1.
Mycobacterium tuberculosis causes tuberculosis. The DNA of M. tuberculosis contains a direct
repeat (DR) region. The DR region consists of 43 different, non-coding base sequences called
spacers. Each spacer is found in a specific place in the DR region.
In different strains of M. tuberculosis, some of these spacers have been lost.
(a)
(i)
The DR region consists of non-coding base sequences.
What is meant by a non-coding base sequence?
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(1)
(ii)
Name the process by which the base sequence of a spacer is lost from a DR region.
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(1)
Scientists investigated the DR regions of different strains of M. tuberculosis. They produced a
DNA probe for each of the 43 spacer sequences. Each probe was:
•
•
labelled with a fluorescent marker that gave off light if the probe attached to its
complementary spacer
attached to a particular square on a slide.
They obtained samples of the DR region from each strain. These were cut into small singlestranded DNA fragments. The fragments from each strain were added to a slide with the DNA
probes attached. The diagram below shows their results for one strain of M. tuberculosis with 20
of the probes.
(b)
The scientists cloned the DR region DNA in vitro before testing for the presence of spacers.
Give the name of the method they used to clone the DNA in vitro.
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(1)
Page 1 of 19
(c)
Explain how the use of DNA probes produced the results in the diagram.
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(3)
(d)
Doctors can use the method with DNA probes to identify the specific strain of M.
tuberculosis infecting a patient. This is very important when there is an outbreak of a
number of cases of tuberculosis in a city.
Suggest and explain why it is important to be able to identify the specific strain of M.
tuberculosis infecting a patient.
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(2)
(Total 8 marks)
Page 2 of 19
2.
(a)
Describe and explain how the polymerase chain reaction (PCR) is used to amplify a DNA
fragment.
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(4)
The figure below shows the number of DNA molecules produced using a PCR.
(b)
Explain the shape of the curve in the figure above.
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(2)
(Total 6 marks)
Page 3 of 19
3.
(a)
What is a DNA probe?
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(2)
DNA probes are used to detect specific base sequences of DNA.
The process is shown in Figure 1.
Figure 1
(b)
Describe how the DNA is broken down into smaller fragments.
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(2)
Page 4 of 19
(c)
The DNA on the nylon membrane is treated to form single strands. Explain why.
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(1)
A scientist used DNA probes and electrophoresis to screen four volunteers for five different viral
DNA fragments.
Figure 2 shows the results the scientist obtained. The lanes numbered 2 to 5 represent the four
volunteers.
Figure 2
(d)
Lane 1 of Figure 2 enabled the size of the different viral fragments to be determined.
Suggest and explain how.
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(2)
Page 5 of 19
The lengths of the viral DNA fragments were:
•
•
•
•
•
600 base pairs
250 base pairs
535 base pairs
300 base pairs
500 base pairs.
(e)
Which volunteers had at least one of the viral DNA fragments with 250 base pairs or 535
base pairs?
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(1)
(Total 8 marks)
4.
(a)
Adrenaline binds to receptors in the plasma membranes of liver cells. Explain how this
causes the blood glucose concentration to increase.
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(2)
(b)
Scientists made an artificial gene which codes for insulin. They put the gene into a virus
which was then injected into rats with type I diabetes. The virus was harmless to the rats
but carried the gene into the cells of the rats.
The treated rats produced insulin for up to 8 months and showed no side-effects. The
scientists measured the blood glucose concentrations of the rats at regular intervals. While
the rats were producing the insulin, their blood glucose concentrations were normal.
(i)
The rats were not fed for at least 6 hours before their blood glucose concentration
was measured. Explain why.
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(1)
Page 6 of 19
(ii)
The rats used in the investigation had type I diabetes. This form of gene therapy may
be less effective in treating rats that have type II diabetes. Explain why.
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(1)
(iii)
Research workers have suggested that treating diabetes in humans by this method of
gene therapy would be better than injecting insulin. Evaluate this suggestion.
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(4)
(Total 8 marks)
5.
SCID is a severe inherited disease. People who are affected have no immunity. Doctors carried
out a trial using gene therapy to treat children with SCID. The doctors who carried out the trial
obtained stem cells from each child’s umbilical cord.
(a)
Give two characteristic features of stem cells.
1. _________________________________________________________________
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2. _________________________________________________________________
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(2)
Page 7 of 19
The doctors mixed the stem cells with viruses. The viruses had been genetically modified to
contain alleles of a gene producing full immunity. The doctors then injected this mixture into the
child’s bone marrow.
The viruses that the doctors used had RNA as their genetic material. When these viruses infect
cells, they pass their RNA and two viral enzymes into the host cells.
(b)
One of the viral enzymes makes a DNA copy of the virus RNA. Name this enzyme.
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(1)
The other viral enzyme is called integrase. Integrase inserts the DNA copy anywhere in the DNA
of the host cell. It may even insert the DNA copy in one of the host cell’s genes.
(c)
(i)
The insertion of the DNA copy in one of the host cell’s genes may cause the cell to
make a non-functional protein. Explain how.
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(2)
(ii)
Some of the children in the trial developed cancer. How might the insertion of the
DNA have caused cancer?
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(2)
(d)
Five out of the 20 children in the trial developed cancer. Although the cancer was treated
successfully, the doctors decided to stop the trial in its early stages. They then reviewed the
situation and decided to continue. Do you agree with their decision to continue? Explain
your answer.
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(2)
(Total 9 marks)
Page 8 of 19
6.
(a)
There are different types of gene mutation.
Put a tick (✓
✓) in the box next to the statement which describes incorrectly the effect of the
mutation in an exon of a gene.
A substitution may not result in a change to the encoded
amino acid.
An inversion will result in a change in the number of
DNA bases.
A deletion will result in a frame shift.
An addition will result in a frame shift.
(1)
(b)
Describe how alterations to tumour suppressor genes can lead to the development of
tumours.
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(3)
Page 9 of 19
(c)
A type of malignant tumour cell divides every 8 hours.
Starting with one of these cells, how many tumour cells will be present after 4 weeks?
Assume none of these cells will die.
Give your answer in standard form.
Answer = ____________________
(2)
(Total 6 marks)
7.
Sickle cell disease (SCD) is a group of inherited disorders. People with SCD have sickle-shaped
red blood cells. A single base substitution mutation can cause one type of SCD. This mutation
causes a change in the structure of the beta polypeptide chains in haemoglobin.
(a)
Explain how a single base substitution causes a change in the structure of this polypeptide.
Do not include details of transcription and translation in your answer.
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(3)
Page 10 of 19
Haematopoietic stem cell transplantation (HSCT) is a long-term treatment for SCD. In HSCT, the
patient receives stem cells from the bone marrow of a person who does not have SCD. The
donor is often the patient’s brother or sister. Before the treatment starts, the patient’s faulty bone
marrow cells have to be destroyed.
(b)
Use this information to explain how HSCT is an effective long-term treatment for SCD.
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(3)
Page 11 of 19
A new long-term treatment for SCD involves the use of gene therapy.
The diagram shows some of the stages involved in this treatment in a child with SCD.
(c)
Some scientists have concluded that this method of gene therapy will be a more effective
long-term treatment for SCD than HSCT. Use all the information provided to evaluate this
conclusion.
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(3)
(Total 9 marks)
Page 12 of 19
Mark schemes
1.
(a)
(i)
Does not code for amino acid/tRNA/rRNA;
Accept ‘does not code for production of protein/polypeptide’
Reject ‘that produces/makes amino acid'
1
(ii)
Deletion mutation;
Accept ‘deletion’
Ignore references to splicing
1
(b)
(The) polymerase chain reaction;
Accept PCR
1
(c)
1.
2.
Probes are single stranded / have a specific base sequence;
Complementary base sequence on (specific) spacer
OR
3.
4.
Complementary/specific to (particular) spacer;
(In white squares probe) binds (to single-stranded spacer) and
glows/produces light/fluoresce;
2. Need idea of complementary to spacer
3. Accept converse for dark squares
3
(d)
1.
2.
To see if strain is resistant to any antibiotics;
So can prescribe effective/right antibiotic;
OR
3.
4.
To see whether (any) vaccine works against this strain/ see
which vaccine to use/ to produce specific vaccine;
(So) can vaccinate potential contacts/to stop spread;
OR
5.
6.
Can test other people to see if they have the same strain/ to
trace where people caught TB;
Allowing control of spread of disease/vaccinate/treat contacts
(of people with same strain) before they get TB;
Do not allow mix and match of points from different alternative pairs
2 max
[8]
Page 13 of 19
2.
(a)
1.
(Requires DNA fragment) DNA polymerase, (DNA) nucleotides and primers;
2.
Heat to 95 °C to break hydrogen bonds (and separate strands);
Accept temperature in range 90 to 95 °C.
3.
Reduce temperature so primers bind to DNA/strands;
Accept temperature in range 40 to 65 °C.
4.
Increase temperature, DNA polymerase joins nucleotides (and repeat method);
Accept Taq polymerase for DNA polymerase.
Accept temperature in range 70 to 75 °C.
4
(b)
1.
(Initially) number (of molecules) doubling is low
OR
Doubles each cycle to produce exponential increase;
First alternative relates to idea of low numbers i.e., 2, 4, 8, 16, 32
etc.
2.
Plateaus as no more nucleotides/primers;
Accept ‘levels out’ or ‘flattens’ for plateaus.
Accept enzyme/polymerase (eventually) denatures.
2
[6]
3.
(a)
1.
(Short) single strand of DNA;
2.
Bases complementary (with DNA/allele/gene);
2
(b)
1.
Restriction endonuclease/enzyme;
2.
(Cuts DNA at specific) base sequence
OR
(Breaks) phosphodiester bonds
OR
(Cuts DNA) at recognition/restriction site;
Accept palindromic sequence.
2
(c)
(So DNA) probe binds/attaches/anneals;
1
(d)
1.
(Lane 1 has DNA fragments) of known sizes/lengths;
2.
Compare (position of viral fragment/s);
2
Page 14 of 19
(e)
3, 4, 5 with these numbers in any sequence;
All three numbers required.
Reject if more than three numbers given.
1
[8]
4.
(a)
1.
Adenylate cyclase activated / cAMP produced / second messenger produced;
2.
Activates enzyme(s) (in cell so) glycogenolysis / gluconeogenesis occurs /
glycogenesis inhibited;
2. Neutral: ‘glucose produced’ as given in the question stem
Accept: correct descriptions of these terms
2
(b)
(i)
1.
Glucose / sugar in food would affect the results;
1. Accept references to starch / carbohydrate
Or
2.
Food / eating would affect blood glucose (level);
Or
3.
(Allows time for) blood glucose (level) to return to normal;
3. Neutral: allows time for insulin to act
1 max
(ii)
Type 2 diabetes is a failure to respond to insulin / still produces insulin / is not
insulin-dependent;
1
Page 15 of 19
(iii)
(For) – 3 max
A maximum of three marks can be awarded for each side of the
argument
1.
Avoids injections / pain of injections;
2.
Long(er) lasting / permanent / (new) cells will contain / express gene;
Ignore references to methodology e.g. sample size not known
3.
Less need to measure blood sugar / avoids the highs and lows in blood
sugar;
4.
Less restriction on diet;
(Against) – 3 max
5.
Rats are different to humans;
6.
May have side effects on humans;
6. Accept: virus may be harmful / disrupt genes / cause cancer
7.
Long(er) term effects (of treatment) not known / may have caused effects
after 8 months;
8.
(Substitute) insulin may be rejected by the body;
4 max
[8]
5.
(a)
Will replace themselves / keep dividing / replicate;
Undifferentiated / can differentiate / develop into other cells / totipotent / multipotent /
pluripotent;
Accept tissues
2
(b)
Reverse transcriptase;
Allow phonetic spelling
1
(c)
(i)
Alters base / nucleotide sequence / causes frame shift;
Different sequence of amino acids in polypeptide / protein / primary structure
alters the tertiary structure;
Accept any reference, such as adding bases, to changing the base
sequence of the gene. Reject deletion / substitution.
Idea of sequence essential so not makes different amino acids.
Accept answers involving stop / start codons and effect on protein.
2
Page 16 of 19
(ii)
Affects tumour suppressor gene;
Inactivates (tumour suppressor) gene;
Rate of cell division increased / tumour cells continue to divide;
Ignore answers relating to oncogenes. May gain third point.
2 max
(d)
Yes
SCID patients unlikely to survive / quality of life poor unless treated;
Cancer that develops is treatable / only affects 25% / five children;
No
Risk of developing cancer is high / 25%;
Cancer may recur / may not be treated successfully in future / only short time scale
so more may develop cancer;
No mark for yes or no. Marks are for supporting argument based on
biological reasoning.
Accept any points
2 max
[9]
6.
(a)
Box 2.
An inversion will result in a change in the number of DNA bases.
Reject if more than one box with tick. Ignore crossed-out ticks
1
(b)
1.
(Increased) methylation (of tumour suppressor genes);
Accept abnormal methylation or hypermethylation
Ignore decreased acetylation of histones
2.
Mutation (in tumour suppressor genes);
3.
Tumour suppressor genes are not transcribed/expressed
OR
Amino acid sequence/primary structure altered;
Accept mRNA for transcription/transcribed
Accept tertiary structure altered
Accept different amino acid
Ignore reference to protein not being formed
4.
(Results in) rapid/uncontrollable cell division;
Accept cell division cannot be regulated
Ignore growth
3 max
Page 17 of 19
(c)
1.
Correct answer of 1.9/1.93 × 1025 = 2 marks;;
Accept 2 × 1025 = 2 marks
Ignore any numbers after 1.93
2.
Incorrect answer but shows 84 = 1 mark
OR
28 × 3 = 1 mark
OR
Incorrect answer but shows 672 divided by 8 = 1 mark;
2
[6]
7.
(a)
1.
Change in (sequence of) amino acid(s)/primary structure;
Reject amino acids are formed.
Reject amino acids code.
2.
Change in hydrogen/ionic/disulfide bonds;
3.
Alters tertiary/30 structure;
Reject active site.
Ignore quaternary.
Ignore 3D.
3
(b)
1.
Produce healthy (red blood) cells
OR
Produce (normal) polypeptide/haemoglobin;
Produce only healthy (red blood) cells is only equivalent
to mark point 1.
Accept produce ‘normal’/non-SCD cells.
Ignore type of stem cell e.g. pluripotent.
2.
No sickle/faulty/SCD (red blood) cells (produced)
OR
No defective polypeptide/haemoglobin;
3.
Stem/marrow cells (continuously) divide/replicate
OR
Less chance of rejection (from brother/sister);
Differentiate is not equivalent to divide/replicate.
Ignore type of stem cell e.g. pluripotent.
3
Page 18 of 19
(c)
Max 2 marks for marking points 1, 2 and 3
(For gene therapy)
1.
No destruction of bone marrow
OR
No destruction of stem cells;
Accept no destruction of faulty bone marrow unless
context indicates this is against gene therapy.
2.
Donors are not required;
Stating ‘only own cells used’ is not equivalent.
3.
Less/no chance of rejection (own stem cells);
(Against gene therapy)
4.
Sickle/faulty (red blood) cells still produced
5.
Immune response against genetically modified cells/virus
OR
Long-term effect not known (as is new treatment)
OR
Virus could cause side effects;
Accept ‘virus could cause problems’ or ‘risk(s) with virus’.
3 max
[9]
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