Section 3 - Nucleic Acids

Question 1

Explain the chemical structure of nucleic acids

Answer 1

• Nucleic acids are polymers of individual nucleotide monomers (sugar + nitrogenous base + phosphate) —> unbranched “polynucleotides”
• nucleotides linked by 3’,5’ phosphodiester bonds
• written in the 5’ to 3’ direction

Question 2

Describe B-DNA

Answer 2

Features of the “B” form: (Fig. 5.7)

• 10 base pairs/turn
• Rise: 0.34nm
• Width: 2nm
• right-handed antiparallel
• the two grooves btwn the sugar-phosphate backbones are not equal in size… bc of arrangement of H bond participants + bases
• major groove & minor groove bc the bonds that connect bases to sugars are not 180deg
• the chemical groups that project from the bases into the grooves are used by proteins to recognize the ds DNA sequence
• no water in interior

Question 3

Describe the important forces that stabilizes the B-DNA structure (the double helix)

Answer 3

1. Base Stacking
   - hydrophobic interactions btwn transient (short-lived) dipoles (LD/dispersion forces) in the rings of stacked bases
   - not sequence-specific (any base can stack on another type)
2. Base pairing
   - H bonds btwn chemical groups on the edges of their rings… maintains size of helix
   - sequence-specific
   - 2 H bonds in AT, 3 in GC
   - aka. Watson-Crick bping

Question 4

Describe the higher order structures formed by RNA, and the important forces that stabilize these structures

Answer 4

• strand can fold back and base pair with itself (intramolecular H bonds) to make stem and loop structure (aka, base modifications) that are antiparallel and right-handed

Question 5

Identify chemical and structural similarities and differences between DNA and
RNA

Answer 5

Similarities: nucleic acids joined in same way; antiparallel H bonding
1. Sugar
DNA: deoxyribose RNA: ribose
- the -OH group at the 2’ position of ribose makes the phosphodiester bond sensitive, and thus RNA is degraded into mononucleotides in basic solution
2. Base
DNA: thymine RNA: uracil
- differ by methyl group
3. Length + Sequence
RNA: sequence-dependent DNA: not (any sequence forms the same essential structure)
RNA is shorter
4. Modification
RNA: many nitrogenous base modifications – can influence the 2dary and tertiary structure (a lot in tRNA)

Question 6

Explain the different levels of structural organization displayed by DNA in
eukaryotic cells.

Answer 6

• must organize its genome in compact form to fit in nucleus, but still must be accessible for transcription —> Chromatin = DNA+ essential packing proteins (histones, etc.)

First level: ~200-bp nucleosomes composed of histone octamers and linker regions

Second level: nucleosomes interact with histone H1 to form 30-nm fibres

Third level: 30-nm fibres form loops attached to a scaffold

DNA—>nucleosome (type of chromatin) —> chromatin fibre (packed nucls) —> chromatin fibre folded into loops —> to become full chromosome

Question 7

Describe the structure of the nucleosome.

Answer 7

beads on a string
Nucleosome core particle (the wrapped DNA + histone octamer core proteins) + linker (joining the core particles).
core histones contains two each of H2A, H2B, H3, and H4.

Question 8

Explain the effect of DNA packaging on gene expression and identify, in general
terms, the key mechanisms cells use to alter DNA packaging to regulate gene
expression.

Answer 8

some sort of reorganization must take place

Question 9

Steps in DNA Replication

Answer 9

1. Initiation
2. Priming
3. DNA synthesis (and proofreading)
4. Ligation

-using template strand = semi-conservative

Question 10

Explain the process of replication initiation,
including the protein and DNA
elements involved.

Answer 10

• starts at replication origins (specialized sequences), which are rich in AT– easy to pull apart by
• initiator proteins recognize sequence and pull apart the two strands
• helicase unwinds helix creating replication bubble, consuming ATP as it goes
• s-s binding protein prevents unwound regions from re-forming by attaching to them

Question 11

Explain the process of DNA priming,
including the
protein and nucleic acid elements involved.

Answer 11

• DNA Polymerase cannot start synthesizing DNA without a primer, bc it needs another strand to attach the nucleotides onto
• enzyme primase synthesizes 10nuc-long RNA primers in the 5’–>3’ on the DNA strands

Question 12

Explain the process of DNA synthesis,
including the
protein and nucleic acid elements involved.

Answer 12

• DNA Polymerase extends primer, synthesizing DNA 5’->3’, extending 3’ end occurring along the two templates at replication forks
• energetically favourable rxns happening – pyrophosphate is byproduct
• helicase travels at the front of replication fork, unwinding… with Polymerase attached behind
• product created from template strand is leading strand – continuous synthesis using one primer
• sliding clamp protein(bagel) keeps Polymerase attached to template strand (trombone model so we can just use same DNA polymerase)
• synthesis on complementary strand makes lagging strand
• synthesis occurs in opposite direction of rep fork direction
• discontinuous – forming Okazaki fragments, requiring more primers

Question 13

Explain the process of DNA proofreading (humans),
including the
protein and nucleic acid elements involved.

Answer 13

• DNA Polymerase (3) has 3’-5’ exonuclease activity in another active site. error occurs–> mononucleotide(base) removed from 3’ end –> synthesis resumes
• catches 99%

Question 14

Explain the process of DNA ligation in bacteria

Answer 14

• joining of Okazaki fragments
• 3’-5’ exonuclease activity degrades RNA primer and replaces it with new DNA (all repair DNA Polymerase 1) —> “nick translation”
• repair polymerase falls off and nicks are sealed by DNA ligase (using NAD+)
• human DNA ligase use ATP

Question 15

Draw a replication fork and a replication bubble, and describe the processes
occurring at each.

Answer 15

note the leading and lagging strands, the 3’ and 5’ ends, the primers, the Okazaki fragments, etc.

Question 16

Describe how common inhibitors of viral DNA replication function.

Answer 16

AZT - Azido thymine

• taken orally
• inhibits reverse transcriptase (the enzyme) of HIV
• converted to triphosphate by our cells
• is incorporated into synthesized DNA strand during reverse transcription, and prevents another nucleotide from being added because it has azide (N3) instead of a hydroxyl group at its 3’ carbon
• therefore preventing viral replication
• doesn’t inhibit our DNA Pol bc we have higher affinity for dTTP than AZT

Acyclovir - inhibits herpes virus DNA Pol
- and causes chain termination if incorporated into growing DNA strand

Question 17

Describe the arrangement of DNA at telomeres, and explain the roles of these
structures.

Answer 17

Telomeres

• found at ends of chromosomes
• DNA: long repetitive ds 6-nucleotide sequences, followed by a 30-200 nuc 3’ overhang that loops back and displaces a section of its strand
• enables the cell to distinguish btwn the end of chromosome and a ds break
• protect coding sequences

Problem: where RNA primer was, now there’s a gap at end of lagging strand

• bacteria resolve this by having circular DNA
• humans use telomerase (a cellular REVERSE TRANSCRIPTASE = copies RNA into DNA) to extend end of lagging strand using its own RNA template, by adding same sort DNA sequence to template (top) strand (Fig 6-22), then DNA Pol synthesizes lagging strand more. still leaving gap there but now the extension accounts for what gets lost
• 2 days after birth, telomeres shorten w/ each cell generation

Question 18

Describe and differentiate between the different kinds of commonly observed
DNA damage

Answer 18

1. Copying Mistakes
   - DNA Pol makes mistake during replication (1 per 10^7 nucleotides), which results in 1+ mismatch base pairs that leads to a permanent mutation in on of the two daughter DNAs (cuz top strand with error will then get the wrong base in that spot, but the bottom one gets the right) (Fig. 6-27)
2. Depurination
   - Acid–> glycosidic bond broken–loss of A or G from phosphate backbone–>abasic site which blocks replication
   - Now translesion DNA Pol comes and synthesizes top and bottom new strands, resulting in one molecule thats unchanged, and the other with the nucleotide deleted, then DNA Pol resumes.
   - occurs spontaneously (-DeltaG)
3. Deamination
   - conversion of amine group to carbonyl (C—>U)
   - Spontaneous (Fig 6-23,25)
   - results in new DNA having a UA pairing instead of CG; the other is unchanged
4. Pyrimidine Dimers
   - UV rad makes the c=c in adjacent thymines makes a ring formation instead with other thymine
   - causes a 4-membered C ring that joins the two adjacent bases together covalently which causes distortion in double helix, and will also cause problems in translation and replication. Need translesion to fix it
5. Base Modification & Strand Breaks
   - ionizing rad (X- & gamma-rays), chemical mutagens; mechanical stress –> ss or ds break in s-p backbone & base changes
   - none of the Pols can synthesize past a break in template —> chromosomal abnormalities or cell death

Question 19

Explain the fundamental mechanisms of the major DNA repair systems

Answer 19

1. Proofreading during DNA Replication
2. Mismatch Repair
   - enzymes detect site of mismatch and then search for the unmethylated strand. This is the newly synthesized strand because it hasn’t had time to be methylated. An endonuclease nicks that backbone and exonuclease removes nucs from the nick to the mismatch site. Then DNA Pol fills the gap
   - limited time + takes a lot of energy
   - Eukaryotic Mechanism of finding site is unknown. But then, same thing: removes it and resynthesizes it with exo
3. Base Excision Repair
   - fixes local issues: ss breaks, modified bases, deaminated bases, basic sites
   - excision of segment of damaged strand by endonuclease. DNA repair Pol resynthesizes using bottom strand as template. Connected by endonuclease and then DNA ligase seals nicks, requiring ATP, which remakes the broken phosphodiester bond btwn the nucs
   - Short-patch BER– synthesis of 1 new nuc. Long-patch BER – synthesizes 2-10
4. Nucleotide Excision Repair
   - fixes (bigger damage) base modifications that distort the helical structure of DNA (e.g. Pyrimidine dimers)
   - entire nucleotides are lost, not just bases
   - endo cuts the damaged backbone strand on either side of damage (2 nicks). Helices unwinds and removes part and new DNA is synthesized using undamaged strand as template. Ligase fixes nicks
   - people with NER defect have xeroderma pigmentosum – UV light (direct sunlight) exposure causes dimer accumulation — increased risk of skin lesions and skin cancer
5. Homologous Recombination
   - fixes ds breaks
   - occurs when a break happens in 1 of the 2 daughter chromosomes but they haven’t been separated yet (shortly after DNA rep, before cell div) so you can use the undamaged helix as template to repair by homo recomb
   - nuclease chews 5’ end of the two broken strands. One of the 3’ ends invades the helix of the undamaged DNA and searches for a solid amount of complementary base pairing. once found, it is elongated using template by a repair DNA pol, then it rejoins its partner.
6. Non-homologous end-joining
   - repairs ds breaks
   - the parts that were broken are cleaned up into nice cuts by a nuclease. Then they’re stitched back up by ligase.
   - nucleotides are lost but accomplishes repair without another DNA

Question 20

Predict which repair mechanism will be used to repair a particular instance of
DNA damage

Answer 20

1. Proofreading – happens all the time
2. Mismatch Repair –for copying mistakes
3. Base Excision Repair – local issues: ss breaks, modified bases, deaminated bases, basic sites
4. Nucleotide Excision Repair – bigger damage: base modifications that distort the helical structure of DNA (e.g. Pyrimidine dimers)
5. Homologous Recombination – ds breaks that happen before cell division
6. Non-homologous End-Joining – ds breaks

Question 21

Predict the consequences of DNA damage that is not repaired before the next
round of replication.
Ex. Suppose the C in the sequence TCGA is deaminated, and the base is only repaired after the DNA undergoes one round of replication. Which sequence would you expect to find after the repair?

Answer 21

Result of unrepaired cytosine deamination:
TTGA
Because Polymerase looks at the U and says it looks like a T, so it’ll put an A there. Fits fine.
But now the cell realizes there’s a mistake there, so it removes U and use other strand as template to synthesize it and put a T in its place because it’ll use the A.

Question 22

Describe the role of translesion DNA polymerases as they relate to DNA damage

Answer 22

Translesion creates higher chance of mutation, but that’s better than leaving cell
Its active site is not as fussy, it’ll put whatever on sometimes and won’t affect it.. but that’s also why we only let it do a couple nucs and then we switch back to Polymerase bc it misincorporates bases a lot (but at least now can resume)