Organization of DNA
• Large DNA molecules must be packaged to fit
inside the cell and still be functional !
• mt DNA and the DNA of most prokaryotes are
closed circular structures. These molecules may
exist as relaxed circles or as supercoiled structures
in which the helix is twisted around itself in three-
• Supercoiling results from strain on the molecule
caused by under- or overwinding the double helix:
• Negatively supercoiled DNA: if the DNA is wound more
loosely than in Watson-Crick DNA. This form is
required for most biologic reactions.
• Positively supercoiled DNA: if the DNA is wound more
tightly than in Watson-Crick DNA.
– can change the amount of supercoiling.
– make transient breaks in DNA strands by alternately breaking
and resealing the sugar-phosphate backbone.
– E.g., in Escherichia coli, DNA gyrase (DNA topoisomerase II)
can introduce negative supercoiling into DNA, whereas DNA
topoisomerase I can relax the supercoils
Nucleosomes and Chromatin
• Nuclear DNA in eukaryotes is found in chromatin
associated with histones and nonhistone proteins.
• The basic packaging unit of chromatin is the nucleosome
• Histones are rich in lys and arg a positive charge.
• Two copies each of histones H2A, H2B, H3, and H4
aggregate to form the histone octamer.
• DNA is wound around the outside of this octamer to
form a nucleosome (a series of nucleosomes is
sometimes called "beads on a string").
Nucleosome and Nucleofilament Structure in Eukaryotic DNA
Organization of human DNA,
illustrating the structure of
Structural organization of eukaryotic DNA.
• Histone H1 is associated with the linker DNA
found between nucleosomes to help package them
into a solenoid-like structure, which is a thick 30-
• Further condensation occurs to eventually form
the chromosome. Each eukaryotic chromosome
contains one linear molecule of DNA.
Electron micrograph of a
showing the residual
scaffold and loops of DNA.
Individual DNA fibers can be
best seen at the edge of the
DNA loops. Bar = 2μ. (From
Laemmli UK  The
structure of histone-depleted
metaphase chromosomes. Cell
Cells in interphase contain two types of chromatin:
• Euchromatin is loosely packaged and
• Heterochromatin is tightly packaged and inactive.
• Euchromatin generally corresponds to looped 30-
• Heterochromatin is more highly condensed.
An Interphase Nucleus
• Gene expression requires that chromatin be opened for
access by transcription complexes (RNA polymerase and
• Chromatin-modifying activities include:
– Histone acetylation
– Histone phosphorylation
• During mitosis, all the DNA is highly condensed to allow
separation of the sister chromatids. This is the only time
in the cell cycle when the chromosome structure is
• Chromosome abnormalities may be assessed on mitotic
chromosomes by karyotype analysis (metaphase
chromosomes) and by banding techniques (metaphase,
prophase or prometaphase), which identify aneuploidy,
translocations, deletions, inversions, and duplications.
1. Cytosine arabinoside (araC) is used as an effective
chemotherapeutic agent for cancer, although resistance to
this drug may eventually develop. In certain cases,
resistance is related to an increase in the enzyme cytidine
deaminase in the tumor cells. This enzyme would
inactivate araC to form
B. cytidylic acid
C. thymidine arabinoside
D. uracil arabinoside
2. A double-stranded RNA genome isolated from a
virus in the stool of a child with gastroenteritis
was found to contain 15% uracil. What is the
percentage of guanine in this genome?
3. Endonuclease activation and chromatin
fragmentation are characteristic features of
eukaryotic cell death by apoptosis. Which of the
following chromosome structures would most
likely be degraded first in an apoptotic cell?
A. Barr body
B. 10nm fiber
C. 30 nm fiber
DNA Replication and Repair
OVERVIEW OF DNA REPLICATION
• Genetic information is transmitted from parent to
progeny by replication of parental DNA
• During DNA replication, the two complementary strands
of parental DNA are pulled apart. Each of these parental
strands is then used as a template for the synthesis of a
new complementary strand (i.e., semiconservative).
• During cell division, each daughter cell receives one of
the two identical DNA molecules.
Replication of Prokaryotic and Eukaryotic
• The bacterial chromosome is a closed, double-
stranded circular DNA molecule having a single
origin of replication.
• Separation of the two parental strands of DNA
creates two replication forks that move away from
each other in opposite directions
• Replication is, thus, a bidirectional process.
• The two replication forks eventually meet,
resulting in the production of two identical
circular molecules of DNA.
• Each eukaryotic chromosome contains one linear
molecule of DNA having multiple origins of
• Bidirectional replication occurs at each origin.
• Completion of the process results in the
production of two identical linear molecules of
• DNA replication occurs in the nucleus during the
S phase of the eukaryotic cell cycle.
• The two identical sister chromatids are separated
COMPARISON OF DNA AND RNA
• The overall process of DNA replication requires
the synthesis of both DNA and RNA.
• These two types of nucleic acids are synthesized
by DNA polymerases and RNA polymerases
3'→5'-Exonuclease activity enables DNA polymerase III to “proofread” the newly synthesized
Polymerases and Nucleases
• Polymerases are enzymes that synthesize nucleic
acids by forming phosphodiester (PDE) bonds.
Nucleases are enzymes that hydrolyze PDE bonds
• Exonucleases remove nucleotides from either the
5' or the3' end of a nucleic acid.
• Endonucleases cut within the nucleic acid and
release nucleic acid fragments.
Table 1-2-1. Comparison of DNA and RNA Polymerases
*Certain DNA and RNA polymerases require RNA templates. These enzymes are
most commonly associated with viruses.
• The newly synthesized strand is made in the 5'3'
• The template strand is scanned in the 3'5' direction.
• The newly synthesized strand is complementary and
antiparallel to the template strand.
• Each new nucleotide is added when the 3' hydroxyl
group of the growing strand reacts with a nucleoside
triphosphate, which is base-paired with the template
strand. Pyrophosphate (PPi, the last two phosphates)
is released during this reaction.
• The substrates for DNA synthesis are the dNTPs, whereas
the substrates for RNA synthesis are the NTPs.
• DNA contains thymine, whereas RNA contains uracil.
• DNA polymerases require a primer, whereas RNA
polymerases do not. That is, DNA polymerases cannot
initiate strand synthesis, whereas RNA polymerases can.
• DNA polymerases can correct mistakes ("proofreading"),
whereas RNA polymerases cannot. DNA polymerases have
3'5' exonuclease activity for proofreading.
Steps of DNA Replication in E. coli
1. The base sequence at the origin of replication is
recognized and bound by the dna A protein. The two
parental strands of DNA are pulled apart to form a
2. Helicase uses energy from ATP to break the hydrogen
bonds holding the base pairs together. This allows the
two parental strands of DNA to begin unwinding and
forms two replication forks.
3. Single-stranded DNA binding protein (SSB) binds to the
single-stranded portion of each DNA strand, preventing
the strands from reassociating and protecting them from
degradation by nucleases.
Elongation of the leading and lagging strands (E. coli)
4. Primase synthesizes a short (about 10
nucleotides) RNA primer in the 5'3' direction,
beginning at the origin on each parental strand.
The parental strand is used as a template for this
process. RNA primers are required because DNA
polymerases are unable to initiate synthesis of
DNA, but can only extend a strand from the 3' end
of a preformed "primer".
5. DNA polymerase III begins synthesizing DNA in the
5'3' direction, beginning at the 3' end of each RNA
primer. The newly synthesized strand is complementary
and antiparallel to the parental strand used as a template.
This strand can be made continuously in one long piece
and is known as the "leading strand".
• The "lagging strand" is synthesized discontinuously as a
series of small fragments (about 1,000 nucleotides long)
known as Okazaki fragments. Each Okazaki fragment is
initiated by the synthesis of an RNA primer by primase,
and then completed by the synthesis of DNA using DNA
polymerase III. Each fragment is made in the 5'3'
• There is a leading and a lagging strand for each of the
two replication forks on the chromosome.
synthesis of DNA.
6. RNA primers are removed by DNA polymerase I.
This enzyme removes the ribonucleotides one at a
time from the 5' end of the primer (5'3'
exonuclease). DNA polymerase I also fills in the
resulting gaps by synthesizing DNA, beginning at
the 3' end of the neighboring Okazaki fragment.
7. Both DNA polymerase I and III have the ability
to "proofread" their work by means of a 3'5'
exonuclease activity. If DNA polymerase makes a
mistake during DNA synthesis, the resulting
unpaired base at the 3' end of the growing strand
is removed before synthesis continues.
Removal of RNA primer and filling of the resulting “gaps” by DNA polymerase-I
8. DNA ligase seals the "nicks" between Okazaki
fragments, converting them to a continuous strand of
• DNA gyrase (DNA topoisomerase II) provides a "swivel"
in front of each replication fork. As helicase unwinds the
DNA at the replication forks, the DNA ahead of it
becomes overwound and positive supercoils form. DNA
gyrase inserts negative supercoils by nicking both strands
of DNA, passing the DNA strands through the nick, and
then resealing both strands again.
• DNA topoisomerase I can relieve supercoiling in DNA
molecules by the transient breaking and resealing of just
one of the strands of DNA.
Replication is completed when the two replication forks
meet each other on the side of the circle opposite the
Formation of a phosphodiester
bond by DNA ligase.
• Quinolones are a family of drugs that block the
action of topoisomerases. Nalidixic acid kills
bacteria by inhibiting DNA gyrase.
• Inhibitors of eukaryotic topoisomerase II
(etoposide, teniposide) are becoming useful as
9. The mechanism of replication in eukaryotes is
believed to be very similar to this. However, the
details have not yet been completely worked out.
Eukaryotic DNA Polymerases
• DNA polymerase δ elongates leading and Okazaki
fragments during replication.
• DNA polymerase α initiates synthesis and has primase
• DNA polymerase γ replicates mitochondrial DNA.
• DNA polymerases β and ε are thought to participate
primarily in DNA repair. DNA polymerase ε may
substitute for DNA polymerase δ in certain cases.
• Telomeres are repetitive sequences at the ends of
linear DNA molecules in eukaryotic
• With each round of replication in most normal
cells, the telomeres are shortened because DNA
polymerase cannot complete synthesis of the 5'
end of each strand.
• This contributes to the aging of cells, because
eventually the telomeres become so short that the
chromosomes cannot function properly and the
• Telomerase is an enzyme in eukaryotes used to maintain
the telomeres. It contains a short RNA template
complementary to the DNA telomere sequence, as well
as telomerase reverse transcriptase activity (hTRT).
Telomerase is thus able to replace telomere sequences
that would otherwise be lost during replication. Normally
telomerase activity is present only in embryonic cells,
germ (reproductive) cells, and stem cells, but not in
• Cancer cells often have relatively high levels of
telomerase, preventing the telomeres from becoming
shortened and contributing to the immortality of
Mechanism of action of telomerase
Telomere Ends are Loops Not “naked” ends
• Telomeres and TRF1 (telomere repeat-binding factor)
protein stabilize ends for two reasons (replication gaps &
• Reverse transcriptase is an RNA-dependent DNA
polymerase that requires an RNA template to
direct the synthesis of new DNA. Retroviruses,
most notably HIV, use this enzyme to replicate
their RNA genomes.
• DNA synthesis by reverse transcriptase in
retroviruses can be inhibited by AZT, ddC, and
Eukaryotic cells also contain reverse transcriptase
• Associated with telomerase (hTRT).
• Encoded by retrotransposons (residual viral
genomes permanently maintained in human
DNA) that play a role in amplifying certain
repetitive sequences in DNA
Bridge to Pharmacology
• Quinolones and DNA Gyrase: Quinolones and
fluoroquinolones inhibit DNA gyrase (prokaryotic
topoisomerase II), preventing DNA replication and
transcription. These drugs, which are most active against
aerobic gram-negative bacteria include:
• Nalidixic acid
• Resistance to the drugs has developed overtime; current
uses include treatment of gonorrhea, and upper and lower
urinary tract infections in both sexes.
Steps and Proteins Involved in DNA Replication
• The structure of DNA can be damaged in a number of
ways through exposure to chemicals or radiation.
• Incorrect bases can also be incorporated during
• Multiple repair systems have evolved, allowing cells to
maintain the sequence stability of their genomes.
• If cells are allowed to replicate their DNA using a
damaged template, there is a high risk of introducing
stable mutations into the new DNA.
• Thus any defect in DNA repair carries an increased risk
• Most DNA repair occurs in the G1 phase of the
eukaryotic cell cycle. Mismatch repair occurs in the G2
phase to correct replication errors.
Repair of Thymine Dimers
• Ultraviolet light induces the formation of dimers
between adjacent thymines in DNA (also
occasionally between other adjacent pyrimidines).
• The formation of thymine dimers interferes with
DNA replication and normal gene expression.
• Thymine dimers are eliminated from DNA by a
nucleotide excision-repair mechanism.
Steps in nucleotide excision repair:
• An excision endonuclease (excinuclease) makes
nicks in the phosphodiester backbone of the
damaged strand on both sides of the thymine
dimer and removes the defective oligonucleotide.
• DNA polymerase fills in the gap by synthesizing
DNA in the 5'3' direction, using the undamaged
strand as a template.
• DNA ligase seals the nick in the repaired strand.
Thymine Dimer Formation and
Nucleotide Excision Repair in Eukaryotes
• People with Xeroderma Pigmentosum (XP) have
defective NER systems
– at least eight different genes for NER
• They can not repair bulky DNA damage (e.g., T-dimers)
• 1000 times more likely to develop skin cancer
• Sun (UV) increases cancer in people with XP (T-dimers)
Global Genome NER Model
Table 1-2-3. DNA Repair
Repair of Deaminated and Missing Bases
• Cytosine can become deaminated spontaneously or by reaction
with nitrous acid to form uracil. This leaves an improper base pair
(G-U), which is eliminated by a base excision repair mechanism.
Failure to repair the improper base pair can convert a normal G-C
pair to an A-T pair.
Steps in base excision repair:
• A uracil glycosylase recognizes and removes the uracil base,
leaving an apyrimidinic (AP) site in the DNA strand.
• An AP endonuclease nicks the backbone of the damaged strand at
the missing base.
• Additional nuclease action removes a few more bases, and the gap
is filled in by DNA polymerase.
• DNA ligase seals the nick in the repaired strand.
Repair of Deaminated and
Base Excision Repair in E. coli
• Removes damaged base (e.g., O6-ethylguanine)
Mismatch Repair Fixes Heteroduplex DNA
• Heteroduplex DNA when two non-pairing bases on
• Involves mut genes in E. coli
• Involves msh genes (mutS homologes) in eukaryotes
• Which strand is correct and which strand has the mutated
• Methylation of A bases provides clues as to which is
strand is correct
Mismatch Repair (MMR)
• Mismatch repair deals with correcting mismatches of the
normal bases. It can enlist the aid of enzymes involved in
both base-excision repair (BER) and nucleotide-excision
repair (NER) as well as using enzymes specialized for this
• Recognition of a mismatch requires several different
proteins including one encoded by MSH2.
• Cutting the mismatch out also requires several proteins,
including one encoded by MLH1.
• Mutations in either of these genes predisposes the person to
an inherited form of colon cancer. So these genes qualify as
tumor suppressor genes.
DNA Is Methylated in E. coli
• Fixes errors during replication!
• “Old” DNA is methylated (CH3) at A in 5’-GATC-3’
• Takes 10-20 min after DNA replication for “new” strand
to be methylated
• Repair systems use “delay” to select old/correct strand
(methylated) from the new/defective strand (not
DNA Mismatch Involves Excision Repair
Repairing Strand Breaks
• Ionizing radiation and certain chemicals and viral
infections can produce both single-strand breaks (SSBs)
and double-strand breaks (DSBs) in the DNA backbone.
• Breaks in a single strand of the DNA molecule are
repaired using the same enzyme systems that are used in
Base-Excision Repair (BER).
Double-Strand Breaks (DSBs)
• There are two mechanisms by which the cell attempts to
repair a complete break in a DNA molecule:
• Direct joining of the broken ends. This requires proteins
that recognize and bind to the exposed ends and bring
them together for ligation. They would prefer to see some
complementary nucleotides but can proceed without
them so this type of joining is also called
Nonhomologous End-Joining (NHEJ).
• Errors in direct joining may be a cause of the various
translocations that are associated with cancers.
• Here the broken ends are repaired using the information
– on the intact sister chromatid (available in G2 after chromosome
duplication), or on the
– homologous chromosome (in G1; that is, before each chromosome
has been duplicated). This requires searching around in the nucleus
for the homolog — a task sufficiently uncertain that G1 cells usually
prefer to mend their DSBs by NHEJ, or on
– the same chromosome if there are duplicate copies of the gene on the
chromosome oriented in opposite directions (head-to-head or back-to-
• Two of the proteins used in homologous recombination are
encoded by the genes BRCA-1 and BRCA-2.
Resolving Holliday Junctions
• 50% chance of resolving as Crossover Recombinant
• 50% chance of resolving as Noncrossover Recombinant
Non-cross over Cross over
Meiotic Recombination in Eukaryotes
- DNA polymerase
Diseases Associated with DNA Repair
• Inherited mutations that result in defective DNA repair
mechanisms are associated with a predisposition to the
development of cancer. Some examples:
• Xeroderma pigmentosum is an autosomal recessive
disorder, characterized by extreme sensitivity to sunlight,
skin freckling and ulcerations, and skin cancer. The most
common deficiency occurs in the excinuclease enzyme.
• Hereditary nonpolyposis colorectal cancer results from a
deficiency in the ability to repair mismatched base pairs
in DNA that are accidentally introduced during
Tumor Suppressor Genes and DNA Repair
• DNA repair may not occur properly when certain tumor
suppressor genes have been inactivated through mutation
• The p53 gene encodes a protein that prevents a cell with
damaged DNA from entering the S phase. Inactivation or
deletion associated with LiFraumeni syndrome and many
• ATM gene encodes a kinase essential for p53 activity.
ATM is inactivated in ataxia telangiectasia, characterized
by hypersensitivity to x-rays and predisposition to
• BRCA-1 (breast, prostate, and ovarian cancer) and
BRCA-2 (breast cancer) required for p53 activity.
1. It is now believed that a substantial proportion of the
single nucleotide substitutions causing human genetic
disease are due to misincorporation of bases during DNA
replication. Which proofreading activity is critical in
determining the accuracy of nuclear DNA replication and
thus the base substitution mutation rate in human
A. 3' to 5' polymerase activity of DNA polymerase δ
B. 3' to 5' exonuclease activity of DNA polymerase γ
C. Primase activity of DNA polymerase α
D. 5' to 3' polymerase activity of DNA polymerase III
E. 3' to 5' exonuclease activity of DNA polymerase δ
2. The proliferation of cytotoxic T-cells is markedly
impaired upon infection with a newly discovered human
immunodeficiency virus, designated HIV-V. The defect
has been traced to the expression of a viral-encoded
enzyme that inactivates a host-cell nuclear protein
required for DNA replication. Which protein is a
potential substrate for the viral enzyme?
A. TATA-box binding protein (TBP)
B. Cap binding protein (CBP)
C. Catabolite activator protein (CAP)
D. Acyl-carrier protein (ACP)
E. Single-strand binding protein (SBP)
3. The deficiency of an excision endonuclease may produce
an exquisite sensitivity to ultraviolet radiation in
Xeroderma pigmentosum. Which of the following
functions would be absent in a patient deficient in this
A. Removal of introns
B. Removal of pyrimidine dimers
C. Protection against DNA viruses
D. Repair of mismatched bases during DNA replication
E. Repair of mismatched bases during transcription
4. The anti-Pseudomonas action of norfloxacin is related to
its ability to inhibit chromosome duplication in rapidly-
dividing cells. Which of the following enzymes
participates in bacterial DNA replication and is directly
inhibited by this antibiotic?
A. DNA polymerase I
B. DNA polymerase II
C. Topoisomerase I
D. Topoisomerase II
E. DNA ligase