Oct 23

Question 1

What are the 4 steps in CPD photolyase breaking down thymine dimer?

Answer 1

1. Photolyase sees kink in DNA, binds to timer with light
   2.TT dimer is flipped into active site, the MTHF absorbs photon, gives electron to FADH
2. FADH gives electron to dimer, breaks it
3. Electron goes back to FADH to give it neg charge

Question 2

What removes O6-methylguanine?

Answer 2

O6-methylgauanine-DNA-methyltransferase has a cys that takes methyl group, then the enzyme is degraded (sacrifice)

Question 3

Do all organisms have mechanism that has enzyme take methyl group from base and sacrifice itself? Y or N

Answer 3

Yes

Question 4

How BER works in prokaryotes?

Answer 4

1. DNA glycosylase recognizes base, hydrolyses it
2. The Apurinic/apyrmidinic (AP endonculease) cleaves the backbone at that site to create a nick.
3. Pol 1 removes some DNA, then puts it back including the missing nuclotide, then ligase seals nick

Question 5

Is DNA glycosylase conserved between species? Y or N

Answer 5

Yes

Question 6

What does it mean by DNA pol 1 in prokaryotes having nick translational activity?

Answer 6

It has exonuclease activity 5’->3’, but has 3’ to 5’ adding nucleotides. It backs up and puts them back and more

Question 7

What glycosylase targets UG and UA BP?

Answer 7

Uracil DNA N-Glycosylase

Question 8

What glycosylase targets methyl groups on purines? What positions on the purines?

Answer 8

It is position 3 and 7 both for adenine and guanine. It is methyl purine DNA glycosylase

Question 9

What glycosylase targets 8-oxo-guanine?

Answer 9

The same name with glycosylase on the end

Question 10

What glycosylase targets AG and A-8-oxo-guanine BPs?

Answer 10

MutY homolog

Question 11

What glycosylase targets pyrimidine glycols?

Answer 11

Endonuclease tree homolog

Question 12

What are the steps of Eukaryotic BER?

Answer 12

1. Glycosylase removes base, forms abasic site
2. AP Endonuclease cut away nucleotide = nick present

Long patch repair: DNA Pol has 3-5’ exonuclase activity in eukaryotes, but not 5-3’, so it just builds onto the 3’ end present in the nick, which displaces the strand in front of it (5’ end facing it). Then flap endonuclase cuts off the 5’ terminus flap
Short patch:simply has polymerase replace missing base, then ligase seals it

Question 13

How does DNA glycosylase recognize damage as the bases are facing inwards?

Answer 13

It scans minor groove, looking for kink. It then flips damaged base into the active site

Question 14

How is NER used to remove bulky damaged bases (E. coli)?

Answer 14

1. Excinuclease cleaves phosphodiester bond on each side of DNA lesion
2. UvrA and B complex scans DNA helix
3. UvrA leaves and UvrB melts the strands and bends the DNA strand, forming bubble
   4.UvrC excinuclease (dimer) attaches to the B subunit, and each part of it on each side of the lesion, and makes a cut on each side.
   5.ssDNA gap treated, then UvrD helicase comes along and removes the cut fragment.
4. DNA pol 1 and ligase fill and seal gap
   KEY POINT is that for NER, nicks on each side

Question 15

How does NER occur in Eukaryotes?

Answer 15

1.XPG recognizes lesion
2. XPB and D recruited to lesion and separate DNA to form ssDNA bubble
3.RPA binds and positions XPF on 5’ side, XPG on the 3’ side, they cleave the DNA
4. Fragement displaced by helicase, then pol and ligase recruited by PCNA clamp

Question 16

How does Transcription coupled repair work?

Answer 16

1. Damage recongized by RNA polymerase, stalls it
2. RNA pol has some proofreading ability, not as much as DNA pol, so it recruits NER proteins

Question 17

What causes Xeroderma pigmentosum (XP)?

Answer 17

It is defective NER system, cells can’t repair UV damage, so early onset of skin cancer, skin very photosensitive

Question 18

Do carriers of chromosomal inversion show a phenotype? What about offspring?

Answer 18

No, have balanced rearrangement. The carriers may produce gametes with unbalanced chromosomes, so can have unbalanced offspring with phenotype

Question 19

What is paracentric inversion?

Answer 19

Imagine have P and Q arm of chromosome (short and long). The P arm has A, the Q arm has BCD. If paracentric inversion happens, it is a loop on itself. If have ABCD and ACBD, sill balanced. If have ABCA (2 centromeres) and DBCD (no centromeres, this is unbalanced. Paracentric produces acentric or dicentric chromosome (no or 2 centromeres)

Question 20

What is pericentric inversion?

Answer 20

It is inversion loop that forms duplicated deficient chromosomes, it involves the centromere

Question 21

How does repair by translesion DNA synthesis work?

Answer 21

It works by DNA pol 3 stalling at lesion in replication fork. Then it leaves and polymerase of Y family is recruited, it has less proofreading ability, is less picky. Once it synthesizes past the lesion, then pol 3 will take over again. Mutations produced, but cell doesn’t die

Question 22

How does non-homologous end joining work, when occur? Do you get mutations from this?

Answer 22

It is when sister chromosome not available to invade to use as a template. It will have ds break, then Ku70/80 will bind on each end, recruit other proteins to join the strands, can get indels if break happens before DNA replication, as during break can get indels put in