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Genetics COPY > Chapter 12 > Flashcards

Chapter 12 Flashcards

1
Q

All DNA polymerases synthesize new DNA by adding nucleotides to the _____ of the growing DNA chain.

a. 3′ OH
b. 5′ OH
c. 3′ phosphate
d. 5′ phosphate
e. nitrogenous base

A

a

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2
Q

Okazaki fragments are found in all of the following EXCEPT

a. leading strand.
b. lagging strand.
c. eukaryotic DNA.
d. bacterial DNA.
e. linear replication models.

A

a

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3
Q

Which of the following does NOT utilize bidirectional replication?

a. Theta model
b. Rolling circle model
c. Linear model
d. Eukaryotes
e. Bacteria

A

b

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4
Q

Which of the following typically only has one origin of replication?

a. Prokaryotes
b. Eukaryotes
c. Linear model of replication
d. Humans
e. Plants

A

a

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5
Q

Meselson and Stahl showed that DNA is replicated by a __________ system.

a. conservative
b. semiconservative
c. dispersive
d. semidispersive
e. conservative in prokaryotes and dispersive in eukaryotes

A

b

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6
Q

What type of synthesis occurs on the leading strand?

a. Conservative
b. Dispersive
c. Continuous
d. Discontinuous
e. Recombination

A

c

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7
Q
  1. In the diagram below, which letter indicates the 5′ end of the leading strand?

a. A
b. B
c. C
d. D
e. C and D

A

c

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8
Q

Suppose that some cells are grown in culture in the presence of radioactive nucleotides for many generations so that both strands of every DNA molecule include radioactive nucleotides. The cells are then harvested and placed in new medium with nucleotides that are not radioactive so that newly synthesized DNA will not be radioactive. What proportion of DNA molecules will contain radioactivity after two rounds of replication?

a. 0
b. 1/8
c. 1/4
d. 1/3
e. 1/2

A

e

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9
Q

You learn that a Mars lander has retrieved a bacterial sample from the polar ice caps. You obtain a sample of these bacteria and perform the same kind of experiment that Meselson and Stahl did to determine how the Mars bacteria replicate their DNA. Based on the following equilibrium density gradient centrifugation results, what type of replication would you propose for these new bacteria?

a. Conservative
b. Semiconservative
c. Dispersive
d. Semiconservative or Dispersive
e. Conservative or Dispersive

A

c

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10
Q

You are studying a new virus with a DNA genome of 12 Kb. It can synthesize DNA at a rate of 400 nucleotides per second. If the virus uses rolling-circle replication, how long will it take to replicate its genome?

a. 7.5 seconds
b. 15 seconds
c. 30 seconds
d. 1 minute
e. 2 minutes

A

c

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11
Q

You are studying a new virus with a DNA genome of 12 Kb. It can synthesize DNA at a rate of 400 nucleotides per second. If the virus uses theta replication, how long will it take to replicate its genome?

a. 7.5 seconds
b. 15 seconds
c. 30 seconds
d. 1 minute
e. 2 minutes

A

b

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12
Q

The nuclear genome of a single human cell (i.e., the entire diploid complement) contains about 6.6 billion (6.6 × 109) base pairs of DNA. If synthesis at each replication fork occurs at an average rate of 50 nucleotides per second, all the DNA is replicated in 5 minutes. Assume that replication is initiated simultaneously at all origins. How many origins of replication exist in a human diploid genome?

a. 220,000
b. 440,000
c. 880,000
d. 2.64 x 107
e. 1.32 x 108

A

a

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13
Q

The nuclear genome of a single human cell (i.e., the entire diploid complement) contains about 6.6 billion (6.6 × 109) base pairs of DNA. If synthesis at each replication fork occurs at an average rate of 50 nucleotides per second, all the DNA is replicated in 5 minutes. Assume that replication is initiated simultaneously at all origins. Assuming that the origins are approximately equally distributed across the chromosomes, what is the average number of origins per human chromosome?

a. 4783
b. 19,130
c. 4.6 x 106
d. 1.21 x 109
e. 2.9 x 1010

A

a

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14
Q

Suppose Meselson and Stahl had obtained the following results in their experiment. These results would be consistent with which model of replication?

a. Conservative replication only
b. Semiconservative and conservative replication
c. Semiconservative replication only
d. Dispersive replication only
e. Semiconservative and dispersive replication

A

e

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15
Q

Which of the following enzymes do NOT aid in unwinding of DNA for replication?

a. Helicase
b. Single-stranded binding proteins
c. Primase
d. Gyrase
e. Topoisomerase

A

c

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16
Q

DNA polymerase I and DNA polymerase III both have ______but only DNA polymerase I has _______.

a. 5′→ 3′ exonuclease activity; 3′→5′ exonuclease activity
b. 5′→ 3′ polymerase activity; 3′→5′ polymerase activity
c. 3′→5′ polymerase activity; 5′→ 3′ polymerase activity
d. 3′→5′ exonuclease activity; 5′→ 3′ exonuclease activity
e. 5′→ 3′ polymerase activity; 3′→5′ exonuclease activity

A

d

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17
Q

What is the function of DNA gyrase?

a. Connects Okazaki fragments by sealing nicks in the sugar- phosphate backbone
b. Unwinds the double helix by breaking the hydrogen bonding between the two strands at the replication fork
c. Reduces the torsional strain that builds up ahead of the replication fork as a result of unwinding
d. Binds to oriC and causes a short section of DNA to unwind
e. Prevents the formation of secondary structures within single-stranded DNA

A

c

18
Q

What is the function of single-strand-binding proteins?

a. Connects Okazaki fragments by sealing nicks in the sugar- phosphate backbone
b. Unwinds the double helix by breaking the hydrogen bonding between the two strands at the replication fork
c. Reduces the torsional strain that builds up ahead of the replication fork as a result of unwinding
d. Binds to oriC and causes a short section of DNA to unwind
e. Prevents the formation of secondary structures within single-stranded DNA

A

e

19
Q

What is the function of DNA ligase?

a. Connects Okazaki fragments by sealing nicks in the sugar- phosphate backbone
b. Unwinds the double helix by breaking the hydrogen bonding between the two strands at the replication fork
c. Reduces the torsional strain that builds up ahead of the replication fork as a result of unwinding
d. Binds to oriC and causes a short section of DNA to unwind
e. Prevents the formation of secondary structures within single-stranded DNA

A

a

20
Q

Which of the following is a protein that facilitates the termination of replication in E. coli?

a. Telomerase
b. DNA gyrase
c. Tus
d. Primase
e. Topoisomerase

A

c

21
Q

DNA synthesis during replication is initiated from

a. a free 5′ OH.
b. DNA primers.
c. RNA primers.
d. telomerase.
e. DNA polymerase I.

A

c

22
Q

DNA polymerases require all of the following for DNA replication, EXCEPT

a. DNA template.
b. primer.
c. free 3′ OH.
d. 3′ to 5′ polymerase activity.
e. dNTPs.

A

d

23
Q

What type of bonds does DNA ligase create between adjacent nucleotides?

a. Hydrogen
b. Phosphodiester
c. Ionic
d. Metallic
e. Ribonucleotide

A

b

24
Q

The proofreading function of DNA polymerases involves

a. 5′ → 3′ exonuclease activity.
b. 3′ → 5′ exonuclease activity.
c. 5′ → 3′ telomerase activity.
d. 3′ → 5′ telomerase activity.
e. 5′ → 3′ gyrase activity.

A

b

25
Q

DNA primase requires a ____ template and _____ nucleotides to initiate primer synthesis.

a. DNA; DNA
b. RNA; RNA
c. DNA; RNA
d. RNA; DNA
e. leading strand; DNA

A

c

26
Q

Which one of the following statements is not true for all E. coli DNA polymerases?

a. They require a primer to initiate synthesis.
b. They use dNTPs to synthesize new DNA.
c. They produce newly synthesized strands that are complementary and antiparallel to the template strands.
d. They possess 5′ 3′ exonuclease activity.
They synthesize in the 5′ 3′ direction by adding nucleotides to a 3′-OH group

A

d

27
Q

Which activity is NOT associated with DNA polymerases?

a. Ability to attach a DNA nucleotide to the 3′ end of previously incorporated DNA nucleotide
b. Ability to excise a newly incorporated nucleotide that does not match the template strand
c. Ability to “read” a template strand 3′ to 5′ and synthesize a complementary strand
d. Ability to synthesize a DNA from scratch without a primer
e. Ability to synthesize new DNA in a 5′ to 3′ direction

A

d

28
Q

If ribonucleotides were depleted from a cell during S phase, how would DNA synthesis be affected? (Ignore energetic considerations.)

a. There would be no effect because ribonucleotides are used in RNA synthesis, not DNA synthesis.
b. DNA synthesis would continue, but at a slower rate.
c. There would only be an effect during M phase, not in S phase.
d. DNA synthesis would not be affected because ribonucleotides are only used during the process of transcription.
e. Replication would cease because ribonucleotides are required to initiate DNA synthesis.

A

e

29
Q

If a deletion occurs in a gene that encodes DNA polymerase I and no functional DNA polymerase I is produced. What will be the most likely consequence of this mutation?

a. The DNA strands would contain pieces of RNA.
b. The DNA would not exist in a supercoiled state.
c. There would be no DNA replication on the leading or lagging strands.
d. There would be no RNA primers laid down.
e. The DNA will not be able to unwind to initiate replication.

A

a

30
Q

Telomerase activity is most likely to be found in which cells in humans?

a. Red blood cells
b. Muscle cells
c. Neurons
d. Germ line
e. Any type of somatic cell

A

d

31
Q

For which of the following is the “end-replication problem” relevant?

a. Circular DNA
b. Linear chromosomes
c. The centromere region of a chromosome
d. Rolling circle model of replication
e. Theta model of replication

A

b

32
Q

Telomerase uses _____________ to synthesize new DNA.

a. exonuclease activity
b. a licensing factor
c. strand invasion
d. a DNA template
e. an RNA template

A

e

33
Q

_______ are tandemly repeated DNA sequences located at the ends of eukaryotic chromosomes.

a. Replication bubbles
b. Telomeres
c. Nucleosomes
d. Licensing factors
e. Holliday junctions

A

b

34
Q

Which of the following is TRUE regarding nucleosome formation during DNA replication?

a. Nucleosomes are only reassembled on the lagging strand.
b. Nucleosome assembly consists entirely of newly synthesized histones.
c. Nucleosome assembly occurs at a faster rate in prokaryotes than in eukaryotes.
d. The addition of newly synthesized histones is a part of nucleosome assembly.
e. Nucleosome assembly does not occur during semiconservative replication.

A

d

35
Q

Which of the following is TRUE of DNA polymerases of eukaryotic cells?

a. The same DNA polymerase replicates mitochondrial, chloroplast, and nuclear DNA.
b. There are only two different DNA polymerases that function in the process of replication.
c. Some DNA polymerases have the ability to function in DNA repair mechanisms.
d. All eukaryotic DNA polymerases have 3′ → 5′ exonuclease activity.
e. Leading and lagging strand synthesis are performed by the same type of DNA polymerase.

A

c

36
Q

What would be a likely result of expressing telomerase in somatic cells?

a. Premature aging
b. Cancer
c. Lower rates of replication
d. Immortality of gametes
e. Early termination of replication

A

b

37
Q

Which of the following best describes heteroduplex DNA?

a. A single-stranded DNA molecule of one chromosome pairs with a single-stranded DNA molecule of another chromosome.
b. A single-stranded DNA molecule of one chromosome pairs with a single-stranded RNA molecule of another chromosome.
c. DNA that consists of sequences from two different species brought together through homologous recombination.
d. DNA that consists of an RNA primer and newly synthesized DNA on the lagging strand.
e. Newly synthesized DNA that has yet to be reassembled into nucleosomes.

A

a

38
Q

Which of the following is a necessary step in the Holliday model of recombination?

a. DNA primase generates an RNA primer.
b. The RNA template of telomerase binds to the telomere.
c. Topoisomerases aid in supercoiling.
d. DNA polymerase α initiates DNA synthesis.
e. A single-strand break occurs in the DNA molecule.

A

e

39
Q

The Holliday model describes which process?

a. Semiconservative replication
b. Homologous recombination
c. End replication
d. RNA primer synthesis
Rolling-circle replication

A

b

40
Q

Why did the Meselson and Stahl experiment require two rounds of replication?

A

The first round of replication, which produced a single band of intermediate density, was enough to rule out the conservative model of replication. However, a second round was necessary to show that now two bands of equal intensity were observed, consistent with semiconservative replication and inconsistent with dispersive replication.

41
Q

Summarize the similarities and differences in rolling-circle replication, theta replication, and linear eukaryotic replication.

A

(1) Rolling-circle replication
• Initiated by cleavage in a single DNA strand
• Uncleaved strand used as template
• New nucleotides added to 3′ end of cleaved strand
• Linear single-strand DNA produced; subsequently circularized
(2) Theta replication
• Two replication forks
• Initiated by DNA denaturation at single origin
• Single, expanding replication bubble
• Bidirectional replication
• Circular DNA molecule produced
(3) Linear eukaryotic replication
• Two replication forks
• Initiated by DNA denaturation at multiple origins
• Multiple, expanding replication bubbles
• Bidirectional replication
• Linear DNA molecule produced

42
Q

Why do eukaryotic chromosomes have multiple origins of replication, whereas prokaryotic chromosomes typically have only one origin?

A

The large genome size of a typical eukaryotic chromosome, compared to that of a prokaryotic chromosome, necessitates multiple origins of replication so that the genetic material can be copied in a reasonable amount of time. Also, replication is much slower in eukaryotes (the average is about 10–100 nucleotides/second) than in prokaryotes (the average is about 1000 nucleotides/second). For example, the human haploid genome complement is estimated to be about 3.3 × 109 nucleotide base pairs. Divide this number by 23 (the haploid number of human chromosomes) to get an average number of 3.3 × 109/23 = 1.4 × 108, or about 140 million nucleotides (nt) for a single human chromosome. Even at 100 nucleotides/second, it would take (140,000,000 nt) × (100 nt/sec)−1 = (1,400,000 sec) × (1 min/60 sec) × (60 min/hr)−1 × (1 day/24 hr) = 16.2 days to replicate one eukaryotic chromosome if it contained only one origin.

Genetics COPY (23 decks)

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