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GGA 2 > Meriems lecture > Flashcards

Meriems lecture Flashcards

1
Q

DNA replication is

A

semi conservative

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2
Q

DNA replication in E.coli has 3 steps

A
  1. Initiation, 2. elongation, 3. termination
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3
Q

Elongation is performed by

A

DNA polymerase III

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4
Q

The alpha subunit of RNApolII

A

is responsible for 5’ to 3’ polymerisation

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5
Q

The epsilon subunit of RNApolII

A

is responsible for proof reading exonuclease, it removes any mismatches via a repair system

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6
Q

To replicate the double helix,

A

the strands need to be separated, and this is an energy dependent process

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7
Q

The DNA polymerase acts like a

A

hand (attach slide pic)

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8
Q

DNA polymerase synthesises DNA strand in

A

5’ to 3’ direction

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9
Q

The lagging strand cannot be replicated

A

continuously, the DNApolII will have to go backwards as the DNA continues to get unwound.

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10
Q

The lagging strand forms a loop because

A

the DNApolII wants to move in the same direction

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11
Q

The lagging strand forms a loop because

A

the DNApolII wants to move in the same direction

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12
Q

RNA primers need to be removed

A

to complete replication

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13
Q

DNA polI is responsible for

A

Maturation of Okazaki fragments (and DNA repair)

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14
Q

Initiation of replication in E.coli occurs at

A

oriC

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15
Q

SSB proteins bind to the separated strands to ensure

A

that both strands don’t re-ligate automatically

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16
Q

SSB proteins bind to the separated strands to ensure

A

that both strands don’t re-ligate automatically

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17
Q

H.Sapiens have

A

60000 origins of replication

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18
Q

Eukaryotic DNA is assembled into

A

chromatin

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19
Q

If there is a free double stranded end for DNA, this would suggest that

A

there has been some DNA damage

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20
Q

End replication problem is basically where

A

the lagging strand will not completely replicate, resulting in a terminal gap

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21
Q

Telomerase extends the 3’ end of

A

the chromosome. This means that overall, no genomic information has been lost from each time DNA replication occurs.

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22
Q

Telomerase is activated in

A

85% of cancers

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23
Q

DNA damage does not necessarily

A

mean mutation

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24
Q

Rate of spontaneous mutations can be

A

5x10^-11 for mamallian systems

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25
Q

Mutations can result from errors in DNA replication

A
  1. failure of epsilon subunit of DNA polymerase to proof- read
  2. Tautomeric shift of bases causing mismatches in base pairing
26
Q

Each of the 4 bases can exist in

A

alternative forms (tautomers)

27
Q

A tautomeric shift can cause a point mutation

A

Where a CG pair turns to a CG+, and causes the G+ to bind with T, and replication to thereby occur, it would result in a CG base pair to convert to a TA pair

28
Q

slippage of the template strand can cause

A

deletion

29
Q

Slippage on the newly synthesised strand can cause

A

insertion

30
Q

Mismatch repair system

A

repairs replication errors

31
Q

Base excision error

A

repairs damaged bases. This occurs via DNA glycosylase which removes the base and leaves a gap. AP endonuclease recognises a sugar with missing base and cuts the phosphodiester backbone of DNA. DNA polymerase I fills in the gap and DNA ligase seals the DNA strand

32
Q

Nucleotide excision repair

A

repairs unusual base structures which distort DNA such as pyrimidine dimers

33
Q

Mechanism of mismatch repair system

A

see attach slide

34
Q

What happens to your DNA upon DNA damage?

A

where there are two thymine bases beside each other, exposure to UV can cause Thymine dimers to form. This can be repaired using CPP photolyase and light. Placental mammals cant do this though so they just cut i.e. NER.

35
Q

Synthesis of RNA from DNA template require

A

RNA polymerase as well as RNTP’s

36
Q

RNA polymerase copies one strand

A

of duplex DNA into RNA

37
Q

In prokaryotes, a single RNA polymerase

A

transcribes all genes

38
Q

The three steps of RNA transcription

A
  1. initiation
  2. elongation
  3. termination
39
Q

Transcription is initiated at specific regions in DNA

A
  • gene promoters
40
Q

There is a signal sequence on RNA which signals to end transcription

A

AAA

41
Q

Formation of charged tRNA

A

occurs via a synthetase enzyme joining amino acyl and tRNA

42
Q

tRNA’s decode the

A

mRNA message

43
Q

Each amino acyl tRNA synthetase charges a tRNA

A

with the correct amino acid

44
Q

Translation also occurs in 3 steps

A
  1. Initiation
  2. Elongation
  3. Termination
45
Q

Prokaryotes have

A

polycistronic mRNA’s,

46
Q

Eukaryotes have

A

monocistronic mRNA’s

47
Q

The initiator tRNA is the only tRNA which can

A

bind to the P- site

48
Q

Why regulate gene expression?

A

a) economical
b) some gene products are required at different times or in different environments
c) gene products are required in different amounts
d) In eukaryotes, differential gene expression occurs during development

49
Q

How is gene expression regulated?

A

At the transcription level, initiation and elongation of the transcript
At the post transcription level, Transcription processing (folding of mRNA prevents translation) and protein modification

50
Q

The E.coli sigma 70 sub unit recognises a specific set of gene promoters

A

and binds to the -35 and -10 regions

51
Q

The initiation of gene transcription is controlled by regulatory proteins

A

These include positive regulators and negative regulators. In prokaryotes, Activator proteins bind to a DNA sequence close to the promoter and help recruit the RNA polymerase to the promoter.

52
Q

Genes can be activated by either

A

binding the activator protein or removing the repressor protein

53
Q

Prokaryotes generally

A

have polycistronic mRNA’s. This gene structure is called the operon.

54
Q

Lactose is the inducer of

A

the lac operon

55
Q

LacI creates the

A

lac repressor which binds to the operator to prevent transcription

56
Q

With no inducer present,

A

the repressor makes contact with the operator

57
Q

When inducer is present it binds

A

to repressor, altering its structure and it cannot bind to the operator

58
Q

The lac operon is positively regulated by

A

CAP

59
Q

Lactose + glucose =

A

low basal lac gene expression

60
Q

Lactose no glucose

A

high lac gene expression

GGA 2 (4 decks)

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