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Biochemistry 2 > Lecture 1 > Flashcards

Lecture 1 Flashcards

1
Q

What did Frederick Griffith study in 1928? And what did he do in order to study it?

A

The Streptococcus pneumoniae, a pathogenic bacterium causing pneumonia.

He infected mice with these strains hoping to understand the difference between the strains

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2
Q

How many strains of Streptococcus are there?

A

There are 2 strains:
1. S strain - virulent
2. R strain - nonvirulent

(virulent - extremely severe or harmful in its effects.)

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3
Q

What was the result of Griffith’s experiment?

A
  • Live R strain cells didn’t kill the mice
  • Live S strain cells killed the mice
  • Heat-killed S strain cells did not kill the mice
  • Heat-killed S strain + live R strain killed the mice
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4
Q

What did Griffith conclude from the mice experiment? And what did he call the result?

A

He concluded that information specifying virulence got passed from the dead S strain to the live R strain cells

He called the transfer of this information Transformation

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5
Q

After using purified cell extracts in the grifith experiment what did they find

A

The transforming material is DNA

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6
Q

DNA structure

A
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7
Q

phosphodiester bond

A

Bond between adjacent
nucleotides. Formed between the phosphate group of one nucleotide and the 3’ –OH of the next nucleotide. Chain contains 5’ to 3’ orientation

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8
Q

Who found the amounts of the four bases on DNA and what are those amounts (Thymine, adenine, guanine and cystosine)

A

Erwin Chargaff
A:30.7%
G:30.7%
C:19.7%
T:19.7%

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9
Q

Who identified the 3D structure of DNA, what is it shape and the diameter and length to form a complete turn of the helix?

A

Franklin Rosalind and Maurice Wilkins
Shape is helical
2nm diameter, 3.4nm for complete turn

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10
Q

Whoo built the first DNA model using Franklin’s x-ray’s?

A

Watson and Crick in 1953

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11
Q

The two strands of DNA are ________
and are one orientated ____ the other ______

The wrap around each other in a ______ shape

A

Anti-parallel
orientated 5’ to 3’the other 3’ to 5’
helical shape

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12
Q

How many base pairs in a complete turn,
Name of left twisted Dna.
Where do hotspots occur

A

10 base pairs
Z-DNA or southpaw
When Right sided and Left sided DNA meet (Forms mutations)

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13
Q

4 requirements for DNA to be genetic material?

A
  1. Must carry information
  2. Must replicate
  3. Must allow for exchange of information
  4. Must govern the expression of phenotype
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14
Q

Information stored in the bases of DNA

A

Only accesible when DNA is unwound (Very rarely can find info in double strand DNA)

Proteins read the DNA sequence as DNA helix unwinds. Either bind to DNA sequence or copies it.

Some proteins recognize the base sequence of DNA without unwinding it (Restriction enzyme)

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15
Q

DNA replication

A

Copied before cell divides during the synthesis phase of interphase. New cells need Identical DNA strands. Replication is extremely accurate.

E.coli replicates at 1000 neucleotides/sec
Eukaryotes are slower in replication for higher accuracy

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16
Q

Purpose of DNA replication

A

Cells need to make a copy of DNA before dividing so each daughter cell has a complete copy of genetic information

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17
Q

What are the 3 proposed methods of DNA replication

A

Semiconservative, conservative and dispersive

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18
Q

Meselson-Stahl experiment. What is it and what method was proven?

A

Semi-conservative method

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19
Q

Basic rules of replication

A

Semi-conservative
Starts at the ‘origin’
Synthesis always in the 5-3’ direction
Can be uni- or bidirectional
Semi-discontinuous
RNA primers required

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20
Q

Who deduced the steps of DNA replication and what are those steps

A

Arthur Kornberg

Initiation
* Proteins bind to DNA and open up double helix
* Prepare DNA for complementary base pairing
Elongation
* Proteins connect the correct sequences of nucleotides into a continuous new strand of DNA
Termination
* Proteins release the replication complex

21
Q

Problem of DNA replication

A

DNA is maintained in a compressed supercoiled state but the basis of replication is the formation of strands based on specific bases pairing with their complementary bases,

DNA must be unwound before it can get replicated.

22
Q

Core proteins at replication fork

A
23
Q

Bacterial DNA replication

A

Initiation: 245 bp in the oriC (single origin replicon); an initiation protein

  • Gyrase (Topoisomerase II) removes supercoiling ahead of the replication fork. Single-stranded DNA is prevented from annealing by single-stranded binding proteins.
  • Unwinding of DNA is performed by Helicase.
  • Primers: an existing group of RNA nucleotides with a 3′- OH group to which a new nucleotide can be added; usually
    10 ~ 12 nucleotides long

*Primase: RNA polyamerase

24
Q

Primer for DNA polyamerase

A

Primer:
- short strand complementary to the template
- contains a 3’-OH to begin the first DNA polymerase-catalyzed reaction
- can be made of DNA or RNA (more common)

24
Q

Functions of key proteins in DNA replication

A

DNA helicase breaks hydrogen bonds in strands

Topoisomerase alleviates positive supercoiling

Single-strand binding proteins keeps parental strands apart

DNA polymerase III synthesizes a daughter strand of DNA

DNA polymerase I excises RNA primers and fills in with DNA

DNA ligase covalently links the Okazaki fragments together

25
Q

What direction does the synthesis proceed

A

5’ to 3’
Leading strand made continuosly as replication fork advances
Lagging strand made discontinuously that are later joined together (Okazaki fragments: 1000 to 2,000 neucleotides long but 150 t0 200 in eukaryotes)

26
Q

Why does DNA (and RNA) only “grow” in the 3’ direction?

A

Growing strand requires a free 3’ OH group.

27
Q

DNA polyamerase structure

A
28
Q

Structure on how leading and lagging strand co-ordinate at replication fork

A
29
Q

Requiremnets of DNA replication in complex organism

A
  • Very low error rate:
    One human cell: 6 billion bp of DNA. A copying error rate of 1 error/million nt→
    6000 errors with every cell division
  • Very fast copy rate
    E. coli –1000 nt per minute → 3 days to replicate (real life: 20 minutes per cell
    cycle; 1000 nt per second)
30
Q

Fidelity of DNA replication

A

Proofreading: DNA polymerase I: 3′→5′ exonuclease activity removes the incorrectly paired nucleotide.

Mismatch repair: correcting errors after replication is complete

31
Q

Eukaryiotic DNA replication

A

Typical human chromosome length: 100 million bp
Time to replicate chromosome: Minutes to hours
Origins: Hundreds per chromosome
Replicon: 20,000-300000bp long
Replication per minute: 500-5000bp/minute at each replication fork

32
Q

How are errors during synthesis corrected

A

3’-5’ exonuclease activity(Proofreads for mismatched base pair)

33
Q

Errors in E.coli
How many times does DNA polyamerase insert wrong base

A

Errors in E.coli: 1/10^9 – 1/10^10 bp(1 per 1000-10000 replications)
DNA polyamerase error:1/10^4–1/10^5
times.(Repair mechanisms fix this)

34
Q

Bacterial DNA replication

A

Elongation:Carried out by DNA polymerase iii

Removing RNA primer: DNA polymerase i

DNA ligase: Connecting nicks after RNA primers are removed

Termination: When replication forks meet or by termination protein

Only one origin on their circular chromosome

35
Q

Functions of the 5 DNA polymerase

A

DNA polymerase i: Abundant but not ideal for DNA replication. Rate is slower, has lower processivity. Primary function is clean up. Moves ahead of 3’-5’ exonuclease activity and hydrolyzes things in its path

DNA polymerase iii: Principle replication polymerase

DNA polymerase ii iv and v: DNA repair

36
Q

Structure of DNA polymerase iii

A

Complex structure with 10 subunits
Two core domains of a, e, and 0 subunits
Domains linked by clamp loader complex

Core domains interact with a dimer of beta subunits that increase processivity of complex
-Form a sliding clamp to prevent
dissociation
-processivity of DNA Pol III is >500,000
bp because of the beta clamps

37
Q

Regulation of Replication Initiation via Methylation

A

After replication, oriC is hemimethylated by Dam(DNA Adenine methylase. Methylates N^6 of A in the palindromic GATC sequences)

Hemimethylated oriC sequences interact with the plasma membrane (uses protein
SeqA

After a period, SeqA dissociates, oriC sequences released from membrane

Dam methylase fully methylates DNA and allows new DnaA to bind

38
Q
A
39
Q

Transitioning between Okazaki fragments

A

Core subunits of DNA Pol III dissociate from one beta clamp and bind to a new one

DNA Polyamerase I or RNase H1 removes RNA primer

DNA Polyamerase I fills in the gap

DNA ligase seals the backbone

40
Q

Describe the termination of replication

A

Replication forks meet at region with 20bp sequences Ter (TerA-TerF)

Ter sites found near each other but in opposite directions create a site that replication forks cannot leave

41
Q

Ter is a binding site for what protein?

A

Tus (terminus utilization sequence)
Causes a replication frok to stop

42
Q

Functions of eukaryotic DNA polyamerase a,d and e

A

DNA polyamerase a- acts like Primase to initiate. Has Primase activity in one subunit and polymerization in the other, No 3’-5’ proofreading

DNA polyamerase d- replicates lagging strand - associated with PCNA (proliferating cell nuclear antigen-protein), highly processive, 3’-5’ proofreading

DNA polyamerase e- replicates leading strand. - associated with PCNA (proliferating cell nuclear antigen-protein), highly processive, 3’-5’ proofreading

43
Q

Replication in eukaryotes

A

More complex than bacteria

Yeast have ~400 well-defined origins
-called autonomously replicating sequences (ARS) or replicators
Entire genome replicated 1X/cycle
-regulation due to cyclin proteins and cyclin-dependent kinases (CDKs)
- cyclins ubiquitinated for proteolytic destruction at the end of the M (mitosis) phase

44
Q

Initiation of Replication

A

Requires a pre-replicative complex (pre-RCs)

The origin recognition complex (ORC) loads a helicase onto the DNA
-ORC functions like bacterial DnaA
-Helicase is a hexamer of mini-chromosome maintenance proteins (MCM2-7). – MCM2-7 function like bacterial DnaB helicase

45
Q

Eukaryotic rate of replication

A

Occurs more slowly than E. coli does
Synthesis ~50 nucleotides/sec
-(~1/20th the rate seen in E. coli)
Compensated by origins every 30–300 kb

46
Q

Termination of replication (Eukaryotes)

A

Synthesis of specialized structures known as telomeres are found at the linear ends of the nuclear chromosomes

47
Q

How telomerase finishes replication of linear chromosomes

A

Biochemistry 2 (5 decks)

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