Solved MCQs(1-27) along with explanation concerned with DNA replication.

1. DNA REPLICATION
SOLVED MCQS WITH EXPLANATION
(1-27)
BY
Dr Ichha Purak
University Professor
Department of Botany
Ranchi Women’s College,Ranchi

2. Which of the following reactions is required for proofreading (i.e. correcting
replication errors) during DNA replication by DNA polymerase III?
a) 3' - 5' exonuclease activity
b) 5' - 3' exonuclease activity
c) 3' - 5' endonuclease activity
d) 5' - 3' endonuclease activity
Answer c) 3' - 5' endonuclease activity
Explanation
The proofreading activity possessed by many DNA polymerases is an
exonuclease activity that degrades mismatched bases that have been wrongly
incorporated into the growing chain. This is, therefore, an exonuclease
activity (exonucleases digest from theend of a DNA chain) and it operates
'backwards' from the 3' growing end, i.e. 3' - 5'.
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3. How does the mismatch repair system distinguish between the parental (i.e. correct)
DNA strand and the newly synthesised strand containing the mismatched base?
a) Thymine in the parental strand of the helix is methylated at GATC.
b) Thymine in the new strand of the helix is methylated at GATC.
c) Guanine in the parental strand of the helix is methylated at GATC.
d) Guanine in the new strand of the helix is methylated at GATC.
Answer d) Guanine in the new strand of the helix is methylated at GATC.
Explanation
Mismatch repair is a system that repairs mismatches that have slipped
evaded proofreading during DNA replication. DNA becomes methylated
at the G of GATC sequences after replication; however, this does not
occur immediately so the new strand, containing the error, can be
distinguished from the methylated parental strand
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4. What is the name of the DNA repair system in E. coli in which dual incisions
are made in the damaged part of the double helix, and a 12-13 base
segment is removed and replaced with new DNA?
a) Mismatch repair
b) Base excision repair
c) Nucleotide excision repair
d) AP site repair
Answer c) Nucleotide excision repair
Explanation
Nucleotide excision repair is an almost universal repair mechanism in which
a section of damaged DNA is removed and replaced with new DNA by a
DNA polymerase. It is used to repair photoproducts caused by UV damage
and bulky DNA lesions caused by a variety of mutagens.
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5. Which of the following is the name of the human genetic disorder resulting from defects in
nucleotide excision repair?
a) Hereditary nonpolyposis colorectal cancer (HNPCC)
b) Xeroderma pigmentosum (XP)
c) Lynch syndrome
d) Diabetes
Answer b) Xeroderma pigmentosum (XP)
Explanation
People born with the disorder, xeroderma pigmentosum, have a mutation
in one of the genes coding for nucleotide excision repair enzymes.
Therefore they are unable to carry out efficient repair on sunlight damage
and they are hypersensitive to sunlight. They have to protect their skin
from daylight or risk getting skin cancer.
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6. In which of the following would you find telomeres?
a) Human mitochondrial DNA
b) Human chromosomes
c) Bacterial chromosomes
d) The influenza virus genome
Answer b) Human chromosomes
Explanation
Telomeres are found at the ends of the linear double-stranded DNA
molecules in human chromosomes. They protect chromosome ends from
nucleases and they also provide a special mechanism for replication of
chromosome ends, using the enzyme telomerase. Circular molecules such
as bacterial chromosomes and most mitochondrial genomes do not need
these specialised DNA ends
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7. Which of the following enzymes are used to join bits of
DNA?
a) DNA ligase
b) DNA polymerase
c) primase
d) Endonuclease
Answer DNA ligase
Explanation
DNA ligase enzyme joins bits of DNA on lagging strand
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8. Why is an RNA primer considered essential during DNA synthesis by
DNA polymerase III?
a) The enzyme requires a free 3'-PO4 group.
b) The enzyme requires a free 5'-PO4 group.
c) The enzyme requires a free 5'-OH group.
d) The enzyme requires a free 3'-OH group.
e) There is no particular reason, that is simply the observation.
Answer d) The enzyme requires a free 3'-OH group
Explanation
Polymerase III requires a free 3'-OH group to begin synthesis of DNA. An
RNA primer provides a 3' hydroxyl group, from which the DNA polymerase
can start synthesis.
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9. Which molecule serves to destabilize the DNA helix in order to open it
up, creating a replication fork?
a)DNA helicase
b) DNA ligase
c) DNA polymerase
d) SSBPs
e) DNA gyrase
Answer a) DNA Helicase
Explanation
The DNA helicase enzyme destabilizes the DNA helix by breaking hydrogen
bonds
For DNA Replicaion, unwinding of DNA is done by
a) Helicase
b) ligase
c) Hexonuclease
d) Topoisomerase
Answer a) Helicase
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10. Semi conservative replication of DNA was first demonstrated in
a) Escherichia coli
b) Streptococcus pneumonae
c) Salmonella typhimuriam
d) Drosophila melanogaster
Answer a) Escherichia coli
Explanation
The experiment was performed by Maselson and Stahl (1959 ) to know about mode of
replication. The experiment was performed using Escherichia coli
Mode of DNA replication in E.coli is
a) Conservative and unidirectional
b) Semiconservative and unidirectional
c) Conservative and bidirectional
d) Semiconservative and bidirectional
Answer d) Semiconservative and bidirectional
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11. During the replication of DNA, the synthesis of DNA on lagging
strand takes place in segments, these segments are called
a) Satellite segments
b) Double helix segments
c) Kornbeg segments
d) Okazaki segments
Answer d) Okazaki segments
Explanation
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12. The elongation of the leading strand during DNA synthesis
a) Progresses away from the replication fork
b) Occur in 3’-5’ direction
c) Produces Okazaki fragment
d) Depend on the action of DNA polymerase
Answer d) Depend on the action of DNA polymerase
Explanation
Takes place with advancement of DNA polymerase on the
template ,On leading strand it is towards the neck of Replication
forlk in 5’----3’ direction
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13. Eukaryotes differ from prokaryote in mechanism of DNA replication
due to:
a) Different enzyme for synthesis of lagging and leading strand
b) Use of DNA primer rather than RNA primer
c) Unidirectional rather than bidirectional replication
d) Discontinuous rather than semidiscontinuous replication
Answer
d) Discontinuous rather than semidiscontinuous replication
Explanation
Okazai fragments are smaller and are formed discontinuously
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14. True replication of DNA is possible due to
a) Hydrogen bonding
b) Phosphate backbone
c) Complementary base pairing rule
d) None of the above
Answer c) Complementary base pairing rule
Explanation
Adenine pairs with Thymine only( A = T ) and Guanine pairs with
Cytosine ( G ≡C ) only ,because of strict base pairing pattern
the new daughter strand is assembled according to bases
present on the template
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15. All of the following are differences between eukaryotic and
prokaryotic DNA replication except __________.
a) the type and number of polymerases involved in DNA synthesis
b) multiple vs. single replication origins
c) the rate of DNA synthesis
d) the ability to form a replication fork
Answer
d) the ability to form a replication fork
Explanation
Both prokaryotes and eukaryotes form replication forks during DNA
replication
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16. Telomeres have all of the following properties or characteristics
except __________.
a) the telomerase enzyme
b) forming "hairpin" loops
c) a link to the aging process in eukaryotic cells
d) being found in both eukaryotic and prokaryotic chromosomes
Answer d) being found in both eukaryotic and prokaryotic
chromosome
Explanation
The telomere sequence occurs at the ends of linear eukaryotic
chromosomes. Prokaryotes have circular chromosomes so donot
have the terminities or telomeres
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17. Assume that in the following cross, d+ x e+, the only gametes formed
consist of de alleles. This may be due to a phenomenon called
__________.
a) gene conversion
b) reciprocal recombination
c) nonreciprocal recombination
d) genetic exchange
Answer a) Gene conversion
Explanation
Gene conversion is a situation in which the products of meiosis do not fit the
normal Mendelian patterns of inheritance. One or more alleles are not
represented, and it is as if one gene has been converted into another
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18. In the Meselson-Stahl experiment, which mode of replication
can be eliminated based on data derived after one generation
of replication?
a)dispersive
b) semiconservative
c) conservative
d) all three modes
Answer c) Conservative
Explanation
Conservative replication theory says that parental strands re-
anneal with parental strands, and daughter strands with daughter
strands after DNA replication. This experiment showed this to be
not true.
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19. The discovery of Okazaki fragments suggested that DNA synthesis
is __________.
a) discontinuous
b) continuous
c) 3' to 5'
d) semiconservative
Answer a) discontinuous
Explanation
DNA must be synthesized on both strands of the double helix at the
same time. However, one strand runs in a 5' to 3' direction and one runs
in the 3' to 5' direction. In order for replication to occur on both strands in
the same direction simultaneously, one strand must be made in a
discontinuous fashion (in short pieces—Okazaki fragments) and
reannealed later.
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20. In vivo synthesis of DNA is __________
a) 3' to 5'
b) 5' to 3'
c) both 3' to 5' and 5' to 3'
d) neither 3' to 5' nor 5' to 3'
Answer b) 5' to 3‘
Explanation
DNA polymerase must add nucleotides in a 5' to 3' direction.
Therefore, DNA synthesis must also occur in this direction
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21. Semiconservative replication of DNA means that
a) Each daughter duplex will have one of the original strand and one new
strand
b) One daughter duplex will be entirely new and other will have both original
parental strands
c) Both daughter duplexes will be entirely new and the parental duplex will be
degraded
d) Each strand of each daughter duplex will have parts of parental strands and
part of new strands
Answer a) Each daughter duplex will have one of the original strand and one
new strand
Explanation
Experiment conducted by Maselson and Stahl (1959) clearly demonstrates that
density of DNA after first generation is intermediate between n N 15 and N14
which supports Semiconservative replication
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22. The theta structure that results from autoradiography of replicating bacterial
chromosome illustrates all the following except
a) The unreplicated portion of the chromosome
b) The pair of daughter molecules in the process of being formed
c) Relication forks
d) The replication origin
Answer d) The replication origin
Explanation
When replication starts from origin it proceeds bidirectionally and when two
replication forks meet each other on other side theta like structure is formed
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23. A replicating prokaryotic chromosome has --------
replicaton forks and a replicating eukaryotic
chromosome has ------------ replication forks.
a) one:many
b) Two :two
c) Two : many
d) Many : many
Answer c) two : many
Explanation
As prokaryotic chromosome (DNA ) is circular ,
it has only one origin (ori) of replication
which consists of two Y shaped replication
whereas eukaryotic chromosome is large and
linear having many orgins each of which consists
of 2 replication forks
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24. Which of the following forms of DNA can serve as a template for
DNA polymerase
a) Partially double stranded DNA
b) Circular double stranded DNA
c) Intact double stranded DNA
d) Circular single stranded DNA
Answer a) Partially double stranded DNA
Explanation
DNA double helix gradually opens by DNA helicse by dissolving H
bonds between base pairs giving single strands which behave
as template for DNA polymerase
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All input of energy is not required for which of the following steps of
replication ?
a) Separation of two stranda
b) Unwinding of DNA duplex
c) Linking together of the individual deoxynucleotides
d) All of the above require energy
Answer d) All of the above require energy
Explanation
Most of the steps of DNA replication require energy which is
generally provided by hydrolysis of energy rich comounds like
ATP

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Why should patients with Xeroderma pigmentosa avoid sunlight
a) The UV wavelengths do irrepairable damage to DNA
b) Sunlight inhibits any residual DNA repair activity of the cell
c) These patients lack pigmentation to protect them from burning
d) Sunlight inhibits DNA polymerases
Answer a) The UV wavelengths do irrepairable damage to
DNA

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The exonuclease activity of the DNA polymerases functions to
a) Remove the RNA primer sequences
b) Proofread the new DNA strand and remove inappropriate
nucleotides
c) Maximize the fidelity of DNA replication
d) All of the above
Anwer d) All of the above
Explanation
By having exonuclease activity it can remove the nucleotides from the
ends of Okazaki fragments and also by moving backwards for proof
reading

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The last DNA to be replicated in the eukaryotic chromosome is
a) Telomeres at the end of the chromosomes
b) Heterochromatin
c) Euchromatin in the arms of the chromosome
d) Facultative heterochromatin
Answer b) heterochromatin
Explanation
As heterochromatin is densely packed it is always late replicating

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