DNA, RNA, Protein & Gene Technology Questions from VCAA
Exams (includes answers)
2009
Question 8
Using gene therapy to treat a disease involves introducing
A. particular proteins into a person with the disease.
B. various types of blood into a person with the disease.
C. viruses that destroy certain cells of a person with the disease.
D. functional alleles into the cells of a person with the disease.
74% got this correct
\
The following information is relevant for Questions 9 and 10.
Examine the following diagram.
Question 9
The diagram is most likely to be of
A. translation.
B. transcription.
C. gene regulation.
D. DNA replication.
64% got this correct
Question 10
The molecule labelled X represents
A. ligase.
B. a ribosome.
C. RNA polymerase.
D. DNA polymerase.
52% got this correct
Question 11
A mutation is
A. a product of natural selection.
B. caused by immigration and emigration.
C. a change in an allele due to a change in DNA.
D. a random change in gene frequencies from one generation to the next.
77% got this correct
Question 13
Tay Sach’s disease is a genetic condition that can be caused by mutations in the hexoseaminidase
gene resulting in an abnormal enzyme.
The table below shows details for part of a normal and abnormal peptide sequence for
hexoseaminidase.
Using this information it is reasonable to say that the abnormal hexoseaminidase enzyme could
have resulted from a change in the mRNA due to
A. a substitution of G for C at the beginning of codon 7.
B. a duplication of ACU at the beginning of codon 4.
C. an insertion of UAC at the beginning of codon 5.
D. deletion of UCU at the beginning of codon 6.
83% got this correct
Question 17
The process of gene expression involves
A. translation followed by transcription.
B. exons carrying information required for protein synthesis.
C. identical action in every cell of the organism containing the particular gene.
D. tRNA molecules, each carrying four nucleotides that specify one amino acid.
56% got this correct
Question 20
Regulatory and structural genes differ in their arrangement in the genomes of prokaryotic and
eukaryotic cells. In prokaryotic cells, regulatory genes are arranged side by side. This
arrangement is known as an operon. In eukaryotic cells, the regulatory genes and structural genes
may be located on different chromosomes.
Therefore, it would be reasonable to say that
A. eukaryotic stem cells have all genes switched on.
B. the environment has no impact on whether a gene is switched on or off.
C. all bacterial operons are located on a large circular chromosome within the cell.
D. mutations in distant regulatory genes will have no effect on their related structural genes in
eukaryotic cells.
46% got this correct
Question 1
Translation occurs within the cytosol of a cell.
d. Outline the steps that normally occur in translation. Use specific terms and names of the
molecules involved. Name the final product of the process.
All of:
• mRNA travels to the ribosomes where its codons are read
• tRNA carries specific amino acids to the ribosomes or complementary base pairing occurs
between the codons and anticodons
• product; protein/polypeptide.
This question was well answered by many students. However, some students failed to name the
final product, which was specifically asked for in the question. Other students used many
irrelevant terms, enzymes and structures which detracted from their answers.
3 marks
39% got 0, 15% got 1, 16% got 2, 30% got 3
Question 5
A Wagyu breeder discovered a small number of individuals in her elite herd that were suffering
from Chediak- Higashi Syndrome (CHS). CHS is an autosomal recessive condition that can
affect species other than cattle. The breeder required further information.
Gene probing was used to target CHS1, the allele responsible for the condition. The genetic
probes for the Wagyu CHS locus were derived from human alleles.
b. Given that the gene probe for a human works for the Wagyu, what can you infer about the
chemical code for this allele?
The chemical code is identical/very similar.
Some students stated that the code was ‘similar’, which was not precise enough. Answers which
made inferences on evolutionary relationships were not awarded any marks as they were not
addressing the question. Some students added ‘due to the universal nature of DNA’; however,
this was incorrect.
1 mark
43% got this correct
The Wagyu CHS1 allele was isolated and given a fluorescent tag. It was introduced into a yeast
cell as a large, independent, cytoplasmic chromosomal segment called a Yeast Artificial
Chromosome (YAC). In addition to the allelic DNA, a YAC includes a centromere and a
replication sequence. The yeast cells are then incubated in the presence of growth stimulants and
given time to replicate.
This procedure is similar to genetic engineering of bacterial plasmids, however the YAC is able
to contain much larger pieces of DNA than a plasmid.
c. i. In the bacterial cell above, draw a plasmid in the blank box.
A circle was to be drawn
ii. Bacterial plasmids lack a centromere. Why are YACs made with a centromere?
Either of:
• the centromere was required for spindles to attach
• to allow mitosis to occur.
iii. What term describes the process of copying a gene?
Gene cloning
Gene replication and DNA replication were also accepted. PCR and transcription were
incorrect answers and replication was deemed too vague
1 + 1 + 1 = 3 marks
18% got 0, 43% got 1, 30% got 2, 9% got 3
A test was developed to identify each of the normal and mutant alleles. Two cows were chosen
for testing.
Cow X – a cow known to have the autosomal recessive disease CHS
Cow Y – a phenotypically normal cow with a family history of CHS
The CHS locus was isolated from each, amplified and then treated with FokI restriction enzyme
which recognizes the nucleotide sequences
5’ – G G A T G – 3’ and 5’ – C A T C C – 3’
The genotypes at the CHS locus for the two cows are shown in the following figure.
d. i. Explain whether Cow Y is heterozygous or homozygous at the CHS locus.
Both of:
• heterozygous
• it has different alleles (sequences).
To gain the mark students had to state ‘heterozygous’ and identify the reason for their choice.
ii. On the electrophoretic gel diagram below, draw in the band(s) that would accurately show a
profile for an unaffected cow Z with no history of CHS in the family.
Two bands needed to be illustrated:
• near but slightly above 100
• near but slightly above 50.
Some students included extra bands or only one band,
indicating that the DNA had not been cut.
1 + 2 = 3 marks
39% got 0, 25% got 1, 16% got 2, 20% got 3
A farmer suspected that one of his cows was a CHS carrier. He sent a sample of the cow’s hair
follicles for testing. A technician ran a gel of DNA sequences from the hair follicles and obtained
the following result.
e. What mistake must the technician have made in his procedures to obtain this result?
The electrode had been connected the wrong way.
1 mark
48% got this correct
Question 7
Many techniques in molecular biology require the use of probes.
a. What is a probe?
A single-stranded segment of DNA which is (radioactively) labeled.
It was important that students’ answers distinguished a probe from a primer. It was clear that
the concept of a probe was not well understood.
1 mark
21% got this correct
DNA microarray technology, also known as DNA chip technology, allows screening to detect
mutations.
A DNA chip is made of glass and can contain thousands of fields. Each field is like a tiny well in
which reactions can occur.
DNA microarray technology has been used to survey the p53 gene because a mutation of this
gene is present in about 60% of all cancers. The position of a mutation in the p53 gene of a
patient, Patient X, who has breast cancer, can be determined. Steps in the screening procedure
are outlined below.
Step 1 Treat a normal allele of p53 to break it down into nucleotide sequences.
Step 2 Each segment must be tested, one nucleotide at a time. Tests for the first segment of the
allele are outlined below.
Consider nucleotide five [G] in the normal sequence. Sequences of this section are manufactured
so that all possible mutants of base five are formed. Each of these sequences is placed in a
different field.
Step 3 Two solutions are added to each of the fields.
These solutions contain
Solution I Complementary normal strand, labelled with green fluorescent dye.
Solution II Complementary strand from DNA of person with breast cancer, labelled with
red fluorescent dye.
Step 4 Allow time for hybridisation of strands and then wash the DNA chip to remove excess
dyes.
Step 5 Examine fields under UV light to distinguish colours remaining. Interpret results.
The results for Patient X are shown below.
b. What is the function of each of the two different fluorescent dyes used?
Both of:
• green – indicates hybridisation with a normal allele
• red – indicates a specific mutation and hybridisation with that allele have occurred.
2 marks
31% got 0, 43% got 1, 27% got 2
c. What does hybridisation mean?
The joining of complementary DNA from different sources.
1 mark
13% got this correct
d. What mutation resulted in Patient X having breast cancer?
G was replaced by A
A point mutation occurred. The specific type of mutation was a base substitution; however, in
light of the data, the best answer was: G was replaced by A.
1 mark
25% got this correct
A daughter of Patient X was also tested for the first segment of the allele.
e. Would you expect the result of her test to be green, red or black? Explain your answer.
Either of:
• green if she inherited a normal allele from her mother
• red if she inherited the defective allele from her mother.
1 mark
20% got this correct
2008
Question 3
The following diagram summarises the steps involved in the production of a cloned sheep.
The chromosomes in the cells of the cloned sheep will be identical with those in the cells of
A. sheep M.
B. sheep N.
C. sheep P.
D. sheep Q.
92% got this correct
Question 5
Mitochondrial DNA
A. is a linear molecule.
B. is inherited only from the female parent.
C. exists in mammalian cells but not plant cells.
D. contains the same amount of uracil bases as adenine bases.
73% got this correct
Question 9
In the following table, the information that is accurately presented is in row
Answer is D
50% got this correct
Question 17
Biologists have sequenced the genomes of many organisms. The number of genes found in
organisms varies greatly. Some examples are listed in the table below.
From this data it can be concluded that
A. larger organisms have larger genomes.
B. puffer f sh show greater genetic variety than E. coli.
C. a nematode and flowering mustard plant have the same number of chromosomes.
D. the larger the genome of an organism, the greater the number of proteins it produces.
76% got this correct
Question 18
β-cells in the human pancreas produce insulin. mRNA transcripts from these cells can be used to
produce complementary (cDNA) copies of the active gene.
The enzyme needed to produce cDNA from an mRNA transcript is
A. restriction endonuclease.
B. reverse transcriptase.
C. ligase.
D. RNA polymerase.
57% got this correct
Question 19
The diagram below shows part of the translation process of protein synthesis.
Structure X in the diagram contains
A. messenger RNA.
B. ribosomal RNA.
C. transfer RNA.
D. nuclear DNA.
62% got this correct
Question 20
The genome of a small virus is depicted below, showing the positions of cutting sites (P and Q)
for two restriction enzymes.
The length of DNA fragments obtained when using these restriction enzymes is shown in the
table below.
If both EcoR1 and BamH1 are used together on this viral DNA, the length of fragments obtained
would be
A. 3, 8, 5, 2
B. 7, 2, 1
C. 3, 5, 2
D. 3, 7, 8, 2
37% got this correct
Question 6
RFLP (Restriction Fragment Length Polymorphism) analysis is commonly used to determine
genetic variation between individuals. The procedure is summarised below.
In this procedure, scientists select a particular restriction enzyme from an available range.
a. Explain the reason for their choice.
Restriction enzymes cut at a particular base sequence/recognition site.
Answers that referred to cutting at ‘a place or gene locus’ were not awarded a mark.
Many students incorrectly thought that specific restriction enzymes cut DNA into specific
lengths.
1 mark
40% got this correct
Electrophoresis uses electrical current to sort DNA fragments.
b. i. Describe one characteristic of this sorting process.
Either of:
• DNA is negatively charged and moves towards the positive electrode
• shorter DNA fragments travel further/more quickly through a gel.
ii. Explain why the DNA of each individual produces a different pattern of fragments after gel
electrophoresis, even when the same restriction enzyme is used.
DNA of individuals is different, therefore will be cut at different places and produce fragments of
different length.
1 + 1 = 2 marks
29% got 0, 42% got 1, 39% got 2
Examine stages Y and Z.
c. Describe, at the molecular level, what is meant by the term ‘hybridised’. Why is it necessary to
carry out hybridisation?
Both of:
• the RNA probe is complementary to the single stranded DNA sample and these join
together/hybridise
• this enables the DNA to be seen/identified.
Many students treated this question as ‘describe DNA hybridisation’ and in doing so, they
ignored the instruction to
‘examine stages Y and Z’.
2 marks
61% got 0, 27% got 1, 12% got 2
2007
The following information relates to Questions 8, 9 and 10.
The following diagram outlines the production of protein in a cell when DNA is activated.
Question 8
At stage I, the DNA molecule involved has the base sequence
A. TACTACGGCCTCCTGCCTGGGGGACTAACT
B. TACTACGGCCTTCTGCCTCCCGGACTAACT
C. TACTACGGCCTTCTGCCTGGGGGACTAACT
D. TACTACGGCCTTCTGTTTGGGGGACTAACT
92% got this correct
Question 9
In stage I
A. step E represents translation.
B. removal of exons occurs at step F.
C. product G is the initial mRNA produced.
D. the enzyme RNA polymerase is active.
45% got this correct
Question 10
In stage II
A. structure K is made of tRNA.
B. the three bases of group H form an anticodon.
C. bond J represents a hydrogen bond.
D. the mRNA shown will code for a protein containing 10 amino acids.
65% got this correct
Question 12
The normal order of development of features in an embryo is determined by gene sequences. The
relevant groups of genes have been studied extensively in insects and chordates. These groups of
genes are responsible for directing the sequential embryonic development of the head, thorax,
limbs and wings of insects and the head, thorax and limbs in chordates.
In the summary below, gene activity starts at the left and continues in sequence to the right. The
action of each gene in the group promotes the action of the subsequent gene.
The above information suggests that if the action of a particular gene in a sequence was blocked
A. an abnormal insect could develop possessing a head but no legs.
B. an abnormal insect could develop possessing a thorax and legs but no head.
C. an abnormal chordate could develop with a thorax but no head.
D. an abnormal chordate could develop possessing digits but no legs.
83% got this correct
Question 2
Victoria Police forensic scientists conduct DNA profiling using samples taken from crime
scenes. Traces of DNA of less than 1 nanogram can be amplified and then profiled.
a. Name the process which is used to amplify the DNA.
PCR (or Polymerase Chain Reaction)
1 mark
71% got this correct
Below is a diagram showing part of this process.
b. What must be done between stages 1 and 2 to separate the strands of the DNA molecule?
Heating:
• to 90°C or greater
• to dissociate the strands
• to break the hydrogen bonds.
Heating had to be mentioned, and the effect it had on the process.
1 mark
52% got this correct
c. Complete and label the diagram at stage 2.
On the diagram or in the space below
the question, students needed to show
primers anneal, one per strand, at
opposite ends (for the first mark), and
show one of:
• primers attach at 3’ end
• temperature reduced to
approximately 50°C
• DNA/Taq polymerase is added
• propagation occurs away from
the primer.
2 marks
62% got 0, 22% got 1, 16% got 2
Small pieces of DNA of differing length can be compared to determine whether or not a sample
could have come from a particular person. In a case, samples of DNA from the victim and the
crime scene were compared with samples from two suspects.
The DNA samples were treated with restriction enzymes, amplified and run through gel
electrophoresis. The results for one gene locus are shown in the diagram below.
d. Draw an arrow on the right-hand side of the diagram to indicate the direction of movement of
the DNA fragments.
The arrow needed to point towards the top of the page.
Students were required to know that DNA has a negative charge and hence would move towards
the positive electrode.
1 mark
70% got this correct
e. What do the standards consist of, and what is their purpose?
Both of:
• the standard consists of fragments of known length
• they are used to estimate the size of the samples.
To gain the two marks, a link had to be established between the comparison of known to
unknown. Too often students made vague statements such as ‘The standard is used to compare to
the sample’.
2 marks
46% got 0, 23% got 1, 31% got 2
f. From these results, give a conclusion which could be drawn about the sample taken from the
crime scene.
A suitable conclusion could be:
• the sample is not from the victim
• the sample could be from either suspect.
To gain the mark here students had to make a valid conclusion rather than a simple observation
such as ‘The sample matched the suspects’. Some incorrect conclusions included ‘both suspects
must be guilty’, ‘both suspects are identical twins’ or even ‘the victim wasn’t at the crime scene’.
1 mark
39% got this correct
g. What further action would you recommend to the forensic scientists investigating this case?
Any one of:
• apply the same process to a different gene locus
• use another suitable DNA technique, such as DNA sequencing
• use another forensic method, such as blood analysis or fingerprinting.
To gain the mark, a forensic, logical and feasible action had to be recommended. Incorrect
suggestions included ‘check alibis’ and ‘interview the victim’.
1 mark
37% got this correct
Question 3
a. Explain what is meant by gene regulation.
Either of:
• genes are only activated/transcribed when required
• required genes are expressed, which can save energy.
To gain the mark the purpose of gene regulation was required. A common answer that lacked
sufficient information was ‘genes are turned on and off’.
1 mark
21% got this correct
Bacteria require amino acids to produce proteins. For example, bacteria in a human intestine may
absorb amino acids from digested food, but at times there may be a deficiency of a particular
amino acid. If this is the case, the bacteria will produce the necessary amino acid themselves.
The diagram below is a regulation system in a bacterial cell involving the production of the
amino acid tryptophan.
Note that there are two pathways (X and Y). Tryptophan is the regulatory compound in these two
pathways and acts as a repressor in both.
b. Describe the immediate outcome when tryptophan activates pathway X.
The transcription of all genes is stopped or suppressed.
1 mark
18% got this correct
c. Describe the immediate outcome when tryptophan activates pathway Y.
Either of:
• enzyme 1 is inhibited
• compound 1 is not produced.
1 mark
17% got this correct
d. Suggest how the action of tryptophan as a repressor in this system could be of selective
advantage to a bacterial cell in the digestive tract.
Both of:
• when tryptophan is available, its presence prevents further tryptophan being produced
• therefore the cell does not waste energy or resources producing tryptophan.
In parts b., c. and d. students were not awarded the mark if they stated that the gene/enzyme was
repressed, as the stem of the question stated that tryptophan acts as a repressor. However,
students could gain the mark if they indicated in their response that they understood the term; for
example, for part c. ‘tryptophan represses enzyme 1 and stops production of compound 1’.
2 marks
74% got 0, 13% got 1, 13% got 2
2006
Question 1
In eukaryotic organisms genes are
A. composed of DNA.
B. alternative forms of an allele.
C. composed of DNA and protein.
D. the same length as a chromosome.
58% got this correct
Question 2
In prokaryotic organisms
A. translation occurs at the ribosome.
B. transcription occurs in the nucleus.
C. chromosomes are usually linear.
D. DNA is only found in plasmids.
35% got this correct
Question 11
Amplification of DNA in the polymerase chain reaction requires
A. nucleotides of uracil.
B. DNA polymerase.
C. amino acids.
D. ribose sugar.
80% got this correct
Question 20
One section of a polypeptide has the amino acid sequence
Ala . Cys . Lys . Ile . Asn
The codons for these amino acids are:
Ala GCA GCC GCG GCU
Cys UGC UGU
Lys AAA AAG
Ile AUU AUC AUA
Asn AAC AAU
The sequence of DNA coding for this section of the polypeptide could be
A. CGTACGTTTTATTTG
B. CGTTCGTTTTATTTG
C. CGTACTTTTTACTTG
D. CGAACATTCTATTTT
20% got this correct
Question 2
The following diagram outlines events associated with the production of a polypeptide chain in a
eukaryotic cell.
a. What is the name of the process at step 1?
Transcription
b. i. Name the product of step 1.
Either of:
• Pre-mRNA
• primary RNA.
mRNA was not an acceptable answer.
ii. Outline what occurs at step 2.
Either of:
• introns are removed
• any other post transcriptional modification.
c. Name the event that occurs at structure 3.
One of:
• translation
• polypeptide synthesis
• protein synthesis.
4 marks for a-c
24% got 0, 13% got 1, 23% got 2, 19% got 3, 21% got 4
d. i. Name the structure at 4.
Transfer RNA (tRNA)
ii. Outline the function of the structure you named in d.i.
Transfer RNA brings a specific amino acid to the ribosome. The anticodons of the tRNA are
complementary to the codons of the mRNA.
Many students incorrectly called tRNA, transport RNA. An incorrect answer was that ‘tRNA is
involved in making an amino acid’.
1 + 1 = 2 marks
47% got 0, 15% got 1, 38% got 2
Question 4
a. Describe the appearance of a bacterial plasmid.
Circular
18% got this correct
A bacterial plasmid was modified in the laboratory so that it contained a gene for an enzyme
which provided resistance to the antibiotic tetracycline. Bacterial cells, which in their natural
environment were sensitive to the antibiotic tetracycline, were mixed with the modified plasmid.
The bacterial cells were treated so that they could take up the plasmid.
b. What is the name of the process in which a bacterial cell takes up a plasmid and expresses the
genes of the plasmid?
Transformation
‘Heat shock’ was not an acceptable answer as this is a process of taking up the plasmid. Neither
was ‘transfection’ as this applies to viral vectors.
Many students were aware that the required answer began with ‘trans’. Examples of common
incorrect answers included transduction, translocation and transgenic. Recombination was also
a common incorrect answer.
1 mark + 1 mark = 2 marks
36% got 0, 54% got 1, 10% got 2
The outcome of an experiment is shown below.
With respect to the growth of bacteria the results of plates A and C are shown. On plate A there
is a continuous growth of bacteria over the surface of the agar. On plate C the colonies are
distinguishable from each other.
c. i. What result would you expect on plate B with respect to the growth of the bacteria?
No growth of bacteria.
ii. Explain your answer to c.i.
The bacteria are sensitive to the antibiotic tetracycline and therefore do not grow.
1 + 1 = 2 marks
34% got 0, 12% got 1, 54% got 2
d. Explain why there is a difference in the way the bacteria have grown on plates A and C.
• Plate A: The bacteria are able to grow as there is no tetracycline present.
• Plate C: Only those bacteria which take up the plasmid can grow in the presence of
tetracycline.
It was important that students distinguished between Plates A and C, recognised the significance
of the presence/absence of the antibiotic and realised that only some of the bacteria were
transformed. One mark was awarded for each of the points above. Some students gave a specific
percentage transformation which could not be deduced from the information. Students need to be
aware that if their answers include incorrect information they may not receive full marks.
2 marks
51% got 0, 27% got 1, 22% got 2
Question 5
‘CC’ for Carbon Copy is the name of the first cloned kitten born in 2001. The nucleus of a cat’s
egg cell was removed. It was replaced by a nucleus from a somatic cell of a donor female cat.
Once development commenced the egg cell was transferred into a surrogate female.
a. What is meant by the term cloning?
Making genetically identical copies (of organisms).
The key point was that the copies are genetically identical. Many students stated that the copies
were identical or phenotypically identical; these responses were not awarded a mark.
1 mark
58% got this correct
The diploid number of a cat is 38.
b. i. How many chromosomes would have been in the nucleus that was removed from the egg
cell?
19
ii. Is CC male or female? Explain.
Female. The donor of the nucleus was from the somatic cell of a female donor.
A common incorrect answer was ‘The cell was an egg cell and this comes from females’.
1 + 1 = 2 marks
15% got 0, 37% got 1, 49% got 2
To determine if CC was in fact a true clone, studies were made of specific variable regions in the
DNA of the donor, CC and surrogate.
The results are shown in the table.
c. For each region of DNA there are two values, for example, 164/164. Suggest a reason for this.
The reason for the two numbers is that this locus is found on a pair of homologous
chromosomes.
This part was not well answered. Many students incorrectly thought the two numbers were
representative of the two strands of DNA.
1 mark
15% got this correct
d. In the case of DNA variable region 4 in the donor DNA, why are the pairs of values different?
The different values indicate that the individual being tested is heterozygous for this locus,
indicating that the alleles are of different lengths.
Most students attempted this question, however, many students again related the numbers
incorrectly to the two strands of DNA or made reference to the occurrence of mutations having
occurred, which was also incorrect.
1 mark
12% got this correct
e. From the data it was concluded that CC was a true clone. Explain the evidence in the table that
supports this claim.
CC is a true clone as the DNA of CC is identical to the donor DNA and not the surrogate.
This was generally well answered, however students who stated the DNA was almost identical
could not be awarded the mark.
1 mark
74% got this correct
Question 7
Organisms can regulate the expression of their genes in a number of ways.
a. Suggest why an organism regulates the expression of its genes.
Genes are only expressed when required to conserve energy or time.
1 mark
42% got this correct
One example in bacteria is the regulation of the expression of a gene which produces an enzyme
(enzyme X) involved in the metabolism of the amino acid tryptophan. Enzyme X is only
produced when tryptophan is in high concentration. This gene regulation involves several genes.
Two of the genes include a gene for the production of enzyme X and an operator gene. If a
protein, called a repressor protein, binds to the operator gene, transcription of the gene for
enzyme X is stopped. If no repressor protein is bound to the operator, transcription of the gene
for enzyme X occurs. A summary of this regulation is shown in Figure 1.
b. The gene coding for enzyme X is not transcribed when the repressor protein binds to the
operator gene. What enzyme is prevented from functioning during this binding?
RNA polymerase
This part was poorly answered as students did not relate their answer to gene expression and the
enzyme needed to bind to allow this process to occur. The most common incorrect answer was
enzyme X.
1 mark
17% got this correct
When tryptophan binds to the repressor protein, the repressor protein can no longer bind to the
operator gene. (See Figure 2.)
c. When tryptophan binds to the repressor protein what will happen to the production of enzyme
X?
Production of enzyme X will occur.
d. Based on Figure 2, suggest how tryptophan prevents repressor protein function.
The binding of the tryptophan to the repressor protein changes its shape (from the diagram) and
this complex can no longer bind to the operator gene (therefore transcription can occur).
1 mark + 1 mark = 2 marks
32% got 0, 29% got 1, 33% got 2
2005
Question 1
The nucleotide found in messenger RNA but not in DNA is
A. uracil.
B. adenine.
C. guanine.
D. cytosine.
97% got this correct
Question 14
DNA was incubated with radioactive nucleotides. After one cycle of replication the distribution
of radioactive and non-radioactive nucleotides in the DNA would be
Answer is B
64% got this correct
Question 15
Translation of the genetic code occurs in the cytosol of a cell. The following diagram is one
representation of translation.
In the model presented
A. M represents a ribosome.
B. N represents messenger RNA.
C. O represents transfer RNA.
D. P represents an amino acid.
79% got this correct
Question 16
The following diagram indicates the cutting sites of three different restriction enzymes on a
particular bacterial plasmid.
If the plasmid was incubated with the restriction enzyme Eco R1, the number of pieces of DNA
obtained would be
A. two.
B. three.
C. four.
D. seven.
79% got this correct
Question 17
An investigation was carried out to determine the mutation rate in DNA exposed to different
wavelengths of UV light. The results are presented graphically in the diagram below.
The data indicates that the wavelength of UV that produces the maximum mutation rate in DNA
is
A. 240 nm.
B. 260 nm.
C. 270 nm.
D. 290 nm.
93% got this correct
Question 18
A restriction enzyme cutting site may be present or absent in a particular 200 kb region of human
chromosome 1.
The DNA of a person heterozygous for this cutting site would have the following gel pattern
after digestion with the enzyme.
Answer is A
52% got this correct
Question 5
Fungi can make their own amino acids and are able to grow on a substance called minimal
media.
An abnormal strain of fungus that could not grow on minimal media was discovered. The
abnormal fungus was able to grow on minimal media to which all of the twenty amino acids
were added. These observations can be summarised as follows.
It was assumed that the deficiency in the abnormal fungus was because of a fault in the genetic
material coding for the production of one of the amino acids.
An experiment was designed to investigate which amino acid was involved. The experimental set
up, involving 22 tubes, is outlined in the following diagram and the results are shown.
a. Why were tubes 1 and 2 included in the experiment?
Tubes 1 and 2 are the control group to which other tubes can be compared.
1 mark
65% got this correct
The diagram shows that the fungus grew in tube 12. The amino acid histidine had been added to
this tube.
b. From the above experiment what two conclusions can be made about the ability of the
abnormal fungus to produce amino acids?
The conclusions which can be drawn are:
• the abnormal fungus cannot produce histidine
• the abnormal fungus can produce the other 19 amino acids.
2 marks
59% got 0, 26% got 1, 14% got 2
Consider the following genetic code table.
c. i. What sequences of nucleotides in DNA code for the amino acid histidine?
(Note: in the table his represents histidine.)
The sequence codings for ‘his’ in RNA are CAU and CAC, which are GTA and GTG respectively
in DNA.
ii. Which of the following DNA sequences could lead to the production of an uninterrupted chain
of
amino acids in the abnormal fungus?
sequence 1 AACGCCTCGGTGCCA
sequence 2 CAAGTAGGTACACTC
sequence 3 TAATGGACCCCCGGT
Sequence 3
iii. Explain why you made this selection in c.ii.
This sequence does not have the DNA triplet for histidine (GTA or GTG) whereas sequences 1
and 2 do, so the polypeptide could be made without interruption.
1 + 1 + 1 = 3 marks
47% got 0, 27% got 1, 8% got 2, 18% got 3
During their work in establishing the structure of DNA, Watson and Crick were interested in the
proportion of nucleotides in skin cells from a particular organism. They considered the results
from three different laboratories.
The result of each of the laboratories was as follows.
d. Watson and Crick used the results of laboratories 2 and 3. What difference would it have
made to the model they constructed of DNA had they used the results of laboratory 1 only?
Watson and Crick would have concluded that Adenine pairs with Guanine and Thymine pairs
with Cytosine because these pairs are in approximately equal numbers.
To gain the mark, specific reference had to be made to the pairing of the bases.
1 mark
23% got this correct
2004
Use the following information to answer Questions 9 and 10.
The following nucleotide sequence forms part of the template strand of a gene coding for protein
X.
Question 9
The complementary base found at the fourth nucleotide (marked *) in a sequence transcribed
from this sequence would be
A. C
B. G
C. T
D. U
60% got this correct
Question 10
In a double-stranded molecule formed from this DNA template strand (shown above) the number
of deoxyribose sugar units you would expect to find is
A. 4
B. 8
C. 16
D. 32
56% got this correct
Question 11
Reverse transcriptase catalyses the production of
A. DNA from an mRNA template.
B. DNA from a protein template.
C. mRNA from a DNA template.
D. tRNA from a DNA template.
62% got this correct
Use the following information to answer Question 12.
The list 1–4 below describes events and outcomes of the replication of DNA within a eukaryotic
cell.
1. Complementary nucleotides bind to each of the two strands.
2. Sugar phosphate bonds form between the nucleotides.
3. The newly formed DNA molecules are semi-conserved.
4. Unwinding of the DNA molecule forms two single strands.
Question 12
The correct order of these events during DNA replication, with the earliest event first, is
A. 1, 2, 3, 4
B. 1, 4, 3, 2
C. 4, 2, 1, 3
D. 4, 1, 2, 3
79% got this correct
Question 4
Below is the DNA sequence from the beginning of a gene coding for an enzyme involved in
photosynthesis. The upper line of bases (in bold) represents the template strand.
a. Write the mRNA sequence that would be transcribed from this DNA sequence.
AUGAAAUUCUCGAAUAGC
1 mark
62% got this correct
Use the table of part of the genetic code (below) to answer part b.
b. i. The 6th base on the template strand of the sequence above is substituted by C.
What type of mutation is this?
This mutation is a point mutation or base substitution.
ii. Explain the effect this mutation will have on the amino acid sequence of the protein produced.
There will be a different amino acid in the amino acid sequence. Asn (asparagine) is replaced by
Lys (lysine).
1 + 1 = 2 marks
35% got 0, 33% got 1, 32% got 2
c. The 11th base pair of the sequence is deleted.
Explain the effect that this mutation will have on the amino acid sequence of the protein
produced.
The amino acid sequence will be shortened as a stop codon has been introduced into the mRNA.
1 mark
22% got this correct
Question 5
A US company has recently released what is believed to be the first designer transgenic pet.
The ‘Glofish’ is derived from the zebra fish, which is normally silver with black stripes. The
zebra fish has been genetically modified by adding a gene found in a sea coral into zebra fish
eggs. The offspring resulting from these modified eggs are reddish in colour and glow or
fluoresce under ultraviolet light.
Although the added gene is from a different species, it is expressed in the same way in both the
sea coral and transgenic zebra fish.
a. What feature of the genetic code makes it possible for a gene to be transferred from one
species to another and to be expressed in the second species?
The genetic code is universal which means that the codons in mRNA code for the same amino
acids irrespective of the species.
1 mark
16% got this correct
b. What steps are required, within the zebra fish cells, for the sea coral gene to be expressed?
The DNA is transcribed into mRNA in the nucleus and then translated in the cytoplasm at the
ribosome into a polypeptide/protein.
The most common error was that students did not relate expression of a gene to the production
of the protein that was coded for by the sea coral gene.
2 marks
85% got 0, 6% got 1, 9% got 2
c. Predict the chance of survival of the Glofish if it were released into the natural environment of
the zebra fish. Explain your answer.
The survival of the Glofish is likely to be poor in the natural environment. The fish would be
more noticeable to predators because it is red or glows in ultraviolet light, so may be captured
more often. Another reason for the decrease in the number of fish is that they are no longer
recognised by the silver fish with black stripes and so have fewer opportunities to mate.
Students often recognised that the fish had less chance of survival but needed to say more than
that the fish would be at a disadvantage to be awarded full marks.
2 marks
32% got 0, 20% got 1, 48% got 2
Question 7
Sickle cell anaemia is a serious inherited blood condition. It leads to tiredness and kidney or
heart failure and without treatment children usually die before the age of 10. Sickle cell anaemia
is due to a change in the gene which codes for beta haemoglobin. There are two alleles for the
beta haemoglobin gene; HbA coding for normal beta haemoglobin and HbS coding for the
changed haemoglobin. An individual with two copies of the HbS allele will develop symptoms
of sickle cell anaemia.
It is now possible to genetically test people to see if they carry the HbS allele. This test uses
PCR, the restriction enzyme MstII and gel electrophoresis.
MstII is a restriction enzyme that recognises the 7-base sequence in DNA,
and cuts it between the C and the T to produce
a. What term is used to describe the ends of the fragments produced by MstII?
Sticky ends are produced by MstII.
1 mark
76% got this correct
Molecular studies have shown that the sickle cell allele differs only by one base pair from the
normal allele. This base change occurs in a 7-base sequence that is recognised by the restriction
enzyme MstII. This is the only MstII site found within the region of the gene that is used in the
genetic test.
The PCR products are digested using MstII. The resulting fragments undergo gel electrophoresis.
b. How is the action of the MstII enzyme affected by the HbS mutation?
The enzyme does not cut the DNA, or the enzyme no longer recognises the sequence.
1 mark
49% got this correct
c. Mark on the picture of the gel below the banding patterns you could expect to see for someone
who has each of the following genotypes.
i. HbA/HbS
ii. HbS/HbS
A person with a copy of the HbA allele will still have the MstII recognition sequence and the
DNA will be cut into two pieces, one piece of 52 base pairs the other 175 base pairs.
A person with a copy of the HbS allele will not have the MstII recognition sequence and the DNA
will not be cut, leaving it as a piece 230 base pairs in length.
A person who has the genotype HbS/HbS will have all pieces of DNA 230 base pairs in length
and thus present one band on the gel.
A person who has the genotype HbA/HbS will have pieces of DNA 230 base pairs in length as
well as pieces of DNA 175 and 52 base pairs in length. There will be three separate bands on the
gel, the first lining up with the band created by the person with the HbS/HbS genotype and the
other two running further down the gel as they are smaller in size.
1 + 1 = 2 marks
52% got 0, 36% got 1, 12% got 2
2003
Question 8
In DNA, the number of
A. phosphate groups equals the number of nitrogen bases.
B. adenine nucleotides equals the number of cytosine nucleotides.
C. phosphate groups equals twice the number of sugar molecules.
D. guanine nucleotides equals the number of uracil nucleotides.
70% got this correct
Question 9
During DNA replication
A. messenger RNA (mRNA) is produced.
B. reverse transcriptase enzymes play an important role.
C. bonds between phosphate and sugar molecules break.
D. each of the DNA strands acts as a template strand.
67% got this correct
Use the following information to answer Question 10.
Figure 3 shows portion of a cell engaged in protein synthesis. The various parts of the cell are
not drawn to scale.
Question 10
It is reasonable to conclude that
A. process N is translation.
B. structure P is made of t-RNA.
C. compound Q is messenger RNA.
D. structure R is the site of protein synthesis.
79% got this correct
Use the following information to answer Questions 13 and 14.
Haemoglobin consists of four polypeptide chains. In normal haemoglobin, two of these chains
are beta chains, each comprising 146 amino acids. Variations exist in the amino acid composition
of these chains and this results in different kinds of haemoglobins. One of these variants is called
haemoglobin S.
The first seven amino acids in the beta chains of these two haemoglobins, and the amino acid at
position 143 are given below. The amino acids at each of the remaining positions are the same
for each haemoglobin.
The genetic code for the amino acids in these sequences, coded for by the template strand of
DNA, is shown in the following table.
Question 13
It is reasonable to conclude that during transcription of normal haemoglobin, the mRNA codon
sequence could be
A. GTT for amino acid 1.
B. ACT for amino acid 4.
C. CTT for amino acid 3.
D. CAC for amino acid 2.
60% got this correct
Question 14
Considering the DNA responsible for the haemoglobins, it is reasonable to conclude that
A. haemoglobin S could be the result of a single base mutation in the DNA of adult haemoglobin.
B. deletion of nucleotide 12 would change the fourth amino acid in the sequence of adult
haemoglobin.
C. a change in the 4th base of the DNA sequence would produce no change in either of the two
haemoglobins.
D. in haemoglobin S, a change in the 10th base of the DNA sequence would produce no change
in the amino acid sequence.
68% got this correct
Question 4
Genes can be transferred from one species to another in different ways. One method is to use
plasmids, circular pieces of DNA found in some bacteria.
In this method, a plasmid is cut and a piece of foreign DNA inserted. The foreign piece of DNA
usually contains more than one gene. The process is shown in Figure 5 below.
Many copies of the new plasmid are then incubated with bacteria.
a. What is the name given to a plasmid that is used to transfer DNA from one organism to
another?
vector or recombinant plasmid
1 mark
49% got this correct
b. What is used to cut the DNA of a plasmid?
restriction enzyme or endonuclease
1 mark
70% got this correct
c. What is used to join the inserted piece of DNA to the plasmid?
DNA ligase
1 mark
54% got this correct
One of the foreign genes inserted into the plasmid, codes for resistance to a particular antibiotic.
d. Explain why it is important to include a gene for antibiotic resistance in the plasmid produced.
The importance of including a gene for antibiotic resistance in the plasmid is in identification of
bacteria that have taken up the plasmid. After the new plasmids are incubated with bacteria, the
bacteria is grown on media containing antibiotics. Only those bacteria that have taken up the
plasmid will be able to grow on the media.
Many students simply stated to make the bacteria resistant to antibiotics and therefore were not
awarded a mark.
1 mark
9% got this correct
Bacteria containing plasmids, that are constructed in the way outlined in Figure 5, can be used in
a variety of settings with plants and animals.
For example, some plants are resistant to attack by insects. The plants produce a protein that
poisons the larval stage of some insects that feed on them. The production of the protein is under
genetic control.
A particular species of crop plant was genetically engineered to contain this gene. Such plants
are referred to as GM (genetically modified) plants.
e. Explain why a farmer might choose to grow a crop that was genetically engineered to be
resistant to insects, rather than spray the crop with insecticide.
Any one of the following responses would have been awarded a mark:
• Insecticide may be harmful to humans and other living things.
• Insecticide may kill useful insects.
• GM crop may be cheaper to grow and manage.
• Genetically engineered resistance may be more specific for a harmful insect.
1 mark
52% got this correct
Some plants are resistant to particular herbicides, chemicals that are used to kill plants. This trait
is also under genetic control.
The gene that confers herbicide resistance has also been incorporated into some GM crop plants.
This enables a farmer to spray his GM crop with a herbicide that will not harm the GM crop but
does kill weed plants growing within the crop.
f. Suggest one advantage for a farmer to be able to spray his crops with a herbicide.
One advantage for a farmer to spray his crops with herbicide would be that there would be less
competition for resources from other plants so a greater yield from his crop. Another correct
response would be that the farmer may reduce the time spent weeding.
1 mark
34% got this correct
Two farmers have properties next door to each other. They grow the same cereal crop.
• Farmer X wishes to grow GM crops that are resistant to herbicide.
• Farmer Y wishes to continue to grow non-GM crops.
Farmer Y was concerned, and suggested to farmer X that pollen from the GM crop could fertilise
the non-GM plants.
g. Explain why farmer Y might be concerned about the possibility of his crop being fertilised by
pollen from farmer X’s crop.
Farmer Y may be concerned that he may lose some of his markets if he cannot state that his
crops are GM free.
Other correct responses included ethical issues raised by Farmer Y.
1 mark
11% got this correct
The farmers agreed to carry out field trials to establish whether leaving a gap between crops
reduced the likelihood of cross-pollination. A number of trials were planted so that the results of
one trial did not interfere in any way with the results of another. The percentage of seeds
produced at various positions as a result of cross-pollination was measured for each trial. The
outline of these trials and the results gathered are shown in the following table.
h. From the data, what conclusions can be drawn about cross-pollination and the gap between
crops?
Conclusions that can be drawn from the experiments are:
• When the crops are adjacent to each other there is cross-pollination.
• Cross-pollination can be reduced by moving crops 5 m apart.
• the difference between separating 5 m or 7 m does not influence cross pollination at the edge,
only 10m into crop.
Many students restated the figures in the table without making conclusions from the data. An
example of a response that could not be awarded a mark included ‘there was 2 per cent crosspollination when there was no gap between the crops and 0.3 per cent when the gap was 7
metres’.
3 marks
44% got 0, 36% got 1, 16% got 2, 4% got 3
2002
Use the following information to answer Question 11.
Figure 3 represents a length of DNA and its cutting sites for the restriction enzymes Spe I, Eco
RI, Bgl II and
Hin dIII.
Question 11
Incubation of this length of DNA in a tube containing
A. Spe I would result in three pieces of DNA.
B. Hin dIII would result in two pieces of DNA.
C. Spe I and Eco RI would result in five pieces of DNA.
D. Bgl II and Hin dIII would result in four pieces of DNA.
61% got this correct
Question 4
Figure 10 represents a piece of DNA from a gene. The template strand for a section of the gene is
shown.
a. In the space provided in Figure 10 write in the complementary bases for the other strand of
DNA.
1 mark
84% got this correct
b. What does the A represent in this DNA sequence?
Adenine
1 mark
79% got this correct
c. The first step in the process of gene expression is transcription. What is produced during
transcription?
mRNA or messenger RNA is produced during transcription. (pre-mRNA) would be correct.
1 mark
69% got this correct
d. The compound produced during transcription attaches to an organelle in the cell.
i. What is the name of this organelle?
Ribosome
ii. What is the name of the step in gene expression which follows transcription and occurs at this
organelle?
Translation
iii. What is produced during this process mentioned in part ii.?
A polypeptide or protein
1 + 1 + 1 = 3 marks
14% got 0, 12% got 1, 22% got 2, 53% got 3
In the section of the gene sequence shown in Figure 10 a base substitution mutation occurred.
The 9th base from the left, Guanine, was replaced by Thymine (at arrow). The genetic code is
provided below (Figure 11).
e. What effect will this mutation have on the sequence of amino acids in the polypeptide?
asp (aspartic acid) is replaced by glu (glutamic acid) or asp to glu
1 mark
40% got this correct
Question 5
There were three suspects in an assault case. A forensic scientist found blood, other than the
victim’s, at the site. DNA was extracted from five blood samples.
• the victim
• the blood at the assault site (not the victim’s)
• the three suspects
Polymerase Chain Reaction (PCR) was used on the extracted DNA.
a. A DNA polymerase enzyme is involved in the PCR process. Explain the role of the
polymerase enzyme in PCR.
The polymerase enzyme catalyses the production of a new strand of DNA or is involved in
making multiple copies of DNA or amplification of DNA
and
DNA polymerase replicates the DNA by extending from the primer or by complementary base
pairing or by using the original DNA as a template.
Some students incorrectly identified the enzyme and discussed the role of another enzyme. Many
other responses gave one part of the expected answer. Students need to be reminded to use the
number of marks allocated to the question as an indication of the depth required in their answer.
2 marks
47% got 0, 38% got 1, 15% got 2
One of the regions used in the forensic analysis was a short tandem repeat (STR) sequence of 4
bases, called HUMTHO1. This sequence, located on chromosome 11, has many alleles which
differ from each other by the number of times the sequence AATG is repeated. It was this region
of chromosome 11 which was amplified using PCR.
The amplified samples were loaded onto a gel and electrophoresis was performed to separate the
fragments of DNA.
b. Name two properties of the DNA fragments which allow them to be separated from each other
during gel electrophoresis.
A DNA fragment will move according to its charge and molecular weight (size) or DNA is
negatively charged and moves to the positive pole; smaller DNA fragments move further or
faster than larger fragments.
2 marks
37% got 0, 31% got 1, 32% got 2
A diagram of the gel is shown below (Figure 12).
c. Why is there only one band in lane 2 but two bands in lanes 3, 4, 5 and 6?
There is only one band in lane 2 because individual 2 is homozygous, the others on the gel are
heterozygous or the two fragments of DNA are the same size or the number of repeats in the two
fragments is the same.
1 mark
30% got this correct
d. How many different alleles at the HUMTHO1 locus are represented on the gel in individuals
2, 3, 4, 5, 6?
There are 5 different alleles at the HUMTHO1 locus represented on the gel.
1 mark
37% got this correct
e. Which piece of DNA, A or B, has the greater number of the 4 base repeat sequence?
DNA piece A has the greater number of the 4 base repeat sequence. The greater the molecular
weight of the sample the smaller distance the sample will move from the loading well.
1 mark
77% got this correct
f. Which of the suspects appears to have committed the assault? Explain.
The bands on the gel for suspect 5 match the sample of blood found on the victim,which was not
the victim’s blood (lane 3).
Students could not be given a mark for the correct identification of suspect 5. The mark was
awarded for the explanation as to why suspect 5 appears to have committed the assault.
1 mark
38% got this correct