DNA Replication and Repair

Question 1

If bacteria from the heavy N media
were transferred to light N medium
and allowed to replicate their DNA
once, the new DNA had an
intermediate density.

Answer 1

Ruled out Conservative Model

Question 2

They subjected the DNA to heat,
which denatured it into single strands
and produced one heavy and one light
band

Answer 2

dispersive model was ruled
out.
Fig. 6.7

Question 3

Suppose the bacteria in the Meselson-Stahl experiment were allowed to replicate their DNA
twice after transfer from heavy N to light N (instead of just once). Their DNA molecules would consist of
what type(s) of density, and which model(s) would this result support?

A. Intermediate density DNA only, semiconservative and dispersive
B. Light density DNA only, conservative and semiconservative
C. Half intermediate density DNA and half heavy density DNA, semiconservative and conservative.
D. Half intermediate density DNA and half light density DNA, semiconservative
E. Half heavy density DNA and half light density DNA, conservative

Answer 3

E. Half heavy density DNA and half light density DNA, conservative

Question 4

Initiator proteins pry
apart the two strands of the
double helix at

Answer 4

replication
origins

Question 5

DNA polymerase synthesizes DNA using

Answer 5

parental strand as template.

Question 6

In DNA polymerase Nucleotides are added in which Direction?

Answer 6

3’ end of
growing strand

Question 7

In DNA polymerase energy of two _______Bonds

Answer 7

phosphoanhydride

Question 8

DNA polymerase
enzyme catalyzes
consecutive
additions while

Answer 8

Remaining Attacthed

Question 9

Chp 6 Q2: What would happen in a DNA replication reaction with a 10-fold excess of normal
dCTP relative to dideoxyCTP (ddCTP), which has only H’s bonded to the 2’ and 3’ carbons of its
deoxyribose (no OH groups at these positions)?
A. DNA replication would proceed without interruption, with incorporation of both dCMP and
ddCMP.
B. DNA replication would proceed until incorporation of the first dCMP is needed, after which it
would stop.
C. DNA replication would proceed to different extents in different molecules, until
incorporation of the first ddCMP, after which it would stop.
D. DNA replication would proceed with incorporation of only dCMP (no ddCMP).
E. ddCMP would be incorporated only at positions in which dCMP is never incorporated (eg, not
opposite a dGMP in the template).

Answer 9

C. DNA replication would proceed to different extents in different molecules, until
incorporation of the first ddCMP, after which it would stop

Question 10

The replication fork is asymmetrical, with one strand growing in the
5’ to 3’ direction and the other in the 3’ to 5’ direction.

Answer 10

But all DNA
polymerases synthesize in the 5’ to 3’ direction

Question 11

The DNA strand growing in the 3’ to
5’ direction is discontinuously
synthesized in small _______in the 5’ to 3 ‘ direction

Answer 11

Okazaki
fragments

Question 12

Polymerase moves backward on
this

Answer 12

“lagging strand.”

Question 13

The______are later stitched
together to make the lagging strand
continuous.

Answer 13

Ozaki Fragments

Question 14

Note that each strand is
synthesized by

Answer 14

both replication
forks and has both leading and
lagging strands

Question 15

Which of the following statements about one newly synthesized strand of a human
chromosome is true?
A. It was synthesized from a single origin solely by leading strand DNA synthesis.
B. It was synthesized from a single origin solely by lagging strand DNA synthesis
C. It was synthesized from a single origin by a mixture of leading and lagging strand DNA
synthesis.
D. It was synthesized from multiple origins solely by leading strand DNA synthesis
E. It was synthesized from multiple origins by a mixture of leading and lagging strand DNA
synthesis.
Chp 6, Q4

Answer 15

E. It was synthesized from multiple origins by a mixture of leading and lagging strand DNA
synthesis.
Chp 6, Q4

Question 16

proofreads during synthesis,
removing incorrectly added nucleotides and
resynthesizing that addition.

Answer 16

DNA polymerase

Question 17

Choose correct order of DNA Polymerase proof reading
A) Correctly paired 3’ end allows for addition of next nucleotide. B) Mispaired nucleotide removed by proof reading. C)Polymerase adds incorrect nucleotide. D) continues synthesis in 5 to3 direction

1) A, C, D, B
2) C, D, A, B
3) C, B, A, D
4) D, B, A, C

Answer 17

3) C, B, A, D

Question 18

A hypothetical
DNA polymerase
synthesizing in the
3’ to 5’ direction
could not
proofread because

Answer 18

the 5’ triphosphate
would be gone
after the first
addition. Hence
the need for
lagging strand
synthesis.
Fig. 6.16

Question 19

Polymerization and proofreading are accomplished by

Answer 19

by
different sites on the enzyme.

Question 20

Short lengths of RNA act as

Answer 20

primers for DNA synthesis

Question 21

RNA polymerases do not

Answer 21

need a base-paired end (PRIMER)

Question 22

Primases are RNA
polymerases

Answer 22

synthesize short RNA
primers from DNA
templates in 5’ to 3’
direction.

Question 23

leading strand synthesis, only

Answer 23

one primer is
needed

Question 24

lagging strand synthesis an ____ needed for each okazaki fragment

Answer 24

RNA primer

Question 25

Steps of Synthesis of Okazaki fragments

Answer 25

A) DNA polymerase adds nucleotides to the 3’ end of new RNA Primer to synthesize OF

B) DNA Polymerase finishes OF

C) Prev RNA primer removed by nucleases and replaced with DNA by Polymerase

D)Nick Sealed by Ligase

Question 26

DNA ligase uses the energy of_____

Answer 26

ATP hydrolysis to
join together Okazaki fragments

Question 27

In DNA ligase ____ is released to seal nick of DNA strand

Answer 27

AMP

Question 28

Sliding clamps keep the
DNA polymerase

Answer 28

fully
attached to DNA template,

Question 29

these detach each time
an Okazaki fragment is
finished

Answer 29

Sliding clamps

Question 30

DNA helicases and
single-stranded
binding proteins

Answer 30

unwind he helix and
keep the DNA in the
unwound state

Question 31

The proteins involved in
DNA replication are held
together in a

Answer 31

large
multienzyme complex.

Question 32

the clamp loader

Answer 32

moves the lagging strand
polymerase to the next primer

Question 33

topoisomerases does what

Answer 33

Relieve tension

Question 34

telomerase, an enzyme
with a bound RNA
template

Answer 34

a reverse
transcriptase) that allows
extra copies of the
telomere repeat to be
added to the telomere
template strand

Question 35

telomerase identifies
A) end of choromsome
B) beginning of chromosome

Answer 35

A

Question 36

Telomerase
A. uses a DNA primer and an RNA template
B. uses an RNA primer and an RNA template
C. does not use a primer and uses a DNA template
D. does not use a primer and uses an RNA template
E. uses a DNA primer and a DNA template

Answer 36

A

Question 37

Addition of deoxyribonucleotides to only the lagging strand DNA at the end of
a chromosome is synthesized during addition of telomere sequences by
A. Telomerase
B. Primase
C. DNA polymerase
D. a nuclease
E. DNA ligase

Answer 37

C

Question 38

Depurination reactions remove _____
from DNA
A) Adenines and Guanines
B) Cytosin and Thymine
C) Thymine dimers
D) Cytosin and Guannines

Answer 38

A

Question 39

Deamination of Cytosine produces

A) Cytosine
B) Guanine
C) Thymine
D) Uracil

Answer 39

D

Question 40

What does ultra Violet light promote?

A) Linkages between Guannines
B) LInkages between Cytosine
C) Linkages between Adinine
D) Linkages between Thymine

Answer 40

D

Question 41

Individuals with Xeroderma pigmintation is caused by ?

A) Depurination
B) UV light
C) Deamination
D) Damage by thymine

Answer 41

B

Question 42

Failure of proof reading will result in
A) Addition of correct nucleotide
B) Incorporation of incorrect nucleotide
C) Deletion of a nucleotide during DNA replication
D) Both B and C
E A, B, C

Answer 42

D

Question 43

Which of the following is NEVER involved in the repair of a single damaged
deoxyribonucleotide base in a single strand of double helical DNA?
A. Recognition of the damaged base by a protein
B. Cutting of the damaged DNA strand on both sides of the damaged base
C. DNA ligase activity
D. Synthesis of the region of DNA damage with a repair DNA polymerase, using the other
undamaged strand as template
E. Excision of the purine or pyrimidine base and direct replacement with the correct base on
the remaining deoxyribose in the DNA strand (no excision of the remaining deoxyribose).

Answer 43

E

Question 44

Mismatched repair removes

A) DNA strand with mod
B) Replication errors caught by proof reading
C) DNA strand with out mod
D) A and B

Answer 44

C

Question 45

Chp 6 Q9: The strand that carries a base WITH a modification prior to incorporation of a
mismatched nucleotide in mismatch repair is
A. the template strand, which is used as a template to repair the newly synthesized strand.
B. the template strand, which is repaired to have a complementary sequence to the newly
synthesized strand.
C. the newly synthesized strand, which is repaired to have a complementary sequence to the
template strand.
D. the newly synthesized strand, which is used a a template to repair the template strand.
E. both the template and the newly synthesized strand.

Answer 45

A

Question 46

: Place the steps of repair of double-stranded DNA breaks as mediated by recombination in the
correct order.
A. Paring of the 3’ ends of damaged DNA with the intact helix, nucleases cut back the 5’ ends of the severed
strands, synthesis of DNA using the intact helix as template, broken helix reformed
B. Paring of the 3’ ends of damaged DNA with the intact helix, nucleases cut back the 5’ ends of the severed
strands, broken helix reformed, synthesis of DNA using the intact helix as template
C. Nucleases cut back the 5’ ends of the severed strands, paring of the 3’ ends of damaged DNA with the
intact helix, synthesis of DNA using the intact helix as template, broken helix reformed
D. Nucleases cut back the 5’ ends of the severed strands, paring of the 3’ ends of damaged DNA with the
intact helix, broken helix reformed, synthesis of DNA using the intact helix as template
E. Broken helix reformed, nucleases cut back the 5’ ends of the severed strands, paring of the 3’ ends of
damaged DNA with the intact helix, synthesis of DNA using the intact helix as template

Answer 46

D

Question 47

In Homologous joining
A) There is RNA Loss
B) DNA loss
C) Both A & B
D) replicated DNAs are still close to each other
E) B and B

Answer 47

B

Question 48

Unlike most cancers, there is a cancer that causes eye tumors in young children rather than
old people. The children are not born with the cancer but develop them as adolescents. This cancer
would most likely be caused by
A. a recessive loss of function cancer mutation in a single gene, with the other copy wild type
B. A dominant cancer mutation in a single gene inherited from one of their parents
C. inheritance of two recessive loss of function cancer mutation genes, one from each parent
D. a recessive loss of function cancer mutation in a single gene, with the other copy originally wild type
but mutating to a recessive loss of function cancer form during childhood
E. mutations of multiple genes to a dominant cancer causing form in childhood

Answer 48

D

Question 49

Which of the following results in the Meselson-Stahl experiment would support the semiconservative rather than the dispersive model of DNA replication?

A. In the first generation of replication after transfer from 15N to 14N medium the DNA was of intermediate density.
B. In the first generation of replication after transfer from 15N to 14N medium the DNA was of heavy density.
You Answered
C. In the first generation of replication after transfer from 15N to 14N medium the DNA, after denaturation, was half heavy density and half light density.
D. In the second generation of replication after transfer from 15N to 14N medium the DNA was half intermediate density and half light density.
Correct Answer
E. C and D both supported the semiconservative model rather than the dispersive model.

Answer 49

E. C and D both supported the semiconservative model rather than the dispersive model.

Question 50

What does the DNA polymerase that fills in the gap between two Okazaki fragments after they are synthesized use as a primer to synthesize the missing DNA?

A. an RNA molecule
B. the 3’ end of the Okazaki fragment synthesized second
C. the 3’ end of the Okazaki fragment synthesized first
D. the 5’ end of the Okazaki fragment synthesized second
E. It does not need a primer.

Answer 50

B

Question 51

The enzyme that seals the nicks in DNA after complete synthesis of two Okazaki fragments and removal of the RNA primer between them is

A. DNA ligase
B, DNA polymerase
C. Primase
D. Topoisomerase
E. helicase

Answer 51

A

Question 52

Which of the following enzymes uses an RNA template?

A. the primase
B. the telomerase
C. repair DNA polymerase
D. the lagging strand DNA polymerase
E. C and D are both correct answers.

Answer 52

B

Question 53

Repair of which one of the following can occur on either the newly synthesized DNA strand or the older template strand?

A. Depurination
B. Deamination
C. Thymine dimers
D. None of the above (A, B or C)
E. All of the above (A, B and C)

Answer 53

E

Question 54

Mismatch repair

A. eliminates depurinated deoxyribonucleotides from DNA.
B. eliminates thymine dimers from DNA.
C. eliminates a mispaired base introduced during DNA replication from the newly synthesized strand.
D. eliminates a mispaired base introduced during DNA replication from the original template strand.
E. C and D are both correct.

Answer 54

C

Question 55

Which of the following repair mechanisms can be used to repair a double-stranded break in DNA?

A. use of the other strand of the same DNA molecule to fix the damage
B. nonhomologous end joining
C. homologous recombination with the other chromosome of the same type
D. B and C are both correct answers
E. A, B and C are all correct answers

Answer 55

D

Question 56

Which of the following results would support either the semiconservative or the dispersive model of DNA
replication (i.e., it ruled out the conservative model but supported either the semiconservative or dispersive
model)?
A. In the first generation of replication after transfer from 15N to 14N medium the DNA was of intermediate
density.
B. In the first generation of replication after transfer from 15N to 14N medium the DNA, after denaturation, was
half heavy density and half light density.
C. In the second generation of replication after transfer from 15N to 14N medium the DNA was half
intermediate density and half light density.
D. A and B both supported the semiconservative model or the dispersive model, but C did not.
E. B and C both supported the semiconservative model or the dispersive model, but A did not

Answer 56

A

Question 57

2) All DNA replication proceeding from one replication origin involves a total of _______(1) DNA polymerases
and ________(2) helicases.
A. Two (1), One (2)
B. Two (1) Two (2)
C. Four (1), Two (2)
D. Four (1), Four (2)
E. Two (1). Four (2

Answer 57

C

Question 58

Which of the following is (are) NOT true about DNA synthesis?
A. The DNA polymerase adds deoxyribonucleotides to the 5’ end of the newly synthesized strand.
B. The DNA polymerase cleaves two phosphates from the 5’ end of the newly added deoxyribonucleotide during
DNA synthesis.
C. The addition of a deoxyribonucleotide to the newly synthesized strand is facilitated if it base pairs correctly
with the template strand.
D. If base pairing is not correct after addition of a deoxyribonucleotide the polymerase usually cleaves the
phosphodiester bond of the added deoxyribonucleotide and repeats the addition.
E. All of the above (A, B, C and D) are correct (no false statements).

Answer 58

A

Question 59

If DNA synthesis occurred in the _____(1), it would pose a problem for proofreading because there
would be no __________ (2) on the newly synthesized strand to allow resynthesis of a mismatched
deoxyribonucleotide after its removal.
A. 3’ to 5’ direction, triphosphate on the 3’ end
B. 3’ to 5’ direction, triphosphate on the 5’ end
C. 5’ to 3’ direction, triphosphate on the 5’ end
D. 5’ to 3’ direction, triphosphate on the 3’ end
E. 3’ to 5’ direction, hydroxyl on the 3’ end

Answer 59

B

Question 60

What does the DNA polymerase that fills in the gap between two Okazaki fragments after they are
synthesized use as a primer to synthesize the missing DNA when the primer between the two is degraded?
A. The 3’ end of the Okazaki fragment synthesized first
B. The 3’ end of the Okazaki fragment synthesized second
C. An RNA molecule
D. The 5’ end of the Okazaki fragment synthesized first
E. It does not need a primer

Answer 60

B

Question 61

Which of the following enzymes does NOT require a primer to begin synthesis?
A. DNA polymerase
B. A primase
C. A telomerase
D. A and C are both correct.
E. B and C are both correct

Answer 61

B

Question 62

When DNA ligase seals the nick in a molecule of DNA, what is released from attachment to the 5’ end of the
DNA nick?
A. ATP
B. AMP
C. A phosphate attached to the 5’ end of a deoxyribonucleotide
D. A phosphate attached to the 3’ end of a deoxyribonucleotide
E. Ribose

Answer 62

B

Question 63

__________ (1) adds DNA to the lagging strand template at the ends of chromosomes to lengthen the ends,
and it uses___________ (2) as a primer for this synthesis.
A. Telomerase (1), the lagging strand (2).
B. Telomerase (1), the lagging strand template (2).
C. DNA polymerase (1), the lagging strand (2)
D. DNA polymerase (1), the lagging strand template (2)
E. Telomerase (1), RNA (2)

Answer 63

B

Question 64

The enzyme that nicks the DNA ahead of a replication fork to relieve tension is
A. DNA polymerase.
B. Helicase.
C. Primase.
D. Topoisomerase
E. DNA ligase

Answer 64

D

Question 65

Which of the following DNA damage events might involve removal of an adenine base from DNA?
A. Deamination
B. Thymine dimer formation
C. Mismatch synthesis
D. a double-stranded break in the DNA
E. Depurination

Answer 65

E

Question 66

Place the following events in correct order during repair of a single strand of DNA, using the other
strand as a template for the repair.
A. Nicking and excision of the damaged strand, recognition of the damaged base or bases, DNA ligase
activity, resynthesis of the damaged region of the strand.
B. Nicking and excision of the damaged strand, recognition of the damaged base or bases, resynthesis
of the damaged region of the strand, DNA ligase activity.
C. Recognition of the damaged base or bases, nicking and excision of the damaged strand, resynthesis
of the damaged region of the strand, DNA ligase activity
D. Resynthesis of the damaged region of the strand, recognition of the damaged base or bases, nicking
and excision of the damaged strand, DNA ligase activity
E. Recognition of the damaged base or bases, DNA ligase activity, nicking and excision of the
damaged strand, resynthesis of the damaged region of the strand.

Answer 66

C

Question 67

Which of the following is NEVER involved in the repair of a damaged base in a single strand of double
helical DNA?
A. Recognition of the damaged base by a protein
B. Cutting of the damaged DNA strand on both sides of the damaged base
C. DNA ligase activity
D. Synthesis of the region of DNA damage with a repair DNA polymerase, using the other undamaged strand as
template.
E. Excision of the purine of pyrimidine base and direct replacement with the correct base on the remaining
deoxyribose in the DNA strand with no excision of the remaining deoxyribose

Answer 67

E

Question 68

n homologous recombination, what is the nature of the strand that provides the initial template for the
resynthesis of the DNA in the region that has the double-stranded break?
A. The 3’ end of a strand of the damaged DNA
B. The 5’ end of a strand of the damaged DNA
C. The DNA in the unbroken chromosome that is complementary to the invading 3’ end of the broken
chromosome.
D. The DNA in the unbroken chromosome that has the same sequence as the invading 3’ end of the broken
chromosome.
E. The 3’ and 5’ ends of a strand of the damaged DNA

Answer 68

C

Question 69

Why is there some positive selection for a mutation in hemoglobin that causes sickle cell anemia?
A. In the heterozygous genotype it provides resistance to malaria
B. In the homozygous genotype it has no effect on blood cells
C. In the homozygous genotype it causes blood cells to be less fragile than in the wild type
D. In the homozygous genotype it causes hemoglobin to be more soluble than in the wild type
E. There is not any positive selection for sickle cell anemia mutations

Answer 69

A

Question 70

Which of the following is NOT true about cancer?
A. The accumulation of single gene cancer mutations typically happens at the SAME rate in old people
as in young people.
B. Most cancers require the accumulation of multiple mutations.
C. Some cancer-causing mutations are recessive.
D. Cancers tend to only develop in old people mostly because they do not have the defense mechanisms
against cancer that younger people have.
E. A few types of cancers are more common in children than in old people.

Answer 70

D