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DNA Replication

and Synthesis

Fundamental Genetics Lecture 10

John Donnie A. Ramos, Ph.D. Dept. of Biological Sciences

College of Science University of Santo Tomas

The Flow of Biological Information

DNA

RNA

Protein

Replication

Transcription

Translation

Modes of DNA Replication Semiconservative Replication

Semiconservative Replication in Prokaryotes

Mathew Messelson and Franklin Stahl (1958)

15N – heavy isotope of N (contains 1 more neutron) compared to 14N

15N has high sedimentation rate in cesium chloride compared to 14N

Semiconservative Replication in Prokaryotes

Expected results of the Messelson-Stahl experiment

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Semiconservative Replication in Eukaryotes

J. Herbert Taylor, Philip Woods, and Walter Hughes (1957)

Used root tip cells from Vicia faba (broad bean)

Monitored replication using 3H-Thymidine to label DNA

Used autoradiography to determine the incorporation of 3H-Thymidine

Arrested cells at metaphase using colchicine

Replication of E. coli Plasmid

Shown by John Cairns (1981) using radioisotopes and radiography

Replication starts in a single OriC – origin of replication (245 bp)

Replication is bidirectional

Replication fork – unwound DNA helix

Replicon – replicated DNA

Ter region – region of replication termination

DNA Synthesis in Microorganisms

DNA polymerase I (928 aa) – catalyses the synthesis of DNA in vitro (A. Kornberg, 1957)

Requirements:

Deoxyribonucleoside triphosphates, dNTPs (dATP, dCTP, dGTP, dTTP)

DNA template

Primer

Chain Elongation

5’ to 3’ direction of DNA synthesis (requires 3’ end of the DNA template)

Each step incorporates free 3’ OH group for further elongation

DNA replication using DNA polymerase is of high fidelity (highly accurate)

With exonuclease activity (proofreading ability)

DNA Polymerases

All 3 types requires a primer

Complex proteins (100,000 Da)

Functions of DNA polymerases in vivo

DNA Pol I – proofreading; removes primers and fills gaps

DNA Pol II - mainly involved in DNA repair from external damage

DNA Pol III – main enzyme involved in DNA synthesis

a holoenzyme (>600,000 Da) – forms replisome when attached to a replication fork.

Replication in Prokaryotes

1. Unwinding of DNA helix

2. Initiation of DNA synthesis

3. DNA synthesis proper (elongation)

4. Sealing gaps

5. Proofreading and error correction

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Unwinding of DNA Helix

Takes place in oriC (245 bp) – repeating 9mers and 13mers

Function of helicases (Dna A, B, C) – requires ATP hydrolysis to break hydrogen bonds

Initiated by Dna A – binds to 9mers

Binding of Dna B and Dna C to unwound helix

Single-stranded binding proteins (SSBPs) – prevents reannealing of replication bubble.

DNA gyrase (a DNA topoisomerase) – relaxes the supercoiling of DNA helix

Initiation of DNA Synthesis

Synthesis of RNA primer – 5 to 15 RNA bases complementary to the DNA template

Catalysed by primase (an RNA polymerase)

Pimase does not require free 3’ end to initiate synthesis (not unlike DNA polymerase III)

Function of primase will be continued by DNA polymerase III.

DNA Synthesis (Elongation)

Function of DNA polymerase III

Requires free 3’ end

Direction of elongation: 5’ to 3’

DNA synthesis is continuous in 3’ to 5’ DNA strand (leading strand) and discontinuous in the 5’ to 3’ DNA strand (lagging strand).

Okazaki fragments – short DNA fragments produced in the lagging strand

Concurrent synthesis of leading and lagging strands occur by using DNA pol dimer and by a looping mechanism for the lagging strand

Sealing of Gaps, Proofreading

and Error Correction

DNA polymerase I removes all RNA bases produced by primase (creates gaps in the lagging strand) and replaces it with DNA bases (U to T).

DNA ligase seals the gaps by forming phosphodiester bonds

Exonuclease proofreading (identification of mismatched bases) is a function of both DNA polemerase I and III (both with 3’-5’ exonuclease activity)

subunit of DNA polymerase III is involved in proofreading.

Assures high fidelity of DNA replication

Mutations Affect Replication Replication in Eukaryotes

Presence of multiple replication origin (faster replication, guarantees replication of a big genome) – 25K replicons in mammalian cells

Autonomously replicating sequences (ARSs) – origin of replication in yeasts (11 bp)

Origin site is AT rich region

Helicase unwinds double stranded DNA and removes histone proteins from DNA

Histones reassociates while DNA synthesis occurs.

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Eukaryotic DNA Polymerases

Pol - initiates nuclear DNA synthesis

4 subunits (2 acts primase – produces RNA primers)

Acts on both leading and lagging strands

2 other subunits continue elongation step (DNA synthesis)

Low processivity (short length of synthesized DNA prior to dissociation)

Pol - replaces Pol (called polymerase switching)

High processivity (during elongation)

With 3’-5’ exonuclease activity (proofreading)

Pol - nuclear DNA synthesis

Pol - DNA repair (the only eukaryotic DNA polymerase with single

subunit)

Pol - DNA repair

Pol - mitochondrial DNA synthesis (encoded by nuclear gene)

Eukaryotes has a high copy number of DNA polymerases (ex. Pol may be up to 50K copies)

Eukaryotic DNA Polymerases

Eukaryotic DNA Replication

Telomeres – linear ends of eukaryotic chromosomes

Problem with lagging strand: no 3’ needed by DNA polymerase I (after removal of RNA primers)

Possible result: chromosome with shorter lagging strand every replication step

Telomerase

Enzyme that adds TTGGGG repeats on the telomeres (first identified in Tetrahymena)

Prevents shortening of chromosomes

Forms a “hairpin loop” on chromosome ends using G-G bonds

Creates a free 3’ on lagging strand that can be used by DNA polymerase I to replaced the removed RNA primer

Telomerase is a ribonucleoprotein and contains RNA sequence (5’ AACCCC 3”- serving as template) – reverse transcriptase

Cleavage of loop after DNA synthesis

Homologous

Recombination

Exchange of genetic material

Directed by specific enzymes:

Endonuclease – introduces single strand nicks

Ligase – seals loose ends (nicks)

Rec A protein promotes the exchange of reciprocal single-stranded DNA molecules and it enhances hydrogen bond formation during strand displacement

Gene Conversion

Exchange of genetic information between non-homologous chromosomes (non-reciprocal genetic exchange)

Type of chromosome mutation (recombination)

First identified in Neurospora (by Mary Mitchell)

Can be repaired but forms recombined genetic material

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Short Quiz

Assume that the sequence of bases given below is present on one nucleotide chain of a DNA duplex and that the chain has opened up at a replication fork. Synthesis of an RNA primer occurs on this template starting at the base that is underlined.

3’……..GGC TAC CTG GAT TCA……..5’

a. If the RNA primer consists of 8 nucleotides, give the sequence of the primer (showing the 3’ and 5’ ends)

b. In the intact RNA primer, which nucleotide has a free 3’ OH terminus?

c. Give the sequence of the complementary strand of the given DNA.

d. How many hydrogen bonds are found in the given double helical DNA strand?

e. Give the sequence of the primer that will be synthesized in the complementary strand if the primer synthesis starts in the complementary base of the underlined base above.

Short Quiz

Assume that the sequence of bases given below is present on one nucleotide chain of a DNA duplex and that the chain has opened up at a replication fork. Synthesis of an RNA primer occurs on this template starting at the base that is underlined.

3’……..GGC TAC CTG GAT TCA……..5’

a. If the RNA primer consists of 8 nucleotides, give the sequence of the primer (showing the 3’ and 5’ ends)

Given DNA strand 3’……..GGC TAC CTG GAT TCA……..5’

RNA Primer 5’………..AC CUA AGU……..3’

Answers

b. In the intact RNA primer, which nucleotide has a free 3’ OH terminus?

Given DNA strand 3’……..GGC TAC CTG GAT TCA……..5’

RNA Primer 5’………..AC CUA AGU……..3’

Uracil

Short Quiz

c. Give the sequence of the complementary strand of the given DNA.

Given DNA strand 3’……..GGC TAC CTG GAT TCA……..5’

Complementary Strand 5’……….CCG ATG GAC CTA AGT………3’

Short Quiz

d. How many hydrogen bonds are found in the given double helical DNA strand?

Given DNA strand 3’……..GGC TAC CTG GAT TCA……..5’

Complementary Strand 5’………CCG ATG GAC CTA AGT………3’

Between C and G = 3 hydrogen bonds = 8 x 3 = 24

Between A and T = 2 hydrogen bonds = 7 x 2 = 14

______

Total hydrogen bonds = 38

Short Quiz

e. Give the sequence of the primer that will be synthesized in the complementary strand if the primer synthesis starts in the complementary base of the underlined base above.

Given DNA strand 3’……..GGC TAC CTG GAT TCA……..5’

Complementary Strand 5’………CCG ATG GAC CTA AGT………3’

RNA Primer 3’……..GGC UACCU……5’

or

5’…..UCC AUC GG…..3’