Download IV. Enzymology of DNA Replication

DNA Replication
I.
SOURCE OF NUCLEOTIDES
A. Salvage pathway
1. Bases or nucleosides are obtained from degraded nucleic acids and are built up
to NTPs
B. de novo pathway
1. Ribonucleoside monophosphates are made from
phosphoribosylpyrophosphate, amino acids, CO2, and NH3
2. Kinases can add phosphates to form di- and tri-ribonucleotides
3. Ribonucleotides can be reduced by reductases to form deoxynucleotides
II.SEMICONSERVATIVE
REPLICATION
A. Models of DNA replication
1. Conservative replication (click for diagram)
a) This model, which was shown to be incorrect, predicted that after
replication, the parent double-stranded DNA would remain intact while the
daughter double-stranded DNA would be entirely newly synthesized
2. Semiconservative replication (click for diagram)
a) This model predicts that both daughter DNA molecules would contain one
strand each from the parent DNA and one entirely newly synthesized strand
B. Meselson and Stahl's investigation (1958) provided evidence that DNA is replication
semiconservatively
1. Background
a) DNA contains nitrogen in its bases (A, T , C, & G)
b) There are different isotopes of nitrogen, either of which can be incorporated
into DNA
(1) One has an atomic mass of 14 (14N) which we will refer to as light nitrogen and
the other with an atomic mass of 15 (15N) which will be referred to as heavy
nitrogen
c) Bacteria grown in media containing only heavy nitrogen (15N) will have DNA
that weighs more than DNA from bacteria grown in light nitrogen (14N)
(1) The weight of DNA can be determined by isolating it from the bacteria and
then ultracentrifuging it in a CsCl gradient
(2) Heavier molecules will be located more toward the bottom after
ultracentrifugation than the lighter DNA molecules
2. The experiment
a) Bacteria were grown in media with light nitrogen and then switched to media
with heavy nitrogen for two generations
b) At each step, DNA was isolated from some bacteria and density of their DNA
measured by ultracentrifugation
3. Predictions
a) Conservative replication model
(1) This model predicts that after one generation on heavy nitrogen two DNA
bands would be detected with different densities
(a) The parent DNA, which the model predicts stays intact, would contain only
light nitrogen
(b) The daughter DNA, which the model predicts in newly synthesized, would
contain only heavy nitrogen
b) Semiconservative replication model
(1) This model predicts that after one generation on heavy nitrogen only one DNA
band would be detected with a density more than DNA from bacteria grown
only on light nitrogen
(a) Each daughter strand would have one strand from the parent DNA (with
light nitrogen) and a newly synthesize strand (with heavy nitrogen)
(2) This model predicts that after two generations on heavy nitrogen two DNA
bands would be formed
(a) During the second generation on heavy nitrogen each daughter DNA
would donate a light DNA chain and a heavy DNA chain as templates for
newly synthesized heavy DNA
4. Results
a) DNA from bacteria grown on light nitrogen showed only one band that was
located near the top of the tube after ultracentrifugation
b) DNA from bacteria grown for only one generation on heavy nitrogen showed
only one band that was located near the middle of the tube after
ultracentrifugation
c) DNA from bacteria grown for two generations on heavy nitrogen showed two
bands after ultracentrifugation; one located in the middle of the tube, the
other at the bottom
5. Interpretation
a) These results support the semiconservative model of DNA replication
C. Taylor, Woods, and Hughes' experiment with eukaryotes
1. Background
a) There are several isotopes of hydrogen, one of which, 3H, is radioactive
b) 3H containing nucleotides will be incorporated into DNA
c) The radioactive DNA can be detected using photographic film
(1) The process, called autoradiography, turns the film dark where it was exposed
to particles released during radioactive decay
2. Experiment
a) Broad beans were grown in media containing 3H-thymidine for
approximately the time it took the actively dividing root cells to undergo one
generation
b) They were then changed to media with only cold (non-radioactive) nutrients
3. Predictions
a) Conservative model of replication
(1) This model would predict that only one chromatid on each chromosome would
be radioactive (expose the film) while the other chromatids would be cold
b) Semiconservative model of replication
(1) This model would predict that radioactivity would be equally distributed
throughout the whole chromosome
4. Results and interpretation
a) The results were compatible with the semiconservative method of DNA
replication
III.TOPOGRAPHY
OF DNA REPLICATION
A. Unwinding of parental DNA during replication causes stress in the unreplicated
portion of DNA, which if not relieved, could prevent the replication fork from moving
upstream
1. Bacterial chromosomes cannot relieve their stress as it is a covalently closed
circle
2. Eukaryotic chromosomes, though linear, are too large to rotate to relieve stress
B. DNA topoisomerases
1. DNA gyrase, a topoisomerase, uses breaking, twisting, and ligating ability to
remove stress
a) DNA gyrase wraps DNA around it
(1) Similar to the way DNA wraps around histones
b) The gyrase cuts both strands of DNA
(1) Staggered cut resulting in a 4 base overhang
(2) The 5' phosphate are covalently attached to DNA gyrase
c) The gyrase passes DNA through the gap of broken strands and reforms the
phosphodiester backbone
2. Sign inversion
a) DNA always undergoes an even number of superhelical twist introductions
3. Inhibition of DNA gyrase
a) Prevents DNA replication and hence growth
b) Courmermycin, nalidixic acid, oxolinic acid, and novobiocin
(1) Resistant strains do occur whose DNA gyrase does not bind these antibiotics
IV.ENZYMOLOGY
OF DNA REPLICATION
A. DNA polymerases
1. Substrates
a) Needs all four deoxynucleoside 5'-triphosphates
(1) 5' mono- and di- phosphates and 3' mono-, di-, or tri-phosphates do not work
b) Needs a DNA template
c) Needs a nucleotide primer with free 3'-OH group
(1) All nucleic acids are synthesized in the 5' to 3' direction
2. Reaction
a) Poly(deoxynucleotide)n-3'-OH + dNTP  Poly(deoxynucleotide)n+1-3'-OH +
PPi
b) Monophosphates will not work
(1) Formation of the phosphodiester bond is extremely endergonic
c) Energy released from breaking pyrophosphate from NTP provides the
energy for polymerization
(1) Reaction is slightly endergonic
(2) Pyrophosphates must be removed or reaction would go into reverse
(a) Pyrophosphatases break down pyrophosphates in an exergonic reaction
3. Prokaryotic DNA polymerases
a) DNA polymerase I
(1) First isolated by Arthur Kornberg in 1957 and was shown to be used to
connect Okazaki fragments
(2) Has strand displacement, nick translation, and endonuclease activity
(3) 3' to 5' exonuclease activity
(a) Occasionally a nucleotide is added to a 3'-OH that cannot H-bond to base
on other strand
(b) DNA polymerase I remove mis-paired base in a process called proof
reading
(i)
(4) 5' to 3' exonuclease activity
(a) Nucleotides are removed from 5'-P terminus
(b) Functions to remove ribonucleotide primers
(5) Nick translation
(a) 5' to 3' exonuclease activity occuring with 5' to 3' polymerization reactions
(6) Strand displacement
(a) 5' to 3' polymerization reactions without exonuclease activity
(7) Endonuclease activity
(a) Can cleave sugar-phosphate backbone two base pairs following a 5'phophate terminated segment of unpaired bases
b) DNA polymerase II
(1) Used in DNA repair
(2) Has 5' - 3' polymerase activity and 3' - 5' exonuclease activity
c) DNA polymerase III
(1) Major enzyme used in replication
(2) Has 5' - 3' polymerase activity and 3' - 5' exonuclease activity
4. Proofreading
a) If an incorrect base is added to the growing DNA chain DNA polymerases
can back up and remove that base and then continue
b) The removal of the incorrect base is referred to as 3' to 5' exonuclease
activity
(1) Exonuclease refers to degrading DNA from its end and 3' to 5' refers the
direction of removal
5. Eukaryotic DNA polymerases
a) DNA polymerases  and 
(1) Main polymerase for nuclear chromosomal replication
b) DNA polymerase 
(1) Nuclear, but function is unknown, possibly used for repair
c) DNA polymerase 
(1) DNA polymerase for mitochondrial DNA replication
B. DNA ligase
1. Function
a) Forms phosphodiester bonds between two segments of DNA
2. Mechanism
a) Joins 3'-OH to a 5'-monophosphate group
(1) Both bases must be properly H-bonded
(2) Energy for bond formation comes from either ATP or NAD
C. DNA Gyrase
1. Function
a) Unwinds DNA helix into single-stranded DNA so that replication can proceed
2. Mechanism
a) DNA wraps around gyrase (similar as to around histones)
b) DNA gyrase breaks both strands of the hind DNA, passes it in front of the
overlapping strand, and reseals bond
(1) Energy released by breaking the phosphodiester bond is stored and is later
used to reform that bond
D. Primase / RNA polymerase
1. Function
a) DNA polymerase must connect nucleotides to 3'-OH group
(1) Cannot lay down first nucleotide
b) RNA polymerase does not have this requirement
(1) A few ribonucleotides laid down by RNA polymerase can serve as a primer for
DNA polymerase
2. RNA polymerase
a) Primes the leading (continuous strand)
b) Sensitive to rifampicin
3. Primase
a) Primes lagging (discontinuous strand)
b) Resistant to rifampicin
4. Primer
a) 1 to 60 bases
(1) Depending on organism
b) Remains H-bonded to DNA
c) Provides 3'-OH group for DNA polymerase III to add a deoxynucleotide
V.
REPLICATION FORK
A. Origin of replication
1. DNA replication begins at specific regions of DNA referred to as 'Origins of
Replication' or ori sites
a) Prokaryotes contain only one ori site
b) Eukaryotes contain multiple ori sites per chromosome
(1) Multiple ori sites are needed due to the larger size of DNA in eukaryotes and
the slower speed of DNA replication of eukaryotic DNA polymerases
B. Replication forks
1. DNA is replicated bi-directionally from each ori site
2. A replication fork is the area of DNA that is being unwound prior to replication
3. There are two replication forks for every one ori
a) As DNA replication begins continuously on one strand, the first Okazaki
fragment produced becomes the leading strand for the other replication fork
C. Advance of the replication fork and unwinding the helix
1. Addition to nucleotides and unwinding of DNA are two different processes
a) DNA polymerase III cannot unwind DNA
2. Energy is required
a) Must break H-bonds and disrupt hydrophobic interactions
b) DNA polymerase I
(1) Can utilize the energy of hydrolysis of a triphosphate and energy of H-bond
formation to unwind parental molecule
(2) No other polymerase can do this
c) Unwinding is catalyzed by enzymes called helicases
(1) Uses energy from ATP hydrolysis to somehow unwind the parental strand
(2) Requires two ATP per bp broken
3. Single-stranded binding proteins
a) DNA polymerase III is not directly behind the helicase
(1) There is therefore some single-stranded DNA in the leading strand
(2) There is a larger gap of single stranded-DNA on the lagging strand
b) Single-stranded binding proteins coats single-stranded DNA so they cannot
reform hydrogen bonds
(1) These single-stranded binding proteins must be displaced by DNA polymerase
III or another enzyme
VI.CONTINUOUS
REPLICATION
A. DNA helicase
1. Unwinds DNA double helix
a) Separates double-stranded DNA into single-stranded sections
b) Starts at ori site
c) Results in topographical stress
B. Single-stranded DNA binding proteins
1. Keeps complimentary strands of DNA from reannealing
C. DNA topoisomerases (e.g., DNA gyrase)
1. Relieves stress caused by helicases
D. Primase (RNA polymerase)
1. Lays down RNA primer
E. DNA Polymerase III
1. Adds nucleotides to 3’ end of primer
2. Adds nucleotides to 3’ end of growing DNA polymer
F. DNA Ligase
1. Seals the ends of the newly created DNA circle
G. Uracil fragments
1. General
a) About half the fragments are Okazaki fragments and half uracil fragements
2. Occurrence of uracil in DNA
a) DNA polymerase cannot distinguish dTTP from dUTP
(1) Both H-bond properly with adenine
(2) Some U is incorporated into DNA
b) Deamination
(1) Cytosine can be deaminated into uracil
(a) G-C  G-U  A-U
3. Prevention
a) dUTPase
(1) Converts dUTP to dUMP, which cannot be incorporated into DNA
4. Treatment
a) Uracil N-glycosylase
(1) Cleaves glycosidic bond
(a) Leaves sugar and phosphate
b) AP endonuclease
(1) Cleaves sugar phosphate backbone at apurinic sites
(2) Deoxyribose and a few adjacent bases are removed
c) DNA polymerase I
(1) Fills the gap
5. Mutants
a) dut- mutants lack dUTPase activity
(1) Uracil fragments are smaller
(2) More uracil is incorporated
b) ung- mutants lack uracil N-glycosylase activity
(1) Only half the radiolabeled DNA was found in the fragments
(2) Could not remove U, so sugar phosphate backbone was not exposed to alkali
VII.DISCONTINUOUS
REPLICATION
A. Replication fork
1. At the replication fork, one strand is synthesized continuously, the other
discontinuously, because the strands are antiparallel
a) All DNA is synthesized in the 5' to 3' direction
B. Steps
1. DNA is unwound
2. RNA primer is made at fork
3. DNA polymerase adds nucleotides to 3’ end
4. DNA polymerase runs into previous primer
5. Cycle starts over again (1 through 4)
6. DNA polymerase I removes RNA primer and replaces
deoxynucleotides
C. Okazaki fragments
it with
1. Detection using a pulse-labeling experiment
a) Pulse experiment
(1) DNA is replicated in the presence of radioactive thymidine
(a) Newly synthesized radioactive DNA (1000 to 2000 nucleotides) did not
sediment as fast as parental DNA (20,000 to 100,000 nucleotides) when
denatured in alkali
b) Pulse-chase experiment
(1) Radioactivity labeled DNA sedimented as fast as parental DNA after time
because Okazaki fragments were connected
2. Size
a) Eukaryotes
(1) 100 - 200 bases
b) Prokaryotes
(1) 1000 - 2000 bases
3. Connecting Okazaki fragments
a) Okazaki fragments are joined to form a continuous DNA containing no
ribonucleotide
b) DNA polymerase I
(1) Removes the primer ribonucleotides
(2) Replaces them with deoxyribonucleotides
c) DNA ligase
(1) Catalyzes formation of phosphoester bond between nucleotides
d) Events
(1) DNA polymerase III extends growing strand until RNA primer is reached
(a) Has little 5' to 3' exonuclease activity, therefor, cannot go further
(b) It cannot join a 5' triphosphate terminus of primer to 3'-OH group
(c) Cannot carry out strand displacement
(d) DNA polymerase moves away from 3'-OH terminus, leaving a nick
(2) DNA ligase cannot seal the nick
(a) There is a triphosphate on the 5' terminus of the RNA primer
(b) DNA ligase will not catalyze a reaction with a ribo-containing nucleotide
(ribonucleotide)
(3) DNA polymerase I
(a) Performs nick translation
(i) Degrades ribonucleotides and replaces them with
deoxyribonucleotides
(a) Continues past RNA primer to replace some deoxyribonucleotides
(4) DNA ligase seals nick
(a) DNA ligase competes with DNA polymerase I and seals the nick
(5) By this time, the next Okazaki fragment will be reaching the RNA primer of this
just joined segment
e) No 3' to 5' polymerase
(1) It would be simpler, however, evolution has not taken this course
VIII.ROLLING
CIRCLE METHOD OF DNA REPLICATION
A. 
B. The other method of replication of circular DNA molecules found in bacteria is
known as the theta  method
IX.METHYLATION
A. DNA adenine methylase
1. Adds methyl groups to adenine
B. Functions
1. Allows cell to determine how to correct mismatched base pairs
a) If the wrong base is incorporated into DNA, the cell must decide which base
to replace
(1) It needs to replace the one on the daughter strand
b) The cell always replaces the base on the strand that is less methylated
(1) The daughter strand would have had less time to become methylated
2. Protects cell's DNA from digestion with its own restriction enzymes
a) Bacteria possess restriction enzymes to degrade viral DNA
C. Mutants
1. DAM- mutants have higher rates of permanent mutations
X.
EUKARYOTES
A. Rate
1. Eukaryotic DNA polymerases are slower than bacterial DNA polymerases
a) Can replicate about 500 - 5,000 bases per minute
b) Bacterial can replicate about 105 (100,000) per minute
(1) ?Average eukaryote replicates at 2,600 bp / min while average prokaryote
replicates at 50,000 bp / min
2. To make up for the slower replication, eukaryotes have more origins of
replication
a) Mammals have around 12,000 ori sites
3. Number of DNA polymerases
a) There are about 50,000 copies of eukaryotic DNA polymerases per cell to
accommodate all the ori sites
(1) This compares to about 15 copies of DNA polymerase III per prokaryotic cell
4. Genome replication
a) With increased number of ori sites and more DNA polymerases, eukaryotic
cells can replicate their genomes faster than prokaryotes
(1) Takes Escherichia coli about 15 minutes to replicate its DNA and Drosophila
about 3 minutes
B. DNA polymerases
1. DNA polymerase 
a) Polymerizes the discontinuous strand
2. DNA polymerase 
a) Is used in DNA repair
3. DNA polymerase 
a) Polymerizes the continuous strand
4. DNA polymerase 
a) Found in mitochondria and chloroplasts
C. Nucleosomes
1. DNA must be unwrapped from histones to be replicated
2. The dissociated histones reform nucleosomes with DNA of the leading strand
3. DNA from the lagging strand form nucleosomes with newly synthesized
histones
D. Telomeres
1. Problems with linearity
a) Eukaryotes have linear chromosomes, the ends of which are called
telomeres
b) After replication, the newly synthesized lagging strand will contain an RNA
primer at its 5' end
(1) DNA polymerase I cannot replace this with deoxynucleotides as there is no 3'OH group available
2. Solution
a) Eukaryotes have an enzyme called telomerase
(1) This enzyme adds the sequence TTGGGG to the ends of linear DNA that
contains that same sequence
(2) Telomeres end in TTAGGG so the polymerase will add more TTAGGGs to the
end
(a) The actual sequence is unique to the particular species
b) This repeat has the ability to form a hairpin loop
(1) This involves the ability of Gs to form hydrogen bonds with each other
c) The 3'-terminal G of the hair pin loop can provide the hydroxyl group for
DNA polymerase I
(1) It can then remove the RNA primer and replace it with deoxynucleotides
(2) DNA ligase can seal the gap in the sugar-phosphate backbone
d) The hairpin loop with the added TTAGGGs is then cleaved off, resulting in a
linear DNA molecule
3. Mechanism
a) The telomerase is a ribonucleoprotein
(1) An enzyme containing RNA and protein
b) The RNA of the enzyme serves as its own template for the added DNA
sequence TTAGGG
(1) The sequence is CCCUAA
(2) The enzyme can be considered a type of reverse transcriptase
(a) It uses an RNA template to make DNA, which is the reverse direction of
transcription
XI.LABELING
DNA
A. Add flourouracil deoxyribose to media lacking thymidine
1. Flourouracil is a derivative of thymidine in which flourine replaces the methyl
group
2. Use mutant bacterial strain lacking the ability to synthesize thymidine
B. Add radioactive thymine
1. Thymine is converted to thymidine via the salvage pathway
a) Used labeled thymine as it is converted only to thymidine and thymidine is
found only in RNA