lecture 8 & 9

Question 1

where are primers needed at each strand

Answer 1

leading strand- primer required at 5’ end of newly synthesized strand

lagging strand- needed at each Okazaki

Question 2

DNA helicase, called ___ in prokaryotes, do what?

Answer 2

DnaB
slides on DNA template for the lagging strand and in the 5’ to 3’ direction, uses ATP to separate strands (breaks hydrogen bonds)

unzips double helix

Question 3

topoisomerase, also called ____ in prokaryotes, does what?

Answer 3

gyrase or topoisomerase II

removes positive supercoils, untwists DNA by cutting one or both strands of DNA to unwind it when helicase is breaking the hydrogen bonds, then reseals it

Question 4

Primases are specialized ____

in E. coli, an RNA primer is synthesized by the ___ primase, this primase must bind ____ for activity

Answer 4

RNA polymerases

DnaG

DNA helicase

Question 5

RNA primers initiate ____

leading strand initiated by…

Answer 5

RNA primers initiate each of the thousands of the Okazaki fragments on lagging strand

leading stranded initiated by primase at replication origin

Question 6

what does DNA polymerase I do

Answer 6

removes RNA primer at the end of each Okazaki fragment and replace with DNA (b/c cannot have RNA) — 5’ to 3’ activity- “nick-translation activity”

(RNaseH can also remove RNA primer by breaking phosphodiester bonds in RNA), but not the last rNMP) — RNaseH is like a backup

Question 7

what does DNA ligase do

Answer 7

seal the nick in the phosphodiester bond after DNA polymerase I goes through

Question 8

SSB’s protect DNA from ___ , it stimulates DNA polymerase activity by melting small ____ in the single-stranded template

Answer 8

endonucleases

DNA hairpin structures

Question 9

which DNA polymerase is the replicative enzyme in E. coli

Answer 9

DNA poly III

Question 10

what do DNA polymerases 1, 2, 4, and 5 do in E. coli

Answer 10

DNA Poly I = okazaki fragment processing and DNA repair

DNA poly III, IV, V = DNA repair

Question 11

name the E. coli DNA polymerases that have 3’ to 5’ exonuclease activity

Answer 11

Pol I
Pol II
Pol III

Question 12

name the E. coli DNA polymerases that have 5’ to 3’ exonuclease activity

Answer 12

DNA poly I

Question 13

E. coli DNA Pol III is called a ____ and has ___ subunits

Answer 13

holoenzyme
22

Question 14

name the components of the E. coli DNA Pol III holoenzyme

Answer 14

3 Pol III cores
Beta sliding clamps attached to each core
clamp loader at bottom — tau proteins attach clamp loader to the 3 cores

Question 15

which of the 3 Pol III core subunits in the holoenzyme has the DNA pol activity

Answer 15

alpha

Question 16

function of beta clamps in holoenzyme
what happens without beta sliding clamp?

Answer 16

help DNA poly III slide across subunit — increase speed and processivity of replication

without beta clamps, DNA poly III added 10 nucleotides/second, with beta — much faster(100,000 nucleotides per binding event)

without beta sliding, DNA poly III synthesizes new DNA through distributive synthesis (associated and deassociates with template)

Question 17

clamp loader complex of holoenzyme also called ___

Answer 17

tau complex

Question 18

tau subunit of clamp loader has ___ activity and binds ___

Answer 18

ATPase

binds Poly III and DnaB (helicase)

Question 19

clamp loader of holoenzyme does what

Answer 19

assembles beta clamps onto DNA

• attaches to beta clamp using energy of ATP to open the beta clamp - when B clamps opens, it is now able to encircle a single-stranded DNA with a primer, ATP hydrolysis ejects the clamp loader

Question 20

what are the components of the E. coli replisome?

Answer 20

DNA Poly III holoenzyme, DnaB helicase, primase

DnaB helicase connects with DNA poly III holoenzyme — increases helicase activity

leading strand - Pol III - beta clamp moves continuously with DnaB helicase

lagging strand - Poly III - beta clamp moves on and off DNA to extend multiple primers

Question 21

describe trombone model

Answer 21

model for replication of lagging strand – forms a loop
- looping template DNA grows and shortens during lagging strand synthesis
Poly III core synthesizes lagging strand, forms loop, okazaki fragments dissociate (ligase and Pol I fill in gaps)

Question 22

beta clamp recycling in e. coli

Answer 22

1- at end of replication, poly III core dissociates from beta after completing okazaki fragment

2- pol I target beta clamp to remove primer

3- dna ligase targets beta clamp to join okazakis

4- unoccupied beta clamp is unloaded

Question 23

replication fork in eukaryotes - enzymes

Answer 23

helicase = CMG complex (moves in 3’ to 5’, opposite to E. coli DnaB)

sliding clamp = PCNA

clamp loader = RFC

SSB’s = RPA

Question 24

describe the DNA polymerases in eukaryotes

Answer 24

3 different enzymes:

1- DNA Pol alpha (primase) - extends each RNA primer

2- DNA Pol epsilon - synthesizes leading strand

3- DNA Pol delta - synthesizes lagging strand

Question 25

initiation of replication- origin is bound by ___

origin = ___ element

Answer 25

initiator protein (trans-acting element)

origin = cis-acting

Question 26

when initiator binds to origin…

Answer 26

leads to DNA melting (denaturation of strands) and recruitment of other proteins — DNA helicase binds to initiator protein, helicase opens helix and binds primase

Question 27

describe prokaryotic initator protein

Answer 27

DnaA, binds to 9 bp repeats with ATP
- leads to melting at 13 bp repeats, required for binding of helicase (DnaB) and fork formation (helicase loader (DnaC))

Question 28

describe open-complex formation after DnaA binds origin

Answer 28

when DnaA and ATP bind origin, DnaA oligomerizes and wraps DNA around oligomer – another protein, HU, facilitates open complex formation at 13 bp

open complex = DnaA-ATP-OriC-HU

Question 29

what happens after open-complex formation

Answer 29

DnaC loads DnaB helicase onto each strand, ATP hydrolysis causes DnaC to be ejected, DnaB helicase expands replication bubble and primase forms RNA primers at each replication fork, RNA primers direct clamp loading and assembly of one Pol III holoenzyme

Question 30

replication initiation in E. coli is controlled at multiple sites…

Answer 30

1- binding of DnaA at OriC

2- DNA methylation

3- nucleotide-bound state of DnaA (ATP)

Question 31

describe control of initiation by methylation

Answer 31

only methylated origins can be replicated

OriC contains many GATC sequences that is a target for adenosine methylation by Dam methylase

there are 11 GATC sequences with 245 bp of OriC

SeqA protein binds hemimethylated DNA, preventing DnaA binding

Question 32

describe control of initiation by ATP

Answer 32

DnaA has to be bound to ATP to oligomerize and form open complex

DnaA-ADP cannot destabilize the A=T rich region to maintain open complex — forms closed complex

Question 33

termination of DNA replication

Answer 33

replication forks meet head on when replication is complete

forks cannot travel past Ter sites

at end – 2 daughter chromosome separated by topoisomerase II

Question 34

mismatch repair

Answer 34

1- first protein, MutS- recognizes mismatched bp and binds to it

2- MutS-MutL - scan bidirectionally along DNA scanning for GATC sites- forms loops
- MutS-MutL recruits helicase II and degrades newly replicated DNA past the
mismatch
3- MutH - site-specific endonuclease that cleaves unmethylated GATC sites