DNA Replication Flashcards
Describe the process of semi-conservative replication
- DNA helicase breaks the hydrogen bonds between the complementary base pairs in replication forks
- DNA binding proteins bind to the strands
- Each exposed polynucleotide strand acts as a template where complementary free activated DNA nucleotides bind to with hydrogen bonds, by the specific base pairing rule (hence why two anti-parallel strands are needed) oh
- Phosphodiester bonds form between the new nucleotides by condensation catalysed by DNA Polymerase to form the replicated DNA strand
- Both DNA molecules contain one original DNA strand and one new identical one
How does ATP release energy?
The energy required to break the bond on the edge is much smaller than the energy given off. Much more energy s gained relative to breaking the other bonds
Replication is an active process
What stops the bases on the exposed polynucleotide strands binding back together?
DNA binding proteins bind to the strand to stop the attraction (caused by the hydrogen bonds) between the two strands
Describe how DNA Polymerase catalyses the formation of phosphodiester bonds
• always catalyses from
5’ -> 3’
• always acts on leading strand
• continuous replication
Why is there no need for an enzyme to catalyse the formation of hydrogen bonds between the bases on the exposed polynucleotide strand and the free DNA nucleotides
Hydrogen bonds are attraction bonds
Describe DNA Ligase
- lagging strand of the replication fork
- 3’ -> 5’
- join adjacent Okazaki fragments
- involved in any process that required sealing of phosphodiester bonds from the DNA backbone
- catalyses the formation of phosphodiester bonds between 3’OH and 5’phosphate on various substances (e.g. DNA nicks, DNA fragments with blunt ends; some DNA/RNA hybrids)
Why are replication forks necessary?
- DNA is unwound at different parts (replication forks) to speed up replication and decrease length of exposure (decreases likelihood of chemical and physical damages which cause mutations)
- helicase enzymes join at the replication forks
Conservative replication
- original DNA molecule remains intact
- separate daughter DNA copy built from new molecules of deoxyribose, phosphate and nitrogenous bases
- one strand contains original material, one completely new DNA strand
Describe the logistics of semi-conservative replication
- needs ATP
* 2 template strands
Describe Meselsohn and Stahl’s experiment
- grew E. coli for many generations in a 15N medium
- bacteria incorporated the 15N into the DNA, making it denser and forming ‘heavy’ DNA
- control culture grown in normal, ‘light’ DNA medium formed normal ‘light’ DNA
- bacteria with ‘heavy’ DNA transferred to a 14N medium and left for periods of time that corresponded to the generation time
- samples of parental, first and second generations taken
- analysed using density gradient centrifugation; suspended in CsCl and centrifuged at a high speed
- bands of DNA visible under UV light
15N
Heavy nitrogen isotope
Meselsohn and Stahl’s results:
Generation 0: 100% 2 heavy
Generation 1: 100% 1 heavy, 1 light
Generation 2: 50% 1 heavy, 1 light and 50% 2 light
Generation 3: 25% 1 heavy, 1 light and 75% 2 light
Explanation of Meselsohn and Stahl’s results:
- at generation 0, the bacteria has only been growing in 15N, so both strands are heavy
- at generation 1, the bacteria has semi-conservatively replicated in 14N, si both molecules form 1 heavy and 1 light strand
- at generation 2, both molecules replicate and form 2 molecules with 1 heavy and 1 light and 2 molecules of both light
- at generation 3, replication continues
NB. There are only ever 2 ‘heavy’ strands
