Practice Multiple-choice Questions

Question 1

Suppose you are provided with an actively dividing
culture of E.coli bacteria to which radioactive thymine
has been added. What would happen if a cell replicates
once in the presence of this radioactive base?
a) One of the daughter cells, but not the other, would have radioactive DNA
b) Neither of the two daughter cells would be radioactive
c) Radioactive thymine would pair with radioactive guanine
d) DNA in both daughter cells would be radioactive

Answer 1

d) DNA in both daughter cells would be radioactive

Question 2

Which of the following help(s) to hold the
DNA strands apart while they are being
replicated?
a) Primase
b) Ligase
c) DNA polymerase
d) Single-strand binding proteins

Answer 2

d) Single-strand binding proteins

Question 3

The leading and the lagging strands differ in that
a) The leading strand is synthesized in the same direction as
the opening of the replication fork, and the lagging
strand is synthesized in the opposite direction
b) The leading strand is synthesized by adding nucleotides
to the 3’ end of the growing strand, and the lagging
strand is synthesized by adding nucleotides to the 5’ end
c) The lagging strand is synthesized continuously, whereas
the leading strand is synthesized in short fragments that
are ultimately stitched together
d) None of the above are true

Answer 3

a) The leading strand is synthesized in the same direction as
the opening of the replication fork, and the lagging
strand is synthesized in the opposite direction

Question 4

Why is the lagging strand of DNA replicated in
short Okazaki fragments?
a) Because DNA is tightly wound
b) Because DNA polymerase can only synthesize
in one direction
c) To help with repair enzymes
d) To help with proof reading

Answer 4

b) Because DNA polymerase can only synthesize

in one direction

Question 5

Which of the following is true regarding PCR?
a) Tm = temperature at which 90% of helical
structure is lost
b) PCR utilises the separation of DNA strands at
high temperature above the TM
c) G and C melts at a lower temperature
d) The orientation of a primer contributes to
the Tm of the primer

Answer 5

b) PCR utilises the separation of DNA strands at

high temperature above the TM

Question 6

If the OD260 reading of a sample of DNA that has been
diluted 100x is 0.64 and 50g of DNA in a volume of 1 ml
of H2O gives an OD260 of 1, what is the original
concentration of the DNA sample?
a) 32 mg/ml
b) 3.2 mg/ml
c) 0.64 mg/ml
d) 64 mg/ml

Answer 6

b) 3.2 mg/ml

Question 7

The human angiotensin-converting enzyme (ACE) gene has an insertion (I) / deletion
(D) polymorphism in the population that can be detected via polymerase chain
reaction. The D polymorphism produces a band of 191 bp and the I polymorphism
produces a band of 478bp. A total of ten individuals were tested to determine their
genotypes with respect to the two alleles in the ACE gene. The results shows that we
obtained 5 individuals that were homozygous for the D allele, 5 individuals there were
homozygous for the I allele and no individuals who were heterozygotes. What is the
frequency of the D allele?
a) 0.10
b) 0.24
c) 0.48
d) 0.50

Answer 7

d) 0.50

Question 8

If the frequency of the D allele is 0.5, what is the frequency of II homozygotes in the
next generation provided the Hardy-Weinberg equilibrium is maintained.
a) 0.24
b) 0.50
c) 0.25
d) 0.36

Answer 8

c) 0.25

Question 9

Individuals with Cockayne syndrome are more
likely to:
a) Have an increase rate of somatic mutation
b)Have an increased rate of germ line mutation
c) Have a decreased rate of spontaneous
mutation
d)Have a decreased rate of induced mutation

Answer 9

a) Have an increase rate of somatic mutation

Question 10

Which DNA repair enzyme is primarily responsible
once apurinic or apyrimidinic sites are detected?
A.Mismatch repair pathway
B.Uracil glycosylase
C. AP endonuclease
D.Homologous recombination

Answer 10

C. AP endonuclease

Question 11

In homologous recombination, DNA
hybridisation creates a region of DNA helix
formed from strands that originate from two
different DNA molecules. This region is known
as a
a) Heterochromatin
b) Heteroduplex
c) Strand invasion
d) homoduplex

Answer 11

b) Heteroduplex

Question 12

If a cell has completed the first meiotic division and is
just beginning meiosis II, which of the following is an
appropriate description of its contents?
a) It has half the amount of DNA as the cell that began meiosis
b) It has the same number of chromosomes but each of them
has different alleles than another cell from the same meiosis
c) It has half the chromosomes but twice the DNA of the
originating cell
d) It is identical in content to another cell from the same
meiosis

Answer 12

a) It has half the amount of DNA as the cell that began meiosis

Question 13

In the Holliday model, branch migration
results in the formation of
a) Double stranded breaks
b)Gene conversion
c) Heteroduplex regions on one chromosome
d)Heteroduplex regions on both chromosomes

Answer 13

d)Heteroduplex regions on both chromosomes

Question 14

If double stranded break recombination
results in gene conversion, how is the original
allele lost?
a) It isn’t, it is merely rearranged
b)It is eliminated by DNA repair enzymes
c) It is prevented from replication due to
selective pressure
d)It is digested away after a break within the
gene

Answer 14

b)It is eliminated by DNA repair enzymes

Question 15

Which of the following is true about new
genes and genome evolution:
a) DNA is made from pre-existing DNA
b) Genetic innovation can arise from intragenic
mutation
c)The basis of evolution of new genes is through
gene duplication
d) All of the above

Answer 15

d) All of the above

Question 16

Which of the following statements regarding the genetic code is correct?

a) Wobble base pairing of the anticodon decreases the number of codons a tRNA can recognise
b) The third base of the codon is the most critical for recognition by the tRNA anticodon
c) More than one nucleotide triplet can serve as a stop codon
d) More than one amino acid can be coded for by a single codon
e) Alignment of codon and anticodon base pairs is in a parallel direction

Answer 16

c) More than one nucleotide triplet can serve as a stop codon

Question 17

All of these features are in most (but not necessarily all) tRNA synthetases except for one that is incorrect:

a) Active site catalysing the covalent attachment of amino acids to tRNAs
b) Binding surface for the mRNA codon
c) Binding surface for the ribosome
d) Active site editing, i.e. the removal of incorrect amino acids
e) Binding site for ATP

Answer 17

b) Binding surface for the mRNA codon

Question 18

In the process of protein synthesis, each mRNA is translated:

a) only once
b) by only one ribosome at a time
c) simultaneously by many ribosomes until it is degraded
d) simultaneously by many ribsomes for the life of the cell
e) only when the ribosomes are attached to the endoplasmic reticulum

Answer 18

c) simultaneously by many ribosomes until it is degraded

Question 19

Which structure is correctly paired with its function in relations to the initiation of protein synthesis in prokaryotes?

a) mRNA - interacts first with the large ribosomal subunit
b) AUG codon - the first three nucleotides at the 5’ end of every mRNA
c) Shine-Dalgarno sequence - acts as a start signal for protein synthesis
d) Initiation Factors - promote the assembly of the large and small rRNA subunits
e) fmet-tRNA - undergoes codon/anticodon base pairing with the 16S rRNA

Answer 19

c) Shine-Dalgarno sequence - acts as a start signal for protein synthesis

Question 20

Emma aims to follow the production of an antiviral factor in infected cells. She detects the mRNA of this factor using microarray analysis but fails to detect the protein by Western blot. This negative result may be due to:

a) Targeting of the antiviral factor to the nucleus
b) Proteosome degradation following polyubiquitination of the protein
c) The protein has failed to fold and is degraded by chaperones
d) The virus induces an arrest of cellular transport

Answer 20

b) Proteosome degradation following polyubiquitination of the protein

Question 21

Deletion of a single base from the coding sequence of a gene is likely to result in any of the following outcomes except one. Which one?

a) A protein whose sequence of amino acids differs markedly from normal
b) A non-functional protein being generated
c) A shorter protein
d) A protein in which a single amino acid is replaced by another amino acid
e) A protein that is still functional depending on the site of the deletion

Answer 21

d) A protein in which a single amino acid is replaced by another amino acid

Question 22

The TrpR repressor:

a) is encoded by the trpE gene
b) controls the trp operon via transcriptional attenuation
c) binds to the trp operon in the presence of co-repressor
d) binds cAMP in the presence of an inducer

Answer 22

c) binds to the trp operon in the presence of co-repressor

Question 23

Where would the TrpR repressor be bound when tryptophan levels are high?

a) TrpR
b) Operator site
c) Promoter site
d) the repressor would not be bound

Answer 23

b) Operator site

Question 24

Where would the TrpR repressor be bound when tryptophan levels are low?

a) trpR
b) operator site
c) promoter site
d) it would not bind

Answer 24

d) it would not bind

Question 25

Tryptophan molecule that binds to and activates the TrpR protein is called:

a) inducer
b) repressor
c) co-repressor
d) transcription activator

Answer 25

a) co-repressor

Question 26

Transcriptional attenuation of the trp operon involves:

a) the CAP protein and TrpR
b) the TrpR repressor molecule
c) translational stalling of a tryptophan rich leader peptide
d) TrpR binding to a tryptophan rich leader peptide

Answer 26

c) translational stalling of a tryptophan rich leader peptide

Question 27

Transcriptional attenuation of the tryptophan biosynthesis operon results in:

a) increased TrpR repressor binding resulting in reduced gene expression
b) reduced TrpR binding resulting in increased gene expression
c) early transcription termination when tryptophan levels are low
d) early transcription termination when tryptophan levels are high

Answer 27

d) early transcription termination when tryptophan levels are high

Question 28

What fraction of the eukaryotic genes are located in the close-conformation?

a) all genes
b) most of the genes
c) none of the genes
d) small minority of the genes

Answer 28

b) most of the genes

Question 29

Deacetylation of histone tails:

a) favours transcription
b) stops condensation of DNA
c) blocks access by RNA polymerase
d) inhibits translation

Answer 29

c) blocks access by RNA polymerase

Question 30

Which statement is correct?

a) some eukaryotic activator protein complexes direct de-acetylation of histones near promoters
b) some eukaryotic repressor protein complexes direct acetylation of histones near promoters
c) some eukaryotic activator protein complexes direct acetylation of histones near promoters
d) eukaryotic activator/repressor proteins do not regulate acetylation of histones

Answer 30

c) some eukaryotic activator protein complexes direct acetylation of histones near promoters

Question 31

An important difference between the transcription in prokaryotes and eukaryotes is that:

a) only prokaryotes use upstream sequences
b) only eukaryotic RNA polymerase possesses a sigma-subunit
c) only eukaryotes use transcription activators
d) only eukaryotes require chromatin-remodelling proteins
e) only prokaryotic genes have promoters

Answer 31

d) only eukaryotes require chromatin-remodelling proteins

Question 32

Termination of transcription in prokaryotes involves:

a) an intrinsic terminator sequence that has an AT-rich inverse repeat sequence
b) a terminator sequence that has a string of Gs which form weak hydrogen bonds with the string of Cs on the template DNA
c) an upstream Rut sequence to which the Rho factor binds
d) an upstream Rut sequence to which RNA polymerase binds
e) the sequence AAUAAA bound by a cleavage factor

Answer 32

c) an upstream Rut sequence to which the Rho factor binds

Question 33

True or False: VDJ recombination uses the same
mechanisms as the recombination events
that cause isotype switching. The 2 events
occur at the same time

Answer 33

False
– Isotype switching does not use Rag genes
– Isotype switching occurs only after B cells have
been activated by contact with antigen. Thus, by
definition, it must occur later than VDJ
recombination

Question 34

True or False: TCR and BCR are both formed as a result of
VDJ recombination. The regions of the
receptors that bind antigen and determine
specificity are small and confined to the 3
CDR regions. Therefore the function of the
TCR and BCR is identical

Answer 34

False, whilst there are several strong
similarities between the biology of the TCR
and BCR they do NOT function identically
– After activation the BCR is secreted (antibody),
The TCR is never secreted
– The TCR recognises only peptide antigen
presented by MHC molecules

Question 35

True or False: Making a mutant mouse by either transgenic
or gene knockout technology uses the same
biological process of homologous
recombination

Answer 35

False.
• A gain of function transgenic mouse
– does not use homologous DNA (think of the
jellyfish DNA into mice or plants).
– The transgene inserts randomly into the
genome….perhaps multiple times.
• In contrast a loss of function knockout mouse
must have homologous DNA.
– This technique works by the endogenous gene
being replaced by the mutant by homologous
recombination

Question 36

True or False: T. brucei switches expression of its VSG genes as a
result of selection pressure from the immune system.
Antibodies recognise the VSG, kill the parastie which
forces the parasite to switch VSG genes

Answer 36

False.
• It is true that the immune system is important in
giving a parasite that has recently switched its VSG
gene a selective advantage over non-switched
counterparts. However, the switching/gene
conversion process itself is random, and occurs
spontaneouly during mitosis.

Question 37

Cells of the immune
system:
A. Originate within the spleen
B. Respond only to contact with
other immune cells
C. Can communicate with one
another
D. Act independently of each other
E. Are generated at the site of need

Answer 37

C. Can communicate with one

another

Question 38

MHC molecules:
A. Are not expressed by immune cells
B. Are an important component of the
innate immune system
C. Regulate the composition of mucus
in the gastrointestinal tract
D Present peptide to the antigen
receptors of B cells
E Present peptide to the antigen
receptors of T cells

Answer 38

E Present peptide to the antigen

receptors of T cells

Question 39

The immune system is said to be
very diverse in terms of cells types.
One such cell, the B cell:
A. Is not necessary any more because of T
cells
B. Is useful because it can identify and kill
virus infected cells
C. Is useful because it can produce toxin
blocking molecules
D. Is restricted to the bone marrow and
protects the haematopoietic stem cells
from destruction
E. Is associated with the innate immune
system

Answer 39

C. Is useful because it can produce toxin

blocking molecules

Question 40

T cell receptors:
A. Bind directly to surface of bacteria
B. Can only bind peptides presented
by MHC class I molecules
C. Interact with MHC and peptide
complexes
D. Expressed by a single cell are
diverse and quite different
E. Once formed can mutate to
produce stronger binding receptors

Answer 40

C. Interact with MHC and peptide

complexes

Question 41

The number of V, D and J regions for the
TCR b-subunit is 50, 2 and 12 respectively
and the number of V, and J regions for the
TCR a-subunit is 45 and 55 respectively.
Based on this information, the number of
different TCR a-chains that can be formed
is:
A. 50 x 2 x 12 x 45 x 55
B. (45 x 55) + (50 x 2 x 12)
C. 45 x 55
D. about 1,000,000,000,000
E. 45 + 55

Answer 41

C. 45 x 55

Question 42

The isotype of an antibody
molecule is:
A. governed by the binding sites
produced
B. indicates what kind of antigen it
will bind to
C. is governed by the constant
domain
D. has no influence on the function
of the antibody
E. is by default IgA

Answer 42

C. governed by the constant domain

Question 43

Knock out mice are produced by
mutating a known gene and thus eliminating
the action of a particular gene. Using this
technology, it has been shown that:
A. RAG genes are not required for TCR
and BCR rearrangement
B. CD4 T cells bind MHC class I
C. Mice lacking MHC class I do not
produce CD8 T cells
D. All genes are required for life
E. Gene repair mechanisms efficiently
replace the deleted gene

Answer 43

C. Mice lacking MHC class I do not
produce CD8 T cells

Question 44

Concentrating on the B cell
receptor, the genes that encode
the receptor:
A. Will be rearranged in all the cells of
the body
B. Are rearranged identically in T
and B cells
C. Will be different in B cells
compared to liver cells.
D. Are rearranged only when
required to produce antibodies
E. Are fixed at birth and does not

Answer 44

C. Will be different in B cells

compared to liver cells.

Question 45

Gene rearrangement or
recombination:
A. Can be associated with disease
B. Is confined to the generation of
antigen receptors in the immune system
C. Is confined to meiosis
D. Results in cell death because of
gene destruction
E. Results in tissue rejection

Answer 45

A. Can be associated with disease

Question 46

A sample of normal double-stranded DNA was found to have a thymine content of 35%. What is the expected proportion of guanine?
Select one:
a. 30%
b. 65%
c. 35%
d. 17.5%
e. 15%

Answer 46

e. 15%

Question 47

A target DNA sequence is 5’…..CGATTCA…3’. Which of the following nucleotide sequences would hybridise to this sequence and therefore could be used as a probe?
Select one:
a. 5’… CGATTCA…3’
b. 5’… ACTTAGC…3’
c. 5’… GCTAAGT…3’
d. 5’… TGAATCG…3’ 
e. None of these is suitable

Answer 47

d. 5’… TGAATCG…3’

Question 48

Cell lysis using a French press creates high shearing forces that also break genomic DNA into smaller fragments. This will affect most strongly the following property of DNA:
Select one:
a. Viscosity 
b. Base pairing
c. Double helix structure
d. Net charge
e. Absorption of UV light

Answer 48

a. Viscosity

Question 49

Normal double-stranded DNA is boiled in preparation for a PCR experiment. Which property of DNA is not affected by the high temperature treatment?
Select one:
a. Absorption of UV light
b. Viscosity
c. Base pairing
d. Covalent bonding 
e. Solubility

Answer 49

d. Covalent bonding

Question 50

DNA replication is semi-conservative. This is because:
Select one:
a. daughter DNA does not contain any parent DNA.
b. a DNA molecule is parallel.
c. each daughter DNA contains two strands of parent DNA.
d. each daughter DNA contains only one strand of the parent DNA.
e. daughter DNA consists of strands containing segments of both parental DNA strands.

Answer 50

d. each daughter DNA contains only one strand of the parent DNA.

Question 51

DNA replication occurs by the cooperation of which key enzymes?
Select one:
a. DNA helicase, RNA polymerase and DNA ligase
b. DNA helicase, RNA polymerase, DNA polymerase and DNA ligase
c. DNA polymerase and DNA ligase
d. DNA polymerase, DNA helicase, DNA topoisomerase, single-stranded DNA binding proteins and DNA ligase
e. None of the above

Answer 51

d. DNA polymerase, DNA helicase, DNA topoisomerase, single-stranded DNA binding proteins and DNA ligase

Question 52

The lagging strand Okazaki fragments begin as a mix of:
Select one:
a. DNA and DNA ligase
b. RNA and RNA primers
c. DNA and RNA primers 
d. RNA and DNA primers
e. The parent and daughter strands

Answer 52

c. DNA and RNA primers

Question 53

Eukaryotic telomeres replicate differently than the rest of the chromosome. This is a consequence of which of the following?
Select one:
a. The evolution of telomerase enzyme
b. DNA polymerase that cannot replicate the leading strand template to its 5’ end
c. Gaps left at the 5’ end of the lagging strand
d. Gaps left at the 3’ end of the lagging strand because of the need for a primer
e. The “no ends’ of a circular chromosome

Answer 53

c. Gaps left at the 5’ end of the lagging strand

Question 54

A typical type II endonuclease:
Select one:
a. does not recognise foreign DNA purified from a different bacterial strain.
b. recognises only the template strand in a 5’ to 3’ direction.
c. forms dimers that bend the DNA to cleave at sites distant from their recognition sequence.
d. may not recognise its target sequence if it contains methylated bases.
e. is only present in bacteria that do not possess the recognition sequence for this endonuclease in their genome.

Answer 54

d. may not recognise its target sequence if it contains methylated bases.

Question 55

Which of the following is correctly paired?
Select one:
a. Sanger method of DNA sequencing – primer extension (replication) replacing all deoxyribonucleotides with dideoxyribonucleotides
b. Southern blotting – transfer of DNA fragments from a gel onto a nitrocellulose membrane
c. DNA microarray – chip presenting a large set of DNA molecules as template for shotgun sequencing
d. Denaturation of DNA – removal of histone proteins associated with DNA
e. Hybridisation – translocation of a DNA sequence to a different region of the chromosome

Answer 55

b. Southern blotting – transfer of DNA fragments from a gel onto a nitrocellulose membrane

Question 56

DNA is amplified in Polymerase Chain Reaction by the power of:
Select one:
a. Stepwise amplification
b. Linear amplification
c. Exponential amplification
d. Partial amplification
e. All of the above

Answer 56

c. Exponential amplification

Question 57

In relation to the usefulness of Variable number of tandem repeats (VNTRs) in forensic science, which statement is incorrect:
Select one:
a. Two unrelated individuals usually do not share the same pair of VNTRs
b. VNTRs are not heritable
c. VNTRs can be easily analysed with PCR
d. VNTRs are visualised as separated fragments on a gel
e. Short repeated sequences of variable length are useful markers

Answer 57

b. VNTRs are not heritable

Question 58

All the following statements regarding chromatin are correct EXCEPT ONE - which one?
Select one:
a. Specific binding sequence motifs on the DNA define the fixed positions occupied by nucleosomes
b. Histone acetylation plays an important part in the control of gene expression by nucleosomes
c. A nucleosome is the smallest DNA packaging unit in heterochromatin and euchromatin
d. DNA wraps around nucleosomes, which are composed of 4 dimers of histones H2A, H2B, H3 and H4.
e. Histones can bind to negatively-charged DNA because they are rich in positively charged residues

Answer 58

a. Specific binding sequence motifs on the DNA define the fixed positions occupied by nucleosomes

Question 59

The telomerase enzyme:
Select one:
a. relies on a bound RNA molecule to ensure full replication of the ends of chromosomes.
b. is encoded by telomeric DNA.
c. is required for sister chromatid separation during mitosis, to dissociate newly replicated homologous telomeres.
d. degrades unreplicated telomeres to ensure accurate replication of chromosomes ends.
e. is required for unwinding the DNA double helix at the onset of replication.

Answer 59

a. relies on a bound RNA molecule to ensure full replication of the ends of chromosomes.

Question 60

DNA synthesis catalysed by DNA polymerases:
Select one:
a. requires discontinuous synthesis of RNA primers on both strands of the replication fork.
b. requires only two RNA primers for both strands of the replication fork.
c. requires a primer providing a free 3’-OH group and proceeds only in the 5’ to 3’ direction.
d. requires a primer providing a free 5’-phosphate group and proceeds only in the 3’ to 5’ direction.
e. proceeds first in the 5’ to 3’ on the leading strand and then in the 3’ to 5’ direction on the lagging strand of the DNA template.

Answer 60

c. requires a primer providing a free 3’-OH group and proceeds only in the 5’ to 3’ direction.

Question 61

Retrotransposons are genetic elements that can increase genome size. They:
Select one:
a. likely evolved from ancestral retroviruses.
b. they encode their own DNA-dependent DNA polymerase.
c. they likely evolved from the integration of genomic material of ancestral DNA viruses into the cellular DNA.
d. are universally found in cellular organisms.
e. represent a smaller fraction of the human genome than coding sequences (<2%)

Answer 61

a. likely evolved from ancestral retroviruses.

Question 62

The pigmentation of kernel in maize is not inherited in a classical Mendelian manner because:
Select one:
a. the pigmentation genes are often inactivated by the movement of DNA transposons.
b. the pigmentation genes are located on the X chromosome.
c. viral infection is required for the expression of the pigmentation genes.
d. the pigmentation genes are located on the mitochondrial genome.
e. the Ac element of maize is located on the chloroplast genome.

Answer 62

a. the pigmentation genes are often inactivated by the movement of DNA transposons.

Question 63

The repeated sequences found at chromosome telomeres are important because they:
Select one:
a. are required for the production of the enzyme telomerase, a critical protein for stabilising chromosome length.
b. are the origin of replication for chromosomes.
c. determine the segregation of sister chromatids during mitosis.
d. bind to an RNA molecule within the active site of telomerase, which allows extension of the 5’ end of the DNA molecule.
e. allow chromatin condensation by interaction with histone proteins.

Answer 63

d. bind to an RNA molecule within the active site of telomerase, which allows extension of the 5’ end of the DNA molecule.

Question 64

The temperate bacteriophages are viruses that infect bacteria and:
Select one:
a. are forced to switch from the lysogenic stage to lytic cycle when the two host cells undergo conjugation.
b. can only undergo lytic cycles.
c. can switch from the lysogenic stage to a lytic cycle upon induction by UV irradiation.
d. can switch from the lytic cycle to a lysogenic stage only in the presence of a helper plasmid.
e. can only undergo lysogenic cycles.

Answer 64

c. can switch from the lysogenic stage to a lytic cycle upon induction by UV irradiation.

Question 65

To jump from one locus of the genome to another, retrotransposons require the activity of the following enzymes:
Select one:
a. Histone deacetylase, reverse transcriptase, RNA polymerase
b. DNAase, Integrase, Phospholipase
c. RNA polymerase, Reverse transcriptase, Integrase
d. Exonuclease, Reverse transcriptase, Integrase
e. RNA-dependent RNA polymerase, Reverse transcriptase, Integrase

Answer 65

c. RNA polymerase, Reverse transcriptase, Integrase

Question 66

Which of the following is a consequence of the addition of a 7-methylguanosine at the 5’ end of pre-mRNA in eukaryotic cells?
Select one:
a. Increase in binding of the C-terminal tail of RNA polymerase to the pre-mRNA
b. Blockade of transcription initiation
c. Increase in binding of transcription factors to the mRNA
d. Increase in binding of of RNA polymerase to the TATA box
e. Protection from hydrolysis by exonucleases

Answer 66

e. Protection from hydrolysis by exonucleases

Question 67

Why is cleavage of nascent mRNA near the sequence AAUAAA a crucial step during termination of transcription in eukaryotic cells?
Select one:
a. After cleavage, poly A polymerase adds a polyA tail to the 3’ end of the transcript
b. After cleavage at 30 nucleotides upstream of this sequence, the mRNA forms a hairpin loop, resulting in dissociation of RNA polymerase.
c. RNA polymerase cleaves within this sequence to release the nascent mRNA
d. After cleavage, a 7-methylguanosine can be added to the 3’ end of the transcript
e. Poly A polymerase cleaves within this sequence to release the nascent mRNA

Answer 67

a. After cleavage, poly A polymerase adds a polyA tail to the 3’ end of the transcript

Question 68

Which of the following is crucial for splicing to occur at the correct positions in pre-mRNA?
Select one:
a. Spliceosomes contain a C-terminal tail domain that recognises the 3’ end of the intron
b. Spliceosomes recognise the sequence known as the TATA box at intron-exon junctions
c. Introns form hairpin loop structures that are recognised by spliceosomes
d. All introns contain the conserved sequence known as the TATA box that is recognised by spliceosomes
e. A small nuclear ribonucleoprotein inside the spliceosome base pairs with the 5’ end of the intron

Answer 68

e. A small nuclear ribonucleoprotein inside the spliceosome base pairs with the 5’ end of the intron

Question 69

Which of the following statements about exons is CORRECT?
Select one:
a. Exons are present only in mRNA molecules that contain intercistronic regions
b. Exons are removed after export of pre-mRNA into the cytoplasm
c. Prokaryotic transcripts always contain shorter exons than eukaryotic transcripts
d. They must be removed by splicing to form a coding mRNA molecule
e. Exons are joined together by a transesterification reaction during splicing

Answer 69

e. Exons are joined together by a transesterification reaction during splicing

Question 70

How many intercistronic regions are there in a polycistronic mRNA that encodes 3 genes?
Select one:
a. 1
b. 2
c. 3
d. 4
e. 5

Answer 70

b. 2

Question 71

A mutation in the sigma factor Sigma 28 (σ28) that abolishes the DNA-binding function of σ28 would prevent successful transcription of many genes because:
Select one:
a. sigma factors enable cleavage at the 3’ end of mRNA.
b. sigma factors catalyze the polymerisation reaction carried out by RNA polymerase during RNA synthesis.
c. sigma factors are important for adding a 3’ polyA tail to nascent RNA molecules.
d. sigma factors bind directly to RNA polymerase and the -35 and -10 regions of promoters to initiate transcription.
e. sigma factors are essential for enhancing the binding of RNA polymerase to the DNA template during transcription elongation.

Answer 71

d. sigma factors bind directly to RNA polymerase and the -35 and -10 regions of promoters to initiate transcription.

Question 72

Where on DNA would the LacI repressor bind in an E. coli cell that is growing in low glucose and high lactose?
Select one:
a. lac promoter
b. lac promoter and lac operator
c. lac operator
d. LacZ gene
e. The repressor would not bind

Answer 72

e. The repressor would not bind

Question 73

In gene regulation, an inducer can
Select one:
a. only control the activator protein.
b. promote premature transcription termination.
c. bind to and inactivate the repressor protein.
d. bind to and cause bending of DNA.
e. bind to and activate the promoter.

Answer 73

c. bind to and inactivate the repressor protein

Question 74

Transcriptional attenuation of the trp operon involves
Select one:
a. the CAP protein and cAMP.
b. translational stalling of a tryptophan-rich leader peptide.
c. TrpR binding to a tryptophan-rich leader peptide.
d. the TrpR repressor molecule.
e. the CAP protein and TrpR.

Answer 74

b. translational stalling of a tryptophan-rich leader peptide.

Question 75

Histone acetylation prevents the interaction between neighbouring nucleosomes because
Select one:
a. it causes nucleosomes to dissociate from the DNA.
b. acetylated nucleosomes are targeted for degradation.
c. it inhibits the SWI-SNF complex.
d. it prevents the histone tails from interacting with the adjacent nucleosomes.
e. it promotes binding of the repressor protein.

Answer 75

d. it prevents the histone tails from interacting with the adjacent nucleosomes.

Question 76

When a synthetic polymer of the dinucleotide GU (i.e. …GUGUGUGUGU…) is translated, it produces the peptide with the amino acid sequence…val – cys – val – cys – val. When a polymer of the trinucleotide GGU (i.e. GGUGGUGGUGG…) is translated, a mixture of three peptides is produced: ….gly – gly – gly – gly (known as poly gly), poly val and poly trp.

The triplet code that specifies cysteine (cys) is therefore:
Select one:
a. GUG
b. UGU 
c. GGU
d. UGG
e. Cannot be determined

Answer 76

b. UGU

Question 77

In relation to the genetic code, which of the following statements is correct?
Select one:
a. Each amino acid is specified by a single codon
b. Apart from the stop codons, each codon codes for a specific amino acid
c. Protein synthesis requires 61 different tRNAs - one for each non-stop codon.
d. The genetic code has more codons in eukaryotes than prokaryotes
e. There are 3 stop codons and 3 start codons

Answer 77

b. Apart from the stop codons, each codon codes for a specific amino acid

Question 78

Together tRNAs and tRNA synthetases activate amino acid in preparation for their incorporation in a polypeptide by ribosome. Which statement is correctly paired?
Select one:
a. tRNA synthetases may charge related amino acids due to a wobble in the interaction with the tRNA
b. tRNA attach to a specific amino acid by base-pair complementarity
c. tRNAs are modified RNAs that code for their cognate tRNA synthetase
d. All tRNA folds into a characteristic L shape allowing recognition by the ribosome
e. All tRNA synthetases have the same fold (i.e. shape) and mode of recognition of tRNAs

Answer 78

d. All tRNA folds into a characteristic L shape allowing recognition by the ribosome

Question 79

Which of the following statements regarding the genetic code is correct?
Select one:
a. Wobble base pairing of the anticodon decreases the number of codons a tRNA can recognise
b. Alignment of codon and anticodon base pairs is in a parallel direction
c. More than one nucleotide triplet can serve as a stop codon
d. More than one amino acid can be coded for by a single codon
e. The 3rd base of the codon is the most critical for recognition by the tRNA anticodon

Answer 79

c. More than one nucleotide triplet can serve as a stop codon

Question 80

All of these features are in most (but not necessarily all) tRNA synthetases except for one that is INCORRECT:
Select one:
a. Binding surface for the ribosome
b. Active site editing, i.e. the removal of incorrect amino acids
c. Active site catalysing the covalent attachment of amino acids to tRNAs
d. Binding surface for the tRNA anticodon
e. Binding site for ATP

Answer 80

a. Binding surface for the ribosome - Aminoacyl tRNA synthetases (aaRS) do not bind the ribosome (nor the mRNA)

Question 81

In the process of eukaryotic protein synthesis, each mRNA is translated:
Select one:
a. Only when the ribosomes are attached to the endoplasmic reticulum
b. Simultaneously by many ribosomes until it is degraded
c. Simultaneously by many ribosomes for the life of the cell
d. By only one ribosome at a time
e. Only once

Answer 81

b. Simultaneously by many ribosomes until it is degraded

Question 82

Which structure is correctly paired with its function in relation to the initiation of protein synthesis in prokaryotes?
Select one:
a. AUG codon – the first three nucleotides at the 5´ end of every mRNA
b. fmet-tRNA – undergoes codon/anticodon base pairing with the 16S rRNA
c. mRNA – interacts first with the large ribosomal subunit
d. Shine-Dalgarno sequence – acts as a start signal for protein synthesis
e. Initiation factors– promote assembly of small and large rRNA subunits

Answer 82

d. Shine-Dalgarno sequence – acts as a start signal for protein synthesis

Question 83

Emma aims to follow the production of an antiviral factor in infected cells. She detects the mRNA of this factor using microarray analysis but fails to detect the protein by Western blot. This negative result may be due to:
Select one:
a. Proteasome degradation following polyubiquitination of the protein
b. The virus induces an arrest of cellular transcription
c. Targeting of the antiviral factor to the nucleus
d. The protein has failed to fold and is degraded by chaperones

Answer 83

a. Proteasome degradation following polyubiquitination of the protein

Other valid answers:

• The protein is activated by proteolytic removal of a region that is recognised by the antibody
• The virus prevents cellular translation (e.g. by inhibiting cap-dependent translation)

Question 84

Deletion of a single base from the coding sequence of a gene is likely to result in any of the following outcomes except one. Which one?
Select one:
a. A non-functional protein being generated
b. A protein whose sequence of amino acids differs markedly from normal
c. A protein that is still functional depending on the site of the deletion
d. A shorter protein
e. A protein in which a single amino acid is replaced by another amino acid

Answer 84

e. A protein in which a single amino acid is replaced by another amino acid

This is could be the result of a substitution not deletion.
Insertions and deletions change the reading frame, which results in a protein sequence that differs markedly from normal after the mutation and usually shorter due to premature stop codons. Depending on the location, the protein may or may not be still functional.

Question 85

The normal sequence of amino acids in a particular peptide is Val-Asn-Pro-Thr-Arg-Cys.  The sequences produced following a number of different mutations are listed below.  Which sequence is correctly matched with the type of mutation most likely to have caused the observed change in the peptide?
Select one:
a. Val-Asn-Pro : Frameshift
b. Val-His-Pro-Thr-Arg-Cys  : Silent
c. Val-Val-His-Leu-Ala : Frameshift
d. Val-Asn-Ala-His-Glu-Ser  : Nonsense
e. Val-Asn-Pro-Thr-Arg-Cys : Missense

Answer 85

c. Val-Val-His-Leu-Ala : Frameshift