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Molecular Genetics > Recombination 1 > Flashcards

Recombination 1 Flashcards

1
Q

What is recombination?

A
  • Allows favorable and unfavorable mutations to be separated and tested as individual units in new assortments
  • Provides a means of escape and spreading for favorable alleles and a means to eliminate an unfavorable allele without bringing down all the other genes it is associated with
  • Basis of natural selection
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2
Q

when does recombination occur?

A

During gamete formation
- Two sister chromosomes divide forming chromatids
- 4 chromatids
- The crossing over occurs with the chromatids, but only with 2 out of the 4
- 50% recombinants

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3
Q

homologous recombination

A
  • Homologous (generalized): in eukaryotes, it occurs at meiosis in males and females in the formation of the gametes
  • Takes 50 bases being identical before this occurs in bacteria
  • Occurs during meiosis
  • Does not change the order of the genes - no change in genome organization
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4
Q

site-specific recombination

A

phage integration; inversion of specific regions of the bacterial chromosome
- Site specific recombination changes the genome organization = change order of genes/configuration; deleted or switch order

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5
Q

what are some examples of site-specific recombination?

A
  • VJ joining in mammalian antibody genes
  • mating type switch
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6
Q

leptotene

A
  • chromosomes first visible, DNA replication has already finished
  • condensed chromosomes become visible, often attached to nuclear envelope
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7
Q

zygotene

A
  • chromosomal pairing begins
  • chromosomes begin pairing in limited region/s
  • pair by the pattern of proteins on the DNA
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8
Q

pachytene

A
  • synaptonemal complex complete; recombination is completed
  • Synaptonemal complex extends along entire length of pair chromosomes
  • Synapsis is chromosomal pairing
  • Thick complex
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9
Q

diplotene

A
  • chiasmata are visible
  • Chromosomes segregated but are held together by chiasmata
  • Chiasmata - formation of chi (x formation)
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10
Q

diakinesis

A
  • chiasmata are visible
  • Chromosomes condense, detach from envelope, chiasmata remain, all 4 chromatids become visible
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11
Q

what are the steps of homologous recombination?

A
  • Homologous recombination occurs between synapse chromosomes of the gametes
  • Chromosomes must synapse (pair) in order for recombinant nodes to form where crossing over is thought to occur
  • Leptotene: chromosomes first visible; DNA replication has already finished
  • Zygotene: chromosomal pairing begins (synapsis)
  • Pachytene: synaptonemal complex complete; recombination is completed
  • Diplotene: chiasmata are visible, chromosomes separated
  • Diakinesis: chiasmata are visible, chromosomes condense and all 4 chromatids are visible
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12
Q

when do ds breaks begin?

A
  • Leptotene
  • Homologous recombination always starts at one of the two chromatids
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13
Q

When is recombination complete?

A

Late pachytene

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14
Q

what are the 5 ways to repair a double stranded break in eukaryotes?

A
  • double-strand break repair
  • synthesis-dependent strand annealing
  • nonhomologous end joining
  • single strand annealing
  • chromosome break induced replication
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15
Q

double-strand break repair

A
  • ds break in one pair of the DNA mediated by GEN1
  • 5- end resection - nuclease chews the 5’ end on either side of the break and happens immediately after break is recognized
  • single strand invasion of intact DNA - the 3’ end produced a single strand with a 3’, protruding into the second set of DNA - this serves as a primer for replication/synthesis
  • synthesis of both strands - single strand invasion into the intact DNA forms a D-loop duplex
  • branch migration expands the D-loop, has to expand past the 5’ end is in order to trigger end capture
  • when the ends join together is called end capture
  • need homology to form the second holiday structure
  • double holliday structure forms
  • When you get the displacement loop, it joins with the top molecule (anneal) so you have to have enough exposed to form with the upper strand to repair the break at the top as well
  • ligation occurs
  • outcome is cutting which can result in DSB crossover or no crossover
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16
Q

when does DSBR occur?

A
  • damage
  • collapsed replication fork
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17
Q

branch migration

A
  • branch sites can migrate in either direction
  • this is an important step in expanding the D-loop to reach the far side of the gap left by the ds break
  • movement probably driven by the recombination machinary
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18
Q

when is the DSBR pathway used?

A
  • ds breaks during accidental breaks during mitosis, outcome is no crossover
  • introduced ds breaks during meiosis, outcome is both crossover or no crossover, with a heavy bias towards crossover
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19
Q

how does DSBR occur in bacteria?

A
  • It is much simplified from the eukaryotic double stranded break homologous recombination model
  • No mechanism to automatically generate the ds break
  • Often occurs in repair of DNA damage during replication
  • When there is a crossover, but there will never be a double holliday structure
  • Closely related to repair of damage
20
Q

synthesis dependent strand annealing

A
  • repair of accidental breaks during meiosis and mitosis
  • only one holliday structure forms
  • recombinant joint that does not mature into holliday structure
  • 3’ end invades but no stable displacement loop because there is no template to form a hybrid with
  • unstable and collapses
  • break repair with no crossover
21
Q

when is the SDSA pathway used?

A
  • Repair of accidental break ins during mitosis or meiosis
  • Gamete formation in early pachytene during meiosis
  • Early recombination events there is no crossover
  • Outcome: never get crossover in either mitosis or meiosis
22
Q

in what organism was SDSA first discovered?

A
  • yeast mating type switch
  • the steps are similar to the dsbr model but there is no second end capture and no holliday structure because the D loop is unstable
  • no crossing over
  • used to switch between “sex” types of alpha and A
23
Q

when does SDSA occur?

A

Early/mid pachytene
- Patch recombination

24
Q

Nonhomologous End Joining (NHEJ)

A
  • Typically used when homologous sequences are not available such as G1 phase of cell cycle. Plays important role in VDJ recombination in the immune system shuffling of the parts of antibody genes.
  • Requires Ku proteins -Ku70 and 80- bring to the ends at the double strand break
  • often imprecise ligation with bases deleted/gaps/insertions
25
Q

allelic conversion

A
  • ascomycetes
  • unusual crossover ratios indicate crossover with the allele
  • if no recombination within the allele, expect a 4:4 ratio
  • ratios such as 3:5 and 2:6 indicate one and two recombination events
26
Q

what are the two causes of allelic conversion?

A
  • gap repair
  • mismatch repair
  • not dependent on crossovers
27
Q

Single Strand Annealing

A
  • The ds break happens to occur between two repeated sequences in the genome
  • At ds breaks, repeated sequences anneal to bring the two ends together wrestling in the elimination of non-repeat sequences between the two repeats, only if in the same orientation
  • 5’ resection, but instead of going to sister chromatid, because it is not during replication, you can anneal with yourself
  • Flapases repair the break with the repeats that is similar to FIN1
  • can lose up to 100 bases
  • no D-loop, strand invasion, or crossing over
28
Q

Chromosome Break-Induced Replication

A
  • When the end of chromosome breaks off, only a single end is left
  • Dsbr occurs right after a sequence that is repeated on two nonhomologous chromosomes
  • Can use the homology between two nonhomologous chromosomes as a template
  • Repair involves annealing of a repeated sequence present on a nonhomologous chromosome
29
Q

does the synaptonemal complex form before or after the double-stranded break?

A
  • Forms after the ds break
  • Synaptonemal complexes begin forming in late leptotene
30
Q

what effect does a mutation with Rad 50 produce?

A

Mutation blocks the formation of the protruding 3’ single-strand ends. This block in recombination also prevents formation of the SC

31
Q

what effect does a mutation with ZIP2 produce?

A
  • Zip mutants still form axial associations
  • Wild type zip2 proteins localizes to the axial associations and requires ds breaks - the axial associations hold the pairs together before SC formation
  • There will still be correct chromosome pairing but will not form type SC
  • same applies for ZIP1
  • ZIP2 localizes to the nodules in rad 50 mutants (no 3’ end processing) but not in spo11 mutants, suggesting that ds breaks are a requirement
  • unable to discriminate between homologous and nonhomologous pairing
32
Q

what effect does a mutation with HOP2 produce?

A
  • Mutants form normal synaptonemal complexes but the complex contain non-homologous chromosomes
  • The initial pairing of the homologs is not base pairing of DNA, it is the cooking together of matching proteins to bring the chromosomes together
    Where homologs can recognize each other
33
Q

what is the function of the synaptonemal complex?

A
  • forms after the double stranded break
  • ZIP2 binds to proteins present in early recombination nodules and initiates formation of the SC, it also recruits proteins that bias towards crossover
  • the break is mediated by Spo11, which attaches to the 5’ end
  • immediately after the cut, a 5’ resection is chewed back by rad 50 to leave 3’ protruding to act as a primer
  • rad 51 coats the 3’ end to mediate strand exchange and strand invasion
  • in phase 1 during leptone, the formation of the SC complex begins, recombination events occur but noncrossover events, phase is marked by SDSA
  • in phase 2, the DSBR model is used, crossover events and holliday structure occur during pachytene
34
Q

chiasma

A
  • thought to be region where recombination event occurred
  • Protein structure left over
  • Hold homologs together after SC dissolves
  • Important in chromosomal separation
35
Q

nodules

A
  • represent chiasma
  • Central element in a recombination nodule - presumed region of the chiasma seen later
36
Q

interaxial bridge formation

A

axis of the chromosome of the homologs

37
Q

what is the early stages of meiosis in human fetal ovaries

A
  • leptotene - double strand breaks and strand exchange, initiates synaptonemal complex
  • zygotene
  • pachytene
  • diplotene - synaptonemal complexes disappear but chiasma remain
38
Q

Patch resolution

A
  • If the cuts are on the same strand when they are joined together, they will keep the original strands that were formed before the D loop, patch recombination will occur
  • There will be a region that contains new DNA synthesis where you copied from the homolog, but the two chromatids will still be connected to the original strand and no crossover occurs
39
Q

crossover resolution

A

When you cut on different strands
Not apart of the invading strand
True crossover recombination

40
Q

recombinant joint

A
  • unstable
  • precursor to holliday junction
41
Q

dissolution of holiday structure

A
  • BLM helicase with a topo3A are in a complex
  • BLM = predisposition for cancel and ends up with alot of recombination events
  • The helicase pushes the two holliday structures together to merge into one and topo continuously removes super helical tension and can cause no crossover
  • if both holliday junctions resolve as a crossover then they cancel each other and there is no crossover between the two dsDNA
42
Q

resolution of holliday structure

A
  • Holliday junctions form but do not merge together
  • Use of GEN1
  • Nuclease SLX4 cut
  • Determine whether there is crossover or not
  • Heavy bias towards crossover in meiosis
43
Q

How are double holliday junctions resolved during mitotic divisions?

A

Always end up in a no crossover

44
Q

How are double holiday junctions resolved during meiotic divisions?

A

Strong bias towards crossover

45
Q

Would recombination between each of the chromosomes still occur in a hop2 mutant?

A

No - since the chromosomes would be mispaired no recombination

46
Q

Would recombination still occur in a rad 50 mutant?

A

No - since the 3’ ends at the ds breaks would not be processed

Molecular Genetics (28 decks)

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