Royal Grammar School
Gene expression
Unit 5
Q1.
Scientists are working to produce a genetically modified bacterium to treat patients
suffering from a disease of the digestive system. They plan to collect mRNA from human
cells. This will be used to produce the DNA of the gene for the protein interleukin. They
will then transfer this human gene into the bacterium Lactococcus. The scientists intend
patients to swallow the genetically modified bacteria. These bacteria will release
interleukin inside the digestive system to treat the disease.
(a)
(i)
Name the type of enzyme which will be used to produce the DNA from the
mRNA.
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(1)
(ii)
It is easier to obtain the interleukin gene from mRNA rather than directly from
the DNA removed from human cells. Explain why.
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(1)
(b)
The scientists propose to put the gene directly into the DNA of Lactococcus.
Describe the role of the enzyme ligase in this process.
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(1)
(Total 3 marks)
Q2.
(a)
CFTR is a transmembrane regulator protein. Its molecules have 1480 amino
acids. People with cystic fibrosis produce defective CFTR protein which is missing
one amino acid from its structure.
(i)
What is the minimum number of bases on DNA which would code for the
normal CFTR protein? Explain your answer.
Number of bases ...............................
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(2)
(ii)
Which type of gene mutation produced the cystic fibrosis allele?
Explain your answer.
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(2)
(b)
The diagram shows part of the process of making normal and defective CFTR in a
cell. A normal CFTR protein molecule has sugar molecules attached to it which
make it functional.
(i)
Describe how the information on mRNA is translated into CFTR at the
ribosome.
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(4)
(ii)
Using information in the diagram and your own knowledge, suggest why
defective CFTR, missing one amino acid, is not functional.
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(2)
(Total 10 marks)
Q3.CREB is a transcription factor in the mitochondria of neurones.
(a)
What is a transcription factor?
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(2)
(b)
CREB leads to the formation of a protein that removes electrons and protons from
reduced NAD in the mitochondrion.
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Huntington’s disease (HD) causes the death of neurones. People with HD produce a
substance called huntingtin. Some scientists have suggested that binding of
huntingtin to CREB may lead to the death of neurones.
Suggest how binding of huntingtin to CREB may lead to the death of neurones.
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(Extra space) .................................................................................................
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(3)
(c)
CREB is a protein synthesised in the cytoplasm of neurones. Transport of CREB
from the cytoplasm into the matrix of a mitochondrion requires two carrier proteins.
Use your knowledge of the structure of a mitochondrion to explain why transport of
CREB requires two carrier proteins.
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(2)
(Total 7 marks)
Q4.
Research scientists can increase the nutritional value of potatoes by genetically
engineering potato plants. A gene which results in increased protein production has been
removed from cells of an amaranth plant and inserted into cells of a potato plant.
(a)
Describe how a gene could be removed from cells of an amaranth plant and
inserted into cells of a potato plant.
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(6)
(b)
Describe the advantages of using vegetative propagation rather than sexual
reproduction to reproduce genetically engineered potato plants.
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(3)
(c)
Whole potato plants can be produced from genetically identical potato cells grown in
a tissue culture. Use your knowledge of genes to suggest how different cells, such
as leaf and root cells, can develop from genetically identical cells.
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(2)
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(Total 11 marks)
Q5.
(a)
(i)
The mRNA codon for the amino acid tyrosine is UAU.
Give the DNA triplet for tyrosine.
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(1)
(ii)
Give the tRNA anticodon for tyrosine.
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(1)
(b)
Give two ways in which the structure of a molecule of tRNA differs from the
structure of a molecule of mRNA.
1 ...................................................................................................................
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2 ...................................................................................................................
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(2)
(Total 4 marks)
Q6.
Lysozyme is an enzyme consisting of a single polypeptide chain of 129 amino acids.
(a)
What is the minimum number of nucleotide bases needed to code for this enzyme?
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(1)
(b)
The diagram shows the sequence of bases in a section of the mRNA strand used to
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synthesise this enzyme.
G G U C U U U C U U A U G G U A G A U A
U
(i)
Give the DNA sequence which would be complementary to the first four bases
in this section of mRNA.
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(1)
(ii)
How many different types of tRNA molecule would attach to the section of
mRNA shown in the diagram?
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(1)
(c)
Give two factors which might increase the frequency at which a mutation in DNA
occurs.
1 ...................................................................................................................
2 ...................................................................................................................
(2)
(d)
Two single base mutations occurred in the DNA coding for this section of mRNA.
These mutations caused an alteration in the sequence of amino acids in the
enzyme. The diagram shows the original and altered sequences of amino acids.
Original amino acid
sequence
Original mRNA
base sequence
Altered amino acid
sequence
Altered mRNA
base sequence
(i)
Gly
Leu
Ser
Tyr
Gly
Arg
Tyr
GGU
CUU
UCU
UAU
GGU
AGA
UAU
Gly
Leu
Tyr
Leu
Trp
Arg
Tyr
GGU
CUU
AGA
UAU
Use the mRNA codons provided in the table to complete the altered mRNA
base sequence in the diagram.
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Amino acid
mRNA codons which can be used
Arg
AGA
Gly
GGU
Leu
CUU or UUA
Ser
UCU
Trp
UGG
Tyr
UAU or UAC
(1)
(ii)
Use the information provided to determine the precise nature of the two single
base mutations in the DNA.
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(3)
(Total 9 marks)
Q7.
(a)
Figure 1 shows the exposed bases (anticodons) of two tRNA molecules
involved in the synthesis of a protein.
Figure 1
Complete the boxes to show the sequence of bases found along the corresponding
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section of the coding DNA strand.
(2)
(b)
Describe the role of tRNA in the process of translation.
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(3)
(c)
Figure 2 shows the sequence of bases in a section of DNA coding for a polypeptide
of seven amino acids.
Figure 2
TACAAGGTCGTCTTTGTCAAG
The polypeptide was hydrolysed. It contained four different amino acids. The
number of each type obtained is shown in the table.
Amino acid
Number present
Phe
2
Met
1
Lys
1
Gln
3
Use the base sequence shown in Figure 2 to work out the order of amino acids in
the polypeptide. Write your answer in the table below.
Met
(2)
(Total 7 marks)
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Q8.
The polymerase chain reaction is a process which can be carried out in a laboratory
to replicate DNA. The diagram shows the main stages involved in the polymerase chain reaction.
(a)
Explain why DNA is heated to 95 °C.
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(1)
(b)
What is the role of
(i)
a primer in this process;
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(1)
(ii)
DNA polymerase?
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(1)
(c)
(i)
How many DNA molecules will have been produced from one molecule of
DNA after 6 complete cycles?
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(1)
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(ii)
Suggest one use of the polymerase chain reaction.
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(1)
(d)
Give two ways in which the polymerase chain reaction differs from the process of
transcription.
1 ...................................................................................................................
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2 ...................................................................................................................
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(2)
(Total 7 marks)
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M1.
(a)
(i)
Reverse transcriptase;
1
(ii)
Idea that mRNA is present in large amounts in cell making
the protein / mRNA has been edited / does not contain
introns / mRNA codes for single protein;
1
(b)
(Ligase) splices / joins two pieces of DNA / “sticky ends”;
1
[3]
M2.
(a)
(i)
number of bases = 4440
allow 4446 if they refer to start / stop
each amino acid coded for by triplet / three bases
(so three times more bases than amino acids);
2
(ii)
deletion;
(deletion) of three bases;
because substitution/addition would change amino acid(s);
2 max
(b)
(i)
codon on mRNA;
specific/complementary base pairing with;
anti-codon on tRNA;
specific tRNA for each amino acid;
protein formed by condensation reactions /
peptide bonds formed;
4 max
(ii)
(loss of amino acid) changes tertiary structures/3D shape;
so sugar molecules cannot be attached (to form glycoprotein/
functional protein);
so (defective) unable to bind to chloride ions/use ATP;
2 max
[10]
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M3.(a)
1.
(Protein / molecule) that moves from cytoplasm to DNA;
Accept ‘it’ as TF.
Accept moves into nucleus
2.
(TF) binds to specific gene / genes / to specific part of / site on DNA /
binds to promoter / RNA polymerase;
Accept regulator / enhancer region
3.
Leads to / blocks (pre)mRNA production / allows / blocks binding of RNA
polymerase (to DNA) / allows RNA polymerase to work;
Ignore translation unless context wrong
Max 1 if refer to oestrogen as a transcription factor
2 max
(b)
1.
(Binding to CREB) prevents transcription / mRNA formation;
Accept that lack of protein leaves NAD reduced
2.
(Binding of huntingtin) prevents production / translation of protein (that
removes electrons / protons from NAD);
3.
Fewer electrons to electron transport chain / electron transport chain
slows / stops / stops / slower oxidative phosphorylation;
4.
Fewer protons for proton gradient;
5.
Not enough ATP produced / energy supplied to keep cells alive /
anaerobic respiration not enough to keep cell alive;
Accept neurones require ATP for active transport of ions
Ignore references to resting potential
3 max
(c)
1.
CREB / protein is too large / is water soluble so cannot cross membrane
/ phospholipid bilayer;
2.
Mitochondrion has two membranes / inner and outer membranes;
Accept cristae for inner membrane
3.
For each (different) membrane a (different) carrier required;
3. Ignore reference to channel proteins
2 max
[7]
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M4.
(a)
(cut out gene using an) endonuclease / restriction enzyme;
reference to specificity / recognition site;
sticky ends;
use the same enzyme to cut;
plasmid / virus / potato DNA;
fixed by ligase;
method of introducing vector e.g. micropipette / virus injects DNA /
remove plant cell wall;
6 max
(b)
introduced gene / characteristic passed to offspring;
rapid process;
larger number of plants produced;
asexual reproduction genetically identical / sexual reproduction
causes variation;
3 max
(c)
different genes are expressed;
producing different enzymes/proteins;
2
[11]
M5.
(a)
(i)
ATA;
1
(ii)
AUA;
1
(b)
tRNA ‘clover leaf’ shape; (allow reference to loop / folded structure)
tRNA standard length;
tRNA has an amino acid binding site;
tRNA has anticodon available / three exposed bases;
tRNA has hydrogen bonds (between base pairs);
2 max
[4]
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M6.
(a)
387;
1
(b)
(i)
CCAG;
1
(ii)
5;
1
(c)
high energy radiation / X rays / ultraviolet light / gamma rays;
high energy particles / alpha particles / beta particles;
named chemical mutagens e.g. benzene / caffeine / pesticide /
mustard gas / tobacco tar / free radicals;
(two named examples of any of the above = 2 marks)
length of time of exposure (to a mutagen);
dosage (of mutagen);
2 max
(d)
(i)
UAC UUA UGG;
1
(ii)
addition and deletion (of bases/nucleotides);
thymine added;
adenine deleted;
(addition of thymine and deletion of adenine = 3 marks)
(allow addition of adenine (RNA) and deletion of uracil (RNA)
= 2 marks)
3
[9]
M7.
(a)
AGC; TTC;
2
(b)
anticodon complementary to codon/reads message on mRNA;
specific amino acid;
carried/transferred (to ribosome);
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correct sequence of amino acids along polypeptide;
3 max
(c)
(Met)
Phe
Gln
Gln
Lys
Gln
Phe
2
(three/four/five correct 1 mark; six correct 2 marks)
[7]
M8.
(a)
to separate the two strands / break hydrogen bonds;
1
(b)
(i)
enables replication/sequencing to start (allow keeps strands
separate);
1
(ii)
joins DNA nucleotides (not complementary bases);
1
(c)
(i)
64;
1
(ii)
replication of DNA from crime scene/tissue sample /
for DNA sequencing / gene cloning;
1
(d)
(transcription uses) RNA polymerase;
RNA nucleotides / uracil;
one (template) strand / PCR both strands;
start / stop codons;
(accept enzyme separates strands)
2 max
[7]
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E1.
Unit 2
(a)
Reverse transcriptase was usually given, but the incorrect DNA polymerase or
endonuclease figured frequently. The explanation in (ii) was elusive with many
thinking the mRNA was more accessible as it was in the cytoplasm or in single
strands.
(b)
Answers ranged here from the insufficient ‘ligation’ to lengthy but incorrect
descriptions of joining DNA by means of hydrogen bonds on the base pairs. Those
who considered more carefully gave a correct answer.
Unit 3
The enzyme in (a)(i) was well known, though a few simply referred to transcriptase. In
(a)(ii), it was common to read that mRNA is easier to obtain because it is in the cytoplasm
rather than the nucleus. However, a fair number gave a valid explanation, such as the
idea that introns have been removed, or that the gene will only code for the protein
required. In (b), there was some confusion with restriction enzymes. Some thought the
ligase would cut the gene. Others saw it as a kind of filler, filling the gap between the
sticky ends, rather than forming bonds.
E2.
(a)
(ii)
(i)
Most candidates calculated the number of bases correctly. Many
obtained a second mark for explaining that three bases on DNA (a triplet)
code for each amino acid. A significant number did not get the second mark,
because they stated or implied that bases are amino acids.
This proved very difficult for many candidates. Quite a large number obtained
one mark for suggesting a deletion mutation but only the best candidates went
on to suggest the deletion of a triplet of bases. Some candidates continued a
misconception evident in (i) and stated that an amino acid was deleted. Credit
was given to candidates who suggested another type of mutation affecting a
terminal amino acid.
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(b)
E3.(a)
(i)
There were many good answers to this question and many candidates
obtained all four marks. Surprisingly few candidates made clear reference to
complementary base-pairing between codon and anti-codon.
(ii)
Many candidates obtained one mark for statements suggesting changes in the
tertiary structure of the CFTR protein. Others obtained a mark for suggesting
that sugar molecules would not be able to attach to the defective CFTR
protein. Relatively few candidates put these ideas together and obtained two
marks. A number of candidates drifted into discussions of active sites and
substrate binding which were not appropriate to the CFTR example.
In this part, many students expressed themselves unsatisfactorily and there were a
number of misconceptions; almost half failed to score. Some simply said that a
transcription factor affects transcription. Others attempted to use oestrogen as an
example of a transcription factor, rather than a hormone that binds to a receptor to
form a transcription factor. Yet others failed to state that a transcription factor binds
to a specific site (or sites) on DNA (however expressed). Only about 20% of
students obtained both marks.
(b)
There were some very good, clear and concise answers to this part that obtained all
three marks. All the points on the mark scheme were seen but perhaps the
commonest observations were that binding of huntingtin to CREB stops production
of the protein (that removes protons and electrons from reduced NAD). This stops /
slows the electron transfer chain. This leads to not enough ATP being produced and
the nerve cells die. A lot of students did not read the stem carefully and thought that
CREB removes electrons and protons. Others made general references to
respiration slowing but did not mention the electron transfer chain, or proton
gradients, or ATP production.
(c)
It was pleasing in this part to find many students connecting the requirement for two
carriers for CREB to the two membranes of a mitochondrion; just over 50% obtained
both marks. Some students had problems expressing themselves and others
seemed genuinely confused about the membranes of the mitochondrion; for
example, some wrote about a membrane round the cytoplasm and then the cristae.
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E4.
E5.
Generally this question was well answered with most candidates obtaining at least
four marks. However, part (c) proved difficult for the majority.
(a)
Answers to this question provided the full range of marks. Better candidates gave a
detailed description of genetic engineering and had little difficulty in obtaining five of
the six marks available. A significant number obtained maximum marks. Weaker
candidates referred to the use of a restriction endonuclease and the production of
‘sticky ends’. Many of these candidates also referred to the use of a plasmid but,
unfortunately, not always in the correct context. There was also some confusion
over the use of the enzyme ligase. Many candidates failed to realise that the gene
was inserted into a plant cell but did not take the cell wall into consideration when
describing the method of transfer. Nevertheless, the quality of answers to this
question was pleasing.
(b)
This question caused few problems. Many candidates gained two marks by
describing two advantages of using vegetative propagation to reproduce genetically
engineered potato plants. These answers usually referred to the offspring being
genetically identical and possessing the desired characteristic. A significant number
of candidates gained maximum marks by referring to the large number of plants
which could be produced by vegetative propagation.
(c)
This proved to be an effective discriminator, with only better candidates suggesting
how different cells could be produced by different genes being expressed. The best
candidates then linked this to the production of different enzymes or proteins. Many
candidates referred to the use of hormones without displaying any understanding of
how these may initiate gene expression.
Although only a few candidates obtained maximum marks on this question, most
candidates were able to gain between two and four marks.
Better candidates had little difficulty in providing the correct responses for parts (i) and (ii).
Weaker candidates often correctly identified the DNA triplet for tyrosine but gave an
incorrect tRNA anticodon for the same amino acid.
The most frequent correct responses referred to the 'clover leaf' or folded structure of
tRNA, the presence of an amino acid binding site or the anticodon. Incorrect responses
included references to tRNA consisting of two strands, possessing three anticodons or
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possessing thymine rather than uracil.
E6.
Most candidates were able to score at last two marks on this question. However, very
few candidates obtained maximum marks, usually because of their responses in part
(d)(ii).
(a)
The majority of candidates correctly gave 387 as the answer. However, a significant
number of candidates divided the number of amino acids by three and gave an
incorrect answer of 43.
(b)
Part (i) caused few difficulties with the majority of candidates correctly giving CCAG
as the complementary DNA sequence. A few candidates incorrectly gave CCTG.
Part (ii) proved to be far more discriminating with many candidates giving seven as
an answer rather than five, having failed to realise that two of the codons appeared
twice.
(c)
Most candidates were able to gain at least one mark by naming a specific mutagen.
Unfortunately, a significant number of candidates were not specific enough, using
vague references to ‘radiation’ and/or ‘chemicals’ which were not credited. Similarly,
tobacco tar was credited but ‘smoking’ was not. Better candidates had no problem
obtaining both marks by naming a chemical mutagen, e.g. benzene, and a named
radiation source, e.g. ultraviolet rays. Answers referring to the length of time of
exposure to a mutagen and to the dosage of mutagen were also credited.
(d)
This proved to be the most difficult question on the paper. Most candidates
incorrectly used a substitution mutation when determining the altered mRNA base
sequence in part (i), rather than an addition and deletion. Consequently they had
difficulty obtaining any marks for part (ii). Although approximately a third of
candidates provided the correct mRNA base sequence in part (i), few of them
gained more than a single mark in part (ii). A common error was to refer to amino
acid mutations rather than to base mutations. Many candidates in part (ii) referred to
RNA nucleotides rather than to DNA nucleotides, which restricted their maximum
mark.
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E7.
E8.
The majority of candidates gained the marks for application of knowledge.
Straightforward recall proved difficult for some, through lack of adequate preparation for
the examination.
(a)
There was a very disappointing number of correct responses. The vast majority of
candidates quoted ‘mRNA codons’ and gained no marks.
(b)
Candidates who had not thoroughly learnt the factual material necessary to
understand protein synthesis gave confused accounts and failed to gain any marks.
Many candidates gained two marks but common omissions were failure to mention
the specific nature of the amino acid carried and also failure to complete the story by
reference to the correct sequence on the completed polypeptide.
(c)
The vast majority of candidates gained both marks.
Although this question produced a wide range of marks, many candidates were able
to obtain at least two of the marks available.
(a)
Most candidates gained this mark by referring to separation of the two strands or to
the breaking of hydrogen bonds. Vague references to ‘uncoiling’ or ‘unwinding’ were
not credited.
(b)
Most candidates obtained the mark in part (i) by referring to the role of primers in
enabling sequencing to start. However, in part (ii), considerably fewer candidates
correctly described the role of DNA polymerase in joining DNA nucleotides together.
Many candidates suggested that this enzyme forms the hydrogen bonds between
complementary bases.
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(c)
Part (i) and part (ii) caused few difficulties for the majority of candidates. In part (i),
most candidates correctly calculated that 64 DNA molecules would have been
produced after six complete cycles. In part (ii), the use of the polymerase chain
reaction was often linked to the amplification of DNA for forensic purposes or for
gene cloning.
(d)
This proved to be an effective discriminator with only the best candidates obtaining
both marks. Most candidates failed to consider the processes involved and simply
referred to the products of PCR and transcription. These answers were not credited.
Better candidates often referred to transcription using RNA polymerase, uracil and
one template strand.
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