18 of 32
(d)
Mammals such as a mouse and a horse are able to maintain a constant body
temperature.
Use your knowledge of surface area to volume ratio to explain the higher metabolic
rate of a mouse compared to a horse.
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(3)
(Total 10 marks)
Q9. (a)
Explain five properties that make water important for organisms.
___________________________________________________________________
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(5)
Page 19 of 32
(b)
Describe the biochemical tests you would use to confirm the presence of lipid,
non-reducing sugar and amylase in a sample.
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
(5)
(c)
Describe the chemical reactions involved in the conversion of polymers to
monomers and monomers to polymers.
Give two named examples of polymers and their associated monomers to illustrate
your answer.
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(5)
(Total 15 marks)
Page 20 of 32
Mark schemes
Q1.
(a)
1.
Hydrogen (bonds);
2.
Phosphodiester (bonds);
Accept ester/covalent bond
2
(b)
Correct answer for 2 marks = 1489/1489.2;;
Incorrect answer but for 1 mark accept:
876
OR
1861 - 1862
2
(c)
1.
tRNA is 'clover leaf shape', mRNA is linear;
Must be a comparison
Reject tRNA is double stranded
Accept tRNA is folded for tRNA is ‘clover leaf shaped’
2.
tRNA has hydrogen bonds, mRNA does not;
3.
tRNA has an amino acid binding site, mRNA does not;
Accept ‘CCA end' for amino acid binding site
4.
tRNA has anticodon, mRNA has codon;
2
(d)
1.
mRNA fewer nucleotides
OR
Pre-mRNA more nucleotides
OR
mRNA has no introns/has (only) exons
OR
Pre-mRNA has (exons and) introns;
Accept mRNA is shorter OR pre-mRNA is longer
2.
(Because of) splicing;
2
[8]
Q2.
(a)
1.
Substrate binds to the active site/enzyme
Page 21 of 32
OR
Enzyme-substrate complex forms;
Accept for ‘binds’, fits
2.
Active site changes shape (slightly) so it is complementary to substrate
OR
Active site changes shape (slightly) so distorting/breaking/forming
bonds in the substrate;
3.
Reduces activation energy;
3
(b)
1.
Adenosine diphosphate;
1
(c)
Mark in pairs, 1 and 2 OR 3 and 4 OR 5 and 6
1.
Boil
OR
Add (strong) acid/alkali;
Accept heat at > 50oC OR at very high temperatures
2.
Denatures the enzyme/ATP synthase;
OR
Accept for 'denatures', a description of denaturation
3.
Put in ice/fridge/freezer;
4.
Lower kinetic energy so no enzyme-substrate complexes form;
OR
Accept ES for enzyme substrate complex
5.
Add high concentration of inhibitor;
6.
Enzyme-substrate complexes do not form;
2
(d)
1.
(With) increasing Pi concentration, more enzyme-substrate
complexes are formed;
2.
At or above 40 (mmol dm-3) all active sites occupied
OR
At or above 40 (mmol dm-3) enzyme concentration is a limiting factor;
2
[8]
Q3.
(a)
1.
Large(r) organisms have a small(er) surface area:volume (ratio);
Page 22 of 32
OR
Small(er) organisms have a large(r) surface area:volume (ratio);
2.
Overcomes long diffusion pathway
OR
Faster diffusion;
Accept short diffusion pathway
Accept for ‘faster’, more
2
(b)
Mark in pairs, 1, and 2 OR 3. and 4.
1.
Water has low(er) oxygen partial pressure/concentration (than air);
2.
So (system on outside) gives large surface area (in contact with water)
OR
So (system on outside) reduces diffusion distance (between water
and blood);
3.
Water is dense(r) (than air);
4.
(So) water supports the systems/gills;
2
(c)
1.
In fish, blood leaving (V) has more oxygen than water leaving (E);
2.
(But) in humans, blood leaving (V) has less oxygen than air leaving (E);
3.
Difference in oxygen (concentration) between artery and vein is
greater in fish than in humans;
4.
(So) fish remove a greater proportion from the oxygen they take in;
2 max
(d)
1.
Blood and water flow in opposite directions;
2.
Diffusion/concentration gradient (maintained) along (length of)
lamella/filament;
Accept for 2 marks, suitably labelled diagram
2
(e)
1.
and 2. Correct answer for 2 marks, 4.3 (times greater);;
Accept for 1 mark,
4.333333333 (correct answer not given to 2 significant figures)
OR
Evidence of 130 (cm3 kg-1) and 30 (cm3 kg-1)
Correct explanation for 1 mark,
3.
Provides more oxygen for respiration;
3
Page 23 of 32
[11]
Q4.
(a)
1.
C12H22O11;
2.
Condensation reaction
OR
With a glycosidic bond;
Reject if any other named reaction or named bond given.
Reject if reaction includes addition of water.
Do not credit answers relating to other carbohydrates.
2
(b)
1.
Y shape showing two long and two short (polypeptide) chains correctly
positioned;
Drawing is nothing like an antibody = 0 marks.
2.
(Alpha-gal) binding site labelled on the end of the branches of the Y of the
antibody;
Accept one or two being labelled, if two both must be correct.
3.
Variable region labelled
OR
Constant region labelled
OR
Disulfide bridge/bond labelled;
Accept description of ‘variable region’.
Ignore labelling of light and heavy chains.
List rule applies.
3
(c)
1.
(Part of tick protein and alpha-gal) have a similar shape/structure;
Accept ‘(Part of tick protein and alpha-gal) have the same
shape/structure.’
Do not credit reference to similar/same tertiary structure’.
Ignore reference to alpha-gal being a protein.
2.
Antibody is complementary to both (tick protein and alpha-gal)
OR
Antigen-binding site is complementary to both (tick protein and alpha-gal)
OR
Antibody can form antigen-antibody complex with both (tick protein and
alpha-gal);
Page 24 of 32
Reject reference to substrates or active sites.
2
(d)
1.
Exposure to tick (protein) is followed by increase in antibody (specific to
alpha-gal);
For 'is followed by' accept 'causes'.
2.
(Later) greater/faster increase in antibody suggests there are memory cells;
Must be in relation to EITHER second exposure to tick
(protein) OR to allergic reaction.
3.
Antibody (specific to alpha-gal) increases during/after allergic reaction;
Accept 'eating red meat' or 'eating/exposure to alpha-gal' for
'allergic reaction'.
4.
During/after allergic reaction, total antibody increases more than alpha-gal
antibody;
Accept ‘eating red meat’ or ‘eating/exposure to alpha-gal’ for
‘allergic reaction’.
5.
(So) may be other antibodies (that are causing allergic reaction);
3 max
[10]
Q5.
(a)
Plant v prokaryote
1.
(Associated with) histones/proteins v no histones/proteins;
2.
Linear v circular;
3.
No plasmids v plasmids;
Do not credit if suggestion that prokaryotic DNA only exists
as plasmids.
4.
Introns v no introns;
5.
Long(er) v short(er);
Alternatives must be written directly opposite one another.
Do not award if only half of a mark point is written.
Reference to prokaryotic DNA being single stranded = max
2.
Reference to prokaryotic DNA not being helical = max 2.
3 max
(b)
1.
DNA that does not code for protein/polypeptides
OR
DNA that does not code for (sequences of) amino acids
OR
DNA that does not code for tRNA/rRNA;
Accept the idea of not transcribed for ‘does not code for’.
Page 25 of 32
Do not credit ‘DNA that does not code for an amino acid’.
Ignore reference to introns.
2.
(Positioned) between genes;
Reject (positioned) ‘in introns’ or ‘between exons’.
Accept ‘(Positioned) at the end of chromosomes’ or
‘(Positioned) in the telomeres’.
2
(c)
Top to bottom C N D I;
OR
Top to bottom N C D I;
1
(d)
1.
(Supported) more similar than with any other species;
2.
(Not supported) high (intraspecific) variation in species N (compared with
variation between N and C);
Accept idea that species N has nearly as much variation as
between N and C.
Accept ‘Low/close similarity in species N (in relation to
similarity between N and C)’
3.
Small sample
OR
Only five (individuals);
2 max
[8]
Q6.
(a)
1.
Reduced surface area;
2.
Increased distance for diffusion;
Accept description of efficient gas exchange in healthy
alveolar epithelium as long as reference made to the
damaged tissue changing this.
3.
Reduced rate of gas exchange;
3
(b)
(No)
EITHER
1.
9 (percent per 5 µg cm–3);
2.
1.42/1.8 (percent per 5 µg cm–3);
Accept any number of significant figures as long as rounding
correct, full answer for mp2 is 1.42105263.
OR
Page 26 of 32
3.
1.8 (percent per 1 µg cm–3);
4.
0.28/0.36 (percent per 1 µg cm–3);
Accept any number of significant figures as long as rounding
correct, full answer for mp4 is 0.28421053.
OR
5.
9% and 36/27% increase here;
6.
(To be linear) 100 (µg cm–3) would be 180/171% (increase)
OR
(To be linear) 5 (µg cm–3) would be 1.8% (increase)
OR
% increase is x4 (0-5 µg cm–3 compared with 0-100 µg cm–3) but 5-100 is more
than x4
OR
% increase is x3 (0-5 µg cm–3 compared with 5-100 µg cm–3) but 5-100 is more
than x3;
OR
7.
(Using y = mx + c) at 5 (µg cm–3) m = 1.8;
8.
(Using y = mx + c) at 100 (µg cm–3) m = 0.36;
9.
At 100 (µg cm–3) y would be 186%;
10.
At 5 (µg cm–3) y would be 7.8%;;
If no correct answers accept for one mark
Evidence of incorrect graph reading but division by 19
OR
Evidence of incorrect graph reading but division by 95
Accept 1 and 2 OR 3 and 4
OR 5 and 6
OR 7 and 8
OR 7 and 9
OR 8 and 10
2
[5]
Q7.
(a)
Water
potential /
Concentration of
sodium chloride
Volume of
1 mol dm−3 sodium
Page 27 of 32
Volume of water
__________________
MPa
−1.95
solution /
mol dm−3
0.04
chloride solution /
cm3
____________/ ____
cm3
__________________
0.8
19.2
___________________
__________________
_
1 mark for each row.
If values do not match the given unit, max 1.
Accept dm3 / mm3 for volume unit.
Accept 0.0008/8 x 10−4 and 0.0192/1.92 x 10−2
Accept 800 and 19200
Ignore units in 2nd row.
Do not accept mm−3/cm−3/dm−3/ ml
2
(b)
Correct answer of 0.07 (mol dm−3) = 2 marks;;
Incorrect answer 1 mark for any evidence of
48.6 to 48.8
OR
0.02
OR
0.7
OR
A final answer between 0.04 and 0.10
OR
A final answer of minus 0.07/−0.07;
Ignore minus signs on other 1 mark options.
2
(c)
Correct answer of 9 (cm2) = 2 marks;;
Incorrect answer 1 mark for evidence of water potential of
between -1.85 and -1.95 (MPa)
OR
growth of 15%
OR
69 (cm2)
OR
A final answer between 8.7 and <9;
Allow 9.0
Accept correct reading labelled on the graph
shown on Figure 1 or Figure 2.
2
(d)
EITHER
1.
Low/slow growth;
2.
Due to smaller number/area of stomata (for gas exchange);
OR
Page 28 of 32
3.
Growth may continue at lower water potentials;
4.
(Due to) adaptations in enzymes involved in photosynthesis/metabolic
reactions;
Mark as pair – 1 and 2 OR 3 and 4.
Reference to stomata must not relate only to water loss.
2 max
(e)
1.
Stomata close;
2.
Less carbon dioxide (uptake) for less photosynthesis/glucose
production;
‘Less’ only required once.
Reject ‘no photosynthesis’ but accept ‘carbon dioxide can’t enter
so less photosynthesis’.
Ignore oxygen for respiration but reject oxygen for photosynthesis.
Ignore less water for photosynthesis.
Accept only correct chemical formulae.
For ‘glucose’ accept named product of photosynthesis eg triose
phosphate, TP, amino acid, lipid.
2
[10]
Q8.
(a)
1.
y axis 0 – 100 in linear scale and x axis minimum 1 to 8 in linear scale
and both axes use at least half size of grid;
If tick marks are used on the axis, they must be accurate to within
± half a small square.
2.
Correct plots for 50% and 25% for both animals;
25% - 1.9, 3.3 and 50% - 3.2 and 6.5
Accept plot ± half a small square.
3.
Both curves levelling off (at higher partial pressures and at percentage
saturations ≤100%);
3
(b)
Correct answer of 15 (times faster) = 2marks ;;
If ≥3sf given, accept answers in the range 15.0 to 15.4 (times faster) =
2marks;;
Incorrect answer 1 mark for evidence of:
23−0.27 divided by 550 000−0.27
OR
0.42888777
OR
0.02819045
OR
Between 27 and 27.1
OR
Between 1.77599861 and 1.8
OR
Page 29 of 32
0.06
Accept any number of significant figures ≤2, if rounding correct.
2
(c)
1.
Mouse haemoglobin/Hb has a lower affinity for oxygen
OR
For the same pO2 the mouse haemoglobin/Hb is less saturated
OR
At oxygen concentrations found in tissue mouse haemoglobin/Hb is less
saturated;
For ‘Hb is less saturated’ accept ‘less oxygen will be bound to Hb’.
2.
More oxygen can be dissociated/released/unloaded (for metabolic
reactions/respiration);
Accept ‘oxygen dissociated/released/unloaded more
readily/easily/quickly’
Reject ‘oxygen loaded more readily/easily/quickly’ or ‘more oxygen
loaded’
2
(d)
Accept converse answers in relation to the horse.
Mouse
1.
(Smaller so) larger surface area to volume ratio;
Accept larger SA:V.
Must be comparative.
2.
More/faster heat loss (per gram/in relation to body size);
Ignore heat lost more easily/readily.
Must be comparative.
3.
(Faster rate of) respiration/metabolism releases heat;
Accept respiration/metabolism replaces heat.
Reject produce/generate heat/energy.
3
[10]
Q9.
(a)
1.
A metabolite in condensation/hydrolysis/ photosynthesis/respiration;
2.
A solvent so (metabolic) reactions can occur
OR
A solvent so allowing transport of substances;
3.
High heat capacity so buffers changes in temperature;
For ‘buffer’ accept ‘resist’.
4.
Large latent heat of vaporisation so provides a cooling effect (through
evaporation);
5.
Cohesion (between water molecules) so supports columns of water (in
plants);
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For ‘columns of water’ accept ‘transpiration stream’.
Do not credit ‘transpiration’ alone but accept description of
‘stream’.
For ‘columns of water’ accept ‘cohesion-tension (theory)’.
For cohesion accept hydrogen bonding
6.
Cohesion (between water molecules) so produces surface tension
supporting (small) organisms;
For cohesion accept hydrogen bonding
Ignore reference to pH.
Allow other suitable properties but must have a valid explanation.
For example
•
ice floating so maintaining aquatic habitat beneath
•
water transparent so allowing light penetration for
photosynthesis
5 max
(b)
4 max if marks gained from only 2 substance tests.
Lipid
1.
Add ethanol/alcohol then add water and shake/mix
OR
Add ethanol/alcohol and shake/mix then pour into/add water;
Reject heating emulsion test.
Accept ‘Add Sudan III and mix’.
2.
White/milky emulsion
OR
emulsion test turns white/milky;
Ignore cloudy.
Reject precipitate.
Accept (for Sudan III) top (layer) red.
Non-reducing sugar
3.
Do Benedict’s test and stays blue/negative;
Ignore details of method for Benedict’s test for this mp.
4.
Boil with acid then neutralise with alkali;
Accept named examples of acids/alkalis.
5.
Heat with Benedict’s and becomes red/orange (precipitate);
Do not credit mp5 if no attempt at mp4.
For ‘heat’ ignore ‘warm’/’heat gently’/’put in a water bath’ but
accept stated temperatures ≥ 60°C.
Heat must be stated again, do not accept using residual heat from
mp4.
Accept ‘do the Benedict’s test’ if full correct method given
elsewhere.
Accept ‘sodium carbonate, sodium citrate and copper sulfate
solution’ for Benedict’s but must have all three if term ‘Benedict’s’
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not used.
Amylase
6.
Add biuret (reagent) and becomes purple/violet/mauve/lilac;
Accept ‘sodium or potassium hydroxide and copper sulfate
solution’ for ‘biuret’.
Reject heating biuret test.
7.
Add starch, (leave for a time), test for reducing sugar/absence of starch;
5 max
(c)
Ignore reference to dimers.
1.
A condensation reaction joins monomers together and forms a
(chemical) bond and releases water;
2.
A hydrolysis reaction breaks a (chemical) bond between monomers and
uses water;
3.
A suitable example of polymers and the monomers from which they are
made;
3. and 4. Polymers must contain many monomers.
3. and 4: suitable examples include
•
amino acid and polypeptide, protein, enzyme, antibody or
specific
example
•
nucleotide and polynucleotide, DNA or RNA
•
Alpha glucose and starch/glycogen
•
Beta glucose and cellulose.
If neither specific carbohydrate example is given, allow
monosaccharide/glucose and polysaccharide.
3. and 4. Reject (once) reference to triglycerides.
4.
A second suitable example of polymers and the monomers from which
they are made;
5.
Reference to a correct bond within a named polymer;
Reject reference to ester bond.
5
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