Exam 3: DNA Replication Flashcards
Explain Meselson and Stahl’s experiment
-N14 and N15 centrifuged separately. Lighter N14 always at the top of tube and heavier N15 at the bottom of tube.
-N15 E.coli grown in medium containing N14.
-N centrifuged and found to be in the middle of N14 and N15 bands which proved that the DNA had replicated semiconservatively and was half of each.
Prokaryotic DNA replication begins at the
ORI (origin of replication)
Replication fork
Site of replication where helix is unwound
Replication is
bidirectional (it starts in one place and goes in two directions); therefore, there are two replication forks
Replicon
Length of DNA replicated
DNA polymerase 1
Enzyme catalyzes directional DNA synthesis: chain elongation occurs 5’ to 3’
Requires DNA template, a primer, and all four deoxyribonucleoside triphosphates (dNTPs)
Exonuclease Activity
3’-5’
All three possess 3’ to 5’ exonuclease activity; proofread newly synthesized DNA, remove/replace incorrect nucleotide.
Dramatically increases the fidelity of the replicating enzyme.
5’-3’
Only DNA polymerase 1
Excises primers– fills in gaps left behind
Which DNA polymerase have the initiation of chain synthesis?
None
What DNA polymerase has 5’ to 3’ exonuclease activity?
DNA polymerase 1
Seven key issues that must be resolved during DNA replication
- Unwinding of the helix
DNAa initiator proteins
SSB proteins
Helicase
- Reduce increased coiling generated during unwinding
Topoisomerase/DNA Gyrase
- Synthesis of primer for initiation
Primase
- Discontinuous synthesis of second strand
DNA Polymerase 3
- Removal of the RNA primers
DNA Polymerase 1
- Going of gap-filling DNA to adjacent strand
DNA Ligase
- Proofreading
DNA polymerase exonuclease activity of 3ʹ–5ʹ allows for excise of nucleotides
Components needed to synthesis molecules
Template
Substrates= dNTP (deoxynucleoside triphosphate)
Proteins/enzymes to coordinate assembly of substrate
DnaA
-Initiator protein encoded by DnaA gene
-Binds to ORI causing conformation to change
-Causes helix to destabilize and open up
-Exposes ssDNA
DNA Helicase
-Made of DnaB polypeptides
– Subsequently recruits holoenzyme to bind
replication fork and initiate replication
– Helicases require energy supplied by hydrolysis of ATP denatures hydrogen bonds and stabilizes double helix
SSBPs
-Single-stranded binding proteins
-Stabilize the open conformation of helix. Bind specifically to single strands of DNA.
How does DNA replication avoid supercoiling?
DNA gyrase
–Enzyme relieves coiled tension from unwinding of helix (DNA supercoiling) by making single- or double-stranded breaks
–Driven by energy released during ATP hydrolysis
Primase: an RNA Polymerase
-Synthesizes RNA primer
-Provides free 3’-OH required by DNA polymerase 3 for elongation
-DNA polymerase 1 removes primer and replaces it with DNA
Contrast continuous and discontinuous DNA synthesis
-Because the two strands are antiparallel, and DNA is only synthesized 5’ to 3’, synthesis occurs in opposite directions
-Continuous DNA synthesis occurs on the leading strand
-Discontinuous DNA synthesis occurs on the lagging strand
Describe the importance of eukaryotic “licensing” factors
Eukaryotic DNA replication initiates at specific sites known as origins of replication. The problem with multiple origins of replication is that you want to start all of them once and only once, otherwise the molecules will not be accurately replicated.
Licensing factors ensure that DNA replication only occurs once per cell cycle. This restriction is vital to prevent over-replication or re-replication of the genome or replication not occurring during S phase of the cell cycle.
One of the central licensing factors in eukaryotes is the Origin Recognition Complex (ORC). ORC binds to specific DNA sequences at origins of replication and marks them for replication initiation.
Describe the problems associated with the ends of linear chromosomes
Unlike bacterial chromosomes, the chromosomes of eukaryotes are linear, meaning that they have ends. These ends pose a problem for DNA replication. The DNA at the very end of the chromosome cannot be fully copied in each round of replication, resulting in a slow, gradual shortening of the chromosome.
During DNA replication, the lagging strand synthesis is discontinuous, leading to the formation of Okazaki fragments. DNA polymerase normally can replace primers with DNA and connect the fragments, but it needs a 3’ OH to add dNTPS to which isn’t available at the end of the lagging strand. So because there is no place for a primer at the extreme end of the chromosome, the last few nucleotides at the 3’ end of each DNA strand cannot be replicated in each round of replication.
Explain how semiconservative replication works
The DNA double helix is separated into two strands. Each strand is then replicated into a complementary new strand. Each DNA molecule has one original strand and one new strand.
Chain elongation by DNA polymerase 1
Nucleotide added, two terminal phosphates cleaved off, providing newly exposed 3ʹ-OH
What is a nucleoside triphosphate?
A nucleoside triphosphate is a nucleoside containing a nitrogenous base bound to a 5-carbon sugar, with three phosphate groups bound to the sugar. They are the molecular precursors of both DNA and RNA.
Holoenzyme
Active form of DNA Pol III which complexes with several other subunits each with separate functions.
What are telomeres?
Inert chromosomal ends that protect intact eukaryotic chromosomes from improper fusion or degradation. Long stretches of short repeating TTAGGG sequences preserve the integrity/stability of chromosomes.
Telomeres of chromosomes shorten with each cell division.
In most eukaryotic somatic cells, telomerase is not active.
Active lymphocytes, certain stem cells, male germ cells, and malignant cells maintain telomerase activity.
Telomerase activity and telomere length linked to aging, cancer, and other diseases.
