chapter 11 part 3 Flashcards

(77 cards)

1
Q

2 mechanisms for carrying out double-stranded break repair

A
  1. non-homologous end joining
  2. synthesis-dependent strand annealing
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2
Q

when is non-homologous end joining used

A

when the double-stranded break occurs during G1 and before replication

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3
Q

is replication resumed after non-homologous end joining?

A

yes, but the process inevitably leads to mutation

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4
Q

4 steps of non-homologous end joining

A
  1. double-stranded break produced in DNA
  2. recognized by proteins PKcs, Ku70, and Ku80 that attach to the broken ends
  3. complex trims back the free ends of the break (loss of genetic information)
  4. blunt ends produced by resection are ligated by ligase IV
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5
Q

when is genetic information lost during non-homologous end joining

A

when the complex trims back the free ends of the break

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6
Q

when does synthesis-dependent strand annealing occur

A

when the double-stranded break occurs after replication

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7
Q

does synthesis-dependent strand annealing create mutations?

A

no - error-free process

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8
Q

another name for synthesis-dependent strand annealing

A

homology dependent repair

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9
Q

process of synthesis-dependent strand annealing (SDSA)

A
  • begins with trimming of broken ends
  • attachment of protein Rad15
  • Rad51 facilitates invasion of intact sister chromatid by the resected end of the broken strand
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10
Q

strand invasion process of SDSA

A

displaces on strand of the DNA duplex on the sister chromatid to form a displacement loop

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11
Q

how do sister chromatids reform

A

by dissociation and annealing of the new strand to the other side of the break
- results in replacement of the excised DNA with a duplex identical to the sister chromatid

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12
Q

homologous recombination

A

exchange of genetic material between homologous DNA molecules

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13
Q

when does homologous recombination occur in bacteria/archaea

A

conjugation - consequence of double-strand break repair

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14
Q

when does homologous recombination occur in eukaryotes

A

prophase I of meiosis

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15
Q

in eukaryotes, homologous recombination is initiated by controlled _______________

A

double-stranded DNA breaks

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16
Q

proper chromosome segregation during meiosis depends on

A

homologous recombination

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17
Q

what is the system of homologous recombination in bacteria

A

RecBCD pathway

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18
Q

RecBCD pathway

A
  1. double-strand breaks attract RecA protein
  2. RecBCD then attaches to region where RecA is bound
  3. leads to single-strand invasion and D loop formation
  4. RuvAB and RuvC proteins bind and complete homologous recombination
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19
Q

what does RecBCD pathway rely on to initiate process

A

DNA double-strand breaks

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20
Q

what is RecA protein a homolog of

A

human Rad51 protein

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21
Q

double-stranded break model of meiotic recombination

A
  1. recombination initiated by Spo11
  2. Spo11 degrades, Mrx resect single strands of cut chromatin
  3. proteins Rad51 and Dmc1 help facilitate strand invasion and D loop formation
  4. 2 strands that appear to cross over one another form Holliday junction
  5. heterduplex forms
  6. invading strand extended, with DNA synthesis guided by intact template strands, assisted by Rad52 and Rad59
  7. 3’ end of invading strand joins to 5’ end of strand segment that was initially part of the invading strand (ligation)
  8. non-sister chromatids now connected by double Holliday junctions
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22
Q

where was Spo11 first discovered

A

yeast

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23
Q

Spo11

A

dimeric protein that generates an asymmetric double-stranded cut in one chromatid (both strands)

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24
Q

what proteins associate with Spo11

A

Mrx and Exol

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25
heteroduplex
doubled-stranded DNA formed from single-stranded pieces of DNA of different homologs
26
2 resolution patterns of Holliday junction resolution
1. same sense resolution 2. opposite sense resolution
27
same sense resolution
when two north-south (NS) resolution cuts or two east-west (EW) resolution cuts occur - flanking markers don't recombine - heteroduplex regions remain only in between junction points
28
opposite sense resolution
one Holliday junction is resolved by a NS cut and the other by an ES cut - resulting chromosomes recombinant - major changes in chromosome segments, ends swap
29
is opposite or same sense resolution more common
opposite
30
transposable genetic elements
DNA sequences that can move within the genome by an enzyme-driven process (transposition)
31
2 methods of movement for transposable elements
1. excision of element from its original location and insertion in new location 2. duplication of element and insertion of copy in new location
32
insertional inactivation
when a transposable element causes a mutation by inserting into a wild-type allele and disrupting its function
33
2 sequence features that transposable elements have in common
1. transposable element contains terminal inverted repeats 2. inserted transposable element is bracketed by flanking direct repeats
34
transposition event steps
1. staggered cuts made in both strands of DNA by transposes at new target site for insertion, leaving short single-stranded overhangs on each end 2. double-stranded transposable element is inserted into new site 3. DNA replicated at new site of insertion to fill single-stranded gaps and producing flanking direct repeats
35
what are the staggered cuts made by
transposase
36
what is transposase produced by
transposable element
37
2 types of transposable elements
1. DNA transposons 2. retrotransposons
38
DNA transposons
class II - transpose as DNA sequences, may be replicative or non-replicative
39
retrotransposons
class I - composed of DNA by transpose through RNA intermediate
40
what enzyme do retrotransposons use
reverse transcriptase - RNA copies back into DNA
41
non-replicative transposition
excises transposable element from one position and inserts into new location - cut and paste
42
does non-replicative transposition increase number of transposable elements?
no
43
replicative transposition
increases number of elements per genome - copy and paste
44
3 categories of bacterial transposons
1. insertion sequences 2. composite transposons 3. non-composite transposons
45
insertion sequences
simple transposable elements containing terminal inverted repeats surrounding a gene encoding tranposase
46
composite transposons
carry transposase gene, two flanking IS elements, 1+ additional genes
47
what are composite genes called in bacteria
Tn
48
are composite transposons smaller or bigger than IS elements
much larger
49
what do composite transposons often contain
resistance genes
50
non-composite transposons
similar to composite but lack IS elements
51
are eukaryotic transposable elements frequent or rare?
frequent - plentiful
52
2 types of eukaryotic transposable elements
1. short sequences with inverted repeats (Ac/Ds, P) 2. retrotransposons (Alu, Ty, copia)
53
nearly ________ of the human genome is composed of transposable DNA
half - about 45%
54
retroviruses
infect eukaryotic cells
55
retroviruses genome
composed of single-stranded RNA
56
what happens when a retrovirus infects a eukaryotic cell
RNA is transcribed into dsDNA by reverse transcriptase, allowing DNA to parasitize host cells
57
the genes on retrotransposons are flanked by what?
long terminal repeats (LTRs)
58
what genes in the retrovirus are needed to produce new retroviral particles
gag and env
59
what does pol encode for
reverse transcriptase
60
what are retrotransposons related to
retroviruses
61
what specific genes to retroviruses carry
gag, env, pol
62
what specific genes to retrotransposons carry
pol, some have gag
63
do retrotransposons encode env?
no
64
because retrotransposons don't encode env...
they can be reverse transcribed and inserted into host DNA but are unable to produce viral particles
65
many types of Ty retrotransposons are found where?
yeast - cause insertional mutations
66
structure of Ty retrotransposon
- central element about 6 kb long - flanked by LTRs of about 330 bp - both LTRs contain porters that direct transcription of dif genes
67
where are multiple forms of copia elements found
Drosophila genome
68
structure of copia elements
- central elements about 5-8.5 kb long - contain gag/pol genes - flanked by 250-600 bap LTRs
69
more than ___% of total Drosophila genome is composed of various types of copia elements
5
70
what does LINE stand for
long interspersed elements
71
what does SINE stand for
short interspersed elements
72
L1 elements are _______
LINEs
73
L1 elements
- 6.5-8 kb - full-length encode protein with nuclease and reverse transcriptase function, may also encode RNA-binding protein - associated with spontaneous human mutations
74
Alu elements are the most common of the ______
SINEs
75
what do Alu elements actively generate
mutation
76
Alu elements
- 100-300 bp - flanked by 7-20 bp direct repeats - human genome contains 1+ million
77
data suggests that Alu elements are active in ___ of ____ people are are responsible for about _____% of human hereditary disease
1, 200, 0.3