1. DNA REPLICATION

2. DNA copying Each cell division cell must copy its entire DNA
So each daughter cell gets a complete copy
Rate of synthesis
Bacteria = 1000 bases per second
Mammals = 100 bases per second
Problem - with a single replication origin in DNA
Bacteria genome is 4 x 10E6. Takes 20 minutes to copy.
Human is 3.2 x 10E9. Would take 10,000 times longer.

3. Chromosomes Chromosomes Genes
Strands of DNA that contain all of the genes an organism needs to survive and reproduce
Genes
Segments of DNA that specify how to build a protein
genes may specify more than one protein in eukaryotes
Chromosome maps are used to show the locus (location) of genes on a chromosome
The E. Coli genome includes approximately 4,000 genes

4. Chromosomes Genetic Variation
Phenotypic variation among organisms is due to genotypic variation (differences in the sequence of their DNA bases)
Differences exist between species and within a species
Different genes (genomes)  different proteins (proteomes)
Different versions of the same gene (alleles)
Differences in gene expression (epigenetics)

5. DNA copying solutions Double helix has to be copied
Semi-conservation solution exists
– Each daughter cells gets one of the original copies

6. 06_04_replic.rounds.jpg

7. DNA copying solutions Double helix has to be copied
Semi-conservation solution exists
– Each daughter cells gets one of the original copies
Unwind at one point and use that as the origin of replication

8. 06_05_replic.origin.jpg

9. 06_09_Replic.forks.jpg

10. DNA copying solutions Double helix has to be copied
Semi-conservation solution exists
– Each daughter cells gets one of the original copies
Unwind at one point and use that as the origin of replication
Region is AT rich to allow easy separation
Eukaryotes have multiple replication origins (Humans = 10,000)

11. 06_11_oppositepolarity.jpg

12. Schematic 1

13. Schematic 2
The addition of dNTPs to the growing DNA strand

14. Schematic 3

15. DNA Replication
The DNA is copied by a replication machine that travels along each replication fork.
One of the most important members of this complex is DNA POLYMERASE
It is able to add DNA subunits, making new DNA

16. DNA Polymerase First discovered in 1956 by Kornberg Bacteria E.coli
Bacteria have 3 types
DNA Pol I, II, and III
DNA Pol III involved in replication of DNA
DNA Pol I involved in repair
Humans have 4 types (you need to know, now)
DNA Pol alpha, beta, delta - nuclear DNA
DNA Pol gamma - mitochondrial DNA

17. In prokaryotes, there are three main types of DNA polymerase
Polymerase Polymerization (5’-3’) Exonuclease (3’-5’) Exonuclease (5’-3’) #Copies I
Yes
400 II No ?
III
10-20
•3’ to 5’ exonuclease activity = ability to remove nucleotides from the
3’ end of the chain
•Important proofreading ability
•Without proofreading error rate (mutation rate) is 1 x 10-6
•With proofreading error rate is 1 x 10-9 (1000-fold decrease)
•5’ to 3’ exonuclease activity functions in DNA replication & repair.

18. Eukaryotic enzymes:
Five common DNA polymerases from mammals.
Polymerase  (alpha): nuclear, DNA replication, no proofreading
Polymerase  (beta): nuclear, DNA repair, no proofreading
Polymerase  (gamma): mitochondria, DNA repl., proofreading
Polymerase  (delta): nuclear, DNA replication, proofreading
Polymerase  (epsilon): nuclear, DNA repair (?), proofreading
Different polymerases for the nucleus and mtDNA
Some polymerases proofread; others do not.
Some polymerases used for replication; others for repair.
Polymerases vary by species.

19. DNA Polymerase… ALL DNA Pol’s have 2 properties
Only synthesize DNA in one direction 5’ to 3’
Only add to the end of existing double stranded DNA
Therefore they CANNOT start synthesis of DNA from scratch.
RNA polymerases can, but not DNA polymerases

20. 06_12_asymmetrical.jpg

21. What happens at the other strand?
3’ to 5’ strand replication
Solved by Reiji Okazaki
He saw - one strand made in a continuous manner (leading strand) and the other from short discontinuous pieces (lagging strand)
The discontinuous pieces are known as Okazaki fragments

22. Trombone model

23. DNA ligase seals the gaps between Okazaki fragments with a phosphodiester bond (Fig. 3.7)

24. Okazaki Growth RNA is added by Primase (3 - 10 bases)
DNA is added by DNA Pol alpha
Primase can start RNA synthesis de novo
We have a RNA/DNA joint, so RNA is involved in eukaryote DNA replication

25. Closing the gap between Okazaki fragments
RNA primer removed by RNase H
DNA Pol delta fills the gap
DNA Ligase - the gap closer

26. Stabilization of the replication machine
DNA Pol delta and epsilon are attached to the DNA and held in place by other proteins
Sliding-clamp proteins (proliferating cell nuclear antigen - PCNA), allow the stable binding of DNA Pol and strand synthesis
Clamp-loading proteins (replication factor C - RFC), aid in attaching the sliding-clamp proteins.

27. DNA Pol loading

28. DNA Pol… …more facts Has a proofreading mechanism built in
Checks for base matching
Removes mismatched bases by going backwards
Reason why it is not able to build DNA in the 3’ to 5’ direction. No energy from ATP hydrolysis.
Makes just one error in 10E8 or 10E9 (billion) bases added

29. Other helpers Unwinding of DNA strands by Helicases
Require ATP
Single-stranded DNA-binding proteins
Bind to single stranded DNA to stabilize structure
RPA (replication protein A - in eukaryotes)
Topoisomerase - helps with prevention of DNA strand twisting - ‘swivels’
Two types
Type I - Break one strand only and then rejoin
Type II - Break both strands and then rejoin

30. Protein aids

31. Overview

32. DNA replication in eukaryotes:
Copying each eukaryotic chromosome during the S phase of the cell cycle presents some challenges:
Major checkpoints in the system
Cells must be large enough, and the environment favorable.
Cell will not enter the mitotic phase unless all the DNA has replicated.
Chromosomes also must be attached to the mitotic spindle for mitosis to complete.
Checkpoints in the system include proteins call cyclins and enzymes called cyclin-dependent kinases (Cdks).
Kinases are enzymes that transfer phosphate groups from donor molecules such as ATP to specific substrates by the process of phophorylation. Important for cell signaling and protein regulation.

33. Each eukaryotic chromosome is one linear DNA double helix
Average ~108 base pairs long
With a replication rate of 2 kb/minute, replicating one human chromosome would require ~35 days.
Solution ---> DNA replication initiates at many different sites simultaneously.
Rates are cell specific!
Fig. 3.14

34. Fig. 3.13 - Replication forks visible in Drosophila

35. Replication of circular DNA in
E. coli (3.10):
Two replication forks result in a theta-like () structure.
As strands separate, positive supercoils form elsewhere in the molecule.
Topoisomerases relieve tensions in the supercoils, allowing the DNA to continue to separate.

36. Rolling circle model of DNA replication (3.11):
Common in several bacteriophages including .
Begins with a nick at the origin of replication.
5’ end of the molecule is displaced and acts as primer for DNA synthesis.
Can result in a DNA molecule many multiples of the genome length (and make multiple copies quickly).
During viral assembly the DNA is cut into individual viral chromosomes.

37. What about the ends (or telomeres) of linear chromosomes?
DNA polymerase/ligase cannot fill gap at end of chromosome after RNA primer is removed. If this gap is not filled, chromosomes would become shorter each round of replication!
Solution:
Eukaryotes have tandemly repeated sequences at the ends of their chromosomes.
Telomerase (composed of protein and RNA complementary to the telomere repeat) binds to the terminal telomere repeat and catalyzes the addition of of new repeats.
Compensates by lengthening the chromosome.
Absence or mutation of telomerase activity results in chromosome shortening and limited cell division.