Brainscape's Knowledge GenomeTM

Browse over 1 million classes created by top students, professors, publishers, and experts.


See full index

Biol 200 > Lecture 26 - Replication fidelity, DNA repair and recombination (pt 2) > Flashcards

Lecture 26 - Replication fidelity, DNA repair and recombination (pt 2) Flashcards

1
Q

What nucleotide excision repair mechanism repairs

A

Chemical changes in DNA that affect normal helical structure

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

What induces chemical changes in DNA that affect normal helical structure

A

UV light

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

How chemical changes in DNA that affect normal helical structure are detected

A

Sensor proteins slide along for DNA looking for kinks in the helical structure

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

Example of chemical changes in DNA due to UV light that affect helical structure

A

Adjacent thymine bases covalent bonding

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

Protein that recognizes thymine dimers

A

XP-C/23B complex (dimer)
XP-C = big protein that surrounds it
23 B = protein that binds XP-C

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

First protein that XP-C/23B complex recruits and what it does

A

Transcription factor TFIIH (is a DNA helicase). Unwinds the DNA

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

2 proteins that further unwind the DNA after TFIIH and how what they exactly do

A

XP-G and RPA (replication fork protein A). Further unwind until a 25 nt bubble is formed

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

How TFIIH, XP-G and RPA are positionned together

A

TFIIH thymine dimer side (figure : top)
RPA on A A strand side (figure : bottom)
XP-G on right

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

2 proteins required/recruited after 25 nt bubble is formed and what they do

A

XP-G (already there) and XP-F = 2 ENDOnucleases
XP-G cuts at 3’ end of bubble (on T dimer strand)
XP-F cuts at 5’ end of bubble (on T dimer strand)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

After endonucleases (XP-G and XP-F) action, what happens to 25nt strand that was cut away

A

is degraded

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

How 25nt strand that was cut away is replaced

A

DNAP and DNA ligase fill the gap

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

Name of cancer syndrome in families that lack genes for proteins in nucleotide excision repair

A

xeroderma pigmenosum (XP)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

What induces double-strand DNA breaks

A

X-rays and gamma-rays

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

2 systems that repair double-strand DNA breaks

A

1) Homologous recombination

2) Nonhomologous end joining

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

Principle of homologous recombination and how perfect repair can be

A

Uses unbroken homologous chromosome as a repair template

Can make a perfect repair

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

Nonhomologous end joining difference with homologous recombination and how perfect repair can be

A

does not use unbroken homologous chromosome as repair template
error-prone repair (always errors)

17
Q

Homologous recombination step 1

A

Generate 3’ ssDNA tails at the break site (on broken chromosome) by using 5’-exonucleases

18
Q

Homologous recombination step 2

A

Use Rad-51 to invade (hybridize with) the template chromosome with one of the two tails (part of a strand on the template chromosome makes a bubble cause not coupled anymore)

19
Q

What hybridization of 3’ ssDNA tail of break site to template chromosome allows

A

perfect alignement of the two chromosomes

20
Q

Homologous recombination step 3

A

3’ end of invading tail is extended until DNA in ‘‘bubble’’ base-pairs with other 3’ ssDNA tail. (bubble is now bigger + one of two 3’ ssDNA tails is now repaired)

21
Q

Homologous recombination step 4

A

Extension of the second (non-invading 3’ ssDNA tail) using the strand of the template chromosome that is not hybridized (that forms a bubble)

22
Q

Homologous recombination step 5

A

DNA ligase ligates the repaired strands

23
Q

Structure left after Homologous recombination + name

A

4-stranded DNA structure, all bases paired, but with a bubble of 2 strands mixing with other chromosome ->HOLLIDAY STRUCTURE = EACH CROSSING AT BOTH ENDS OF BUBBLE (so there are 2)

24
Q

When nonhomologous end joining is used

A

when homologous chromosome that can act as a template is not nearby

25
Q

First step of nonhomologous end joining/what happens at broken ends

A

A complex of Ku heterodimer + a DNA-dependent protein kinase binds each broken end of the chromosome AND FORM A SYNAPSE

26
Q

Second step of nonhomologous end joining

A

Several nts are removed from broken end by nucleases

27
Q

Third step of nonhomologous end joining

A

Ligation of the chromosome fragments together

28
Q

2 reasons why nonhomologous end joining is an error-prone repair

A

1) Small deletions at junction point

2) Could join wrong chromosome fragments (if, for example, chrom 6 and 10 are being repaired at the same time)

29
Q

What is metaphase

A

Step in mitosis where chromosomes aligned in the middle of the cell

30
Q

What happens to chromosomes in cells grown in presence of drug that induces DNA double strand breaks

A

Due to the lot of homologous recombination occuring, the chromosomes’ strands are recombinated

31
Q

What method allowed the detection of homologous recombination and what the method does

A

Bodycote and Wolff method : Differentially stains sister chromatids at metaphase

32
Q

why genetic recombination leads to different looks within the population of humans

A

Because of alleles (gene variants)

33
Q

consequence of homologous recombination if no gene variants

A

Homologous recombination would lead to identical chromosomes

34
Q

Hypothesis for why sexual reproduction is prevalant (as opposed to asexual reproduction)

A

Allelic diversity generates phenotyping diversity

35
Q

What happens during metaphase 1 of meiosis

A

strand breaks are induced

36
Q

why strand breaks are induced during metaphase 1 of meiosis (WHICH OF THE TWO TYPES OF dsDNA BREAKAGE REPAIR OCCURS)

A

half of the time, they will result in HOMOLOGOUS RECOMBINATION so advantage for progeny diversity

37
Q

How many of the gametes have a recombined version of the chromosome

A

2 of the 4

38
Q

2 ways of solving a Holliday structure

A

1) Nicking one pair of strands restores original chromosomes

2) Nicking other pair of strands generates recombinant chromosomes

39
Q

How many possibilities for solving the 2 Holliday structures after homologous recombination and how many lead to recombinant DNA

A

4 possibilities. 2 lead to recombinant DNA

Biol 200 (30 decks)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2024 Bold Learning Solutions. Terms and Conditions