Tommy was born small and developed unusual rashes and frequent infections as a child. He was diagnosed with intestinal cancer at age 22 and multiple unrelated tumors appeared over the next 10 years. Genetic testing revealed he had Bloom syndrome, a rare disorder caused by a defect in a DNA helicase enzyme involved in DNA replication. This leads to a high mutation rate and predisposition to many types of cancer. The passage discusses DNA replication and how defects can lead to diseases like Bloom syndrome characterized by genomic instability and cancer predisposition.
2. Tommy was a full-term baby but weighed only
4.5 pounds (2 kg) at birth. At about 9 months of age,
an unusual and persistent rash appeared on his face,
and he frequently caught colds and infections. The
illnesses caused no serious problems; so his parents
were not concerned.
Throughout childhood, Tommy remained small;
by age 18, he was only 4 feet 6 inches (137 cm) in
height. Tommy’s first major health problem arose
shortly after he turned 22—he was diagnosed with
intestinal cancer. The tumor was surgically removed
but additional, unrelated tumors appeared
spontaneously over the next 10 years.
Their appearance startled Tommy’s doctors;
the chance of multiple, independent cancers arising
in the same person is generally remote.
3. The propensity of Tommy’s cells to become
cancerous hinted at a high mutation rate in his genes.
Indeed, when pathologists studied Tommy’s
chromosomes, they observed a wide range of
abnormalities. Tommy had inherited BLOOM
SYNDROME.
Bloom syndrome is a rare autosomal recessive
condition characterized by short stature, a facial
rash induced by sun exposure, a small narrow head,
and a predisposition to cancers of all types.
The disorder is extremely rare; only several
hundred cases have been reported worldwide. Cells
from persons with Bloom syndrome exhibit excessive
mutations in all genes, and numerous gaps and breaks
occur in chromosomes that lead to extensive genetic
exchange in cell division. Rates of DNA synthesis are
retarded.
4. The characteristics of Bloom syndrome suggest
that its underlying cause is a defect in DNA replication.
In 1995, researchers at the New York Blood Center
traced Bloom syndrome to a gene on chromosome 15 that
encodes an enzyme called DNA helicase. A variety of
helicase enzymes are responsible for unwinding double-
stranded DNA during replication and repair.
The cells of a person with Bloom syndrome carry
two mutated copies of the gene and possess little or no
activity for a particular helicase. Normal DNA
replication is disrupted, leading to chromosome breaks
and numerous mutations. The genetic damage resulting
from faulty DNA replication leads to tumors.
It is not yet clear whether the basic defect in
DNA synthesis is associated with replication or DNA
repair or both.
5. To understand Tommy’s case, we need to answer
the following questions:
What models of DNA replication exist among life
forms?
Where is the origin of replication in the DNA
strand?
What is the direction of replication at this site?
How does the chain grow in length?
How does the chain terminate?
What is the enzymology behind DNA replication?
Are there other protein factors that must be
present?
What is the role of DNA replication in the
expression of disease?
6. MODELS OF DNA REPLICATION
These models may differ with respect to the
initiation and progression of replication, but all
produce new DNA molecules by semi-
conservative replication.
7. http://highered.mcgraw-
THETA REPLICATION: E. coli
hill.com/olcweb/cgi/pluginpop.cgi?it=swf::535::535::/sites
/dl/free/0072437316/120073/micro03.swf::Bidirectional%
20Replication%20of%20DNA
11. FUNDAMENTAL RULES OF DNA REPLICATION
1. Replication is semi-conservative
Meselson and Stahl convincingly demonstrated that
each E. coli DNA strand serves as a template for the
synthesis of a new DNA molecule.
http://highered.mcgraw-
hill.com/olc/dl/120076/bio22.swf
12.
13. 2. Replication
begins at an
origin- the
replication fork
http://highered.mcgraw-hill.com/olc/dl/120076/micro04.swf
14.
15.
16.
17. 3. DNA replication is bi-directional,
and proceeds in a 5’-3’ direction
DNA synthesis takes place simultaneously but
in opposite directions on the 2 template strands.
18. 4. DNA Replication is Semi-discontinuous
Replication fork
lagging strand Replication fork
Leading strand
19.
20. The polarity
of DNA
synthesis
creates an
asymmetry
between the
leading
strand and
the lagging
strand at the
replication
fork
21. Requirements of Replication
Although the process of replication includes
many components, they can be combined into
three major groups:
1. a template consisting of single-stranded
DNA,
2. raw materials (substrates) to be assembled
into a new nucleotide strand, and
3. enzymes and other proteins that “read” the
template and assemble the substrates into a
DNA molecule.
22. New DNA is synthesized from deoxyribonucleotide triphosphates
(dNTPs). Since the 5’ end does not get added to and the 3’ end
repeatedly does, the DNA strand is said to grow in a 5’- 3’ manner.
27. DNA helicase
unwinds
the DNA duplex
ahead of DNA
polymerase
creating single
stranded DNA
that can be used
as a template
28. ssDNA binding proteins bind to the sugar phosphate backbone
leaving the bases exposed for DNA polymerase. The binding of
SSB to newly formed ssDNA prevents reassociation of the
single strands and “iron out” the unwound DNA.
29. Since DNA polymerase
requires a template and a
free 3’ OH group to add
nucleotides on to, RNA
primers are required to
initiate DNA
polymerization. Primase,
an enzyme which is part
of a large complex of
proteins called the
primosome, synthesizes
a small stretch of RNA
(the primer) of 3-10
nucleotide in length,
which will act as a
starting site for the DNA
polymerase.
30. DNA polymerase
falls off the DNA
easily. A “sliding
clamp” is required to
keep DNA polymerase
on and allow
duplication of long
stretches of DNA
31. A “clamp
loader:”
complex is
required
to get the
clamp onto
the DNA
32.
33.
34. Ahead of the
replication
fork the DNA
becomes
supercoiled
The supercoiling needs
to be relieved or tension
would build up (like
coiling as spring) and
block fork progression.
35. Supercoiling is relieved by the action of
Topoisomerases.
1. Type I topoisomerases:
Make nicks in one DNA strands
Can relieve supercoiling
2. Type II topoisomersases or DNA gyrase
Make nicks in both DNA strands (double
strand break)
Can relieve supercoiling and untangle
linked DNA helices
Both types of enzyme form covalent
intermediates with the DNA
38. Topoisomerases as drug targets
1. Dividing cells require greater
topoisomerase activity due to increased
DNA synthesis
2. Topoisomerase inhibitors which act by
stablilizing the DNA-topoisomerase
complex are used as chemotherapeutic
agents:
camptothecin -Topo I inhibitor
doxorubicin -- Topo II inhibitor
Some antibiotics are inhibitors of the
bacterial-specific topoisomerase DNA
gyrase: e.g. ciprofloxacin
43. , with less than one error per
billion nucleotides. This accuracy results from the processes
of nucleotide selection, proofreading, and mismatch repair.
44. DNA mismatch repair corrects errors made during DNA
replication.(A) If uncorrected, the mismatch will lead to a
permanent mutation in one of the two DNA molecules produced
by the next round of DNA replication. (B) If the mismatch is
“repaired” using the newly synthesized DNA strand as the
template, both DNA molecules produced by the next round of
DNA replication will contain a mutation. (C) If the mismatch is
corrected using the original template (old) strand as the
template, the possibility of a mutation is eliminated. The scheme
shown in (C) is used by cells to repair mismatches.
45. Chemical modifications of nucleotides, if left unrepaired, produce mutations.
(A) Deamination of cytosine, if uncorrected, results in the substitution of one base for
another when the DNA is replicated. Deamination of cytosine produces uracil. Uracil
differs from cytosine in its base-pairing properties and preferentially base-pairs with
adenine. The DNA replication machinery therefore inserts an adenine when it encounters
a U on the template strand. (B) Depurination, if uncorrected, can lead to the loss of a
nucleotide pair. When the replication machinery encounters a missing purine on the
template strand, it can skip to the next complete nucleotide, thus producing a nucleotide
deletion in the newly synthesized strand. In other cases, the replication machinery places
an incorrect nucleotide across from the missing base, again resulting in a mutation.
46. 1. Helicase enzyme unwinds DNA. This reaction needs ATP. At each replicating
fork, the exposed single-stranded DNA is protected by single-strand binding
proteins (ssb). Primase enzyme binds, preparing to make RNA primers.
2. Primase enzyme makes RNA primer molecules. Each primer hybridizes (base
pairs) with DNA, at the origin of replication. The 3' OH end will attach new deoxy
nucleotides (dNTPs). The primers will each start a leading strand,
3. DNA polymerase III attaches new dNTPs to the 3' OH end of the growing chain of
the leading strand, which elongates toward the replicating fork, 5' to 3'. (For each
origin, there are TWO leading strands) For each NTP, a pyrophosphate (PP) is
released, providing the necessary energy.
4. More primers hybridize to the opposite strand of DNA. Pol III starts elongating 5'
to 3' but it keeps running into the back of an RNA primer. This is the lagging
strand. There are TWO lagging strands.
5. DNA polymerase I (Pol I) starts at “nicks” in the growing strands. It edits the
strand by removing bases ahead of it (5' end), including RNA and mismatched
bases, while elongating the strand "behind" 5' to 3'. It replaces all RNA
nucleotides with dNTPs.
6. Ligase seals the phosphate bonds at all “nicks” in the DNA.
7. Editing endonucleases excise mismatched nucleotides, replacing with the proper
match. How do they know which is old DNA vs. new DNA? The old DNA contains
methyl groups on some of its cytosine bases.
8. Gyrase restores negative superturns in DNA. ATP is needed.
9. Methylases add methyl groups to the new DNA, at the same positions as the
original strands. Now the two daughter helices are indistinguishable from each
other, and from the original helix.
47. THE EUKARYOTIC REPLICON
Time for DNA replication is limited in the S phase of
eukaryotes (6-8 hrs in mammals. Such RFs move only
about 1/10th of the prokaryotic forks, and
chromosomes can be in excess of 108 bp. Completion
of replication at the allotted time requires multiple
RFs called replicons. The Origin Recognition
Complex (ORC) is a complex of 6 ATPases which is
the functional equivalent of DnaA.
50. How does a
linear
chromosome
close
replication
at its two
ends?
51. As DNA synthesis
requires a RNA
primer that will
eventually be
digested away,
standard DNA
replication would
result in linear
chromosomes that
would shrink with
every round of
replication. This is
resolved in bacteria
by the circular
genome which does
not have an end. In
linear chromosomes,
the telomere solves
the DNA end
replication problem.
52. Telomeres have
highly repeated DNA
sequences 5'-
TTAGGG-3'.
Human
chromosomes have
between 100 and
1500 copies of this
sequence.
Telomerase, a
special DNA
polymerase, can add
additional copies of
the 5'-TTAGGG-3' to
the end of a
chromosome.
The telomerase
enzyme is actually a
complex containing
protein and RNA (a
"ribozyme").
53. The RNA portion has a 5'-CCCTAA-3' region that acts
as a template for adding the DNA repeat to the
chromosome ends.
The telomerase enzyme is found mostly in the germ
cells of multicellular organisms.
In somatic cells, the absence of telomerase results in
shorter chromosomal ends with each division and may
be the limiting factor in an organism's life span.
55. Errors of DNA Replication and Disease
Origins or replication are strictly controlled
so that they “fire” only once per cell cycle
Errors lead to over-replication of specific
chromosomal regions = gene amplification
This is commonly seen in cancer cells
and can be an important prognostic
indicator.
It can also contribute to acquired drug
resistance, e.g. Methotrexate induces
amplification of the dihydrofolate
reductase locus.
56. The rate of misincorporation of bases by DNA
polymerase is extremely low, however repeated
sequences can cause problems.
In particular, trinucleotide repeats cause difficulties
which can lead to expansion of these sequences.
Depending where the repeat is located, expansion of
the sequence can have severe effects on the
expression of a gene or the function of a protein.
Looping out of repeats
before replication.
57. Several inherited diseases are associated with
expansion of trinucleotide repeat sequences.
Very different disorders, but they share the
characteristic of becoming more severe in
succeeding generations due to progressive
expansion of the repeats
