biochem 501 exam 2

why is the structure of DNA important?

1. it provides the energetic basis for the molecule to make an exact copy of itself to pass on to the next generation

2. structure can directly “report” out when it has been damaged or mutated, allowing it to signal “repair me!” to the cell

3. information contained in the polymer can be accessed from the inside (by separating the strands) as well as the outside (through large grooves in its structure)

nucleic acids are _______ built from nucleotides

polymers (building blocks)

nucleotides are composed of a ____, ______, and ________

base, sugar and phosphate

purine

short word, bigger structure

pyrimidine

longer word, smaller structure

_ and _ are key positions of the sugar involved in building nucleic acid polymers

5’ and 3’

its the presence or absence of the ___ that is the defining chemical difference between RNA and DNA, not which base is attached to it

2’ OH

N-glycosidic linkage/glycosidic bond

the connection between the base (N) and the sugar (1’)

nucleoside monophosphates

what we see in final nucleic acid polymers

nucleoside triphosphates

high energy building blocks used by cells to synthesize the nucleic acid polymer

purines

A & G

pyrimidines

C, T, U

nucleotides are joined together by _______ bonds

phosphodiester

phosphodiester bonds form by the ____ nucleophilic attack on activated __

3’ OH, 5’ P

activated means there is a LG attached to the __ phosphate to drive the reaction forward

5’

nucleic acids can also be broken down by cleavage of __________

phosphodiester bonds

adenine forms ___ H bonds with thymine, vs guanine forms ___ H bonds with cytosine

two, three

base stacking interactions (stabilization)

• van der Waals interactions between hydrophobic nucleobase faces (steric effects)

• π-stacking (electronic effects)

RNA vs DNA nucleotide preferance

C3’ endo and C2’ endo

B-form _____ allows proteins access through its wide major groove

dsDNA

because RNA prefers the C3’ endo…..

RNA/DNA hybrids and dsRNA adopt A form structures (more tightly packed together)

C2’ endo with DNA

phosphates are farther apart

only purines can adopt the ___ conformation

syn

Hoogsteen base pairs

• form syn and anti pairs

• allow 3-4 strands to be present in a helix

• found in damaged DNA

W-C-F pairs are

anti-anti

why is the genome the principal storage and not DNA/RNA?

RNA is more susceptible to degradation than DNA. The 2’OH can lead to breakdown of the chain

why is the genome the principal storage and not DNA/RNA? (other)

B-form (compared to A-form for dsRNA) allows easier access information in DNA in the major groove without need to unwind helix; keeps it safe!

What sets the relative stability of the two strands being annealed together versus separated as free strands?

The annealed form is great for long-term storage but also need to separate the strands to be able to make use of base-pairing for replication, transcription, etc

B-form DNA is stable, but the two DNA strands will need to be _____ to make use of Watson-Crick-Franklin base-pairing for copying

separated

Genomic DNA consists of two single strands annealed together

1. DNA is denatured during genome replication

2. DNA melts at the 5’ end of a gene before it is expressed

Which base pairs form a stronger bond? AT or CG

CG (less easy to separate dsDNA)

Melting temperature (breaks H bonds)

temp at which the helix is half ds and half ss (50% denatured)

Wound-up materials have _____________ that will constrain how they can be deformed

topological properties

topology

the number of cross-overs in the material

topology of a DNA fragment is described by the ________

linking number (Lk)

linking number

cannot be changed by deforming the structure

Tw (twist)

the number of times each of the curves rotates around the central axis C of the double helix

Wr (writhe)

the number of times the intact B-form helix twists about itself (like in the bottom picture to the right).

DNA in the favorable B-form configuration has an ________ to it

intrinsic twist

If ΔLk<0 the DNA is underwound and _____-supercoiling will result.

negative

If ΔLk>0, the DNA is overwound and ____-supercoiling will result

positive

If ΔLk=0 the DNA is _____

relaxed

Cells use _______ enzymes to change DNA topology

topoisomerase

general rule

most DNA inside the cell is negatively supercoiled

Topoisomerases are also needed to resolve supercoiling that develops when the genome is locally ______ for DNA replication or transcription

unwound

DNA must be unpackaged or ______________ before a gene can be transcribed

unwound from histones

nucleosome

The DNA + histone nucleoprotein complex

histones

1. Eukaryotic genomic DNA has associated basic proteins called histones.

2. Five types of histones: H1, H2A, H2B, H3, H4.

3. Genomic DNA wraps around the histones

heterochromatin proteins

ex. HP1

• bind across methylated histones to promote chromatin compaction

Euchromatin (accessible)

Transcription of genes can occur here. Marked by histone acetylation

Heterochromatin (inaccessible)

Transcription of genes cannot occur here. Marked by histone methylation and binding of heterochromatin proteins

packing DNA into chromatic makes it ….

inaccessible

base activation of 3’OH allows ___________ on 5’ phosphate

nucleophilic attack

DNA polymerase catalyzing nucleotide addition

(a) Uses deoxyribonucleotide 5'-triphosphates (dNTPs) as substrates.

(b) Adds dNTPs that base pair with nucleotides in the original, “template” DNA strand.

(c) Only adds dNTPs to the 3' end of the last nucleotide in the polynucleotide strand.

(d) Synthesis can only proceed off an already existing short double-stranded fragment

Mg2+ coordinated by DNA polymerase facilitates chemistry:

(a) activates 3'-OH attack on dNTP phosphate: makes dNTP Palpha more electrophilic

(b) stabilizes phosphates’ negative charge

Why does DNA polymerase require a primer?

1. Base-pairing provides only a portion of the energy that stabilizes dsDNA structure

2. Base-stacking interactions provide steric and electronic interactions that re-enforce base-pairing to stabilize the structure.

so…

3. Base-pairing interactions are more specific in the context of an existing double-stranded structure than on their own (takes advantage of energy)

tautomerize

position of the H-bond donor/acceptors are different.

wobbles

They occurs because the DNA double helix is flexible enough to accommodate slightly misshaped pairings

-nontautomeric chemical forms of bases (base with extra H⁺)

-bases bond inappropriately

DNA polymerase has a 3’->5’ proofreading activity to correct mistakes

1. DNA polymerase adds nucleotides via its polymerase domain

2. Removes mismatched nucleotides via its exonuclease domain

exonuclease domain

-Exonuclease removes nucleotide(s) from the end of a DNA strand

-The exonuclease domain of DNA pol removes mismatched nucleotide(s) during DNA synthesis

How is the incorrect base detected and corrected?

• Once the 3’end enters the exonuclease site, the terminal nucleotide is removed.

• The polymerase now has a second chance to try to add the correct nucleotide instead.

• This proofreading and error correction mechanism improves the fidelity of DNA replication by 2-3 orders of magnitude

polymerase error rate

1 in every 10⁴ - 10⁶ additions

DNA Pol III

• The “workhorse” polymerase

1. built for speed, need to copy the whole genome!

2. built for processivity, once it gets going its going to run for as long as it can

DNA Pol I

• A “handy-person polymerase”

1. not built for processivity, Polymerizes small stretches of DNA as part of ‘cleanup’ or ‘repair’ jobs and then hops off the DNA

2. built for accuracy over speed, for small jobs, want to be as high-fidelity as possible

oris

genome replication process starts at origins of replication where DNA is. melted and unwound by helicases to produce “replication bubbles” that give ssDNA templates for synthesis

primase

synthesizes a short complementary RNA sequence to provide a primer that DNA polymerase can initiate DNA synthesis on

-DNA/RNA hybrid = A-form (can be detected and eliminated later)

Because DNA synthesis occurs in the 5’->3’ direction, there is one “_____ strand” (continuous synthesis) and one “_____ strand” (discontinuous synthesis) at each fork

leading, lagging

synthesis of lagging strand

1. Addition of primer by primase

2. Extension by DNA pol III until it encounters the 5’ end of another Okazaki fragment and falls off

3. Primase and Pol III continue to work tirelessly to make Okazaki fragments at the head of the replication fork

4. DNA Pol I recognizes the RNA primers and removes them and replaces them with DNA sequence. However, nicks between fragments remain.

5. Nicked DNA fragments are sealed together by DNA ligase.

Okazaki fragments of lagging strand are joined together by ….

DNA ligase

how does DNA ligase catalyze reaction to join lagging strand fragments?

1. DNA ligase uses an ATP to adenylate itself on a lysine residue (forming Ligase-AMP)

2. AMP is transferred from DNA ligase to the 5’ phosphate of the nicked DNA strand

3. base catalyzes nucleophilic attack by 3’ OH seals the strand

topoisomerases

deal with supercoiling from unwinding

single stranded binding proteins (SSBs)

help keep ssDNA from reannealing

telomerase

uses the RNA fragment to template DNA synthesis at the end of telomeres on the parental strand of DNA so it can extend the lagging strand, preventing overhang and shorter chromosomes

two main paths to mutations in the genome

1. polymerase makes a mistake

2. chemistry takes a toll

mistmatches disrupt ideal _-form structure

B

parental strand is used to fix mismatches

cells specifically mark the parental strands of DNA with methylation marks

for a brief period following replication

the parental strand can be distinguished from the daughter strand on the basis of methylation

mismatch is detected by…

MutL-MutS

mismatch repair pathway

1. DNA is threaded through this complex until it encounters a MutH protein bound to a nearby methylated GATC sequence

2. The MutL-MutS-MutH complex stimulates cleavage of the unmodified daughter strand to create a nick

3. Once the DNA is nicked, a complex of nucleases and helicases degrade the unmethylated DNA strand from the point of nick towards the mismatch

4. The resulting gap is filled in by DNA pol III and the nick is sealed by DNA ligase

other times the methylation occurs at a site not involved in base-pairing and is thus not _____

mutagenic

alkylation adds ___ or ___ to the nucleobase

methyl, ethyl

deamination

replaces a nucleobase amine with a carbonyl group

depurination

is loss of the entire purine base from the nucleotide

UV light can cause….

adjacent, stacked thymines and cytosines to become cross-linked into “pyrimidine dimers”

Pyrimidine dimers produce a bulged, ______ structure that is often ________ and results in replication errors.

distorted DNA, misread by DNA polymerase

alkylation: direct repair by methyltransferase

-The aklyl groups can be transferred from the nucleobase using an alkyltransferase.

-In most cases the alkyl group will become permanently attached to the cysteine of the alkyltransferase.

-This means an entire protein is sacrificed to fix one alkylated base. This is a large price to pay, but keeping the genome safe is important.

deamination: base-excision repair

-Simply convert the offending nucleobases into an abasic site

using a damage-specific glycosylase enzyme

-Abasic site with base-excision repair

any process that generates an abasic site can be used to trigger base excision repair. As long as there’s an _____ for that there’s a path to repair.

enzyme

UV dimers: nucleotide excision repair

-Excinuclease makes two cuts in the strand with the UV-dimers: one on each side of the lesion.

-The resulting fragment is ~13nt long with the UV-dimers in roughly the middle.

-The fragment is then removed, and the gaps

can be filled in and sealed using Pol I and Ligase as we’ve seen many times before to complete the repair

alternate base pairing between ______ bases is possible

deaminated

depurination: base-excision repair

-AP nuclease recognizes abasic sites and nicks the offending backbone at the site of damage

-Once nicked, you can see there is a stretch of dsDNA with a free 3’OH, and single nucleotide of ssDNA. This provides a template for DNA polymerase to extend

-Depending on the circumstances, the polymerase can extend past the abasic site (the “long patch” repair pathway); or the 5’ dRP can be removed by lyase and the single abasic site filled in by polymerase and ligase (short patch)

Homologous recombination

this is a process that can recombine any DNA fragments that have “homology” to one another – this just means the sequences match somewhere, only sites of homology need to match

-used to repair broken chromosomes, help with genetic diversity and engineer cells

Sequence specific recombination

this is a process that recombines DNA sequences only at exactly one specific sequence (for example: do it only at AGGGGAGAGAGAGA)

-used by viruses and parasites, highly controllable

Transposons

these are ‘genetic parasites’, DNA sequences that have inserted themselves in our genomes, freeloading off our dna replication processes to propagate their own existence

-insert into genome, they can change how our genes are regulated, create diversity

homologous recombination repairing double strand breaks

1. A 5’->3’ exonuclease generates free 3’ overhangs (useful for repair because DNA polymerase can extend off of it to copy the matching target)

2. The free 3’ overhang forms base pairing interactions with the target, displacing one DNA strand in a process called “strand invasion”

3. Polymerase extends off of the free 3’ end using the homology target as template

4. eventually the displaced strand can base pair with the other 3’ overhang and also get extended

Could a 3’->5’ exonuclease be used to generate functional overhangs instead?

Nope. It could generate overhangs, but they would have 5’P ends instead of 3’OH ends. Remember: DNA polymerase needs a 3’OH to extend off of!

Holliday junctions

If the DNA nicks are sealed by DNA ligase, then the cell is left with a four-way DNA junction (that is: the 4 strands of DNA from the two chromosomes are entangled together)

resolving Holliday junctions

• You could cut “horizontally” and seal the red and blue strands together.

• You could cut “vertically” and seal the green and yellow strands together.

• result: exchange DNA between mom and dad chromosomes!

The key conceptual difference distinguishing “sequence specific recombination” from “homologous recombination”

the enzymes that perform the recombination can ONLY perform the reaction at ONE HIGHLY SPECIFIC sequence

tyrosine recombinase

1. Tyrosine (Y) –OH attacks phosphate in DNA backbone. This nicks the backbone and forms a phosphodiester bond between tyrosine and DNA

2. The DNA –OH now attacks the phosphodiester bond between tyrosine and DNA, kicking off the tyrosine. This results in the formation of a 4-way Holliday Junction between the DNA fragments

3. To resolve the Holliday junction, the tyrosine –OH in recombinase once again attacks phosphate in DNA backbone

4. And this phosphodiester bond between tyrosine –OH and DNA is once again broken when DNA –OH attacks

5. final product