1. EcoR1 is a restriction enzyme that will cut DNA at a certain
base sequence to make DNA fragments suitable for gel
electrophoresis. (a) Below is a diagram showing part of a DNA
molecule that has been cut with EcoR1 into two sections called P
and Q. The four bases at the cut (sticky) end of section P have
been labelled. (i) Name a component of the DNA molecule found in
the part labelled X.
................................................................................................................................
(1) (ii) State the letter of the four complementary bases for the
sticky end of section P.
................................................................................................................................
................................................................................................................................(2)
(iii) Name the base T.
................................................................................................................................(1)(b)
EcoR1 cut a piece of DNA which is shown below.
(i) The letters bp stand for base pairs when referring to DNA.
Name the type of bond that joins two bases together to form a base
pair.
................................................................................................................................(1)
(ii) A gel electrophoresis study was undertaken with 5 samples of
DNA. Each sample was made up of a 5500bp section of DNA that had
been mixed with two restriction enzymes. These enzymes cut the DNA
section into smaller fragments. The restriction enzymes were
different in each of the five samples. The results of the study are
shown in the diagram of the gel electrophoresis plate below.
In one sample the restriction enzymes used were Hind and Bam HI.
These enzymes cut the DNA section as shown below.
Use both pieces of information to choose the sample that
correctly represents the DNA mixed with Hind and Bam HI. Sample
number: ................................
(1) (Total 6 marks)Jun 2008 6131Ans6(a)(i) {phosphate /
phosphoric acid } / {deoxyribose / pentose / eq} (sugar) ; (1)
6(a)(ii) AA ; TT ; (2) 6(a)(iii) thymine ; (1) 6(b)(i) hydrogen / H
(bond) ; (1) 6(b)(ii) (sample) 4 ; (1)
2. 8 Huntingtons disease is a genetic condition that leads to a
loss in brain function. The gene involved contains a section of DNA
with many repeats of the base sequence CAG. The number of these
repeats determines whether or not an allele of this gene will cause
Huntingtons disease.l An allele with 40 or more CAG repeats will
cause Huntingtons disease.l An allele with 36 39 CAG repeats may
cause Huntingtons disease.l An allele with fewer than 36 CAG
repeats will not cause Huntingtons disease.The graph shows the age
at which a sample of patients with Huntingtons disease first
developed symptoms and the number of CAG repeats in the allele
causing Huntingtons disease in each patient.
8 (a) (i) People can be tested to see whether they have an
allele for this gene with more than 36 CAG repeats. Some doctors
suggest that the results can be used to predict the age at which
someone will develop Huntingtons disease.Use information in the
graph to evaluate this
suggestion.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3
marks)(Extra space)
.....................................................................................................................................................................................................................................................................................................................................................................................8
(a) (ii) Huntingtons disease is always fatal. Despite this, the
allele is passed on in human populations. Use information in the
graph to suggest
why.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2
marks)
8 (b) Scientists took DNA samples from three people, J, K and L.
They used the polymerase chain reaction (PCR) to produce many
copies of the piece of DNA containing the CAG repeats obtained from
each person. They separated the DNA fragments by gel
electrophoresis. A radioactively labelled probe was then used to
detect the fragments. The diagram shows the appearance of part of
the gel after an X-ray was taken. The bands show the DNA fragments
that contain the CAG repeats.
8 (b) (i) Only one of these people tested positive for
Huntingtons disease. Which person was this? Explain your
answer.Person
...............................................................................................................................Explanation
........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2
marks)8 (b) (ii) The diagram only shows part of the gel. Suggest
how the scientists found the number of CAG repeats in the bands
shown on the
gel.................................................................................................................................................................................................................................................................................................................................................................................................(1
mark)8 (b) (iii) Two bands are usually seen for each person tested.
Suggest why only one band was seen for Person
L.................................................................................................................................................................................................................................................................................................................................................................................................(1
mark)Total 9 marks Aqa Biology BIOL5Unit 5 Control in cells and in
organismsFriday 22 June 2012 9.00 am to 11.15 am
3. The black mamba is a poisonous snake. Its poison contains a
toxin.The table shows the base sequence of mRNA that codes for the
first two amino acids of this toxin.
Complete the table to show1 (a) (i) the base sequence of the
anticodon on the first tRNA molecule that would bind to this mRNA
sequence (1 mark)1 (a) (ii) the base sequence of the DNA from which
this mRNA was transcribed. (1 mark)1 (b) The length of the section
of DNA that codes for the complete toxin is longer than the mRNA
used for translation. Explain
why.................................................................................................................................................................................................................................................................................................................................................................................................(1
mark)1 (c) A mutation in the base sequence of the DNA that codes
for the toxin would change the base sequence of the mRNA.Explain
how a change in the base sequence of the mRNA could lead to a
change in the tertiary structure of the
toxin.................................................................................................................................................................................................................................................................................................................................................................................................(1
mark)1 (d) The black mambas toxin kills prey by preventing their
breathing. It does this by inhibiting the enzyme
acetylcholinesterase at neuromuscular junctions. Explain how this
prevents
breathing.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3
marks)(Extra space)
.....................................................................................................................................................................................................................................................................................................................................................................................Answer
4. Only one strand of the DNA of a gene (the sense strand) is
normally transcribed tomRNA. The complementary strand of DNA is the
antisense strand, which is notnormally transcribed. By inserting a
promoter at the end of the antisense sequence, this forming an
antisense gene, RNA transcription can occur from it. This anti-mRNA
bonds to normal mRNA to form double-stranded RNA (duplex RNA),
which cannot be translated by ribosomes. This technique can be used
to suppress specific genes. The sequence of events is shown in Fig.
2.1.
(a) Explain(i) The bonding of anti-mRNA with normal mRNA to form
duplex
RNA..........................................................................................................................................................................................................................................................................................................................................................................................[3](ii)
Why the resulting duplex RNA cannot be
translated..........................................................................................................................................................................................................................................................................................................................................................................................[3]A
promoter sequence was inserted into the antisense strand of DNA.(b)
Outline how the promoter sequence may have been inserted into the
antisense strand of
DNA.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................[5]A
recently identified gene (pTOM13) in tomato plants is expressed
during the ripening of tomato fruits.Tomato plants were transformed
by adding an antisense gene to pTOM13 to the normal genome.
Antisense genes, once introduced, are inherited. When transformed
plants that each contain one antisense gene are crossed, the
offspring include plants with no antisense gene, one antisense gene
and two antisense genes in a 1:2:1 ratio.Ethene production during
fruit ripening was measured in plants with one and with two
antisense genes, and in normal plants. The results are shown in
Fig. 2.2.
(c) Suggest:(i) a function of the protein encoded by
pTOM13........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................[2](ii)
a reason for the difference in ethene production between
transformed plantswith one antisense gene and those with two
antisense
genes.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................[2](d)
Explain how two plants, each with an antisense gene added to the
normal genome, can produce offspring with no antisense gene, one
antisense gene and two antisense genes in a 1:2:1
ratio.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................[3](e)
Suggest one benefit and one hazard of the genetic engineering of
plants.benefit
............................................................................................................................................................................................................................................................hazard..................................................................................................................................................................................................................................................[2][Total
: 20]5. 5 All the living affected individuals in a family were
found to have a mutation in the gene locus coding for a kinase
enzyme. DNA profiles of the same part of the normal and mutant
alleles of the gene are shown in Fig. 5.1. Such profiles could form
the basis of genetic screening for the condition.
(a) Explain how such DNA profiles are produced.(In this
question, 1 mark is available for the quality of written
communication.)..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
[9](a) (i) two mRNAs are complementary / shown in diagram;bases
H-bond;base pairing;A-U and C-G [3 marks max](ii) no exposed bases
/ binding sites blocked;cannot bind to, ribosomes/rRNA;cannot bind
tRNA;no start signal;so no protein produced [3 marks
max]www.XtremePapers.netSpecimen Materials OCR 2000Biology Oxford,
Cambridge and RSA Examinations145(b) restriction enzyme / named
restriction enzyme;both DNAs cut with same enzyme;sticky ends;DNAs
join by complementary sticky ends;nucleotides added to make sticky
ends;phosphate-sugar backbones sealed;ligase [5 marks max](c) (i)
enzyme;involved in ethene production / other sensible suggestion;[2
marks max](ii) antisense gene blocks pTOM13;antisense gene not
expressed as rapidly as normal gene;with 1 copy not all mRNA
duplexed;greater chance that with 2 copies of antisense gene;if
even a small quantity of enzyme produced it will have a
detectableeffect;ref partial/incomplete/co-dominance [2 marks
max](d) gene added to one (strand of DNA of) one chromosome;plant
(effectively) heterozygote;chromosomes segregate in meiosis;two
types of gamete;shown in diagram [3 marks max](e) one sensible
benefit e.g. slows ripening / increases keeping time;one sensible
hazard e.g. inherited so passed to other tomatoes and ruinripening
/ very difficult to contain [2 marks max][Total: 20]5 (a) Quality
of written communication assessed in this answer.source of DNA
sample;gene/chromosome/DNA, cut up;by restriction
enzyme/endonuclease;separate fragments of different sizes;by
electrophoresis; samples in wells at, one / cathode end;fragments
move different distances;migrate to anode;shortest, furthest
distance; ref to chargetransfer via Southern
Blotting;description;bands invisible;(incubate) with radioactive
probe;or use stain;autoradiograph/expose to photographic
plate;final banding pattern / ref to bands in Fig. 5.1 [8 marks
max]Q clear, well organised answer using specialist terms [1
mark][9 marks max]
6. The main protein in yoghurt is casein. The synthesis of
casein in cells is shown in the diagram below.
(i) At each stage, one or more nucleic acids is involved. Give
the nucleic acid or nucleic acids involved at each stage on the
dotted line next to the box. The first one has been done for
you.
(4) (ii) Name the site within a cell where stage 3 occurs.
................................................................................................................................(1)7.
(a) Scientists can use restriction enzymes to cut a sample of DNA
into shorter pieces.The lengths of the DNA pieces can be measured
in units called base pairs.Explain why a base pair is a suitable
unit for measuring the length of a piece of
DNA.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
(2 marks)8 (b) Sickle cell anaemia is caused by a point mutation in
the haemoglobin gene. Doctors have developed a test for the allele
causing sickle cell anaemia using a restriction enzyme called Ddel.
The allele for normal haemoglobin, HA, can be cut by Ddel.The
allele for sickle-cell haemoglobin, HS, cannot be cut by Ddel.The
method used by the doctors involves the following steps.1. Take
samples of cells from a person and extract the DNA.2. Treat the DNA
with restriction enzymes other than Ddel to cut the DNA into many
fragments.3. Carry out PCR (the polymerase chain reaction) using
primers that are specific for part of the haemoglobin gene.4. Add
Ddel enzyme to the DNA produced by the PCR.5. Find the lengths of
the resulting DNA pieces.8 (b) (i) What is a
primer?................................................................................................................................................................................................................................................................(1
mark)8 (b) (ii) How can a primer be specific to part of the
haemoglobin gene (step
3)?................................................................................................................................................................................................................................................................(1
mark)8 (b) (iii) Why is it necessary to carry out PCR on the
extracted DNA (step
3)?................................................................................................................................................................................................................................................................(1
mark)8 (c) Mustapha and Shahira are both carriers of sickle cell
anaemia. Shahira is pregnant and they want to know if their baby
will have sickle cell anaemia.The doctors tested samples of DNA
from Mustapha, from Shahira and from the fetus.They used the method
described in part (b) to test for the allele responsible for sickle
cell anaemia. The primers used in this method were specific for the
base sequences that occur 55 base pairs before the point of
mutation and 55 base pairs after the point of mutation in the
haemoglobin gene.The table shows the results.
8 (c) (i) Explain the results of the tests on Mustaphas and
Shahiras
DNA.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
(3 marks)Total 10 marks
AqaHuman Biology HBIO4Unit 4 Bodies and cells in and out of
controlFriday 15 June 2012 9.00 am to 11.00 am7.The polymerase
chain reaction (PCR) is a means of generating large numbers of
identical copies of a sample of DNA. PCR technology allows
scientists to remove tiny samples of DNA from a single hair
follicle or white blood cells in a drop of blood and then make
numerous copies of it. This process can be very useful in forensic
science. The steps involved in PCR technology are shown in the
diagram below.
(a) Suggest how heating of the original DNA sample to 94C (step
1) would bring about separation of the two strands of DNA.
..........................................................................................................................................................................................................................................................................................................................................................................................
[1] (b) In step 2, small lengths of single stranded DNA known as
primers are added to the sample to stop the two sides of the sample
DNA molecule from rebinding with each other. Why is it important
that the sample DNA now remains as two separate
strands?............................................................................................................................................................................................................................................................[1]
(c) In step 3, describe how the enzyme DNA polymerase would produce
new double stranded copies using the single DNA strands and the
pool of nucleotides provided.
........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
[2] (d) Suggest why it is important that some of the DNA copies
manufactured are recycled in step 4.
............................................................................................................................................................................................................................................................(1)In
criminal cases, DNA found on the victim can undergo PCR and
eventually allow the production of a DNA fingerprint. The figure
below illustrates the DNA fingerprint from a victim, the DNA
specimen found on the victim and the DNA fingerprints of three
potential suspects.
(e) (i) Use the DNA fingerprints to identify the guilty suspect.
..............................................................................................................................(1)(ii)
Why is it important to include the victims DNA fingerprint for
analysis?
..............................................................................................................................(1)12.
2 The diagram below shows a stage in the synthesis of a part of a
polypeptide. (a) Name the organelle illustrated in the diagram. [1]
(b) Using the information in the diagram above determine: (i) the
m-RNA codon at 1 (ii) the t-RNA anticodon at 2 (iii) the type of
bond formed at 4 [3]
[7] (a) Heating breaks down hydrogen bonds/bonds holding the two
strands together/bonds between the bases; (b) Each strand acts as a
template against which free nucleotides combine (as in
semi-conservative model); (c) Any two from nucleotides are arranged
in place opposite the exposed bases on each strand according to the
complementary base pairings (A-T, C-G, T-A, G-C) the complementary
bases are held together by hydrogen bonds the sugar of one
nucleotide is joined to the phosphate group of the next nucleotide
a condensation reaction forms the bond (d) This ensures that
multiple copies of the DNA are manufactured; (e) (i) Suspect one;
(ii) To ensure that the DNA specimen is not that of the victim; 12
answer(a) ribosome; (b) (i) CCU; (ii) UUC; (iii) peptide bond;