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Genetics > Exam 3: DNA Replication > Flashcards

Exam 3: DNA Replication Flashcards

1
Q

Explain Meselson and Stahl’s experiment

A

-N14 and N15 centrifuged separately. Lighter N14 always at the top of tube and heavier N15 at the bottom of tube.

-N15 E.coli grown in medium containing N14.

-N centrifuged and found to be in the middle of N14 and N15 bands which proved that the DNA had replicated semiconservatively and was half of each.

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2
Q

Prokaryotic DNA replication begins at the

A

ORI (origin of replication)

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3
Q

Replication fork

A

Site of replication where helix is unwound

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4
Q

Replication is

A

bidirectional (it starts in one place and goes in two directions); therefore, there are two replication forks

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5
Q

Replicon

A

Length of DNA replicated

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6
Q

DNA polymerase 1

A

Enzyme catalyzes directional DNA synthesis: chain elongation occurs 5’ to 3’

Requires DNA template, a primer, and all four deoxyribonucleoside triphosphates (dNTPs)

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7
Q

Exonuclease Activity

A

3’-5’
All three possess 3’ to 5’ exonuclease activity; proofread newly synthesized DNA, remove/replace incorrect nucleotide.

Dramatically increases the fidelity of the replicating enzyme.

5’-3’
Only DNA polymerase 1
Excises primers– fills in gaps left behind

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8
Q

Which DNA polymerase have the initiation of chain synthesis?

A

None

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9
Q

What DNA polymerase has 5’ to 3’ exonuclease activity?

A

DNA polymerase 1

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10
Q

Seven key issues that must be resolved during DNA replication

A
  1. Unwinding of the helix

DNAa initiator proteins
SSB proteins
Helicase

  1. Reduce increased coiling generated during unwinding

Topoisomerase/DNA Gyrase

  1. Synthesis of primer for initiation

Primase

  1. Discontinuous synthesis of second strand

DNA Polymerase 3

  1. Removal of the RNA primers

DNA Polymerase 1

  1. Going of gap-filling DNA to adjacent strand

DNA Ligase

  1. Proofreading

DNA polymerase exonuclease activity of 3ʹ–5ʹ allows for excise of nucleotides

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11
Q

Components needed to synthesis molecules

A

Template

Substrates= dNTP (deoxynucleoside triphosphate)

Proteins/enzymes to coordinate assembly of substrate

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12
Q

DnaA

A

-Initiator protein encoded by DnaA gene

-Binds to ORI causing conformation to change

-Causes helix to destabilize and open up

-Exposes ssDNA

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13
Q

DNA Helicase

A

-Made of DnaB polypeptides

– Subsequently recruits holoenzyme to bind
replication fork and initiate replication

– Helicases require energy supplied by hydrolysis of ATP denatures hydrogen bonds and stabilizes double helix

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14
Q

SSBPs

A

-Single-stranded binding proteins

-Stabilize the open conformation of helix. Bind specifically to single strands of DNA.

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15
Q

How does DNA replication avoid supercoiling?

A

DNA gyrase

–Enzyme relieves coiled tension from unwinding of helix (DNA supercoiling) by making single- or double-stranded breaks

–Driven by energy released during ATP hydrolysis

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16
Q

Primase: an RNA Polymerase

A

-Synthesizes RNA primer

-Provides free 3’-OH required by DNA polymerase 3 for elongation

-DNA polymerase 1 removes primer and replaces it with DNA

17
Q

Contrast continuous and discontinuous DNA synthesis

A

-Because the two strands are antiparallel, and DNA is only synthesized 5’ to 3’, synthesis occurs in opposite directions

-Continuous DNA synthesis occurs on the leading strand

-Discontinuous DNA synthesis occurs on the lagging strand

18
Q

Describe the importance of eukaryotic “licensing” factors

A

Eukaryotic DNA replication initiates at specific sites known as origins of replication. The problem with multiple origins of replication is that you want to start all of them once and only once, otherwise the molecules will not be accurately replicated.

Licensing factors ensure that DNA replication only occurs once per cell cycle. This restriction is vital to prevent over-replication or re-replication of the genome or replication not occurring during S phase of the cell cycle.

One of the central licensing factors in eukaryotes is the Origin Recognition Complex (ORC). ORC binds to specific DNA sequences at origins of replication and marks them for replication initiation.

19
Q

Describe the problems associated with the ends of linear chromosomes

A

Unlike bacterial chromosomes, the chromosomes of eukaryotes are linear, meaning that they have ends. These ends pose a problem for DNA replication. The DNA at the very end of the chromosome cannot be fully copied in each round of replication, resulting in a slow, gradual shortening of the chromosome.

During DNA replication, the lagging strand synthesis is discontinuous, leading to the formation of Okazaki fragments. DNA polymerase normally can replace primers with DNA and connect the fragments, but it needs a 3’ OH to add dNTPS to which isn’t available at the end of the lagging strand. So because there is no place for a primer at the extreme end of the chromosome, the last few nucleotides at the 3’ end of each DNA strand cannot be replicated in each round of replication.

20
Q

Explain how semiconservative replication works

A

The DNA double helix is separated into two strands. Each strand is then replicated into a complementary new strand. Each DNA molecule has one original strand and one new strand.

21
Q

Chain elongation by DNA polymerase 1

A

Nucleotide added, two terminal phosphates cleaved off, providing newly exposed 3ʹ-OH

22
Q

What is a nucleoside triphosphate?

A

A nucleoside triphosphate is a nucleoside containing a nitrogenous base bound to a 5-carbon sugar, with three phosphate groups bound to the sugar. They are the molecular precursors of both DNA and RNA.

23
Q

Holoenzyme

A

Active form of DNA Pol III which complexes with several other subunits each with separate functions.

24
Q

What are telomeres?

A

Inert chromosomal ends that protect intact eukaryotic chromosomes from improper fusion or degradation. Long stretches of short repeating TTAGGG sequences preserve the integrity/stability of chromosomes.

Telomeres of chromosomes shorten with each cell division.

In most eukaryotic somatic cells, telomerase is not active.

Active lymphocytes, certain stem cells, male germ cells, and malignant cells maintain telomerase activity.

Telomerase activity and telomere length linked to aging, cancer, and other diseases.

25
Q

Replication at the telomere

A

Telomerase is a eukaryotic enzyme that makes replication at the telomere possible.

-Telomerase has an associated RNA that complements the 3’ overhang at the end of the chromosome, and the RNA template is used to synthesize the complementary strand in what is known as reverse transcriptase.

–Telomerase adds repeats of six-nucleotide sequence to 3ʹ end to fill gaps.

-Primase and DNA polymerase 1 are then free to synthesize the lagging strand.

26
Q

Okazaki Fragments

A

The lagging strand is synthesized discontinuously by DNA polymerase in sections called Okazaki fragments.

DNA polymerase I removes primers on lagging strand

DNA ligase catalyzes formation of phosphodiester bonds and seals nicks and joins fragments.

27
Q

Concurrent Synthesis

A

Both DNA strands synthesized concurrently

Concurrent DNA synthesis achieved on both strands at single replication fork

Lagging strand is looped

Inverts physical but not biochemical direction

DNA clamp prevents core enzyme dissociation from template

28
Q

What are some similarities between eukaryotic and prokaryotic DNA replication?

A

-dsDNA unwound at the ORI
-Replication forks created
- Bidirectional synthesis creates leading and lagging strands

29
Q

How is eukaryotic DNA replication more complex?

A

-More DNA than prokaryotic cells
– Linear chromosomes
– DNA complexed with nucleosomes
– Eukaryotic chromosomes contain multiple ORIs (Facilitates rapid synthesis of large quantity of DNA)

30
Q

Replication through chromatin

A

Eukaryotic DNA complexed with chromatin

200 base pair nucleosomes wrap around eight histone proteins

Nucleosomes must be stripped away before polymerase can begin synthesis

31
Q

Describe the importance of eukaryotic multiple origins of replication

A

Multiple origins of replication are distributed throughout the genome, ensuring that DNA replication proceeds quickly and efficiently. This multiplicity of origins helps to complete replication within the limited time frame of the cell cycle.

The utilization of multiple origins also minimizes the chances of DNA damage or incomplete replication at a single origin causing catastrophic consequences.

32
Q

T/F: DNA polymerase 3 has 5’ to 3’ exonuclease activity

A

False

33
Q

T/F: DNA polymerase 1 has 5’ to 3’ exonuclease activity

A

True

34
Q

Discontinuous replication is the result of what property of DNA?

A

The antiparallel nature of the nucleotide strands.

35
Q

What would be the result if an organism’s telomerase were mutated and nonfunctional?

A

Chromosomes would shorten with each generation.

36
Q

Differentiate semiconvervative replication from the conservative and dispersive replication hypotheses

A

In semiconservative replication, the original DNA molecule splits into two separate strands, and each of these strands serves as a template for the synthesis of a new complementary strand. The end result is two DNA molecules, each consisting of one original (parental) strand and one newly synthesized (daughter) strand. This model was experimentally supported by the famous Meselson-Stahl experiment in 1958.

In conservative replication, the parental strands and the newly synthesized strands come back together, preserving the original double helix.

In dispersive replication, the parental strands are dispersed into two new double helixes following replication. Hence, each strand consists of both old and new DNA, but this mode involves cleavage of the parental strands creating a patchwork.

37
Q

Which the following best describe bidirectional synthesis?

A

From the point of initiation, replication occurs in both directions along the DNA

Genetics (18 decks)

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