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genetics > chapter 11 part 3 > Flashcards

chapter 11 part 3 Flashcards

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Biology 101 flashcards
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
Q

heteroduplex

A

doubled-stranded DNA formed from single-stranded pieces of DNA of different homologs

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26
Q

2 resolution patterns of Holliday junction resolution

A
  1. same sense resolution
  2. opposite sense resolution
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27
Q

same sense resolution

A

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

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28
Q

opposite sense resolution

A

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

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29
Q

is opposite or same sense resolution more common

A

opposite

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30
Q

transposable genetic elements

A

DNA sequences that can move within the genome by an enzyme-driven process (transposition)

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31
Q

2 methods of movement for transposable elements

A
  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
Q

insertional inactivation

A

when a transposable element causes a mutation by inserting into a wild-type allele and disrupting its function

33
Q

2 sequence features that transposable elements have in common

A
  1. transposable element contains terminal inverted repeats
  2. inserted transposable element is bracketed by flanking direct repeats
34
Q

transposition event steps

A
  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
Q

what are the staggered cuts made by

A

transposase

36
Q

what is transposase produced by

A

transposable element

37
Q

2 types of transposable elements

A
  1. DNA transposons
  2. retrotransposons
38
Q

DNA transposons

A

class II - transpose as DNA sequences, may be replicative or non-replicative

39
Q

retrotransposons

A

class I - composed of DNA by transpose through RNA intermediate

40
Q

what enzyme do retrotransposons use

A

reverse transcriptase - RNA copies back into DNA

41
Q

non-replicative transposition

A

excises transposable element from one position and inserts into new location
- cut and paste

42
Q

does non-replicative transposition increase number of transposable elements?

A

no

43
Q

replicative transposition

A

increases number of elements per genome
- copy and paste

44
Q

3 categories of bacterial transposons

A
  1. insertion sequences
  2. composite transposons
  3. non-composite transposons
45
Q

insertion sequences

A

simple transposable elements containing terminal inverted repeats surrounding a gene encoding tranposase

46
Q

composite transposons

A

carry transposase gene, two flanking IS elements, 1+ additional genes

47
Q

what are composite genes called in bacteria

A

Tn

48
Q

are composite transposons smaller or bigger than IS elements

A

much larger

49
Q

what do composite transposons often contain

A

resistance genes

50
Q

non-composite transposons

A

similar to composite but lack IS elements

51
Q

are eukaryotic transposable elements frequent or rare?

A

frequent - plentiful

52
Q

2 types of eukaryotic transposable elements

A
  1. short sequences with inverted repeats (Ac/Ds, P)
  2. retrotransposons (Alu, Ty, copia)
53
Q

nearly ________ of the human genome is composed of transposable DNA

A

half - about 45%

54
Q

retroviruses

A

infect eukaryotic cells

55
Q

retroviruses genome

A

composed of single-stranded RNA

56
Q

what happens when a retrovirus infects a eukaryotic cell

A

RNA is transcribed into dsDNA by reverse transcriptase, allowing DNA to parasitize host cells

57
Q

the genes on retrotransposons are flanked by what?

A

long terminal repeats (LTRs)

58
Q

what genes in the retrovirus are needed to produce new retroviral particles

A

gag and env

59
Q

what does pol encode for

A

reverse transcriptase

60
Q

what are retrotransposons related to

A

retroviruses

61
Q

what specific genes to retroviruses carry

A

gag, env, pol

62
Q

what specific genes to retrotransposons carry

A

pol, some have gag

63
Q

do retrotransposons encode env?

A

no

64
Q

because retrotransposons don’t encode env…

A

they can be reverse transcribed and inserted into host DNA but are unable to produce viral particles

65
Q

many types of Ty retrotransposons are found where?

A

yeast - cause insertional mutations

66
Q

structure of Ty retrotransposon

A
  • central element about 6 kb long
  • flanked by LTRs of about 330 bp
  • both LTRs contain porters that direct transcription of dif genes
67
Q

where are multiple forms of copia elements found

A

Drosophila genome

68
Q

structure of copia elements

A
  • central elements about 5-8.5 kb long
  • contain gag/pol genes
  • flanked by 250-600 bap LTRs
69
Q

more than ___% of total Drosophila genome is composed of various types of copia elements

A

5

70
Q

what does LINE stand for

A

long interspersed elements

71
Q

what does SINE stand for

A

short interspersed elements

72
Q

L1 elements are _______

A

LINEs

73
Q

L1 elements

A
  • 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
Q

Alu elements are the most common of the ______

A

SINEs

75
Q

what do Alu elements actively generate

A

mutation

76
Q

Alu elements

A
  • 100-300 bp
  • flanked by 7-20 bp direct repeats
  • human genome contains 1+ million
77
Q

data suggests that Alu elements are active in ___ of ____ people are are responsible for about _____% of human hereditary disease

A

1, 200, 0.3

genetics (37 decks)

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