IFYBI003 Biology
THE NCUK INTERNATIONAL FOUNDATION YEAR
IFYBI003 Biology
EXEMPLAR EXAMINATION
MARK SCHEME
Exemplar
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IFYBI003 Biology
This mark scheme should be used in conjunction with the Centre Marking and
Recording Results section of the NCUK Academic Handbook. Contact your Principal/
Academic Manager if you do not have access to this.
NB – Credit should be awarded for any pertinent answers not included in the mark
scheme, not exceeding the total mark allowed for the question.
Notice to markers.
If a student has answered more than the required number of questions, credit should
only be given for the first n answers, in the order that they are written in the student’s
answer booklet (n being the number of questions required for the examination).
Markers should not select answers based on the combination that will give the student
the highest mark. If a student has crossed out an answer, it should be disregarded.
Copyright: Please note that, unless otherwise stated, all diagrams have been produced
from DK Publishing (Dorling Kindersley); Har/Com/Ps edition (30 Mar. 2009)
Exemplar
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IFYBI003 Biology
Questions 1-9
Answer ALL questions.
These questions carry 40 marks in total.
Question
Answer
1
2
3
4
5
6(a)
6(b)
6(c)
6(d)
6(e)
B
C
D
A
D
C
C
C
D
A
Assessment
Objective
AO1
AO1
AO1
AO1
AO1
AO2
AO2
AO2
AO2
AO2
Question 7
(a)
Mitochondrion/mitochondria;
[1]
(b)
P = outer membrane;
Q = cristae;
R = matrix.
Award 1 mark for each of the following points [to Max 2 marks]:
[3]
(c)
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(d)
[2]
muscle requires large amounts of energy;
mitochondrion/mitochondria site of respiration;
respiration releases energy
Examples may include:
[1]
liver cell or sperm cell.
(e)
Note: accept any cell that is very active.
Chloroplast diagram showing correct locations of:
stroma ;
thylakoid membranes / granum ;
inner and outer membrane ;
[3]
AO1 - 7 marks
AO2 – 3 marks
Exemplar
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IFYBI003 Biology
Question 8
(a)
(b)
(c)
(d)
Award 1 mark for each correct answer.
[4]
W = ATP ;
X = reduced NAD/NADH ;
Y = reduced FAD/FADH2 ;
Z = CO2/carbon dioxide ;
matrix of mitochondrion/mitochondria ;
Note: Answer must include matrix
pyruvate ;
production of reduced NAD ;
release of CO2 ;
any two comparative descriptions from:
aerobic
uses oxygen
produces H2O + CO2
more energy / ATP
[1]
[3]
[2]
anaerobic
no oxygen
produces lactic acid
less energy / ATP
AO1 - 8 marks
AO2 – 2 marks
Question 9
(a)
[2]
to identify anomalous results;
to find an average value at each measurement/check
accuracy of the measurements.
(b)
Graph should look like the one below:
[4]
70
60
50
40
30
20
10
0
0
5
10
15
20
25
30
Time in minutes
The y axis units should be distance glucose solution moved (in mm)
Exemplar
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IFYBI003 Biology
Award 1 mark for each of the following points:
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(c)
Correctly labelled x and y axes with units;
Suitable scale used on BOTH axes;
Correct plotting of data;
Line of best fit drawn.
Award 1 mark for each of the following points:
[3]
glucose lowers water potential (makes it more negative);
water moves in from beaker/from higher water potential;
by osmosis.
(d)
suitable line drawn above the on graph;
[1]
AO1 - 0 marks
AO2 – 10 marks
Exemplar
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IFYBI003 Biology
Questions 10-14
Answer 3 questions.
These questions carry 60 marks in total.
Question 10
(a)
Award 1 mark for each of the following points [to Max 10 marks]:
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(b)
DNA is a polymer of nucleotides that forms a double helix;
The four nucleotides are adenine/A, Thymine/T, Guanine/G and
Cytosine/C;
Nucleotides show complimentary base paring, A-T C-G;
With/using hydrogen bonding;
Topoisomerase breaks, swivels and rejoins the DNA molecule
ahead of the replication fork;
This relieves the strain on the molecule during unwinding;
”unwindase”/helicase unwinds the DNA double helix by breaking
hydrogen/H bonds forming a replication fork;
This gives two single stranded molecules/forms a leading strand
and a lagging strand;
These single stranded molecules act as templates for the synthesis
of new DNA molecules;
An RNA primer binds to the leading strand;
DNA polymerase adds nucleotides to the RNA primer;
Nucleotides are added continuously in a complementary manner to
the original leading strand template;
In 5’ to 3’ direction forming a continuous double strand/new DNA;
The lagging strand elongates/adds nucleotides discontinuously;
Multiple primers are added to the lagging strand;
Nucleotides are added and short DNA fragments formed between
primers;
These short segments of DNA are called Okazaki fragments;
DNA ligase then joins the Okazaki fragments into one continuous
DNA molecule;
Result is two new DNA molecules are synthesised, with each new
double helix molecule having one strand running in the 5’ to 3’
direction and one in the 3’ to 5’ direction.
Award 1 mark for each of the following points [to Max 10 marks]:
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Exemplar
[ 10 ]
[ 10 ]
They grew/cultured bacteria;
in a medium containing heavy nitrogen/15N;
15
N was incorporated into the DNA of the growing cells;
They switched/transferred some of the cells to a medium
containing naturally occurring (light) nitrogen/14N;
These cells were allowed to replicate/grow;
DNA was extracted from cells after the first replication and another
DNA sample was taken/extracted after the second replication;
Bacterial DNA ‘weighed’/centrifuged after each replication;
100% heavy became 100% intermediate/the first replication in the
14
N medium produced a band of hybrid /14N-15N DNA;
Then 50% intermediate and 50% light/the second replication
produced both light and hybrid DNA;
This experiment allowed Meselson and Stahl to prove that
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IFYBI003 Biology
replication of DNA was semi conservative/replicated by the semi
conservative method.
AO1 - 10 marks
AO2 – 10 marks
Exemplar
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IFYBI003 Biology
Question 11
(a)
Award 1 mark for each of the following points [to Max 10 marks]:
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(b)
[ 10 ]
Gene is isolated from DNA in cell of interest/isolation of gene from
(usually) eukaryotic/ “foreign” DNA;
Using restriction (endonuclease) enzymes;
These enzymes cut the DNA at specific points/restriction sites;
Plasmid is removed from a bacterial cell/plasmid removed to act as
a cloning vector;
The plasmid is cut/treated using the same restriction enzyme
(used to cut the eukaryotic (“foreign”) DNA at similar restriction
sites;
Complementary sticky ends are formed in both plasmid and
eukaryotic (“foreign”) DNA/complementary plasmid to sticky ends
of isolated gene;
The plasmid DNA and eukaryotic (“foreign”) DNA are then mixed
together (under suitable conditions)/cut plasmid is added to
isolated gene;
Base pairing of plasmid and eukaryotic (“foreign”) DNA produces
recombinant DNA/binding of eukaryotic gene to plasmid
By the formation of hydrogen/H-bonding between complementary
bases/base pairs;
The recombinant plasmid placed/inserted into suitable bacterium;
The transformed bacterium/recombinant bacterium is placed into
suitable growth medium;
The bacterium reproduces/grows and in doing so multiplies the
recombinant DNA (containing the eukaryotic (“foreign”)
gene/DNA;
Reference to selection of transformed bacteria.
Award 1 mark for each of the following points [to Max 5 marks]:
[5]
Medical:
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Used to produce safer drugs;
Used to repair defective genes;
Examples may include discussion/explanation of one or more of
the following:
Anti-thrombin (prevents clottng in surgery);
Factor VIII (haemophiliac treatment);
HGH (stimulates growth);
Penicillin (antibiotic);
Gene therapy;
Insulin production;
Industrial/agricultural:
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(c)
BST (increased yield in cattle);
Rennin production (cheese making);
Pesticide production (crop yield);
Delayed ripening.
Award 1 mark for each of the following points [to Max 5 marks]:
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Exemplar
[5]
Caused by a recessive allele;
Loci on the X-chromosome;
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IFYBI003 Biology
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More common in males due to presence of Y chromosome / lack of
another X chromosome;
Reference to females haemophiliacs being homozygous recessive;
Results in lack of clotting factor VIII or IX;
Blood unable to clot effectively;
Leads to internal bleeding or bleeding out after trauma/surgery.
Note: Allow marks for any other correct points made if another example
of a disease is given in the student’s answer.
AO1 - 10 marks
AO2 – 10 marks
Exemplar
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IFYBI003 Biology
Question 12
(a)
Award 1 mark for each of the following points [to Max 10 marks]:
[ 10 ]
Formation of tertiary structure:
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Proteins are formed from amino acids;
Amino acids have specific R-groups;
Amino acids are joined by peptide bonds;
Description of primary structure as a specific sequence of amino
acids;
Description of secondary structure as alpha helix or beta pleated
sheet folding;
Tertiary structure involves 3D folding of polypeptide chain;
Tertiary structure involves interactions between the side chains/R
groups of the various amino acids;
Tertiary structure is maintained by:
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Hydrogen bonding;
Ionic bonding;
Covalent bonding;
Description of cysteine residues covalent bonding to form
disulphide bridges;
Hydrophobic/hydrophilic interactions;
Van der Walls forces.
Importance/Significance of tertiary structure:
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(b)
Allows protein stability;
Description of effect of temperature change and idea of optimum
temperatures;
Description of effect of pH range and idea of optimum pH values;
Allows active site formation for enzymes;
Binding site for hormones/ligands etc;
Reference to specificity.
Award 1 mark for each of the following points [to Max 10 marks]:
[ 10 ]
Proteins can be:
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Exemplar
Structural/supporting, for example keratin, horn, feathers;
Collagen and elastin proteins provide a framework around
animal/human connective tissues/cytoskeleton;
May function in movement; for example myosin and actin used in
the contraction of muscle tissue;
Proteins function in membrane transport processes, for example in
ion channels/pumps;
Specific proteins play a role in active transport/facilitated diffusion;
Transport for oxygen, for example haemoglobin carries oxygen;
Act as storage molecules, for example ovalbumin of egg white,
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IFYBI003 Biology

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used as an amino acid source for the developing embryo;
Proteins in DNA folding/formation of chromosomes;
Proteins act as hormones in the co-ordination of metabolic activities
of an organism, for example insulin;
Function to protect against disease/defensive proteins, for example
antibodies;
Act as receptors, for example signal transduction in membranes of
nerve cells;
Cell-cell recognition.
AO1 - 10 marks
AO2 – 10 marks
Exemplar
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IFYBI003 Biology
Question 13
(a)
Award 1 mark for each of the following points [to Max 10 marks]:
[ 10 ]
Interphase:
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Alternates with the mitotic phase of the cell cycle;
Consists of a number of different phases including the G1, S and G2
phases;Interphase accounts for about 90% of the cell cycle;
G1/first gap Phase:
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Intense metabolic activity occurs in the cell during this phase/
normal cellular activity occurs;
Cellular components such as mitochondria. Endoplasmic
reticulum/ER and other organelles are duplicated during this
phase/cellullar components are duplicated;
S phase or synthesis phase:
DNA replication/duplication of the chromosomes occurs entirely in
this phase;
G2/second gap phase:
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Cell growth continues/fair amount of metabolic activity takes place;
Prepares for mitosis/DNA checking before the M phase of the cycle;
M phase/mitotic phase:
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Mitosis takes place during this phase;
Consists of prophase, metaphase , anaphase, telophase, and
cytokinesis;
Cleavage furrow/cell plate forms/pinches cell into two;
Importance of regulation by cell cycle:
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(b)
Prevents uncontrolled cell division;
Prevents cancer/tumour formation.
Differences:
[5]
mitosis
1 cell division
2 daughter cells
Diploid cells formed
Genetically identical daughter cells
No crossing over
Forms somatic cells
Exemplar
meiosis
2 cell divisions
4 daughter cells
Haploid
(Genetic) variation
Crossing over
Forms gametes
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IFYBI003 Biology
(c)
[5]
spermatogenesis
Forms spermatocytes
Occurs in testes
Cells divide equally
Four gametes produced
Continuous process
Starts at puberty
Continues throughout life
oogenesis
Forms ova
ovaries
Unequal cell division
One gamete/ovum and 2-3 polar
bodies produced
Interrupted process
Starts in foetus
Stops at menopause
AO1 - 10 marks
AO2 – 10 marks
Exemplar
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IFYBI003 Biology
Question 14
(a)
Award 1 mark for each of the following points [to Max 15 marks]:
 Mutations generally affect one or a few nucleotide pairs;
 Mutations may be referred to as point mutations;
 Two categories of point mutations, which are single nucleotide
substitutions and nucleotide pair insertions or deletions;
 Insertions and deletions can involve more than one nucleotide pair;
[ 15 ]
Substitutions:
 One base pair is replaced by another/a different base pair;
 Some substitutions have no effect on the reading frame/no effect on
the encoded protein;
 This is because of the redundancy of the genetic code;
Silent mutation (a substitution mutation):
 A change in a nucleotide pair may transform/alter one codon for
another that translates the same amino acid;
 This would have no effect on the phenotype of the organism;
Mis-sense mutation (a substitution mutation):
 May be similar to a silent mutation in that such a mutation may have
little effect on the protein function;
 Idea of degenerate code;
 Different codons code for same amino acid;
Non-sense mutation (a substitution mutation):
 This is where there is the insertion of a stop codon;
 Prematurely shortens protein during synthesis/causes translation to
be terminated prematurely/early;
 May have a marked effect on the function of a protein/nearly all nonsense mutations lead to non-functional proteins;
Nucleotide pair insertions and deletions:
 These are where there are additions or losses of nucleotide pairs (in
a gene/protein);These mutations tend to have a marked/disastrous
effect on the resulting gene/protein (compared to substitution
mutations);
Deletion mutation:
 Removes a base pair/base pairs from a DNA/gene sequence;
 Causes a frame shift to occur in triplet codons/alters the normal DNA
nucleotide sequence;
 Leads to abnormal translation of the protein/gene;
 (Generally leading) to a huge effect on protein structure;
 Generally occurs if deletion of base pairs is not in a multiple of three
bases;
 Same/similar effects are seen with insertion mutations;
Note: Allow similar points if made for insertion mutations, but only allow a
mark once for a similar or the same mark point.
Suitable examples of gene mutation:
Exemplar
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IFYBI003 Biology
e.g. sickle cell anaemia, haemophilia, any other correct example.
Note: Allow 1 mark for one or two examples of human disease caused by
mutations.
(b)
Award 1 mark for each of the following points [to Max 5 marks]:
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[5]
Trisomy is a chromosome aberration in which the chromosomes fail
to separate / nondisjunction;
This occurs during meiosis;
The non-disjunction can take place either in the formation of the egg
or the sperm;
The result is an extra chromosome present in all of the body cells /
has a total of 47 chromosomes instead of 46 in each cell;
Trisomy often leads to miscarriage (eg trisomy 16);
An example of trisomy is Down/Edward/Patau syndrome;
Down syndrome occurs due to trisomy 21 (Edwards trisomy 18 and
Patau trisomy 13);
Downs syndrome is associated with physical growth disruption and
learning difficulties;
AO1 - 10 marks
AO2 – 10 marks
Exemplar
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