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genetics > chapter 11 part 3 > Flashcards

chapter 11 part 3 Flashcards

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Biology 101 flashcards
1
Q

when do we do non-homologous end joining?

A

when there is a DSB that occurs during G1 (before replication) which prevents completion of DNA replication

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2
Q

what does non-homologous end joining allow for?

A

cells to regain the ability to complete DNA replication, though we will have mutations for sure

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3
Q

non-homologous end joining process

A

4 steps
-DSB happens
-DSB are recognized by the proteins PKcs, Ku70, and Ku80 that attach to the broken ends of the DNA
-the complex trims back the free ends of the break- LOSS OF GENETIC INFO
-the blunt ends produced by resection are ligates by ligase IV

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4
Q

when does synthesis-dependent strand annealing SDSA occur?

A

we do it when there are DSBs that occur after replication by an error-free process

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5
Q

what is synthesis-dependent strand annealing also called?

A

homology dependent repair (HDR)

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6
Q

what does the strand invasion process do?

A

displaces one strand of the DNA duplex on the sister chromatid, forming a displacement (D) loop
-replication within the loop synthesizes new DNA from the intact template strand

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7
Q

what is the process of SDSA

A

-SDSA begins with trimming the broken ends, followed by attachment of protein Rad51 (a homolog of bacterial RecA protein)
-Rad51 facilitates the invasion of the intact sister chromatid by the resected end of the broken strand
-stand invasion process
-sister chromatids refrom

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8
Q

how do sister chromatids reform in SDSA?

A

by dissociation & annealing of the new strand to the other side of the break, resulting in replacement of the excised DNA with a duplex identical to the sister chromatid

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9
Q

homologous recombination

A

the exchange of genetic info between homologous DNA molecules

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10
Q

homologous recombination in bacteria

A

occurs during conjugation & as a consequence of DSB repair (archaea as well)

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11
Q

homologous recombination in eukaryotes

A

takes place in prophase 1 of meiosis

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12
Q

what is homologous recombination initiated by in eukaryotes?

A

double-strand DNA breaks

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13
Q

what does proper chromosome segregation in meiosis depend on?

A

homologous recombination
-without it, errors like nondisjunction are likely to occur

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14
Q

what is the system of homologous recombination in bacteria?

A

RecBCD pathway

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15
Q

what does RecBCD rely on?

A

DNA double-strand breaks to innitiate the process
-this attracts the RecA protein (homolog of human Rad51 protein)
-RecBCD then attaches to the region where RecA is bound, leading to single-strand invasion & D loop formation (similar to SDSA)
-RuvAB & RuvC proteins bind & complete homologous recombination

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16
Q

DSB model of meiotic recombination simplified

A
  1. recombination initiated by Spo11
  2. Spo11 degrades, Mrx cuts single strand of cut chromatin
  3. Rad51 & Dmc1 facilitate strand invasion & D loop formation
  4. holliday junction formed between 2 strands that appear to cross over
  5. heteroduplex forned
  6. invading strand in extended with DNA synthesis guided by intact template strands assisted with Rad52 & Rad59
  7. 3’ end of invading strand joins 5’ end of a strand segment that was initially invading strand (ligation)
  8. non-sister chromatids are connected by double holliday junctions (DHJs)
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17
Q

heterodulex

A

a double-stranded DNA formed from single stranded pieces of DNA of different homologs

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18
Q

how many ways can holliday disjunction be resolved

A

2

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19
Q

what are the way holliday disjunction can be resolved

A

-same sense resolutions
-opposite sense resolution

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20
Q

same sense resolution

A

when two north-south (NS) resolution cuts or two east-west (EW) resolution cuts occur
-flanking markers DO NOT recombine though heteroduplex regions remain only in between junction points

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21
Q

opposite sense resolution

A

a resolution in which one Holliday junction is resolved by a NS cut & the other by an EW cut, is much more common
-resulting chromosomes are recombinant, & lead to production of recombinant progeny = major changes in chromosome segments
-ends swap

22
Q

what of the two holliday junction resolutions is more commin?

A

opposite sense resolution

23
Q

what are transposable genetic elements?

A

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

24
Q

what do transposable genetic elements vary in?

A

length, sequence composition, number

25
Q

what ways does movement occur in transposable genetic elements?

A
  1. excision of the elements from its original location & insertion in a new location
  2. duplication of the element & insertion of the copy in a new location
26
Q

insertional activation

A

when a transposable element can cause a mutation if it inserts into a wild-type allele & disrupts its function

27
Q

what do all transposable elements have in common?

A
  1. transposable element contains terminal inverted repeats on its ends
  2. the inserted transposable element is bracketed by flanking direct repeats
28
Q

typical transpostion event

A

-staggered cuts are made in both strands of DNA at the new target site for transposable element insertion, leaving short single stranded overhangs on each end of cut
-staggered cuts are made by transposase, produced by the transposable element
-the double-stranded transposable element is inserted into its new site
-dna is replicated at the new site of insertion to fill the single-stranded gaps, producing flanking direct repeats

29
Q

what are the categories of transposable elements?

A

DNA transposons
Retrotransposons

30
Q

DNA transposons

A

(class II transposable elements)
-transpose as DNA sequences & may be replicative or non-replicative

31
Q

Retrotransposons

A

-(Class I transposable elements)
-composed of DNA but transpose through an RNA intermediate
-rna is coped back into DNA by reverse transcriptase
-the reverse-transcribed DNA is then inserted into a new location where flanking directs repeats are formed

32
Q

non-replicative transposition

A

excises a transposable elements from one position & inserts it into a new location
-cut & paste !!!!
-the process moves transposable elements around the genome but results in NO increase in the number of transposable elements

33
Q

replicative transposition

A

increase the number of elements per genome
-COPY & paste !!!!!
-a recombination-like process resolves the cointegrate, & leaves both plasmids with a copy of the element

34
Q

what are the three categories of bacterial transposons

A

-insertion sequences (ISs)
-composite transposons (Tn in bacteria)
-non-composite transposons

35
Q

insertion sequences

A

simple transposable elements containing terminal inverted repeats surrounding a gene (sometimes two genes) encoding transposase

36
Q

composite transposons (Tn in bacteria)

A

carry a transposase gene, two flanking IS elements, & one or more additional genes
-larger than IS element
-often contain resistance genes

37
Q

non-composite transposons

A

-similar to composite transposons but lack the IS element
-typically have additional genes like composite transp.

38
Q

are there a lot of eukaroytic transposable elements?

A

yes, they are highly varied with two types
-short sequences with inverted repeats
-retrotransposons

39
Q

how much of the human genome is composed of transposable DNA?

A

nearly half

40
Q

what do retroviruses do?

A

infect eukaryotic cells & have genomes composed of single stranded RNA (ssRNA)
-on infection the RNA is transcribed into double stranded DNA (dsDNA) by reverse transcriptase, allowing the DNA to parasitize the host cells
-the gene(s) carried on retrotransposons are flanked by long terminal repeats (LTRs)

41
Q

what genes are needed to produce new retroviral particles

A

gag & env encoded by the integrated virus

42
Q

what gene encodes reverse transcriptase?

A

pol

43
Q

do all viral genes contain gag?

A

no, but all contain pol

44
Q

are retrotransposons & retroviruses related?

A

yes

45
Q

what types of retrotranposons are found in yeast?

A

Ty
-these can cause insertional mutations
-central element is about 6kb long & is flanked by long terminal repeats of about 330 bp
-both LTRs contain promoters that direct transcription fo different genes in the central region

46
Q

multiple forms of ______ elements are found in the drosophila genome

A

copia
-have a central element of 5-8.5 kb that contain gag & pol genes & are flanked by 250-600 bp LTRs

*more than 5% of the total genome is comprised of various types of copia elements

47
Q

how much of the human genome is composed of transposable DNA?

A

more than 45%

48
Q

LINEs

A
  • Long interspersed elements
49
Q

SINEs

A

short interspersed elements

50
Q

what do LINEs & SINEs do

A

they are relatively abundant & can cause mutations in humans

51
Q

what is a particularly common LINE

A

L1 elements (600,000/genome)
-associated with spontaneous human mutations
-encode proteins w nuclease & reverse transcriptase function & RNA-binding protein
-6.5-8.0 kb

52
Q

what is the most common SINE

A

Alu elements - they actively generate mutations
-vary in length from 100-300 bp & are flanked by 7-20 bp direct repeats
-more than 1 million of these elements
-active in 1 in 200 ppl & responsible for 0.3% of human hereditary disease

genetics (23 decks)

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