1.
(a)
The inheritance of the ability to produce hydrogen cyanide is controlled by two genes
which are located on different chromosomes. The dominant allele of one gene, G, controls
the production of enzyme G which converts a precursor to linamarin. The dominant allele
of the other gene, E, controls the production of enzyme E which converts linamarin to
hydrogen cyanide. This is summarised in the diagram.
Precursor
Allele G
Enzyme G
Linamarin
Allele A
Enzyme E
Hydrogen cyanide
Explain why plants homozygous for the allele g will not produce hydrogen cyanide when
their tissues are damaged.
(2)
Recently a strain of genetically engineered clover has been developed which has a high
concentration of proteins rich in sulphur-containing amino-acids. A piece of DNA was
prepared which contained the three different genes. This was inserted into a clover plant.
Gene 1 obtained from sunflower seeds. This gene codes for
a protein rich in sulphur-containing amino acids.
Gene 2 ensures that the protein rich in sulphur-containing
amino acids is produced in leaf cells.
Gene 3 prevents this protein being digested in the rumen
of sheep.
(b)
The copy of Gene 1 used in this experiment was obtained from the mRNA of the sunflower
seeds.
(i)
Explain how enzymes could be used to obtain the gene from the mRNA.
(3)
(ii)
Explain why it would be an advantage to obtain the gene from the mRNA rather than
from the DNA of the sunflower seeds. (2)
(Total 7 marks)
2.
One technique used to determine the sequence of nucleotides in a sample of DNA is the Sanger
procedure. This requires four sequencing reactions to be carried out at the same time.
The sequencing reactions occur in four separate tubes. Each tube contains

a large quantity of the sample DNA

a large quantity of the four nucleotides containing thymine, cytosine, guanine and adenine

DNA polymerase

radioactive primers
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A modified nucleotide is also added to each tube, as shown in Figure 1.
Tube 1
A small quantity of
nucleotides containing
modified thymine
Tube 2
A small quantity of
nucleotides containing
modified cytosine
Tube 3
A small quantity of
nucleotides containing
modified guanine
Tube 4
A small quantity of
nucleotides containing
modified adenine
Figure 1
(a)
A large quantity of the DNA sample is required for this procedure. Name the reaction used
to amplify small amounts of DNA into quantities large enough for this procedure.
......................................................................................................................................(1)
(b)
Explain the reason for adding each of the following to the tubes.
(i)
DNA polymerase
.............................................................................................................................
.............................................................................................................................(1)
(ii)
Primers
.............................................................................................................................
.............................................................................................................................(1)
(c)
(i)
When a modified nucleotide is used to form a complementary DNA strand, the
sequencing reaction is terminated. Suggest how this sequencing reaction is
terminated.
.............................................................................................................................
.............................................................................................................................(1)
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(ii)
A sample of DNA analysed by this technique had the following nucleotide base
sequence.
T
G G
T
C
A
C
G A
Give the base sequence of the shortest DNA fragment which would be produced in
Tube 2.
.............................................................................................................................(1)
(d)
A different sample of DNA was then analysed. The DNA fragments from the four tubes
were separated in a gel by electrophoresis and analysed by autoradiography. Figure 2
shows the banding pattern produced.
–ve
T
C
G
A
Direction of
movement of
fragments
+ve
(i)
Explain why the DNA fragments move different distances in the gel.
...........................................................................................................................
...........................................................................................................................(1)
(ii)
What makes the DNA fragments visible on the autoradiograph?
...........................................................................................................................
...........................................................................................................................(1)
(iii)
Use Figure 2 to determine the sequence of nucleotides in this sample of DNA.
........................................................................................................................... (1)
(Total 8 marks)
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3.
Read the following passage.
One aim of cancer therapy is to find a magic bullet that seeks out and kills tumour cells but
leaves normal cells unharmed. For this to work, the bullet needs to be able to recognise a
difference between the two types of cell.
5
10
Some tumours grow so fast that they outgrow their blood supply and the oxygen concentration
in their cells falls. Drugs are being developed that are only effective once they reach the low
oxygen conditions inside a tumour cell. Here enzymes called reductase enzymes activate the
drug which then kills the cell.
Professor Stratford and his colleagues at Manchester are taking advantage of the fact that the
P450 reductase gene is only switched on in an environment which is low in oxygen. His team
have constructed the piece of DNA which is shown in the diagram.
Region of DNA which switches
gene on in low oxygen
concentrations
P450 reductase gene
Gene coding for protein which
acts as a marker on plasma
membrane
This piece of DNA was injected into breast cancer cells and the cells were grown in the
laboratory. The marker protein was used to identify cells with the injected gene. When the oxygen
concentration was reduced, the concentration of P450 reductase increased.
Use information from the passage and your own knowledge to answer the following questions.
(a)
Apart from the rates at which they grow, give one way in which tumour cells differ from
normal cells.
.....................................................................................................................................
.....................................................................................................................................
(1)
(b)
Explain why the oxygen concentration in tumour cells may fall (lines 4 - 5).
.....................................................................................................................................
.....................................................................................................................................(1)
(c)
Explain why the drugs mentioned in this passage do not kill normal cells.
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................(2)
(d) Name the type of enzyme used to:
(i)
remove the P450 reductase gene from a length of DNA;
...........................................................................................................................1)
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(ii)
join the three pieces of DNA together.
...........................................................................................................................(1)
(e)
(i)
The investigators added the gene coding for the protein which acts as a marker on the
plasma membrane to their specially constructed piece of DNA. Explain why.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................(2)
(ii) Some antibodies fluoresce when illuminated with ultraviolet light. Suggest why
these antibodies could be used to identify the cells which had the marker protein on
their plasma membranes.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................(2)
(f)
Describe the parts played by mRNA and tRNA in producing a molecule of a protein such as
P450 reductase.
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................(5)
(Total 15 marks)
4.
(a)
Plasmids are often used as vectors in genetic engineering.
(i)
What is the role of a vector?
...........................................................................................................................
...........................................................................................................................(1)
(ii) Describe the role of restriction endonucleases in the formation of plasmids that
contain donor DNA.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................(2)
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(iii)
Describe the role of DNA ligase in the production of plasmids containing donor
DNA.
...........................................................................................................................(1)
(b)
There are many different restriction endonucleases. Each type cuts the DNA of a plasmid at
a specific base sequence called a restriction site. The diagram shows the position of four
restriction sites, J, K, L and M, for four different enzymes on a single plasmid. The
distances between these sites is measured in kilobases of DNA.
4 kb
10 kb
J K
L
50 kb
M
36 kb
1 kb = 1 kilobase
The plasmid was cut using only two restriction endonucleases. The resulting fragments
were separated by gel electrophoresis. The positions of the fragments are shown in the
chart below.
Position of fragment
100
Direction
90
80
70
60
Length of
50
fragment / kb
40
30
20
10
0
(i)
Which of the restriction sites were cut?
...........................................................................................................................(1)
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(ii)
Explain your answer.
...........................................................................................................................
...........................................................................................................................(1)
(Total 6 marks)
5.
The diagram shows how insulin can be made using genetically modified bacteria.
Human cell which
secretes insulin
Bacterium
Plasmid
Plasmid cut
DNA in
nucleus
mRNA extracted
from cell
Human insulin
gene (DNA)
Marker gene
Bacterial plasmid
containing human
insulin gene
Plasmid
re-introduced
into bacterium
(a)
(i)
The human insulin gene is obtained from mRNA, rather than DNA. Suggest why.
...........................................................................................................................
...........................................................................................................................(1)
(ii)
Name the enzyme used to make a single-stranded DNA copy of the mRNA coding
for insulin.
........................................................................................................................... (1)
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(iii)
The table shows a sequence of bases from the mRNA coding for insulin.
Complete the table to show the sequence of bases you would expect in the
single-stranded DNA copy.
mRNA base sequence
U
C
A
A
C
C
DNA base sequence
(1)
(b) What is the role of DNA ligase in producing genetically-modified bacteria?
.....................................................................................................................................
.....................................................................................................................................(1)
(c)
The plasmid contains a marker gene coding for antibiotic resistance. Explain the
importance of this marker gene.
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................(2)
(Total 6 marks)
6.
(a)
Describe how a particular gene can be removed from the DNA of an animal cell.
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................(2)
(b) Describe how this gene can then be inserted into the genetic material of a bacterium.
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................(4)
(Total 6 marks)
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7.
Bacteria can be genetically modified to produce insulin for human use. To achieve this, human
insulin genes are transferred into bacteria. Plasmids containing two antibiotic resistance genes,
one coding for resistance to tetracycline and one for resistance to ampicillin, are used to carry out
this transfer.
A restriction enzyme was used to cut up the human DNA and plasmids. Figure 1 shows the
different fragments of human DNA and the type of cut plasmid that was produced.
Ampicillin
resistance
gene
Insulin gene
Tetracycline
resistance
gene
Human DNA
Cut
Human DNA
fragments
Cut plasmid
Figure 1
(a)
Describe a plasmid.
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................(2)
(b) Suggest why the restriction enzyme has cut the human DNA in many places but has cut the
plasmid DNA only once.
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................(2)
The fragments of human DNA and the cut plasmids were mixed together with DNA ligase.
Several types of plasmid were formed. Some contained human DNA in the centre of the gene
coding for resistance to tetracycline. The different types of plasmid are shown in Figure 2.
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Ampicillin
resistance
gene
Tetracycline
resistance gene
Plasmid containing
no extra DNA
Ampicillin
resistance
gene
Ampicillin
resistance
gene
Insulin
gene
Plasmid containing
the insulin gene
Human
DNA
Plasmid containing human DNA
fragment without insulin gene
Figure 2
(c)
Explain what causes several types of plasmid to be formed.
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................(2)
(d)
The plasmids are mixed with the bacteria. Some bacteria take up the plasmids.
(i)
Explain how it is possible to distinguish between bacteria which have taken up a
plasmid with human DNA and those which have taken up a plasmid without any
extra DNA.
..........................................................................................................................
..........................................................................................................................
..........................................................................................................................
..........................................................................................................................
..........................................................................................................................
..........................................................................................................................
..........................................................................................................................
..........................................................................................................................(4)
(ii) How is it possible to determine which bacteria have taken up the human insulin
gene?
..........................................................................................................................
....................................................................................................(1)(Total 11 marks)
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8.
Read the following passage.
5
Large numbers of possums in New Zealand are eating crops and spreading disease between
cattle. The use of shotguns and poisons by farmers has not greatly reduced possum numbers.
A better solution to the possum problem may have been found. A crop of carrots has been
genetically modified to produce a ‘sterility protein’. This sterility protein prevents possums
producing offspring.
10
First, scientists identified the gene that codes for this sterility protein. Several copies of the
sterility gene were cut out from long sections of DNA using a special enzyme. The same
enzyme was also used to cut open plasmids which had been removed from bacterial cells. A
different enzyme joined together the ‘sticky ends’ of a plasmid and of a sterility gene to
produce a recombinant plasmid.
15
The scientists then tried to put these plasmids back into the bacteria. Each plasmid also
contained a gene, giving resistance to an antibiotic which normally kills bacteria. Because of
this resistance gene, the scientists could identify bacteria containing the sterility gene and
isolate them from bacteria which had not taken up this gene. Finally, carrot seedlings were
sprayed with bacterial cells. The plasmids entered the carrot seedlings and carried copies of
the sterility gene into the DNA of carrot cells.
The genetically modified crop will be harvested and the carrots scattered across land populated by
possums.
Use information from the passage and your own knowledge to answer the following questions.
(a)
Name the type of enzyme used to
(i)
cut out the sterility gene (line 7)
...........................................................................................................................
...........................................................................................................................(1)
(ii)
join together the plasmid and the sterility gene (lines 9 - 10)
...........................................................................................................................
...........................................................................................................................(1)
(b)
Explain the meaning of the term ‘sticky ends’ (line 9)
.....................................................................................................................................
.....................................................................................................................................(2)
(c)
In this procedure the bacterial plasmids acted as vectors. Explain the function of a vector.
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................(2)
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(d)
Explain how the presence of the antibiotic resistance gene allowed scientists to identify and
isolate the bacteria which contain the sterility gene (lines 12 - 14)
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................(3)
(e)
Explain the arguments for and against using the genetically modified carrots to reduce the
population of possums
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................(6)
(Total 15 marks)
9.
A gene was broken into fragments using enzyme Z. The mixture of fragments produced was then
separated by electrophoresis.
(a)
What type of enzyme is enzyme Z?
............………..............................................................................................................(1)
The table shows the number of base pairs present in the fragments.
Fragment
Number of base pairs (× 103)
1
4.65
2
5.72
3
10.71
4
2.39
5
5.35
6
7.53
The diagram shows the electrophoresis gel used. The mixture of fragments was placed at the start
point marked S and the process started. The boxes indicate the positions reached by the different
fragments.
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Direction of movement of fragments
S
(b)
Explain why base pairs are a suitable way of measuring the length of a piece of DNA.
............………..............................................................................................................
............………..............................................................................................................
............………..............................................................................................................
............………..............................................................................................................(2)
(c)
(i)
Write 6 above the appropriate box on the diagram to show the position you would
expect fragment 6 to have reached.
(1)
(ii)
Explain how you arrived at your answer.
...........................................................................................................................
...........................................................................................................................(1)
(d)
Enzyme Z recognises a particular sequence of bases in the gene. How many times does this
sequence appear in the DNA of this gene?
............………..............................................................................................................(1)
(Total 6 marks)
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10.
-thalassaemia is a genetic condition in which abnormal haemoglobin is produced. In one form,
the recessive allele for -thalassaemia, t, differs from the normal allele, T, by a single base-pair.
A radioactive DNA probe was used to investigate the genotypes of four members of one family.
The flowchart summarises the technique involved.
DNA samples extracted and cut into fragments using a
restriction enzyme
Fragments separated from each other by electrophoresis
One region of the resulting gel was blotted with two
pieces of filter paper. The first was soaked in a solution
containing a radioactive DNA probe for the normal allele.
The second was soaked in a solution containing a
radioactive DNA probe for the -thalassaemia allele.
Surplus probe washed off
The diagram below shows the appearance of the two pieces of filter paper which resulted from the
investigation.
Father
Mother
First
child
Second
child
Filter paper soaked
with probe for
normal allele
Filter paper soaked
with probe for
-thalassaemia
allele
(a)
What is the probability that the next child that this couple have is a girl who has
-thalassaemia? Explain your answer.
............………..............................................................................................................
............………..............................................................................................................
............………..............................................................................................................
............………..............................................................................................................
............………..............................................................................................................
............………..............................................................................................................(3)
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(b)
(i)
The fragment of DNA containing the normal allele and the fragment with the
-thalassaemia allele moved the same distance on the gel. Explain why.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................(2)
(ii)
The allele for -thalassaemia differs from the normal allele by only one base-pair.
Explain why the probe used to identify these alleles consists of a piece of DNA
twenty bases in length and not just one base.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................(2)
(Total 7 marks)
11.
Genetic engineering has made it possible to transfer genes from one species to another. For
example, a gene that gives resistance to herbicide and another gene which gives resistance to
insect attack have been transferred into maize. Some people think that there will be great
advantages in growing maize with these genes. Others are equally convinced that there are
long-term dangers in growing crops of this maize.
Evaluate both of these viewpoints.
...............................................................................................................................................
...............................................................................................................................................
...............................................................................................................................................
...............................................................................................................................................
...............................................................................................................................................
...............................................................................................................................................
...............................................................................................................................................
...............................................................................................................................................
...............................................................................................................................................
...............................................................................................................................................
...............................................................................................................................................
...............................................................................................................................................
...............................................................................................................................................
(Total 6 marks)
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12.
DNA sequencing is a technique used to find the sequence of nucleotides in a sample of DNA.
Enzymes are used to cut the DNA sample into fragments of different lengths.
The ends of these fragments are then labelled using radioactive probes.
Four different probes attach to the end of DNA fragments in which the terminal nucleotide is
adenine, cytosine, thymine and guanine respectively. The labelled fragments are then separated.
The diagram shows apparatus used to separate the DNA fragments and the end result of the
process.
Negative
terminal
Four samples
of radioactively
labelled DNA
fragments are
placed in the
four wells
TCGA
DNA fragments
visible
Gel
Positive
terminal
(a)
(i)
Use the information from the diagram and your knowledge of DNA sequencing
techniques to draw a flow chart for the process. The first box has been completed for
you.
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Four samples of radioactively labelled DNA
fragments are placed in the four wells
(4)
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(ii)
Why do the DNA fragments move different distances in the gel?
...........................................................................................................................
...........................................................................................................................(1)
(iii) This technique is being used to find the DNA sequences of human chromosomes.
Give one advantage to humans of determining this DNA sequence.
...........................................................................................................................
...........................................................................................................................(1)
(b)
Describe how genetic engineering is used to produce alpha-1-antitrypsin which is used to
treat cystic fibrosis.
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
(6)
(Total 12 marks)
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13.
Liposomes are tiny droplets made of lipid molecules. They are sometimes used in gene therapy to
treat people suffering from cystic fibrosis.
(a)
Explain the meaning of gene therapy.
....................................................................................................................................
....................................................................................................................................(1)
(b)
A defective protein causes the symptoms associated with cystic fibrosis.
How does the sequence of amino acids in the defective protein differ from the sequence in
the protein of a healthy person?
....................................................................................................................................
....................................................................................................................................1)
(c)
(i)
Explain how it may be possible for a person with a genetic abnormality to be cured
using gene therapy.
...........................................................................................................................
...........................................................................................................................(1)
(ii)
Explain why a person cured by gene therapy may still have children who suffer from
the abnormality.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................(2)
(Total 5 marks)
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14.
The diagram illustrates the polymerase chain reaction.
Original DNA
sample
Step 1
Step 2
Cycle 1
Step 3
Cycle 2
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(a)
(i)
What method is used to split the original DNA sample into two strands during Step
1?
...........................................................................................................................
........................................................................................................................... (1)
(ii)
Describe what happens during Steps 2 and 3 in order to produce a new DNA strand.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................(3)
(b)
On the diagram, draw the DNA molecules that would be present at the end of Cycle 2. Use
the same method of shading to distinguish between original DNA strands and the new
DNA strands.
(2)
(c)
Describe one example of the use of the polymerase chain reaction.
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................(2)
(Total 8 marks)
Q1. (a)
Do not produce enzyme G/produce non-functional enzyme G;
No linamarin formed;
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21
(b)
(i)
(ii)
Reverse transcriptase;
makes single strand of DNA/cDNA from RNA;
Double strand then formed;
DNA polymerase;
max 3
If from DNA all genes are present in cell;
mRNA from activated genes only/codes for one protein;
DNA has introns/non-coding /junk DNA;
Have been edited out in mRNA;
max 2
[7]
Q2. (a)
(b)
(c)
polymerase chain reaction / PCR;
1
(i)
joins nucleotide together; (not complementary bases)
1
(ii)
enables replication / sequencing to start / keeps strands separate;
1
(i)
(modified nucleotide) does not form bonds/react with other nucleotides;
does not “fit” DNA polymerase/enzyme/active site;
1
(ii)
AC;
1 max
(accept reading from right hand side i.e. TC)
(d)
(i)
different lengths / sizes / mass;
1
(ii)
radioactive primer;
1
(iii)
GAAGTCTCAG;
(accept reading from autoradiogram i.e. CTTCAGAGTC)
1
8]
Q3(a) Lower oxygen concentration/more reductase/
relatively large nucleus/difference in metabolism/
continue to divide when not in contact with another surface;
(b)
(c)
(d)
(e)
High respiratory rate associated with rapid growth and
insufficient blood supply to supply enough oxygen;
They have enough oxygen present;
therefore reductase enzymes not produced;
inactive drug is harmless;
1
Max 2
(i)
Restriction enzyme/endonuclease;
1
(ii)
Ligase;
1
(i)
The marker could be identified easily;
so it would be known which cells contain the piece of DNA;
2
(ii)
(f)
1
Antibodies are specific;
proteins;
will only attach to marker protein;
correspond in shape;
Max 2
mRNA makes a copy of the DNA code;
during transcription;
chains of DNA separate and mRNA nucleotides;
line up with complementary bases;
tRNA carries specific amino acid;
to correct position on mRNA;
anticodon and codon match;
Max 5
[15]
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Q4. (a)
(b)
(i)
transfer/carry genes from one organism to another/into bacteria/cells;
(ii)
cut open plasmid;
cut donor DNA, to remove gene/length of DNA;
cut donor DNA and plasmid with the same enzyme/enzyme
that cuts at the same base sequence;
sticky ends/(overhanging) ends with, single strand/bases exposed;
association/attachment/pairing of complementary strand;
1
2 max
(iii)
annealing/splicing/backbones joined/phosphodiester bonds;
1
(i)
L and M;
1
(ii)
fragments 64 and 36(kilobases obtained)
1
[6]
Q5.
(a)
(i)
Difficulty of finding one gene among all the genes in the nucleus / large
amounts of mRNA coding for insulin will be present in insulin producing
cells / idea that mRNA will be ‘edited’
1
(ii)
Reverse transcriptase;
1
(iii)
AGTTGG;
1
(b)
Joins the gene for insulin into the plasmid;
1
(c)
Allows transformed bacteria to be separated from non-transformed;
Further detail e.g. transformed bacteria survive
when antibiotic applied to medium;
2
[6]
Q6
(a)
(b)
Locate with the use of a gene probe;
Use restriction enzymes / endonucleases;
To cut at specific base sequence;
By hydrolysing / breaking sugar-phosphate bonds;
2
Same restriction enzymes;
Cut at same base sequence in bacterial DNA;
Leaving sticky ends / hydrogen bonds break;
Join / splice with ligase;
Use of plasmid;
max
max 4
[6]
Q7. (a)
(b)
circular DNA;
separate from main bacteria] DNA;
contains only a few genes;
max
enzymes only cut DNA at specific base sequence/recognition site/specific
point;
sequence of bases/recognition site/specific point (on which enzyme acts)
occurs once in plasmid and many times in human DNA;
(max 1 if no reference to base sequence or recognition site)
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2
23
(c)
(d)
all cut DNA have same/complementary base sequence at ends or
same/complementary sticky ends;
random process by which sticky ends join;
(i)
(ii)
2
replica plating;
use of pad/velvet surface to transfer bacteria;
use of agar plate containing ampicillin/no tetracycline and agar
plate containing tetracycline;
in bacteria with human DNA tetracycline gene no longer functional/
not resistant to tetracycline;
bacteria with human DNA grow on plate with ampicillin/no
tetracycline but are killed by tetracycline;
bacteria with no extra DNA in plasmid not killed,
4 max
use of gene probes;
bacteria with insulin gene produce insulin;
1 max
[11]
Q8.
Quality of written communication should be considered in crediting points in the marking
scheme. In order to gain credit, answers must be expressed logically in clear, scientific terms.
(a)
(b)
(c)
(d)
(e)
(i)
Restriction;
(ii)
Ligase / ligating;
I endonuclease only
Cut ends of DNA;
one strand longer than the other / staggered cut / unpaired bases;
Can attach to complementary DNA / bases;
Transfers / carries (foreign) gene / DNA;
Into bacteria / carrot cell / host cell;
1
1
max 2
2
Acts as a marker;
Both genes on same plasmid / bacteria take up both genes;
Reference to antibiotic;
Kills bacteria without (either) gene / plasmid /
only those with gene/plasmid will grow / survive;
max 3
For
More effective than other methods;
More humane method than shooting;
Poisons may harm other animals;
Prevent spread of disease / destruction of crops;
Economic benefit to farmer / consumer;
max 3
Against
Plasmid / (either) gene may enter another species;
May eradicate / greatly reduce possums;
May sterilise other species;
Disruption of food chain / change in numbers of a species;
Not immediate / only affects next generation;
Reference to antibiotic resistance in bacteria;
Possums resistant to sterility protein;
max 3
[15]
q.9(a) Endonuclease / restriction enzyme;
(b)
DNA made of base pairs;
Each base pair is same length / occupies same distance along backbone;
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2
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(c)
(d)
(i)
Second blank box from left labelled 6;
1
(ii)
Distance moved depends on length / number of base pairs /
second longest fragment / second shortest distance identified;
1
5;
1
[6]
Q10(a)
(b)
Mother and father both heterozygotes / Tt / carriers;
Probability of thalassaemia 1/4 and female 1/2;
Probability of both 1/8;
(i)
(ii)
Cut at same base sequence as same enzyme used;
Fragments are same length / size / have same charge;
Only differs by a single base;
Single base occurs many times;
Sequence of 20 unlikely to occur elsewhere;
Allow one mark for establishing the principle where neither marking
point clearly made.
3
max 2
2
[7]
Q11 positive:
less crops lost to insect damage/ diseases spread by insects;
can spray herbicide with no loss to crop/reduce competition from weeds;
more saleable product;
less use of insecticide;
possibly cheaper food;
max 3
negative:
gene transfer to non-crop species;
consumer resistance to “un-natural” products;
transfer of genes into food chains/effect of food chains/examples;
creation of “plague” weeds/uneconomic plants;
excessive use of herbicides;
Reject disadvantages of selective breeding
max 3
[6]
Q12(a)
(b)
(i)
current switched on / fragments move due to electrical attraction;
several hours to run;
DNA transferred to nylon membrane / ‘southern blot’;
(wrapped) photographic film placed on gel;
film developed / radioactivity darkens film
max 4
(ii)
different lengths / mass;
1
(iii)
e.g. progress towards cure for genetic diseases
1
Quality of written communication.
The answer to this part of the question requires continuous prose.
To gain one mark for Quality of Written Communication these answers should be
presented in clear, scientific English. Technical terminology should have been used
effectively and should usually be accurate.
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maximum 4 marks for generic genetic engineering techniques:
human gene identified
removed from human DNA using endonuclease;
detail e.g. sticky ends;
same endonuclease;
used to cut plasmid;
role of ligase;
max 4
combined with promoter sequence;
gene / DNA (+ promoter sequence) injected into nucleus of fertilised
sheep egg;
detail e.g. micropipette;
embryo inserted into sheep uterus;
enzyme obtained from sheep milk.
Overall max 6
[12]
Q13(a)
replace defective genes/treat genetic diseases with (healthy) genes;
(b)
one amino acid missing/different/changed;
(c)
(i)
(ii)
gene is expressed;
healthy genes replicated with cells so not lost;
gamete cells are not affected/do not take up the healthy gene;
still able to pass on the defective gene;
1
1
1 max
2
[5]
Q14(a)
(b)
(c)
(i)
heat (to about 90C)
1
(ii)
primers / short nucleotide chains / RNA added
individual (DNA) nucleotides then added;
by (DNA) polymerase
3
2 double stranded molecules with original (white) and new (black) DNA
strands;
2 double stranded molecules with new (black) DNA strands
2
provides multiple copies of a DNA fragment;
e.g. to analyse in forensic detection
2
[8]
Q15
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