replication and protein synthesis

Question 1

state some features and constraints of DNA polymerase

Answer 1

Template directed
High fecundity- very low error rate
~1 mistake per 10^9 nucleotides
Must be able to replicate massive amounts of DNA in a few hours.

Question 2

DNA polymerase I

Discovery (A. korberg, 1958) relied on:

Answer 2

observation that deoxyribonucleoside 5’ triphosphates (dATP, dCTP, dGTP, dTTP) are heat-stable precursors of DNA synthesis.

Question 3

How is DNA polymerase 1 isolated

Answer 3

Break ecoli cells
centrifuge into layers
test layers with dTTP cooperation assay

Question 4

DNA polymerase Requirements for synthesis:

Answer 4

Single stranded template
dNTP building blocks
Primer can be RNA or DNA

Question 5

which Direction do DNA polymerases work on a Template strand

Answer 5

5’–>3’

Question 6

Describe the consequences of unidirectionality of DNApol on DNA Synthesis

Answer 6

As a consequence of this one strand of DNA is called the lagging strand asn is done in small fragments called ozeki fragments.
Newly formed DNA will be found in ~1000 fragments which are incorporated with time.

Question 7

how are RNA primers removed?

Answer 7

Rnase H removes RNA primers

Question 8

what is the function of DNA ligase

Answer 8

DNA ligase joins ozeki fragments

Question 9

what are the functions of DNA polymerases

Answer 9

• > chromosome replication
• > DNA repair
• > DNA recombination

Question 10

when is the new DNA strand checked

Answer 10

Before adding another nucleotide previous additions are checked if one is mismatched phosphodiester bond is cut and nucleotide replaced.

Question 11

how does proof reading affect DNA polymerase acuracy? how is it done?

Answer 11

Replication DNA polymerases have built in proofreading ability; thus can remove mis-incorporated nucleotides and try again.
For eukaryotic DNA replication: error rate without proofreading: 1 in ~10^5
with proofreading: 1 in ~10^7
with proofreading and mismatch repair: 1 in ~10^9
Proofreading active site is the endonuclease active site
Before adding another nucleotide previous additions are checked if one is mismatched phosphodiester bond is cut and nucleotide replaced.

Question 12

compare DNA replication in eukaryotes vs bacteria

Answer 12

replication can be continuous in bacteria where as happens in ‘s’ phase in eukaryotes

single origin in bacteria multiple in euk

circular DNA no telomeres, euk telomeres special end replication mech

Question 13

what is the problem with replicating the lagging strand end of a chromosome and how is this problem resolved?

Answer 13

No 3’ hydroxyl group to replicate laging end
Without tylermorase chromosomes would get progressively shorter.
Telomeres repete themselves 500-3000 times at end of chromosome

Question 14

why is having multiple origins beneficial for a genome?

Answer 14

allow replication of large genomes in a short time

Question 15

what do telomerases do?

Answer 15

Telomerases add additional bases to the end of the chromosome
DNA polymerase Alpha completes the new strand thus completely copying the chromosome

Question 16

what does miniture RNA (miRNA) do

Answer 16

miniature RNA regulates RNA

Question 17

Why is thymine used not uracil in DNA as well as RNA

Answer 17

Use of t important because at pH7 C will occasionally spontaneously deaminate into a U so our body would have a hard time recognizing damaged deaminated DNA if T was not used.

Question 18

What direction do RNA polymerases work in?

Answer 18

RNA polymerases like DNA polymerases work in the 5’ to 3’ direction

Question 19

RNA polymerase requires

Answer 19

NTPs- nucleotide triphosphate.

Double stranded helix but no primer.

Question 20

why is their another highly conserved region of DNA upstream of the promoter region in bacteria

Answer 20

found to be a result of the structure of RNA pol as there are two spaced out binding sites

Question 21

how is excessive tightening of the DNA double helix prevented before strands are seperated

Answer 21

To prevent DNA helicase tightening adjacent areas of the double helix too much to unwind accessory proteins must go along and loosen the super helical structure of the DNA before transcription.

Question 22

Eucaryotic poll 2 cannot initiate transcription without presence of general transcription factors:

Answer 22

TFIID- (14 subunit structure) dictates start of transcription

TFIIH-(10 subunits) includes helicase DNA unwinding and includes helicase activity.

Question 23

How did Early insight from mutagen studies in tobacco mosaic virus demonstrate that the code was non overlapping. Used as early evidence for triplet code

Answer 23

Addition of a base was shown to scramble code
Removal of base was demonstrated to return phenotype to wildtype
3/+ mutations shown to have no (little) effect on the organism’s phenotype
(Crick, F., Brenner, S., et al. (1961). Nature 192, 1227-1232)

Question 24

why does RNA extend beyond 5’ and 3’ stop and start codons?

Answer 24

enables regulatory functions

Question 25

Bacterial requirements for protein synthesis:

Answer 25

mRNA
tRNA deliver aminoacid
GTP
Additional bacterial proteins (mostly GTPases)
Elongation factors 
Termination factors

Question 26

how are stop codons recognized in ribosomes

Answer 26

No tRNA can read stop codons instead they are recognised by protein ‘release factors’

Question 27

comment on the difference between eukaryotic and prokaryotic ribosomes

Answer 27

pro- 60s
euk- 80s both subunits larger

the ‘s’ is about sedimentation speed the higher the quicker the speed