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Bcem 341 > Lecture 1-2: DNA structure and replication > Flashcards

Lecture 1-2: DNA structure and replication Flashcards

1
Q

RNA and DNA are polymers, each repeating unit has a (3)

A

sugar, phosphate and base

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2
Q

The base in DNA all contain

A

nitrogen

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3
Q

Purines:

includes+shape

A
  • Guanine and adenine
  • double ring structure
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4
Q

Pyrimidines

include+shape

A
  • thymine and cytosine, uracil
  • single ring structure
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5
Q

nucleotide has at least

A

1 phosphate

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6
Q

Nucleoside

A
  • Base bonded to a sugar
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7
Q

General feature of DNA structure: (4)

A
  • Two strands in opposite directions from a right handed helix (one strand 3’-5’ and one is 5’-3’)
  • The sugar phosphate backbone is on the outside of the helix and the purine and pyrimidine bases face the middle
    Bases are perpendicular to the helical axis, with bases seperated by 3.4 A (consitent spacing in helical ladder)
  • Helix is approx. 20 A wide
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8
Q

Lot more —– are more stable then —–

A
  • C-G
    A-T
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9
Q

Deoxyadenosine contains:

A

Sugar+adenine

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10
Q

In the deoxyribose sugar, a base attaches on the—– carbon and the phosphate attach on the —- carbon. The —- carbon has a —-

A
  1. 1’
  2. 5’
  3. 3’
  4. OH group
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11
Q

DNA Base pairing (3):

A
  1. Same amount of purine nucleotides as pyrimidine nucleotides (in dsDNA)
  2. Same amount of A and T; Same amount of G and C
  3. A+T isnt necessarily equal to G+C
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12
Q

There are — base pair in a full twist of DNA and
adjacent bases are separated by approximately —-

A
  • 10
  • 3.4 A
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13
Q

The melting temperature is defined as the —

A

The melting temperature of DNA refers to the temperature at which 50% of DNA in a sample has denatured from double-stranded DNA (dsDNA) to single-stranded DNA

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14
Q

What are the chemical forces that stabilize the double helix?

A

Hydrogen bonds between base pairs and van der Waals interactions among the bases. The van der Waals interactions come into play because of the hydrophobic effect, which forces the bases to the interior of the helix.

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15
Q

Nucleotides contain….. and are

A
  • phosphate
  • sugar
  • base
  • are the repeating unit in nucleic acids
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16
Q

Nucleotides form DNA strands by

A

phosphodiester linkages

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17
Q

Phosphodiester linkages (2)

what it does+what it includes

A
  • Phosphate/sugar backbone
  • links the 3’ carbon of one sugar to the 5’ carbon of aother via a phosphate
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18
Q

The 2 sugar phosphate backbones are

A

antiparallel

one is 5’-3’ nd other is 3’-5’

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19
Q

DNA strands are held together by — hydrogen bonds. AT has — and CG has—-

A

2 or 3
2
3

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20
Q

Semi conservative replication (3)

what happens+ results in

A
  1. 2 DNA strands unwinds from eachother
  2. Each DNA strand acts as a template for the synthesis of a new complementary strand
  3. Results in 2 double helices that are identical to the original
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21
Q

Replication is catalyzed by—- which is—-

A

DNA polymerase which is an enzyme that adds nucleotides into a growing DNA chain

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22
Q

What drives DNA synthesis or DNA polymerase?

A

Cleaving off pyrophohphate produces energy to help drie DNA synthesis

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23
Q

In DNA replication, what does the OH hydroxyl group on the 3’ carbon of growing strand do?

A

Forms a nucleophillic attack on alpha phosphorus. Cleaves off the 2 pyrophosphate which provides energy to drive the rxn.

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24
Q

New nucleotides can only be added on the —- because it has to ——

A
  1. 3’ end
  2. have a free OH group to form the nucleophillic attack for new nucleotides
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25
Q

DNA is synthesized in the

A

5’-3’ direction

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26
Q

Replication is catalyzed by

A

DNA polymerase

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27
Q

What drives the DNA synthesis?

A

Cleaving off the pyrophosphate (2) on the triphosphate base by 3’oh produces energy to help drive DNA synthesis

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28
Q

New nucleotides can only be added on the

A

3’ end

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29
Q

Where is the base attached in dna

A

1’ carbon of the deoxyribose sugar

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30
Q

—– in purines
and —– in pyrimidines form a bond with the sugar.

A

N-9
N-1

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31
Q

Where is the phosphate attached on the deoxy ribose sugar

A

the 5’ carbon

32
Q

DNA pol 3

A
  • The main enzyme that catalyzes DNA synthesis by adding nucleotides. Can extend a chain bit cant start a chain (5’-3’)
33
Q

Helicase

A

Distrupts the H-bonds between strands using ATP to seperate strands

34
Q

The Double Helix Is Stabilized by (2)

A

Hydrogen Bonds and the Hydrophobic Effect

35
Q

4 steps for DNA replication

A
  1. Primase synthesizes short RNA primers using DNA as a template
  2. DNA polymerase III synthesizes DNA starting at the 3’ end of RNA primers
  3. DNA poly 1 removes RNA primers and fills the gap
  4. DNA ligase connects adjacent DNA fragments
36
Q

What type of bond does DNA ligase catalyze?

A

Phosphodiester bond

37
Q

Compare the sugar in RNA and DNA

A
38
Q

Semi-conservative Replication+ configration

A

Each daughter strand remains paired with its complimentary parental strand.

one strand daughter, one strand parent

39
Q

Conservative

A

after replication, both daughter strands pair up parent strand stay together

40
Q

Meselsin Stahl Experiment tell me the procedure (3)

A
  1. Took normal coli cell DNA and made them divide in heavy nitrogen (N15) so all the DNA were heavy.
  2. DNA switched to light nitrogen and dna made after the switch would have to be made up of N14 as this would have been the only nitrogen available for DNA synthesis.
  3. measured the density of the DNA (and, indirectly, N14/N15) using density gradient centrifugation with cesium chloride.
41
Q

generation 0 meselson-Stahl (3)

  • contents
    -Band
    -time
A
  • All heavy N15 E coli DNA
  • before the switch to light nitrogen
  • produced a single band after centrifugation.
42
Q

Generation 1 Meselson Stahl (3)

content
band
fit/reject

A
  • 100% hybrid of light and heavy nitrogen
    -fits with dispersive and semi conservative but not conservative
    -single band middle
43
Q

Generation 2 Meselson Stahl

A
  • Two bands, 1 middle and 1 light band
  • Know that it was conservative as in dispersive replication, all the molecules should have bits of old and new DNA, making it impossible to get a “purely light” molecule.
  • 2 hybrid, 2 light
44
Q

what do you expect over Generation 3/4 of the Meselson Stahl experiment?

A

over the third and fourth generations, we’d expect the hybrid band to become progressively fainter (because it would represent a smaller fraction of the total DNA) and the light band to become progressively stronger (because it would represent a larger fraction).

45
Q

Why is the buffer important in the Meselson and Stahl experiment and what is it? (2)

A

Caesium Chloride solution forms a density gradient when centrifuged, the Cs ions will form an increasing concentration going down the tube and the DNA molecules will float to the density zones that match their own density. This allows separating similar molecules with small differences in density.

Since proteins are heavier they sediment in the heaviest, bottom-most layer while RNA being lighter is present at the topmost layer with the lowest density

46
Q

The greater catalytic prowess of polymerase III is largely due to its …

A

processivity; no time is lost in repeatedly stepping on and off the template.

47
Q

Nucleotides can only be added to the new strand at the____ and thus DNA is

A

3’-OH end and thus, Dna is synthesized in a 5’-3’ direction

48
Q

What end of the daughter cell is growing?

A

3’ end of daughter strand is growing

49
Q

What provides energy for the formation of new phosphodiester bond

A

hydrolysis of pyrophosphate provides energy for the formation of new phosphodiester bond

During strand elongation, two phosphates are cleaved from the incoming nucleotide triphosphate and the resulting nucleotide monophosphate is added to the DNA strand. The subsequent hydrolysis of pyrophosphate to yield two ions of orthophosphate by pyrophosphatase drives the polymerization forward.

50
Q

DNA polymerases (6)

synthesize daughter in+can only add+reads in+cannot synthesize+is an enzyme that +promote the formation of

A
  • synthesizes daughter in 5’-3’
  • can only add bases on the 3’ OH end
  • reads template in a 3’-5’ end
  • cannot synthesize from scratch, it required a RNA primer that has a 3-OH group end untouched for DNA polymerases to bind and replicate.
  • Is an enzyme so it has an active site. Active site can catalyze 4 different reactions with the different DNTP. It binds to whatever the complimentary base is as that is the most energetically favoured.
  • promote the formation of phosphodiester linkages between incomming deoxyribonucleotide triphosphates (dNTPs) and an existing DNA strand
51
Q

Helicase

A

Unwinds the Double Helix by breaking hydrogen bonds btwn the 2 parental strands

52
Q

Primase (3)

What it does+ what it provides+ what it is

A
  • Synthesizes RNA primers (10 nucleotides) complementary to the template strand for DNA polymerase
  • provides the hydroxyl group
  • An RNA polymerase
53
Q

DNA polymerase 3

A

synthesizes DNA by adding nucleotides to the new DNA strand

54
Q

DNA polymerase 1

A

removes RNA primer and fills in the gap between with DNA

55
Q

Ligase+ present in… (2)

A

Joins the ends of DNA segments that DNA pol 1 synthesized after removing the RNA primer by forming phosphodiester bonds.

present in leading strand and in lagging strand.

56
Q

How are the two strands different in DNA Replication?

A

one strand is synthesized continuously (leading)

one strand is synthesized discontinuously (lagging) in small DNA fragments: Okazaki fragments

57
Q

Why is there a lagging strand?

A

Because DNA polymerase can only synthesize in 5’-3’ direction

58
Q

DNA replication steps (7)

A
  1. DNA helices unwinds double helix
  2. RNA primate lay down RNA primer
  3. Topoisomerase prevents dna twisting ahead of replication fork during unwinding
  4. dna poly 3 extends rna primer
  5. dna helicase continue to unwind resulting in leading and lagging strands synthesis by dna pol 3
  6. dna pol 1 removes rna primer of ozakaki fragments and fills in gaps with dNTPs
  7. DNA ligase seals nicks by reforming the phophodiester bond
59
Q

supercoiling

A

The axis of the double helix
can itself be twisted into a superhelix

60
Q

Histones constitute —— of a eukaryotic chromosome. The entire complex of a cell’s DNA and associated protein is called —-

A
  • half the mass
  • chromatin
61
Q

Histones (2)

TYPE+ PROPERTY+what they are

A
  • Five major histones are present in chromatin: four histones, called H2A, H2B, H3, and H4, associate with one another; the other histone is called H1.
  • Histones have strikingly basic properties
    because a quarter of the residues in each histone are either arginine or lysine.
  • Small + charged proteins
62
Q

nucleosomes

A
  • a section of DNA that is wrapped around a core of proteins
  • containing 200 bp of DNA and two copies each of H2A, H2B, H3, and H4, called the histone octamer
63
Q

Linker DNA

A

double-stranded DNA (38-53 base pairs long) in between two nucleosome cores that, in association with histone H1, holds the cores together.

64
Q

Prokaryote (2)

Lack+ DNA

A
  • An organism lacking a nucleus and membrane-bound compartments
  • Single circular double stranded DNA molecule
65
Q

Eukaryote (2)

Lacks+DNA

A
  • An organism whose cells contain a nucleus and other membrane-bound compartments
  • DNA is in multiple linear molecules (chromosomes)
66
Q

To help fit bacterial chromosomes inside a cell ….

A

double helix twists on itself, aka supercoiling

67
Q

Humans have —- different chromosomes

A

23 (2 of each- one maternal, one paternal)

68
Q

Chromosomes are —– that compact DNA in eukaryotes. Talk about how its compacted into chromosomes:

A
  • DNA-protein complexes
  • Nucleosomes are arranged in 30nm fibres, folding of the fibres into loops further compacts the DNA and afterwards more compaction generates the chromosome
69
Q

DNA polymerase structure and shape selectivity

A
  • Induced fit—the change in the structure of the enzyme when it binds the correct nucleotide.
  • DNA polymerases close down around the incoming nucleoside triphosphate (dNTP). The binding of a dNTP into the active site of a DNA polymerase triggers a conformational change: the finger domain rotates to form a tight pocket into which only a properly shaped base pair will readily fit. Such a conformational change is possible only when the dNTP corresponds to the Watson–Crick partner of the template base.
70
Q

DNA pol 1 and pol 3 both have —- which is ——. This is because —–

A
  • 3’-5’ exonuclease activity
  • the ability to remove mismatched nucleotides from the 3’ end of DNA by hydrolysis
  • Mismatched strand caused pausing by the polymerase during which mismatched strand is likely to flop about because of the weaker hydrogen bonding and to find itself in the exonuclease active site, where the trespassing nucleotide is removed
71
Q

DNA polymerase thumb, finger, palm

A

The finger and thumb domains wrap around DNA—much as your own fingers wrap around a baseball bat—and hold it across the enzyme’s active site, which is located in the palm domain.

72
Q

The RNA primer is removed by a —-
exonuclease

A

5’-3’

73
Q

. At a replication fork, both strands are synthesized in the —– direction

A
  • 5’-3’
74
Q

The trombone model

A

The replication of the leading and lagging strands is coordinated by the looping out of the lagging strand to form a structure that acts somewhat as a trombone slide does, growing as the replication fork moves forward. When the polymerase on the lagging strand reaches a region that has been replicated, the sliding clamp is released and a new loop is formed.

75
Q

DNA Synthesis initiation Is More Complex in Eukaryotes Than in Bacteria (2):

A
  1. Proteins, called licensing factors because they permit (license) the formation of the DNA synthesis initiation complex, bind to the origin of replication. These proteins ensure that each replicon is replicated only once in each round of DNA synthesis. After the licensing factors have
    established the initiation complex, these factors are subsequently destroyed.
  2. An initiator polymerase called polymerase begins replication. This enzyme includes a primase subunit, used to synthesize the RNA primer, as well as an active DNA polymerase. After this polymerase has added a stretch of about 20 deoxynucleotides to the primer, it is replaced by DNA polymerase a more processive enzyme and the principal replicative polymerase in eukaryotes.
76
Q
A

Bcem 341 (32 decks)

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