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Molecular Biology > DNA replication > Flashcards

DNA replication Flashcards

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AP® Biology flashcards Biology 101 flashcards
1
Q

Why is DNA replication important?

A

DNA needs to double to pass on equal amounts of DNA when cells divide. Therefore replication is essential for reproduction.

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2
Q

Name the 3 possible methods of DNA replication

A

conservative, semi-conservative and dispersive.

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3
Q

Describe conservative replication

A

Semi conservative replication occurs after separating parent strands, then the new strands detach and the old strands reassociate. A hypothetical mode of DNA replication in which one daughter double helix is made up of the two parental polynucleotides and the other is made up of two newly synthesized polynucleotides.

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4
Q

Describe dispersive replication

A

Daughter strands consist of sections of new and parent DNA. Formed by repeated template switching. A hypothetical mode of DNA replication in which both polynucleotides of each daughter double helix are made up partly of parental DNA and partly of newly synthesized DNA

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5
Q

Why was semi-conservative replication thought impossible?

A

DNA is plectonemic; the strands can’t be separated without unwinding the double helix. In humans, this would require 22.5 million rotations. It is impossible for circular DNA to unwind by rotation this way and since they replicate every 20 minutes, the speed of required rotation would be so high that the cell would heat up and explode.

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6
Q

Describe the first step of Meselson and Stahl’s experiment.

A

They grew a culture of E.coli in NH4Cl with heavy N-15 so that the bacteria all had DNA labelled with heavy nitrogen after a few generations.

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7
Q

Describe the second step of Meselson and Stahl’s experiment.

A

Centrifuged the heavy culture, discarded heavy medium and placed bacteria in a culture with light NH4Cl. Thus light N was incorporated into new DNA polynuceotides.

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8
Q

How was the heavy DNA differentiated from light DNA?

A

Meselson and Stahl used a density gradient centrifuge with 6M CsCl solution. Cellular components migrate to the bottom. Protein and DNA sink or rise until they reach a position where their buoyant density = density of the solution.

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9
Q

Define buoyant density.

A

The density possessed by a molecule or particle when suspended in an aqueous salt or sugar solution.

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10
Q

Define semi-conservative replication.

A

The mode of DNA replication in which each daughter double helix is made up of one polynucleotide from the parent and one newly synthesized polynucleotide.

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11
Q

What was the banding pattern seen after the first cell division in light NH4Cl? Conclusion?

A

One band was seen at an intermediate point (between where the heavy and light bands would have been). This would have been compatible with both semi and dispersive replication.

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12
Q

What was the banding pattern seen after 2 cell divisions in light NH4Cl? Conclusion?

A

One band at intermediate point, another thicker band above it. This lighter band would have no heavy N. This means that semi-conservative replication is the correct hypothesis as dispersive replication would have resulted in only an intermediate band for several divisions.

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13
Q

What is the other topological problem presented by the unwinding of the DNA double helix during replication if the problem of rotation is overcome?

A

Supercoiling; as the double helix unwinds through rotation, tension is produced as the rest of the molecule becomes wound tight. Supercoiling occurs to relieve this tension.

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14
Q

What is the role of DNA topoisomerases?

A

These are enzymes that enable the strands to be separated without unwinding by cutting one or both the strands of the double helix to form a replication bubble. This means that rotation does not occur and so supercoiling is avoided. The double helix can be “unzipped” with the two strands pulled sideways.

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15
Q

Describe the mode of action of DNA topoisomerase type 1.

A

One strand is nicked and the other strand is pulled through. The nick is resealed to form a ‘replication bubble’.

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16
Q

Describe the mode of action of DNA topoisomerase type 2.

A

Cuts both strands. A 2nd segment f DNA is passed through the gap and the 2 cut strands are rejoined.

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17
Q

Why do the toposiomerases not fail to rejoin the cut ends and accidentally break the chromosome into 2 sections?

A

The possibility is greatly reduced because each cut end is attached covalently to a tyrosine amino acid at the active site of the enzyme and is held tightly in place while the free ends are manipulated.

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18
Q

Why is having a template strand important?

A

DNA can be copied exactly with no errors.

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19
Q

What can mutations in DNA polymerases lead to?

A

colorectal cancer

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20
Q

When does DNA replication occur?

A

In the s phase of the cell cycle

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21
Q

How can sites of DNA replication be visualised?

A

Modified nucleotides (e.g. Br instead of CH3 in thymidine) are incorporated into DNA during replication, then detected by using fluorescently labelled antibodies which visualise the sites of DNA replication.

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22
Q

5 patterns of DNA replication were identified when visualised. Give one example.

A

Euchromatin is replicated first, then heterochromatin.

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23
Q

What is the process by which a new DNA strand is synthesised?

A

Template dependent DNA synthesis. In this case, DNA dependent DNA synthesis since DNA is the template.

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24
Q

What are the limitations of DNA polymerase?

A

Can only synthesise in 5’ to 3’ direction, so template has to be 3’ to 5’ for continuous synthesis; new nucleotides must be added to 3’ end. Synthesis is not possible without a primer.

25
Q

What is the main problem posed by the limitations of DNA polymerase?

A

The lagging strand is the 5’ to 3’ strand and thus has to be replicated discontinuously via Okazaki fragments.

26
Q

Define 3’ to 5 exonuclease

A

An enzyme that sequentially removes DNA from 3’ end of a strand. Allows for proofreading during replication.

27
Q

Define 5’ to 3’ exonuclease

A

An enzyme which sequentially removes DNA from 5’ end of a strand. It can remove DNA already attached to the template and displace nucleotides.

28
Q

How does DNA polymerase improve the accuracy of replication?

A

It monitors complementary base pairing closely; only adds new nucleotides after checking if the one just added is correct. When match is correct, small structural rearrangement occurs which allows it to catalyse the addition of a new nucleotide.
It proofreads simultaneously while replicating. Proofreading and polymerisation are tightly coordinated and occur on different catalytic domains.

29
Q

Describe the role of DNA polymerase alpha.

A

Present in eukaryotes with no exonuclease activity in either direction. It extends the RNA primer by adding 20 nucleotides before DNA polymerase delta takes over.

30
Q

Describe the role of DNA polymerase delta

A

Present in eukaryotes. It has 3’ to 5’ exonuclease activity. It is the main replicative enzyme. It takes over after alpha extends the primer. Proofreads what alpha did, then finishes the job.

31
Q

Describe the roles of DNA polymerases I and III in bacterial DNA replication.

A

I: shows both types of exonuclease activity, is involved in DNA repair and replication.
III: main replicative enzyme. Synthesis new strand by extending RNA primer. Only has 3’ to 5’ activity.

32
Q

Which enzyme makes the RNA primer in bacteria?

A

primase

33
Q

What is the typical length of RNA primer?

A

4-15 nucleotides

34
Q

Why must separated single strands of DNA at the replication fork be protected?

A

They may re-attach or be attacked by exonucleases that usually target single stranded DNA viruses

35
Q

How are separated single strands of DNA protected?

A

Single strand binding proteins (SSBs) bind to the newly exposed strand to prevent complementary base pairing and attack by exonuclease. They also keep the strands elongated so that they serve as effective templates.

36
Q

Give an example of an SSB in eukaryotes.

A

Replication protein A

37
Q

Define euchromatin.

A

Regions of a eukaryotic chromosome that are relatively uncondensed, thought to contain active genes.

38
Q

Define heterochromatin

A

Chromatin that is relatively condensed and is thought to contain DNA that is not being transcribed.

39
Q

What is the s phase of the cell cycle?

A

The phase of the cell cycle in which DNA is replicated, occurring between G1 phase and G2 phase

40
Q

Why is an RNA primer required for DNA polymerase to synthesise a new DNA strand?

A

DNA polymerases cannot initiate DNA synthesis unless there is already a short double-stranded region to act as a primer. To solve this problem, the first few nucleotides that are attached are ribonucleotides that are put in place by an RNA polymerase enzyme.

41
Q

What is the replication fork?

A

The region of a double-stranded DNA molecule that is being opened up to enable DNA replication to occur.

42
Q

How is replication of the lagging strand completed in bacteria?

A

DNA polymerase III stops when it reaches the primer. DNA polymerase I continues bc it has exonuclease activity in the 5’-3’ direction. It replaces the RNA primer with DNA. The fragments are connected by DNA ligase.

43
Q

How are Okazaki fragments joined up in eukaryotes?

A

DNA polymerase delta and helicase push aside the primer. FEN1 cuts at the branch point. DNA ligase links the 2 fragments w/ a phosphodiester bond.

44
Q

How come DNA molecules in eukaryotes don’t get progressively shorter with each replication?

A

Telomerase extends the parent DNA strand by adding TTAGGG several times. The final Okazaki fragment can be primed.

45
Q

Which cells have telomerase?

A

Only stem cells have telomerase. Most cells die after 50 divisions- senescence. In senescent cells, the ends are shortened.

46
Q

Define FEN1.

A

The ‘flap endonuclease’ involved in replication of the lagging strand in eukaryotes.

47
Q

Define Okazaki fragment.

A

One of the short segments of RNA-primed DNA synthesized during replication of the lagging strand of the double helix.

48
Q

Define telomerase.

A

The enzyme that maintains the ends of eukaryotic chromosomes by synthesizing telomeric repeat sequences.

49
Q

Define replication origin.

A

A site on a DNA molecule where replication initiates.

50
Q

What is a primosome?

A

Once primases put down RNA primers, the convert the pre-priming complex to a primosome.

51
Q

What is the role of DnaA proteins?

A

They induce the opening of the helix by binding to the replication origin and ‘melting’ the helix through torsional stress.

52
Q

What feature of the sequence in the replication origin makes it easier for DnaA to break the base pairs?

A

It is A-T rich; 3 A-T rich nucleotide repeats at one end of the origin. A-T is easier to break than C-G because it has only 2 hydrogen bonds, while C-G has 3.

53
Q

What does DnaB do?

A

It is a helicase which breaks more bps at origin. Increases the size of the opening at the origin so that a nascent replication fork forms on either end of origin. The forks move away from origin.This forms a pre-priming complex.

54
Q

What controls the attachment and detachment of DNA polymerase III from the lagging strand?

A

the gamma complex, by interacting with beta subunit of each polymerase.

55
Q

What keeps DNA polymerase delta attached to the lagging strand?

A

proliferating cell nuclear antigen (PCNA)

56
Q

What is the purpose of the six terminator sequences in the E.coli genome?

A

they act as binding sites for Tus proteins. The sequences are oriented in a certain way, so the tus proteins can only bind in a certain orientation, allowing the replication fork to only proceed in one direction. This makes sure both replication forks stop and don’t overshoot the halfway point.

57
Q

How does the structure of the Tus protein allow it to determine directionality?

A

It has permissive and non-permissive faces. when approaching from one direction, DnaB encounters a wall of beta strands, causing it to stop, but from the other side, it reaches a less rigid face so it can push aside the protein and continue.

58
Q

How does replication fork arrest occur?

A

A specific interaction between the Ter C6 base and the Tus lock domain results in arrest- the helix won’t open up.

59
Q

Why are there multiple terminator sequences?

A

In case the fork doesn’t stop.

Molecular Biology (2 decks)

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