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Genetics > Module 5 lecture 5-9 > Flashcards

Module 5 lecture 5-9 Flashcards

(110 cards)

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1
Q

in watson and cricks report in the journal they stated

A

the pairing proposed does suggest a possible copying mechanism for genetic material; replication

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2
Q

watson and crick favored what model of DNA Replication

A

Semi-conservative

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3
Q

what models where there of DNA replication

A

semi-conservative model, conservative model, dispersive model

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4
Q

Semiconservative replication

A

one strand is used as a copy for 2 new molecules

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5
Q

conservative replication

A

he parental double helix remains
intact; both strands of the daughter double helix are newly
synthesized

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6
Q

dispersive replication

A

At completion, both strands of both
double helices contain both original and newly synthesized
material.

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7
Q

mehelson stahl experiment was used to determine

A

which type of replication occured

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8
Q

describe mehelson stahl experiment

A

-started out with “normal” DNA
-They grew it for one
generation with the heavy
isotope 15N instead of normal
14N
-

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9
Q

weight differences between N15 and N14 allowed tracking

A

N15 able to track

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10
Q

What DNA-“weight” would
you expect for any of the
three suggested models in
the first generation

A

Conservative: Two distinct bands (heavy and light).
Semi-Conservative: One intermediate band.
Dispersive: One intermediate band

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11
Q

mehelson stahl experiment control

A

e. coli grown in N14-> dna extracted from cells-> centrifuge

results N14N14 single band towards top of tube

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12
Q

mehelson stahl experiment first part

A

e. coli grown for many generation in N15 medium->Extract DNA->Centrifuge

results: 1 band of dna N15 towards the bottom

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13
Q

mehelson stahl experiment second part

A

N15 cells transfered to N14 medium->cells replicate to form first generation of daughter cells-> extract DNA-> centrifuge

Results: single band in middle of tube in between n15 and N14. meant DNA strand had both n15 and N14

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14
Q

mehelson stahl experiment third part

A

cells replicate to form second generation of daughter cells-> extract DNA-> centrifuge

results: band of hybrid N15N14, and band of N14N14

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15
Q

After growing Escherichia coli for several generations in a
medium containing 15N, they found that the DNA of the cells
was heavier than normal because

A

N15 atoms are heavier

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16
Q

what determines where it accumulates in the tube after centrifuge

A

density of DNA

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17
Q

semiconservative replication detailed model shows

A

-original double helix
-strands seperate
-complementary bases align opposite templates (A-T, C-G)
-enzymes link sugar phosphate backbone of aligned nucleotide into continuous strand

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18
Q

end result of semiconservative replication

A

2 molecules of DNA each has old strand and new strand. if layed on top of each other they should be identical.

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19
Q

Complementary base pairing produces

A

semi-conservative
replication

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20
Q

summary of semiconservative replication

A

-Double helix unwinds
-Strands separate
-Each strand acts as template for new strand
- Complementary base pairing ensures that T signals addition of A on new strand, and G signals addition of C
-Two daughter helices produced after replication

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21
Q

New daughter strands have to be synthesized from

A

free nucleotides

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22
Q

5-3 synthesis is _____ then 3-5

A

easier

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23
Q

in 5-3 there is

A

a bond made between hydroxal group on 3 carbon and incoming nucleotide phosphate group

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24
Q

DNA polymerases only works

A

5’ to 3’

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25
DNA polymerase enzyme can only connect a
new nucleotide to a free third carbon atom of a preceding nucleotide
26
what different abou RNA as appose to DNA
RNA polymerases can start from scratch
27
in dna replication there are _______ and ______ strands
leading and lagging
28
Leading strand is made
continuously in 5’->3’ direction following the moving replication fork
29
in leading strand The initial free 3’ end is provided by
a short RNA prime
30
primer is made by ________
Primase
31
primase=
rna polymerase
32
lagging strand is made
discontinuously in form of smaller fragments
33
what are the smaller fragments in Lagging strand
1000 nucleotides and okazaki fragments
34
in lagging strand there are _____ fragments
shorter
35
Each shorter fragment is started by an
RNA PRIMER
36
how many okazaki fragments are there
2
37
After two Okazaki fragments are finished they are connected by a
DNA Ligase
38
what are some key enzymes involved in dna replication (6)
DNA helicase Pol III Pol I Primase Ligase
39
Dna helicase
unwinds double helix
40
polymerase III
produces new strands of complementary DNA
41
Polymerase I
fills in gaps between newly synthesized Okazaki segments
42
Primase
creates RNA primers to initiate synthesis
43
ligase
welds together Okazaki fragments
44
how many key enzymes are involved in dna replication
6
45
Single-stranded binding proteins
keep helix open
46
primary enzyme of dna synthesis
POL III
47
what enzymes re implicated in DNA repair
Pol I and Pol II
48
All DNA polymerase (I, II, III) have
3-5 prime exonuclease activity
49
whats special about DNA pol i as apposed to Pol I and II.
Polymerase I participates in 5-3 prime exonucelase activity.
50
3-5 prime exonuclease activity is involved in
proofreading
51
DNA replication starts at
origin of replication
52
replication bubble made using 2
replication forks
53
lagging strand is tied together by
ligase enzyme
54
in eukaryotes okazaki fragments are smaller. how small?
150 nucleotides
55
how many “classic” DNA polymerases are there
5
56
what are the 5 “classic” DNA polymerases
Pol a pol s pol e pol y pol b
57
pol a (alpha)
extends RNA primer with 20 nucleotides
58
pol s
main polymerase, when reaching next Okazaki fragment it displaces fragment and keeps synthesizing DNA
59
what is the main polymerase
pol s
60
Pol y
replicates mitochondrial genome
61
pol B (beta)
does dna repair
62
pol y, s, e have
exonucelase activity and (proofreading)
63
in eukarotes Replication forks move in
opposite directions
64
is the replication fork moves in opposite directions then
Replication is bidirectional
65
Eukaryotic chromosomes have several hundred
origins of replication
66
In circular chromosomes, such as E. coli, there is only how many ORIC
one origin of replication
67
in linear chromosomes, telomeres ensure the
maintenance and accurate replication of chromosome ends
68
Unwinding of chromosomes causes
supercoiling
69
Errors during replication are ____
rare
70
after errors during replication are made they are
immidientely corrected
71
Enzymes repair ________ to DNA
chemical change
72
Complementary strands are ______
redundant
73
telomeres solve the
end problem
74
unwinding of chromosomes at one location causes another location to
wind tightly (supercoil)
75
solution to supercoil is
topoisomerases
76
topoisomerase works by
cutting dna, rotate strands so not winded, then strand is rejoined by ligase.
77
topoisomerase is in
eukaryotes and prokaryotes
78
important topoisomerase in bacteria is known as
girase
79
since complementay strands are redundant
basis to repair errors correctly either during or after replication
80
what generates diversity
mutations and New combinations of already existing alleles
81
how can New combinations of already existing alleles occur (3)
-Independent assortment of homologous chromosomes -Crossing-over -recombination
82
Molecular Mechanism of Crossing-over first step is
initial cut, by spo11
83
what enzyme is involved in cutting strand of chromatids
spo11
84
non sister chromatids involved in
crossing over
85
Molecular Mechanism of Crossing-over second step
exonuclease chew away from intial cuts and 3 prim single stranded tails are formed
86
one of the tails from cut made will ______ another chromatid
invade, and takes its place
87
when chromatid is invaded and strands place is taken the previous strand that was there before is
displaced
88
displacement of strand during strand invasion forms
D loop
89
Molecular Mechanism of Crossing-over third step
formation of holiday junction
90
formation of holiday junction
2 crosses between strands of 2 different chromatids
91
holiday junctions can
migrate.
92
when holiday junctions migrate we have a
heteroduplex (in between holiday junctions)
93
holiday junction migration moves towards
holiday junctions
94
Whether Crossing over Occurs Depends on
the Resolution of Both Holliday Junctions
95
if resolution of intermediates is the same there will be ______. if it is the same there is_____
gene conversion; crossing over
96
in heteroduplex there are
2 chromosomes, 4 chromatids, 4 alleles
97
after heteroduplex form and repair there are
Mismatch repair no. 1 and 2 Mismatch repair no. 3 and 4
98
Mismatch repair no. 3 and 4
leads to allele frequency changes
99
Mismatch repair no. 1 and 2
regenerate original allele combinations
100
how many chromatids are involved in crossovers
2
101
what happens to the 2 unused chromatids in crossovers
remain same
102
If original was AA BB after mismatch repair, no. 3 and 4 result in genotypes ABBB or AAAB
Allele ratios change from 2:2 to 3:1 or 1:3
103
when allelic changes occur as a result of heteroduplex resolution we get
gene conversion
104
What Happens in Heteroduplexes?
-If the two strands are not identical we end up with a base pair mismatch in the heteroduplex
105
The mismatch will be fixed by _______
cellular enzymes
106
how do cellular enzymes fix mismatches
cut out the wrong base and replace it
107
But how does the enzyme know which base has to be replaced
enzymes recognize pairing isnt correct
108
how does heteroduplex work allele A- ATATCG. allele C- ATCTCG AATAGC TAGAGC after reombination what would happened
the strands are similar so there could be a mixmatch in the dna bases
109
mismatches can be repairs to
A/A or B/B repairs. original A allele. and B/B means original B allele
110
heteroduplex can make
mismatch pairs.

Genetics (29 decks)

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