DNA Replication and Recombination

Question 1

Describe the Meselston Stahl experiment briefly.

Answer 1

• DNA extracted and centrifuged
• Labelled with heavy nitrogen
• Amount of heavy nitrogen, shown by ‘band’ after centrifugation shows how DNA must have hybridised rather than recombining after replication
• Evidence for semi-conservative replication
• Cesium chloride used, when spun in a centrifuge, causes a concentration gradient and allows the DNA to be separated by density in the caesium gradient

Question 2

To which end of the growing strand are nucleotides added by DNA polymerase?

Answer 2

Added to the 3’ end.

Question 3

What is the purpose of the primer? (Hint: what does the primer provide that is necessary for the addition of nucleotides?)

Answer 3

Primer provides a 3’ OH group to which nucleotides can be added.

Question 4

How are the RNA primers in the okazaki fragments dealt with?

Answer 4

DNA polymerase has a 5’-3’ exonuclease which is responsible for degrading the RNA primers

Question 5

What is the function of DnaB helicase?

Answer 5

It opens the replication fork during DNA replication and unwinds it as DNA moves.

Question 6

What are RNA primers synthesised by?

Answer 6

RNA primase

Question 7

What is DnaG?

Answer 7

A primase that generates oligoribonucleotides
(RNA) using the DNA as
a template for copying

Question 8

How are RNA primers removed?

Answer 8

5’ to 3’ exonuclease activity by DNA polymerase I.

Question 9

In E. coli, which DNA polymerase is used for chromosome replication? Why?

Answer 9

DNA polymerase III because DNA polymerase I has a low processivity- only synthesises 3-200 nucleotides before it falls off and can only add 10-20 nucleotides per second.

Question 10

What is the origin of replication called in E. coli?

Answer 10

oriC

Question 11

What is a topoisomerase? Give an example. How does this example work?

Answer 11

An enzyme which alters the supercoiled form of a DNA molecule.

DNA gyrase. It cuts two strands of DNA and passes another through the break before releasing the cut strands.

Question 12

Which two bending proteins aid the initiation reaction of initiation in DNA replication?

Answer 12

Fis and IHF. They bind to the DNA and bend it which causes strand opening.

Question 13

Which two bending proteins aid the initiation reaction of initiation in DNA replication? How do they work.

Answer 13

Fis and IHF. They bind to the DNA and bend it which causes strand opening.

Question 14

What is DUE (DNA unwinding element)?

Answer 14

An enzyme complex which is the initiation site for the opening of the double helix structure. It is good at this because it is A-T rich so strands can be easily separated here.

Question 15

What is DnaA?

Answer 15

A DNA unwinding protein that promotes unwinding at oriC.

Question 16

What structure does DnaB take and how is it loaded onto DNA?

Answer 16

It is a ring structure so must be loaded onto DNA by DnaC at the origin opened by DnaA.

Question 17

How is the DnaABC complex released?

Answer 17

ATP hydrolysis releases DnaC and leaves DnaB bound for its helicase actions to occur.

Question 18

What is SSB and what does it do?

Question 19

What roles do DNA polymerase I, II and III play in replication?

Answer 19

I- removes RNA primers and replaces with DNA
II- fills gaps following the repair of DNA
III- chromosome replication

Question 20

What is the gamma subunit in DNA polymeraseIII responsible for doing?

Answer 20

Responsible for loading the beta clamp onto parental strands.

Beta clamp holds the polymerase onto the DNA and makes sure it doesn’t fall off.

Question 21

What does the epsilon subunit of DNA polymerase III do?

Answer 21

Is a proofreading exonuclease (3’ to 5’)

Question 22

Why does the lagging strand form a loop?

Answer 22

Because the DNA polymerase remains bound at the loop in order for both DNA polymerases on the leading and lagging strands to add nucleotides at the same rate. Once the okazaki fragment is complete, the polymerase dissociates and the loop straightens.

Question 23

What is the beta sliding clamp?

Answer 23

A subunit in DNA polymerase III. It encircles DNA and ensures the polymerase does not fall off. Loaded by the gamma subunit.

Question 24

What do the:
α 
τ 
γ 
β
Subunits of pol III do?

Answer 24

α subunits synthesize DNA on leading and lagging strands
τ subunits ensure dimerization of core polymerase subunits
γ complex loads β clamp at each Okazaki fragment
β clamp encircles DNA and binds complex to fork

Question 25

How is a Holliday junction formed?

Answer 25

A D loop forms as one strand invades the other strand. Cleavage of the D loop and subsequent ligation forms a Holliday junction.

Question 26

How is a Holliday junction resolved?

Answer 26

Enzyme called resolves cleaves either vertically or horizontally.

Question 27

What are the products of horizontal and vertical cleavage of a Holliday junction?

Answer 27

Vertical- recombination products

Horizontal- non recombination products.

Question 28

Why does homologous recombination occur?

Answer 28

To repair DNA, especially important in fixing double stranded breaks

Question 29

What is meant by ‘gene conversion’?

Answer 29

Where there is a mutation in a base pair which is then resolved using homologous recombination (giving a mutant on one strand and the normal base on the complimentary strand). Depending on which strand is used as the template in replication, either the mutation will be inherited or the wild type will be inherited. This is gene conversion.

Question 30

How can asymmetry be explained in the Holliday model? (Hint: another model explains this).

Answer 30

Meselson Radding model explains asymmetry:

• Non sister chromatids are used
• Endonuclease nicks one strand of a duplex
• DNA synthesis follows, displacing the nicked strand
• This displaced nicked strand invades the non sister chromatid, looping out the homologous stretch of DNA (which is then excised).
• Remaining free ends are ligated, resulting in a Holliday junction.
• Only one of the molecules has a heteroduplex region (asymmetry) if branch migration does not occur.

Question 31

What is synthesis dependent strand annealing?

Answer 31

-Otherwise known as homologous recombination repair

• Double stranded break in DNA causes both strands to undergo degradation, leaving 3’ ended tails
• These tails invade another piece of DNA at a region of homology, forming a D loop
• 3’ end of the strand acts as a primer and DNA synthesis
• D loop replication bubble tracks along the template strand
• Newly synthesised strand dissociates from D loop and is captured by the 3’ single stranded tail from the other side of the break in the duplex.
• DNA replication across the break to fill in the gap

Question 32

Describe the three stages of synaptic recombination.

Answer 32

Presynapsis
-Formation of gaps, single or double stranded breaks and exposure of single stranded regions.

Synapsis
-Homologous strands are paired up and strand exchange takes place.

Post synapsis
-Holliday junctions are cut to form cross over and non cross over products.

Question 33

What is the role of RecA in homologous recombination?

Answer 33

Assists the 3’ overhang to invade the homologous strand

Question 34

What is RuvABC and what are the respective purposes of A, B and C.

Answer 34

RuvABC is a protein complex that mediates branch migration and resolution of Holliday junctions.

RuvAB mediates the branch migration through helicase activity. RuvA loads B ring

RuvC is the cleavage enzyme.

Question 35

What is RecQ?

Answer 35

A helicase that recognises replication forks and untangles any complications they encounter.

Question 36

What occurs in Holliday junction dissolution?

Answer 36

Double Holliday junctions migrate together and ‘cancel each other out’ so non recombination products form. This uses RecQ and a topoisomerase.

Question 37

During DNA cloning, how can you ensure that the piece of DNA inserts itself in the correct orientation?

Answer 37

Cut one end of the DNA with one enzyme and another end with another and do the same to the plasmid to ensure that it inserts itself in the correct orientation

Question 38

Briefly describe blue- white screening (alpha complementation)

Answer 38

• LacZ gene encodes 2 peptides, when expressed, peptides come together and form blue colour
• If gene interrupted,