Q1. (a)
Describe the structure of DNA.
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(5)
(b)
Name and describe five ways substances can move across the cell-surface
membrane into a cell.
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(5)
Page 1 of 21
The figure below shows transmission electron micrographs of two cells, one animal cell
and one prokaryotic cell.
(c)
Contrast the structure of the two cells visible in the electron micrographs shown in
the figure above.
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(5)
(Total 15 marks)
Page 2 of 21
Q2. (a)
Describe how mRNA is formed by transcription in eukaryotes.
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(5)
(b)
Describe how a polypeptide is formed by translation of mRNA.
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(6)
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(c)
Define ‘gene mutation’ and explain how a gene mutation can have:
•
•
no effect on an individual
a positive effect on an individual.
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(4)
(Total 15 marks)
Page 4 of 21
Q3. (a)
Explain five properties that make water important for organisms.
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(5)
(b)
Describe the biochemical tests you would use to confirm the presence of lipid, nonreducing sugar and amylase in a sample.
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(5)
Page 5 of 21
(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 6 of 21
Q4. (a)
Describe the gross structure of the human gas exchange system and how we breathe in
and out.
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(6)
(b)
Mucus produced by epithelial cells in the human gas exchange system contains
triglycerides and phospholipids.
Compare and contrast the structure and properties of triglycerides and
phospholipids.
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(5)
Page 7 of 21
(c)
Mucus also contains glycoproteins. One of these glycoproteins is a polypeptide with
the sugar, lactose, attached.
Describe how lactose is formed and where in the cell it would be attached to a
polypeptide to form a glycoprotein.
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(4)
(Total 15 marks)
Page 8 of 21
Q5. (a)
Contrast how an optical microscope and a transmission electron microscope work and
contrast the limitations of their use when studying cells.
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(6)
(b)
The diagram shows an image from an optical microscope of meiosis occurring in a
flower bud of a flowering plant. W and Z are undergoing meiosis.
Page 9 of 21
Explain the appearance of W and Z.
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(4)
(c)
An environmental scientist investigated a possible relationship between air pollution
and the size of seeds produced by one species of tree.
He was provided with a very large number of seeds collected from a population of
trees in the centre of a city and also a very large number of seeds collected from a
population of trees in the countryside.
Describe how he should collect and process data from these seeds to investigate
whether there is a difference in seed size between these two populations of trees.
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(5)
(Total 15 marks)
Page 10 of 21
Mark schemes
Q1. (a)
1.
Polymer of nucleotides;
Accept ‘Polynucleotide’ Accept for ‘phosphate’. phosphoric
acid
2.
Each nucleotide formed from deoxyribose, a phosphate (group) and an
organic/nitrogenous base;
3.
Phosphodiester bonds (between nucleotides);
4.
Double helix/2 strands held by hydrogen bonds;
5.
(Hydrogen bonds/pairing) between adenine, thymine and cytosine,
guanine;
5
(b)
1.
(Simple) diffusion of small/non-polar molecules down a
concentration gradient;
2.
Facilitated diffusion down a concentration gradient via protein
carrier/channel; Reject if active rather than passive
3.
Osmosis of water down a water potential gradient;
4.
Active transport against a concentration gradient via protein carrier
using ATP;
5.
Co-transport of 2 different substances using a carrier protein;
For any answer accept a correct example
For ‘carrier protein’ accept symport OR cotransport protein
5
(c)
1.
Magnification (figures) show A is bigger than B;
2.
A has a nucleus whereas B has free DNA;
3.
A has mitochondria whereas B does not;
4.
A has Golgi body/endoplasmic reticulum whereas B does not;
5.
A has no cell wall whereas B has a murein/glycoprotein cell wall;
6.
A has no capsule whereas B has a capsule;
7.
A has DNA is bound to histones/proteins whereas B has DNA not
associated with histones/proteins
OR
A has linear DNA whereas B has circular DNA;
8.
A has larger ribosomes;
Accept in all marking points, animal/eukaryote for A and
prokaryote/ bacterium for B
5 max
[15]
Page 11 of 21
Q2. (a)
1.
Hydrogen bonds (between DNA bases) break;
Ignore DNA helicase.
Reject hydrolysing hydrogen bonds.
2.
(Only) one DNA strand acts as a template;
3.
(Free) RNA nucleotides align by complementary base pairing;
For ‘align by complementary base pairing’, accept ‘align to
complementary bases’ or ‘align by base pairing’.
4.
(In RNA) Uracil base pairs with adenine (on DNA)
OR
(In RNA) Uracil is used in place of thymine;
Do not credit use of letters alone for bases.
5.
RNA polymerase joins (adjacent RNA) nucleotides;
Reject suggestions that RNA polymerase forms hydrogen
bonds or joins complementary bases.
6.
(By) phosphodiester bonds (between adjacent nucleotides);
7.
Pre-mRNA is spliced (to form mRNA)
OR
Introns are removed (to form mRNA);
5 max
(b)
1.
(mRNA attaches) to ribosomes
OR
(mRNA attaches) to rough endoplasmic reticulum;
2.
(tRNA) anticodons (bind to) complementary (mRNA) codons;
3.
tRNA brings a specific amino acid;
4.
Amino acids join by peptide bonds;
5.
(Amino acids join together) with the use of ATP;
6.
tRNA released (after amino acid joined to polypeptide);
7.
The ribosome moves along the mRNA to form the polypeptide;
6 max
Page 12 of 21
(c)
(Definition of gene mutation)
1.
Change in the base/nucleotide (sequence of chromosomes/DNA);
For 4 marks at least one mark must be scored in each
section of the answer.
Accept named mutation for ‘change’.
2.
Results in the formation of new allele;
(Has no effect because)
3.
Genetic code is degenerate (so amino acid sequence may not change);
OR
Mutation is in an intron (so amino acid sequence may not change);
Accept description of ‘degenerate’, eg some amino acids
have more than one triplet/codon.
4.
Does change amino acid but no effect on tertiary structure;
5.
(New allele) is recessive so does not influence phenotype;
(Has positive effect because)
6.
Results in change in polypeptide that positively changes the properties (of the
protein)
OR
Results in change in polypeptide that positively changes a named protein;
For ‘polypeptide’ accept ‘amino acid sequence’ or ‘protein’.
7.
May result in increased reproductive success
OR
May result in increased survival (chances);
4 max
[15]
Page 13 of 21
Q3. (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);
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.
Page 14 of 21
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’
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)
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
[15]
Page 15 of 21
Q4. (a)
1.
Named structures – trachea, bronchi, bronchioles, alveoli;
Reject mp1 if structures from other physiological systems are
named but award mp2 if the correct structures are in the
correct order.
2.
Above structures named in correct order
OR
Above structures labelled in correct positions on a diagram;
Reject mp1 if structures from other physiological systems are
named but award mp2 if the correct structures are in the
correct order.
3.
Breathing in – diaphragm contracts and external intercostal muscles contract;
4.
(Causes) volume increase and pressure decrease in thoracic cavity (to below
atmospheric, resulting in air moving in);
For thoracic cavity accept ‘lungs’ or ‘thorax’.
Reference to ‘thoracic cavity’ only required once.
5.
Breathing out - Diaphragm relaxes and internal intercostal muscles contract;
Accept diaphragm relaxes and (external) intercostal muscles
relax and lung tissue elastic (so recoils).
6.
(Causes) volume decrease and pressure increase in thoracic cavity (to above
atmospheric, resulting in air moving out);
For thoracic cavity accept ‘lungs’ or ‘thorax’.
Reference to ‘thoracic cavity’ only required once.
If idea of thoracic cavity is missing or incorrect, allow ECF for
mark point 6.
6
(b)
1.
Both contain ester bonds (between glycerol and fatty acid);
All statements must be clearly comparative or linked by the
candidate, not inferred from separate statements.
Accept mark points shown on adjacent annotated diagrams.
2.
Both contain glycerol;
3.
Fatty acids on both may be saturated or unsaturated;
4.
Both are insoluble in water;
5.
Both contain C, H and O but phospholipids also contain P; Must relate to
element.
6.
Triglyceride has three fatty acids and phospholipid has two fatty acids plus
phosphate group;
7.
Triglycerides are hydrophobic/non-polar and phospholipids have hydrophilic
and hydrophobic region;
Accept ‘non-polar’ for hydrophobic and ‘polar’ for hydrophilic.
8.
Phospholipids form monolayer (on surface)/micelle/bilayer (in water) but
triglycerides don’t;
Page 16 of 21
5 max
(c)
1.
Glucose and galactose;
Ignore α or β for glucose
2.
Joined by condensation (reaction);
3.
Joined by glycosidic bond;
4.
Added to polypeptide in Golgi (apparatus);;
4
[15]
Q5. (a)
1.
2.
3.
4.
5.
6.
7.
8.
TEM use electrons and optical use light;
TEM allows a greater resolution;
(So with TEM) smaller organelles / named cell structure can be observed
OR
greater detail in organelles / named cell structure can be observed;
TEM view only dead / dehydrated specimens and optical (can) view live
specimens;
TEM does not show colour and optical (can);
TEM requires thinner specimens;
TEM requires a more complex/time consuming preparation;
TEM focuses using magnets and optical uses (glass) lenses;
3.
‘clearer’ is not equivalent to ‘detail’
4.
Accept ‘Only optical can view live specimens’
5.
Accept ‘Only optical can show colour’
7.
Accept ‘TEM requires a more difficult preparation’
Ignore references to artefacts
6 max
(b)
1.
2.
3.
4.
W has 4 cells / nuclei since it is at the (end of) 2nd division (of meiosis);
Z has 2 cells/nuclei since it is at the (end of) 1st division (of meiosis);
W shows haploid cells / cells containing n chromosomes;
(Cells in) W contain half the (mass of) DNA of (Cells in) Z;
OR
(between Z and W) chromatids have separated;
OR
In Z homologous chromosomes have separated;
Ignore ‘Z shows diploid cells / contains 2n chromosomes’
4.
Accept ‘W contains half the amount of DNA of Z’
4
(c)
1.
2.
3.
4.
5.
6.
Page 17 of 21
Use random sample of seeds (from each population);
Use (large enough) sample to be representative of whole population;
Indication of what size was measured e.g. mass;
Calculate a mean and standard deviation (for each population);
Use the (Student’s) t-test;
Analyse whether there is a significant difference between (the means of)
the two populations;
1.
Accept described, suitable method of random sampling.
1.
Reject description of inappropriate method of random
sampling (e.g. random coordinates in the field / use of
2.
2.
5.
6.
quadrats)
Accept ‘running mean does not change’
For representative accept ‘reliable, reproducible,
repeatable’ OR a mean close to the true value.
Accept ‘Use 95% confidence limits’
Reject unqualified references to results being significant
5 max
[15]
Page 18 of 21
Examiner reports
Q1.
In part (a), most students showed good knowledge of DNA structure. This was less true of
their descriptions of transport processes across cell-surface membranes in part (b). Many
students failed to include the relevant information about channel or carrier proteins in their
descriptions and often referred incorrectly to movement from high gradient to low gradient.
Detail about the co-transport process was not well-known. In part (c), a common
misconception when considering prokaryote structure was to confuse a bacterial capsule
and a viral capsid.
Q2.
(a) Students should be careful to refer to RNA nucleotides/polymerase when describing
transcription. As with question (a) about DNA polymerase, there were students who could
not be awarded marking point 5 and/or 6 for suggesting that RNA polymerase catalyses
formation of hydrogen bonds, forms complementary base pairs or forms phosphodiester
bonds between bases.
(b) Some confusion was demonstrated between tRNA and amino acids, about what joins
with the mRNA and about what is joined together to form the polypeptide. The use of ATP
in the formation of the peptide bonds was rarely seen.
(c) Definition of a gene mutation was generally well done, although some stated that it
always resulted in a change in amino acid sequence. Most who scored marking point 2
did so when discussing the positive effect and the formation of a new allele that would
then be passed on as a result of that individual’s increased reproductive success. Few
students demonstrated the understanding that any mutation would result in a new allele
being formed. The mark for the mutation having no effect was most commonly awarded
for the idea of the genetic code being degenerate, although this was often very poorly
expressed. Many students gave low-level answers for the positive effect, with generalised
suggestions of how it would change an animal’s/human’s appearance or ability to find
food/avoid predators, without expanding to the idea of increased survival chances or
reproductive success. The idea of a mutation leading to the individual having a selective
or competitive advantage was awarded marking point 7 as an equivalent/better
expression of the individual having increased survival chances.
Q3.
(a)
Some excellent answers were seen here, but many students demonstrated some
knowledge of either the property or the importance to organisms but did not give
both. Very few students referred to water as a metabolite. Many students described
water as a solvent (a significant number thought it is a solute), but then did not go on
to describe the importance of this aspect. There was some confusion shown
between heat capacity and latent heat of vaporisation and these terms were often
accompanied by incomplete explanations of why they were important. Many
students referred to the cohesive properties of water but did not then gain credit
because they completed the statement by simply stating “aids transpiration” rather
than an explanation of maintenance of the transpiration stream.
(b)
For a straightforward recall of food tests, it was a little surprising that fewer than a
quarter of students achieved 4 or more marks and fewer than half scored 3 and
above. Many students were able to describe the emulsion test although some are
still using the word cloudy, which has not been accepted for some time. Many
Page 19 of 21
students were able to correctly describe a positive Benedict’s test, but fewer realised
that when testing for non-reducing sugar a Benedict’s test must be carried out first
and shown to be negative. Hardly any students referred to boiling acid and therefore
very few were awarded this marking point. Many students could correctly describe
the biuret test but a significant number confused amylase with amylose, underlining
the need for students to read the questions slowly and carefully before answering.
(c)
This was the most successfully answered part of this question, with nearly 40%
achieving 4 or 5 marks. It was, though, surprising that a large number of students
thought dimers were polymers and only gave disaccharides/dipeptides as their
examples of polymers. Most students understood the role of water in hydrolysis and
condensation reactions, but a smaller number appreciated that these reactions
involve bond breakage or formation and therefore did not always gain credit.
Q4.
Question (a) demanded recall from section 3.3.2 of the specification; 25.6% of students
gained five or six marks. Many, however, omitted any reference to the structure at all and
it was surprising that very few students elected to draw a labelled sketch to show the
gross structure. Many lengthy answers were seen detailing exchange of gases and the
features of the alveolar epithelium, neither of which was required by the wording in the
question. Many answers included tracheoles as a part of the human gas exchange
system. Inhalation tended to be described in the best detail. Marking points 3 and 5 were
often not awarded because of a lack of precision in describing the role of the pair of
antagonistic intercostal muscles. Many students conflated the two, and referred to them
generically. The relaxation and contraction of the diaphragm and its corresponding shape
were frequently confused. For example, students referred to the 'flattened' or 'domed'
shape of the diaphragm without stating how that occurred. This question had the highest
discrimination index of the paper.
In question (b), although many students could demonstrate knowledge of phospholipids
and triglycerides individually, they struggled to complete the required ‘compare and
contrast’ command. When this command is used, every marking point requires a
comparative statement that must be clearly made by the student: examiners will not infer
links between separate statements – in this case, separate descriptions of phospholipids
and of triglycerides. Many students did not include glycerol in their structure of the
phospholipid. Some students were distracted into discussions of applications for the
molecules, and their energetic values.
Question (c) showed that most students had learned the components of lactose and knew
that it would be formed in a condensation reaction, although some omitted that this would
result in the formation of a glycosidic bond. Fewer students knew that the lactose would
be joined to a polypeptide in the Golgi apparatus. Many described that it would be found
on the cell-surface membrane, and some tried to describe where on the polypeptide the
lactose would be attached, rather than where in the cell as required by the question.
Nearly half of the students scored at least two of the four available marks.
Q5.
(a)
The command word here was contrast and so statements showing clear differences
between the use of the two microscopes were required to gain credit. Most students
demonstrated sound knowledge of the optical and electron microscopes, but few
managed to gain all six marks for relevant contrasting statements. Many suggested
that no organelles could be seen with an optical microscope, rather than only larger
organelles being visible. Some referred to SEMs and 3D images; neither was
relevant here.
Page 20 of 21
(b)
There were many good answers here, with over 70% of students scoring 2 marks
and over 40% gaining 3 marks. Mark point 4 was the least often awarded. Students
often failed to gain mark point 3 because they stated that W contained half the
genetic material of Z, rather than specifying half the DNA. Several students
discussed these cells as if they were human cells rather than plant cells, mentioning
23 pairs of chromosomes originally, or 23 chromosomes in cells in W. Some
students named stages of meiosis in their answer. These were ignored as this is not
expected knowledge; we were only looking for the outcomes of 1st and 2nd meiotic
divisions.
(c)
The full range of 0 – 5 marks was seen in similar proportions here (although slightly
fewer with full marks). The main reason for the discrimination was how fully each
student used the information from the question stem. The two groups of seeds to be
investigated had been collected already, and the investigation that was needed was
simply to find out if there was a difference in size of these seeds. Many students
gave extensive answers relating to how to collect the seeds from the environment,
and how to measure the pollution – neither of which was relevant to the question.
Many students selected an inappropriate statistical test or listed several options,
which could not be given credit, even if t-test was amongst the list of possibilities
(see note about the ‘list rule’ in the final paragraph of “General Comments” above).
Page 21 of 21