Jan 22, 24, 29

Question 1

What is semiconservative, conservative, and dispersion strands s

Answer 1

1.Semiconservative: each daughter strand remains paired with its complementary parental strand
2. Conservative: after replication, both daughter strands pair up
3. Dispersion: daughter strands will have a mixture of parental and newly-synthesized DNA

Question 2

Evidence for Semiconservative DNA replication: Meselson and Stahl

Answer 2

Track parental and newly-synthesized DNA strands over several generations with nitrogen isotopes
- 14N is the most abundant (“normal”) nitrogen isotopes
- 15N has an additional neuron and is therefore heavier than 14N
-E.coli was grown for several generations in a medium that only contained 15N
-The nucleotides contained 100% 15N in their nitrogenous bases
-DNA was isolated and centrifuged in a CsCl gradient
-DNA settles in a position based on its density

Question 3

Evidence for semiconservative DNA replication: Meselson and Stahl (2) cont

Answer 3

• they transferred the E.coli into a medium that only contained 14N
  -New nucleotides contained 100% 14N in their nitrogenous bases
  -They sampled the DNA after a single round of DNA replication
  -They sampled the DNA after a second round of DNA replication

Question 4

Evidence for Semiconservative DNA replication Meselson and Stahl predictions

Answer 4

Semiconservative : so beginning we have both N15 strands, 15N medium -> one replication, the 2 original strands and N15 strands separate, new nucleotides are in N14, and so we will call these hybrid double helices that contain a mixture of N15 and N14. (We will have one line at the middle therefore here ) Second round -> one molecule will be hybrid, another one over there that is hybrid(on the opposite side) , then 1 molecule and the other 1 molecule of the opposite side that is going to be ONLY N14. (1 N14 line at the top, and that one hybrid will be at the middle like before)

Conservative -> so beginning we are having both N15 strands-> after 1 round of replication we will see that we have conserved our N15(that should form at the bottom of the tube) the other strand should be 100% N14, and so it should form a a band up at the top of the tube. -> second round of replication we conserve our N15DNA, so we should see a band at the very bottom. But then we got 3 molecules that are completely N14(there at the top and should still be there and be thicker compared to the N15 band)

Dispersive model - 1st replication, we should start out with N15, N15(down at the bottom line) ->. One replication will have 1 hybrid strand (in the middle again), so likely we will have 50-50 mix of N15 , N15 and N14, N14 -> 2nd round of replication we are going to get a band this time around the top but still kinda in the middle(bcs we have our New N14 nucleotides and just 25% N15) , but still hybrid,

Question 5

Evidence for Semiconservative DNA replication: Meselson and Stahl results and conclusions

Answer 5

Results : Meselson and Stahl obtained the following result: CHECK THE PIC
DNA from 15N medium = 15N - 15N (heavy) DNA -> DNA after one replication in 14N (15N - 14N hybrid DNA) ->DNA after 2 replications in 14N (14N-14N (light) DNA, 15N- 14N hybrid DNA)
Conclusion : The predicted DNA banding patterns for the three DNA replication models shown in Figure 11.8 were

Question 6

DNA polymerization reaction:

Answer 6

• The enzyme - DNA Polymerase - that
adds nucleotides to a growing chain can
only add them to the new strand at the
3’-OH end
• Therefore, DNA synthesis occurs in a
5’-3’ direction
• Incoming nucleotides base pair with
the complementary base on the
template (original/old) strand
• Hydrolysis of pyrophosphate provides
energy for the formation of the new
phosphodiester bond

Question 7

Properties of DNA polymerases

Answer 7

Synthesizes the new strand only in the
5’-3’ direction (i.e. can only add new bases
to the 3-OH’ end of existing strands)
•Cannot synthesize a new strand de novo
– requires a RNA primer with a 3’-OH for
synthesis
•Single active site that can catalyze four
different reactions (incorporation of dATP,
dCTP, dGTP and dTTP)

Question 8

How is DNA organized in prokaryotes?

Answer 8

prokaryotes?
•Prokaryotes typically have a single, circular DNA molecule found in the cytosol (no
nucleus)

POINT OF CONFUSION!!!!
Prokaryotes have ONE , CIRCULAR, DOUBLE-STRANDED DNA MOLECULE(PLUS PLASMIDS)
SINGLE-STRANDED GENOMES ARE ONLY FOUND IN VIRUSES

Question 9

Origins of Replication (ori)

Answer 9

Origins of Replication (ori)
• In Eukaryotes, DNA replication starts at multiple points along the length of each DNA molecule
• Makes replication more efficient
• Each starting site is referred to as an origin of replication (ori)
• As the 2 parental strands separate, a
• Each replication bubble has 2 replication forks
• The replication forks are the sites of DNA
polymerization
• Each fork “points” in the opposite direction
• As DNA synthesis continues, each replication
bubble gets longer as the replication forks move away from each other
• Eventually, two replication bubbles (moving in
opposite directions) merge with each other

Question 10

Getting the Party Started -
Helicase-
Single stranded binding protein (SSB)
Topoisomerase

Answer 10

Getting the Party Started
Helicase
• This enzyme at each replication fork
• Separates the two “old” template strands
The separated strands would simply re-anneal after helicase moved past
Single-stranded binding protein (SSB)
• These proteins bind the ssDNA and prevents them from re-annealing before replication starts
Topoisomerase
• Release tension ( supercoiling ) caused by
helicase ahead of the replication fork

Question 11

Primase

Answer 11

Primase
• DNA polymerase can’t add nucleotides to a single-stranded template
• It can only add nucleotides to a free 3’-OH of a free 3’-Oh of a double- stranded molecule
• Forms a phosphodiester bond between the 3’-OH of the previous nucleotide and the 5’-PO4 of the next nucleotide
So when DNA replication starts at each replication fork, there is no 3’-OH!
• Primase synthesizes a short at ori
• This enzyme can add RNA nucleotides without an existing 3’-OH to a single-stranded template
• Adds 10-20 RNA nucleotides base paired with the template strand forming the primer
-A DNA/RNA hybrid

Question 12

The leading Strand

Answer 12

• DNA Polymerase III adds dNTPs (deoxynucleotide triphosphates)
• Starts at the 3’-OH at the end of the RNA primer
• “Reads” the template DNA in the 3’ to 5’ direction
• Adds dNTPs to the new strand in the 5’ to 3’ direction

• A DNA sliding clamp stabilizes the DNA! polymerase so it does not fall off the template strand
• Only 1 of the template strands at each replication fork can be read continuously in the 3’ to 5’ direction
• This leading strand elongates continuously (and tends to “lead” the other strand)

Question 13

The lagging strand (2)cont.

Answer 13

• The other template strand is antiparallel (oriented the other way)
• Therefore, the replication fork must
advance before primase and DNA
Polymerase III can start adding
nucleotides
• This lagging strand is replicated
away from the replication fork
(towards ori) and tends to “lag” behind
the leading strand
• The lagging strand elongates DISCONTINUOUSLY
in a series of short segments called OKAZAKI FRAGMENTS

Question 14

When Replication Bubble Collide

Answer 14

• After the replication bubbles merge into each other, each new double-stranded
DNA molecule contains:
• One parental (old/original) template strand
• One daughter (new) strand that is a mix of DNA plus the RNA primers from the start of
each leading strand and throughout each lagging strand
• DNA Polymerase I: removes the RNA nucleotides and replaces them with DNA nucleotides
• The last dNTP added by DNA Polymerase I will be next to the first dNTP added by DNA
Polymerase III
• But is can’t join these adjacent nucleotides together
• DNA Ligase: seals the “nicks” between these two dNTPs by forming a phosphodiester bond between them

Question 15

The replisome:

Answer 15

The Replisome
• The cartoons on the previous slides show leading and lagging strand synthesis as
somewhat separate events for simplicity
• In reality, the enzymes of DNA replication or organized together into a replisome
• Increases the efficiency of replication

Question 16

The end of replication problem

Answer 16

Requirement for RNA primer to initiate all new
DNA synthesis presents a problem in fully
replicating 3’ ends of linear chromosomes
No DNA polymerase can fill in the gaps at the 5’ ends of the daughter strands

-Therefore, there will be additive loss at the chromosome ends for every round of DNAreplication/cell division

Genes at the end of the chromosome can be deleted leading to death of the organism

Question 17

Telomeres: solution to the end replication problem

Answer 17

Telomeres: solution to the end replication problem
-Noncoding DNA at both ends of our
linear chromosomes
-Usually repeats of 5-8 G’s and T’s
(human: 5’-TTAGGG-3’)
Human telomeres are approximately 10000(10k) bp long
~60 bp of the telomere will be worn away
after each DNA replication/cell division

When the telomere region is gone, the CELL STOPS DIVIDING / if it was to continue to decide it will keep losing 60bp pairs every cell division but cause telomere is gone, then we would be cutting into the coding gene and if u start cutting into those, we can get all sorts of disease , at that point it enters a stage of senescence? Where it stops dividing forever

Question 18

Aging, cancer and the telomeres hypothesis

Answer 18

•Telomerase is an enzyme that restores shortened telomeres
•NOT present in most eukaryotic cells
•Telomeres in older individuals are short
•Leading cause of natural death?
•Present in gametes and stem cells
•Gives newborn cells long telomeres
•Companies (TeloYears) offer testing the length of your telomeres and supplements or
lifestyle advice that promote telomere lengths
•Human telomerase (hTERT) mutations used as a biomarker in cancer
•Many cancers acquire mutations that activate telomerase to negate the limitations of
rapid cell division
•Anticancer therapeutics: telomerase inhibitors (Imetelstat) or telomerase vaccine
(GV1001: peptide of hTERT active site)

Question 19

DNA replication of prokaryote chromosomes

Answer 19

Follows the semi-conservative model where “old” parental DNA is the template for “new”
daughter DNA
- The double helix must be unwound and the template strands are separated
- The circular genome of prokaryotes has only one origin of replication
-The replication bubble that forms has two replication forks that synthesize DNA in opposite directions
-Each replication fork has a leading and a lagging strand
-RNA primers are required and DNA can only be added in the 5’ to 3’ direction
-The replication forks will eventually meet at the opposite side of the DNA molecule and ends TERMINUS

Question 20

Replication of BActeriophage

Answer 20

Lytic Cycle
•The viral genome is replicated using host cell macromolecules and machinery
•Progeny virus are released when the host cell undergoes lysis

Lysogenic Cycle
•Viral DNA is inserted into the host chromosome
•Host cell is called a lysogen
•Viral DNA is called prophage
•As the host cell replicates, a population of lysogen is formed
•Eventually, the prophage comes out of the host chromosome and switches to the
lytic cycle.

Question 21

High fidelity of DNA replication

Answer 21

High fidelity of DNA replication
•DNA must be replicated fully and be devoid of errors
•Failure to maintain high fidelity of DNA replication causes defective genomes
possibly resulting in disease (cancer ) and death of the organism
•DNA repair mechanisms mediated by enzyme complexes ensure that replication error rate is 1/1,000,000,000 nucleotides
• There are three main repair mechanisms:
1. Proofreading
2. Mismatch repair
3. Excision repair

Question 22

Proofreading activities of DNA polymerases

Answer 22

DNA Polymerase III synthesizes new strand in a 5’ to 3’ direction
- Optimum conformation of the active site and incoming nucleotide allows catalysis of the correct base pair, but mistakes still happen (1/10,000)
- DNA polymerase III detects mistakes and uses its 3’ to 5’ exonuclease activity to remove the most recently added nucleotides (including the
mismatched one)
-DNA polymerase III replaces correct nucleotide and resumes synthesis of the new DNA strand (5’ to 3’)
- Polymerases with proofreading exonucleases reduce mismatches to 1/10,000,00

Question 23

Genomics

Answer 23

Genomics
The study of whole genomes
• Includes structures, sequences, functions, and evolution
• The field involves:
1. Sequencing - determining the base to base sequence of nucleotides
in a genome
2. Annotation - determining coding/non-coding parts of a genome
3. functional analysis - determining the function of the coding/non-coding
parts of a genome
4. Evolutionary analysis - comparing genomes to determine the relationships
among organisms

Question 24

-____ covers for _____ ____ not corrected by ____________
-Recognition of mismatch damage by DNA binding protein _____ and _____.
-MutH endonuclease _____ ______ strand several _________ away from the _______
- _____ ______ exonuclease excises region of ________ strand surrounding the ________
-DNA ________ ___ fills the ___ and ___—— the mismatch
- The _____ left after the ____ is sealed by ___ ______

Answer 24

-MMR covers for replication errors not corrected by proofreading
-Recognition of mismatch damage by DNA binding protein MutS and MutL.
-MutH endonuclease nicks daughter strand several nucleotides away from the mismatch
- Exo 1 5’ - 3’ exonuclease excises region of daughter strand surrounding the mismatch
-DNA Polymerase III fills the gap and repairs the mismatch
- The nick left after the gap is sealed by DNA ligase

Question 25

-A mechanism to ______ damage made to nitrogenous bases that are damaged by chemical modifications such as _______, ______ or __________ - DNA glycosylases remove the modified base forming an __ _____ ( a spot on the ______-_______ backbone with a base) - __ _________ removes AP site and _________ repairs the damage Nucleotide-excision repair - A mechanism to correct damage made from exposure to __ ________or _______ mutagens -_____enzymes recognize the damage, ______ the _______ segment, and ______ it

Answer 25

- A mechanism to correct damage made to nitrogenous bases that are damaged by chemical modifications such as oxidation, alkylation or deamination - DNA glycosylases remove the modified base forming an Ap site ( a spot on the pentose-phosphate backbone with a base) - AP endonuclease removes AP site and polymerase repairs the damage Nucleotide-excision repair - A mechanism to correct damage made from exposure to UV radiation or chemical mutagens -Repair enzymes recognize the damage, excise the distorted segment, and replace it

Question 27

Polymerase easy way to think about it

Answer 27

What is Pyrophosphate Hydrolysis? 1. DNA polymerase is an enzyme that builds new DNA strands by adding nucleotides. 2. Each nucleotide comes with three phosphate groups (like a battery with three charges). 3. DNA polymerase removes two of these phosphates when attaching the nucleotide to the DNA strand. 4. These two phosphates together are called pyrophosphate (PPi). 5. Another enzyme breaks pyrophosphate apart into two separate phosphate molecules. 6. This process releases energy, which helps DNA polymerase keep building the DNA strand. Why Does This Matter? • The energy release makes DNA copying fast and efficient. • It also makes sure DNA keeps growing in one direction and doesn’t fall apart. Think of it like gluing puzzle pieces together: • DNA polymerase adds a piece (nucleotide). • Breaking pyrophosphate (PPi) is like pressing down the glue to lock it in place. • The energy from breaking PPi helps add the next piece.