Apricio - Lecture 1-5

Question 1

Arthur Kornberg

Answer 1

1. isolated a protein enzyme; DNA polymerase I (Pol I)
2. dNTP are required precursors (building blocks) of DNA
3. template DNA required for DNA synthesis
4. DNA Pol I and dNTP and template required
5. He showed that newly synthesized DNA molecules had a similar base composition = using templates with different base composition

Question 2

Three models for the replication of DNA

Answer 2

1. distributive
2. semiconservative
3. conservative

Question 3

Kornberg; mechanism by which daughter DNA molecules were assembled

Answer 3

Kornberg’s experiment established that strand of DNA served as a template for DNA synthesis; but not yet revealed the mechanism how daughter DNA molecules were assembled

Question 4

DNA Polymerase I

Answer 4

Enzymatic Synthesis of DNA

1. DNA polymerase removes pyrophosphate
2. pyrophosphatase breaks the bond between phosphate groups

Question 5

Requirements for DNA synthesis

Answer 5

1. single stranded DNA template
2. annealed (complementary) primer with a 3’-OH
3. dNTPs
4. DNA Polymerase

Question 6

Mechanisms of DNA synthesis

Answer 6

1. Correct base pairing allows the reaction to take place (by properly positioning reactive groups in the polymerase active site)
2. Hydrolysis of pyrophospate drives the reaction
   * 3’OH on primer attacks (catalysis) the link between phosphate alpha and beta
   * Structure dictates function

Question 7

DNA Polymerase resembles a RIGHT HAND

Answer 7

1. only close on correct base-pair (two purine is too big, two pyrimidine is too small)
2. catalyze the reaction when correct base-pair (closely adjust)
3. structure allows enzyme to repeat the catalytic cycle (hold on to the DNA)
4. Metal ions used to shield negative charges of dNTPs
5. Metal ions used to activate the reactive 3’ OH

Question 8

DNA Polymerase = processive enzymes

other enzymes work distributed

Answer 8

PROCESSIVITY: ability of DNA polymerases to caarry out continuous DNA synthesis without dissociating from the template
== greatly increases the rate of DNA synthesis
* DNA polymerase binding is slow
* DNA synthesis is fast

Question 9

DNA Polymerase & proofreading exonuclease (“delete” key)

Answer 9

DNA Polymerases contain proofreading exonulease

1. incorrect nucleotide polymerased into DNA (incorrect structure) will inhibit further synthesis - very rare (1/100000)
2. misshapen DNA moves from (polymerase active site) to (second enzymatic site) on protein that contains “3’ to 5’ exonuclease activity”
3. proofreading exonuclease removes the terminal 3’ base = restore correct geometry of primer-template junction
4. DNA moves back to polymerase site

Question 10

Replication Fork :: Opposite Directions of DNA Synthesis

Answer 10

DNA unwinding creates a fork structure

1. Fork structure moves as more DNA is unwound
2. Both (two single strands) DNA strands act as templates for DNA synthesis
3. DNA strands are antiparallel & DNA synthesis ALWAYS occurs 5’ to 3’ ==> two strands replicated in opposite directions relative to the movement of the replication fork
   * leading strand and overall direction of DNA matches
   * lagging strand polymerase move opposite direction (form Okazaki fragments)

Question 11

Leading and Lagging strand DNA synthesis

Answer 11

• Leading strand: 1. replicated in same direction as the fork 2. one long continuous strand of nascent DNA
• Lagging strand: 1. replicated in opposite direction 2. synthesized discontinuously as Okazaki fragment (100s ~1000s)

Question 12

Primase: primer is synthesized by primase

Answer 12

Primase:

1. RNA polymerase; can begin synthesis without a primer DNA polymerase requires a primer
2. begins DNA replication by synthesizing 5-10 base RNA molecule complementary to the template
3. DNA polymerase use 3’OH of RNA primer to continue DNA synthesis

Question 13

dNTPs

Answer 13

deoxyribonucleocide triphosphates

doxyriboNocleocide TriPhosphates

Question 14

template primer junction

Answer 14

template DNA and annealed primer

Question 15

gap in DNA strand

Answer 15

missing a base

Question 16

nick in DNA strand

Answer 16

break of phosphate bond between bases

Question 17

For Konberg to verify his findings, he was lucky to have DNA that are not pure DNA single strands

Answer 17

DNA he threw in had “nicks” in it

Question 18

E. coli has 5 polymerases (10 in human)

Answer 18

Its chromosone is duplicated in 30 minutes

Question 19

DNA synthesis direction and DNA reading direction

Answer 19

DNA synthesis: 5’ to 3’

DNA reading: 3’ to 5’ (direction of polymerase movement)

Question 20

Removal of RNA primer

Answer 20

1. RNAseH: (endonuclease) remove RNA backbone; recognizes DNA-RNA hybrids
2. 5’ exonuclease: removes last ribonucleotide of RNA primer
3. DNA polymerase: extend 3’OH (at new primer-template junction)
4. DNA ligase: joins the DNA using 5’PO_4 and 3’OH at the nick

Question 21

DNA ligase

Answer 21

1. works similar to topoisomerase 1

2. use ATP since no E is stored in DNA

Question 22

5’ exonuclease is not same as proofreading exonuclease

Answer 22

1. 5’ exonuclease removes 5’ end

2. proofreading exonuclease (or 3’ to 5’ exonuclease) removes 3’ end

Question 23

DNA Helicase

Answer 23

1. multi subunit protein - hexamers form donut shape; encircle DNA(single strand); keep bound (processive) and move in one direction (polarity; 5’ to 3’)
2. unwind DNA for replication (and/or transcription) - use energy of ATP (a lot; 1 ATP per 2 bases)

Question 24

DNA unwinding causes DNA supercoiling

Answer 24

1. helicase–> positive supercoil ahead, negative supercoil behind
2. Toposiomerase type 1 and 2 remove positive supercoils; type 2 is necessary

Question 25

Single-Stranded Binding Protein (SSB)

Answer 25

1. keep DNA unwound for DNA synthesis; required for lagging (slow) strand; to prevent intramolecular base-pairing);
2. binding is sequence independent;
3. binding happens immediately
4. cooperative manner (binding of one protein facilitates the binding of the next)
5. SSB binding prevents intramolecular base-pairing
6. SSB bidning keeps the DNA in extended conformation with bases exposed to facilitate synthesis

Question 26

Different polymerases removing RNA primer

Answer 26

Pol III: cannot act at a nick; dissociates so that Pol I can come in
Pol I: perform either 1. strand displacement 2. Nick translation
* Nick in the DNA remains

Question 27

Strand displacement

Answer 27

performed by Pol I

“remove 5’ nicked end intact” as Pol I simultaneously synthesizes DNA on exposed template

Question 28

Nick translation

Answer 28

involves “degradation of nicked strand from its 5’ end - 5’ to 3’ exonuclease” with simultaneous synthesis of DNA on exposed template
* Eukaryotic polymerases do not carry out nick translation

Question 29

Processing lagging strand in Eukaryotes (dealing with RNA primers in Eukaryotes)

Answer 29

Pol-delta displaces RNA primer & some DNA
===> produce flap ==> flap cleaved by Fen1 (flap endonuclease 1)
*Fen1 only works on DNA; so some of DNA will also be cut off
** Nick in DNA is still not removed

Question 30

Prokaryotes: Multiple DNA Polymerases function in DNA replication

Answer 30

Pol III: carries out most synthesis(leading & lagging)

Pol I : completes lagging strand synthesis

Question 31

Eukaryotes: Multiple DNA Polymerases function in DNA replication

Answer 31

1. Pol alpha: takes over from primase; has limited processivity; less accurate than Pol delta, Pol epsilon
2. Pol delta & Pol epsilon: replicate most of the DNA; has sliding clamp
   a. Pol delta : lagging strand
   b. Pol epsilon: leading strand

Question 32

sliding clamps

Answer 32

1. increase processivity of DNA polymerases
   a. beta: prokaryotic clamp; 2 subunit
   b. PCNA: eukaryotic clamp; 3 subunit
2. opened up by “Clamp-loader” protein complex between subunits; requires ATP binding and hydrolysis (coupled to the last step; no energy wastes) to load the clamp
3. DNA polymerase released from sliding clamp in the absence of a primer.. (after completion of an Okazaki fragment)

Question 33

DNA sequencing and Dideoxynucleotides

Answer 33

• Dideoxynucleotide lack 2’OH and 3’OH

- dideoxynucleotide in DNA blocks further polymerization because it lacks 3’OH

Question 34

DNA sequencing by Chain Termination

Answer 34

• use Dideoxynucleotides
  1. (radioactive) primer is annealed to DNA molecule of interest (preceding sequence must be known)
  2. presence of all four dNTPs & DNA polymerase, primer will be extended to end of the template
  3. additional presence of tiny bit of didexoynucleotide, chains will stop growing when dideoxynucleotide is incorporated (rarely happens)
  4. then position of complementary base can be predicted

Question 35

DNA sequencing gel

Answer 35

1. Four different reactions are performed (each with different dideoxynucleotide: ddA, ddC, ddT, ddG)
2. reaction run on a denaturing polyacrylamide gel; allows analysis of the single synthesized strand with single base resolution
3. primer is radioactive; visualized on the gel
4. position of band indicates where each base was incorporated

Question 36

DNA sequencing by High-Throughput DNA Sequencing

Answer 36

• sequencing reactions are carried out in large batches
• sequencing carried out all together in a single reaction with all four normal dNTPs and low concentration of each dideoxy-NTPs (each labeled with different fluorescent-colored dye)
• colored DNA molecules detected as passing the end of column, can read out the sequence

Question 37

Molecular requirements for PCR

Answer 37

PCR: polymerase chain reaction

1. template DNA
2. primer pair complementary to before and after sequences of the target
3. DNA polymerase (thermostable)
4. dNTPs

Question 38

Use of PCR

Answer 38

• amplification of large amounts of specific or random DNA from a small sample
  1. cloning (mole. level)
  2. making probes (for southern/northern blots)
  3. gene mapping
  4. mutagenesis
  5. diagnostics/forensics
• primers become part of the product