Nucleic Acids, DNA Replication, DNA Repair, Transcription, Translation

Question 1

Distinguish purines and pyrimidine bases, ribose and deoxyribose, ribo-and deoxyribonucleosides and nucleotides, and di- and tri- phosphates.

Answer 1

Purines = A and G. Pyrimidines = C and T (and U). Major difference between ribose and deoxyribose is the –OH group on the 2’ hydroxyl (ribonucleotides have it, deoxyribonucleotides do not). Nucleoside = base + sugar. Nucleotide = base + sugar + a phosphate group (ex: deoxyguanosine vs. deoxyguanylate). Diphosphate = 2 phosphate groups on the sugar; triphosphate = 3 groups on the sugar.

Question 2

What are the relative solubility of the components of nucleotides?

Answer 2

Purines are less soluble than pyrimidines because they are more hydrophobic. Solubility ranking = pyrimidine > purine > nucleotide > nucleosides > bases.

Question 3

What diseases related to nucleotide insolubility?

Answer 3

Gout and Lesch-Nyhan disease lead to accumulation of purines of low solubility in tissues.

Question 4

How does a phosphodiester bond form and what does this mean for polarity of DNA?

Answer 4

In the formation of a phosphodiester linkage, phosphate group on 5’C of Nt2 attacks 3’OH group on Nt1 → forms a 5’→3’ linkage, which is why DNA is polarized in this way. (Phosphate group on the first nucleotide in the chain never gets attacked.)

Question 5

What did Avery, McCloud, and McCarty do to establish DNA as the genetic material?

Answer 5

Avery, McCloud, and McCarty showed, using virulent and non-virulent bacteria, that DNA is the transforming material → aka, the genetic material.

Question 6

What did Chargaff do to establish DNA as the genetic material?

Answer 6

Chargaff showed that % purines = % pyrimidines and that %G=%C, and %A = %T.

Question 7

What did Rosalind Franklin and Maurice Wilkins do to establish DNA as the genetic material?

Answer 7

Rosalind Franklin and Maurice Wilkins used x-ray diffraction to show helical structure and 3.4 nm repeat.

Question 8

What did Watson and Crick do to establish DNA as the genetic material?

Answer 8

Watson and Crick put together the double helix structure.

Question 9

What are the implications of Chargaff’s rule?

Answer 9

Chargaff showed that % purines = % pyrimidines and that %G=%C, and %A = %T. So even though the ratio of purines to pyrimidines differs in every organism, within an organism itself, these rules will hold true.

Question 10

What is the Watson-Crick 3D model for DNA structure?

Answer 10

B-form helix DNA is most common, with antiparallel strands in a right-handed helix. Sugar-phosphate backbone is on the outside with bases paired and stacked on the inside (AT and GC favored geometrically). 3.4 nm repeat = 10 base pairs per turn in ladder? Major groove = where backbones are further apart; minor groove = where backbones are closer together.

Question 11

What is the chemical basis for the stability of the double helix DNA in solution?

Answer 11

Hydrophobic interactions/stacking energies between adjacent base pairs and H-bonding between complementary bases help for stability and compensate for negatively charged phosphate groups.
Increased Tm associated with: increased GC content (because joined by 3 H-bonds = harder to tear apart), increased chain length, and increased salt concentration (because cations in salt help neutralize DNA’s negative charge).
pH extremes can also alter stability.

Question 12

What’s the difference between linear and circular DNA?

Answer 12

Linear vs. circular = obvious. Linear = nuclear DNA; circular = mitochondrial or bacterial.

Question 13

What’s the difference between relaxed and supercoiled DNA?

Answer 13

Relaxed vs. supercoiled depends on topoisomerases? Relaxed is underwound; supercoiled happens when DNA becomes under tension, overwound.

Question 14

Describe the chemical modifications of bases in DNA during methylation and its significance to disease.

Answer 14

Methylation = occurs at C of CpG sequences → involved in gene regulation and mutagenesis (covalent modification). Can cause mutations and replication issues.

Question 15

Describe the chemical modifications of bases in DNA during deamination and its significance to disease.

Answer 15

Deamination = generally from nitrous acids, which speeds up deamination (nitrosamine is from cig smoke). Changes a cytosine to a thymine (hard to catch) or uracil (easy to catch since it shouldn’t even be in DNA). If C → T, then the daughter strand will get an A instead of a G. Can cause mutations and replication issues.

Question 16

Describe the chemical modifications of bases in DNA during depurination and its significance to disease.

Answer 16

Depurination = purines are susceptible to attack by water, which causes phosphate backbone to be sensitive to breakage (really bad if on both sides). Can cause mutations and replication issues.

Question 17

Describe the chemical modifications of bases in DNA during UV crosslinking and its significance to disease.

Answer 17

UV cross linking = causes thymines to become covalently linked (happens between stacked bases), which can create a weird kink/bulge in DNA which will make it hard to replicate. Can cause mutations and replication issues.

Question 18

Explain the chemistry of DNA polymerization and how nucleoside analogues are used as drugs.

Answer 18

Can use molecules that look like nucleosides to block replication of virally infected cells because replicating DNA will incorporate analogues.

Question 19

What is DNA melting temperature and what determines this?

Answer 19

It’s the temperature when 50% of the strand pair molecules are separated. Determined by amount of G:C content and A:T content due to H-bonds between the pairs. G:C has three, so requires more heat to melt and increases stability of DNA. A:T only has 2 H-bonds.

Question 20

What diagnostic technique can distinguish the presence of a particular unique sequence (base mutation) in an unknown DNA sample?

Answer 20

You can test for single mismatch between a tagged DNA probe and an unknown DNA sample because perfectly complementary strands pair with higher stability and thus would have a higher melting temp. Do this through “hybridization” properties of DNA.

Question 21

What are the major similarities and differences between DNA and RNA?

Answer 21

DNA = double stranded, uses C,G,T,A, and does not have a OH group at 2’C. 
RNA = normally single stranded (that’s the conformation that gives it the most freedom to carry out function, but it can form hairpin loops and be double stranded with itself), uses C,G,U,A, and DOES have an OH group at the 2’C (makes it more susceptible to nucleophilic attack, which is good because RNA has to be degraded more often).

Question 22

What are the 3 classes of RNA in human cells?

Answer 22

Structural RNA = rRNA, tRNA, snRNA (small nuclear), and snoRNA (small nucleolar). 
Regulatory RNA (control gene expression) = miRNA, siRNA (small interfering)
Information-containing RNA = mRNA

Question 23

How does puromycin mimic amino-acyl tRNA to terminate translation.

Answer 23

Puromycin = nucleotide analogue that mimics the tRNA acceptor region, allowing peptide transfer and termination. Puromycin will bind the ribosome and react with growing peptide chain to terminate translation (example of mimicry in antibiotics).

Question 24

What is the chemical reaction catalyzed by RNA polymerase and why it is unidirectional.

Answer 24

Unidirectional because synthesized in the 5’ → 3’ direction, just like DNA. Phosphodiester bond synthesized by a nucleophilic attack on 3’C by incoming NTP and a pyrophosphate comes off.

Question 25

Define: semi-conservative

Answer 25

each parent strand serves as a template for the daughter strand = conserved in subsequent copies.

Question 26

Define: bidirectional

Answer 26

happening in both directions on the DNA strand at same time

Question 27

Define: Okazaki fragments

Answer 27

pieces of DNA synthesized on the lagging strand in chunks that have to be ligated together

Question 28

Define: origin of replication

Answer 28

where DNA replication begins. Unique segments with multiple short repeats (generally AT rich for easy strand separation) and recognized by origin-binding proteins

Question 29

Define: replication fork

Answer 29

place at which DNA helicase is unwinding the double helix; where replication of new strand is taking place

Question 30

What are the functions of origin binding proteins?

Answer 30

Recognize origin and open up DNA at that sequence; recruits Pol III

Question 31

What are the functions of helicases?

Answer 31

unwinds DNA at the replication fork by breaking H-bonds between base pairs

Question 32

What are the functions of single-strand binding proteins?

Answer 32

bind to ssDNA to prevent it from re-annealing or tangling with each other/itself

Question 33

What are the functions of primase?

Answer 33

lays down an RNA primer for DNA Pol III to synthesize new DNA off of

Question 34

What are the functions of DNA Polymerase I?

Answer 34

‘Distributive’ polymerase that replaces the RNA primers. Has 3’→5’ and 5’→3’ exonuclease proof reading activity. No sliding clamp, slower.

Question 35

What are the functions of DNA Polymerase III?

Answer 35

‘Processive’ polymerase that synthesizes new DNA off template strand. Only has 3’→5’ exonuclease proof reading activity. High processivity because has a sliding clamp that holds it to the DNA.

Question 36

What are the functions of DNA ligase?

Answer 36

seals the nicks between Okazaki fragments, ligating them together

Question 37

What are the functions of the sliding clamp?

Answer 37

present in Pol III that increases processivity by keeping polymerase and DNA closely associated

Question 38

What are the functions of topoisomerase/gyrase?

Answer 38

relieves torsional strain generated by DNA unwinding. Topoisomerase II cuts through both strands of DNA, loop strand through gap, and re-anneals them. Gyrase is a topoisomerase inhibited by quinolones.

Question 39

What are the functions of Telomerase/reverse transcriptase?

Answer 39

provides an RNA template for addition of DNA onto ends of telomeres since chromosomes get shorter after every replication. Heavily used in immortal cells like germ cells. Can act as reverse transcriptases in this way, which make DNA from RNA.

Question 40

How does DNA polymerase create the phosphodiester bond during addition of dNTPs?

Answer 40

3’OH of newly added base attacks the alpha phosphate on 5’C of the incoming base, cleaving off a disphosphate group which provides the energy to create the phosphodiester bond.

Question 41

What are the order of events that occur during, the differences between, and the coordination of DNA synthesis on the leading strand?

Answer 41

On the leading strand, DNA replication can happen continuously (after an RNA primer has been laid down by primase) because replication is happening in the direction of the fork, which keeps on unwinding. Origin binding proteins will bind, DNA will be unwound by helicase, Pol III will elongate new strand.

Question 42

What are the order of events that occur during, the differences between, and the coordination of DNA synthesis on the lagging strand?

Answer 42

On the lagging strand, synthesis has to happen in chunks because it is occurring in the direction opposite of that of the replication fork so it has to keep waiting for more DNA to be unraveled. Primase lays down primers from which Pol III extends DNA, and later, RNA primer is replaced with DNA by Pol I, and chunks are ligated together with DNA ligase.

Question 43

How does telomerase function/solve end replication problem?

Answer 43

Like Reverse transcriptase - copies from RNA
Telomere contains G rich sequence. Telomerase recognizes sequence and using an RNA template within the enzyme itself, it extends the telomere 5’ to 3’, and can extend as much as needed. Lagging strand is then completed by DNA polymerase alpha, which has DNA primase as a subunit to lay down the primer. The extension of telomere ensures that original DNA is conserved and not shortened.

Question 44

What is the End Replication Problem?

Answer 44

Very first primer is degraded because can’t be converted to DNA due to lack of primer for DNA conversion, which shortens the chromosome. This is a MAJOR cause of aging and doesn’t happen in stem cells or cancer cells because of Telomerase

Question 45

What is the role of Ribosomes during translation?

Answer 45

made of rRNA and protein; the complex that brings mRNA, tRNA, and synthetases together to make a peptide. 50S and 30S = 70S in bacteria, and 60S + 40S = 80S in eukaryotes. A site = where tRNAs enter, P = where they deliver their AA (to that in the A site), and E = exit point.

Question 46

What is the role of mRNA during translation?

Answer 46

the RNA copy of DNA that contains the nucleotide sequence that encodes the protein (bacteria have no cap, A tail, or UTR’s, and are polycistronic)

Question 47

What is the role of tRNA during translation?

Answer 47

adapters that “read” the message (codon on mRNA) on mRNA (with their anticodon loop that reads mRNA) and deliver the correct AA (which attaches to acceptor stem)

Question 48

What is the role of aminoacyl tRNA synthetases during translation?

Answer 48

protein enzymes that put/”charge” the right amino acid on the right tRNA. Burns ATP to attach AA to the tRNA.

Question 49

What is the role of initiation factors during translation?

Answer 49

proteins that bring the ribosome to the message (mRNA)

Question 50

What is the role of elongation factors during translation?

Answer 50

proteins that deliver tRNAs to the ribosome and move the ribosome down the mRNA

Question 51

What is the role of release factors during translation?

Answer 51

proteins that end the process and cause dissociation of ribosomal subunits

Question 52

Be able to explain the nature of the genetic code, how it is read. Name the start codon and what amino acid it encodes.

Answer 52

Genetic code = 64 triplet codons that code for 20 AA’s (indicates degeneracy of the code). Start codon = AUG (f-Met) and stop codons = UAA, UAG, UGA. Codons are read by anticodons on the tRNAs that then pair them with the correct AA. Because of the degeneracy/redundancy, the same AA sequence can be encoded by different mRNA sequences. There is some wobble in the 3rd base of the codon (3’ position of codon, 5’ position of anticodon) that adds to the looseness of the code.

Question 53

What are the effects of mutations to the mRNA

Answer 53

mRNA mutations can have some pretty severe consequences for peptide:

• missense mutation = a change in a nt of the mRNA sequence that results in an AA change (because now you have a different codon)
• silent mutation = when mutation of a single nt has no effect due to the redundancy of the code (ex: GAG and GAA wit both code for Glu)
• Frame-shift mutation = if a nt gets inserted or deleted, the whole reading frame as demarcated by AUG get shifted and will result in an entirely different sequence
• Nonsense mutation = mutation that results from a premature stop codon. If a single nt gets changed such that it results in an UGA, UAA, UAG → your peptide is prematurely truncated
• Sense mutation = mutation resulting form a removed stop codon = a mutation that changes one of the stop codons into a codon that actually codes for a protein → you’ll get an elongated peptide

Question 54

Describe initiation in Eukaryotes.

Answer 54

• in eukaryotes → translation begins with a cap-dependent scanning mechanism that uses a lot more IFs, burns more GTP, and involves scanning of mRNA. Relies on the Kozak sequence to position ribosome at start sequence. 5’ UTR can affect rate of scanning process, depending on how convoluted it is since ribosome has to get through it to start codon
• eIF4E = binds to 5’ cap
  
  • eIF4G = recruited by eIF4E (scaffolding)
  • eIF3 = binds 40S
  • eIF2 – binds tRNA charged with Met to start synthesis
  • eIF4A – helicase that melts RNA structures

Question 55

Describe initiation in bacteria.

Answer 55

Initiation in bacteria: IF1 and IF3 bind 30S, and mRNA binds 30S using Shine-Dalgarno sequence (which helps set the reading frame; located upstream of AUG, it can base pair with 3’ end of rRNA of small subunit to position AUG correctly at start sequence. SD sequence allows different AUGs to be considered start site = polycistronic). IF2 delivers initiator f-Met to P site, and GTP hydrolysis leads to IF2 release (and of all IFs) and to binding with 50S.

Question 56

Describe elongation.

Answer 56

Elongation = the peptide bond is catalyzed by the petidyl transferase center (PTC); energy comes from the ATP used in tRNA charging → nascent polypeptide transfers from P-site to A-site. EF2 and GTP, after peptide bond is formed, triggers movement of mRNA and tRNA’s one codon (3 nt’s) in the 3’ direction so tRNA with peptide chain moves to P site, and the other one moves to E site to be kicked out.
-requires 4 high-energy bonds: 2 ATP to charge tRNA, 1 GTP to deliver aa-tRNA to A site, and 1 GTP for translocation

Question 57

Describe termination/recycling.

Answer 57

Termination/Recycling = release factors binding to stop codon terminate polypeptide synthesis. Stop codons not recognized by tRNAs but by release factors that enter A site and cause ribosome to fall apart so the pieces can be recycled

Question 58

Explain the significance and effects of eIF2-alpha phosphorylation and Interferon stimulation.

Answer 58

if a virus attacks the cell, the cell can make interferon, which binds to neighboring cells to activate them so they produce antiviral proteins so they’re protected against the spreading virus. Interferon can activate production of 2-5A synthase → activates an endonuclease that cleaves viral mRNA (prevents translation). Interferon can also trigger phosphorylation of eIF2-alpha to inactivate it → prevents f-Met from being recruited to start protein synthesis

Question 59

Explain the significance and effects of Cap-independent initiation.

Answer 59

this is IRES-driven translation in eukaryotes. Some viruses interfere with the 5’ cap and the eIF4E and eIF4G protein to inhibit translation. Instead then, ribosomes bind the viral 5’ UTR (the IRES; internal ribosomal entry site) so that viral mRNA gets translated instead of host mRNA. Obvious implications for disease.

Question 60

Explain the significance and effects of mRNA editing

Answer 60

mRNA code can be altered after its been made. Example = human apoB mRNA can be edited by cytosine deamination → makes a stop codon instead → truncated protein. Happens in certain tissues to regulate how much protein is made.

Question 61

Explain the significance and effects of Rapamycin treatment

Answer 61

for cancer = phosphorylates 4E-BP so that initiation complex can’t be formed, downregulating translation (not quite? Dephosphorylated 4EBP’s sequester EIF4E and prevent cap-dependent translation, and inhibited when phosphorylated?)

Question 62

Identify antibiotics that operate by affecting translation.

Answer 62

Some antibiotics target the ribosome → inhibit translation by interfering with 30S and 50S subunits through tRNA binding, elongation, and peptidyl transferase. Good at generally just targeting bacteria, but some eukaryotic cells can also suffer.

Question 63

Understand how intracellular levels of iron can be regulated by translation, using this as an example of how protein-mRNA interactions regulate translation.

Answer 63

Fe is necessary for biological activity, but high levels are toxic so it must be kept in balance.
Under high Fe conditions, the IRE-BP’s (iron response binding proteins) are bound to Fe and can’t bind the IRE RNA (iron response element). Under low Fe conditions, IRE-BPs are not bound to Fe and can bind the IRE RNA.
With the transferrin receptor (the protein required to transfer Fe into the cell when levels are low), protein level goes up with low iron and down with high iron.
- low iron concentration = IRE-BP binds to the IRE of transferrin receptor mRNA → protects mRNA from degradation → synthesis of transferrin receptor proceeds
- high iron concentration = IRE-BP cannot bind to IRE → mRNA is degraded and synthesis of transferrin receptor is inhibited
Ferritin binds excess levels of iron and sequesters it in a non-toxic storage form. Protein levels go down with low iron, and up with high iron.
- Low iron concentration = IRE-BP binds to IRE so translation of ferritin mRNA is inhibited
- High iron concentration = IRE-BP cannot bind to IRE so translation of ferritin mRNA proceeds

Question 64

Describe the relationship between mutations, DNA repair and cancer. Give examples of heritable human diseases that are caused by defective DNA repair pathways.

Answer 64

DNA has to be repaired because it cannot be replaced. If it goes un-repaired, it can cause mutations that lead to cancer. HERITABLE HUMAN DISEASES? Xeroderma, Cockayne Syndrome, Hereditary non-polyposis colorectal cancer

Question 65

Generally describe the sources and nature of damage to DNA and the molecular consequences of failure in DNA repair

Answer 65

• Exogenous damage
  ○ (UV and other radiation, chemicals) or
• endogenous
  ○ (ROS, alkylation, hydrolysis).
• These things can either change the base
  ○ (such as base loss, deamination, UV-induced thymine dimers, alkylation, or oxidation)
• or change the DNA structure
  ○ (causing bulges in the helix, strand breaks, or stalled replication forks).
• Consequences include
  ○ blocked DNA replication or base changes that then get copied to subsequent daughter strands (point mutations). Ex: deamination and alkylation changes C:G to T:A.

Question 66

Describe in more detail the types of damages to DNA.

Answer 66

• Deamination = loss of an amine group, CG to UA. Depurination = loss of a purine group. Alkylation = addition of an alkyl group. UV = thymine dimers. ROS (generated via metabolism) can change bases, block replication (via covalent bonds b/w bp that block replication machinery?).
  
  • Inter and intra-strand cross links change DNA structure and interfere with replication/transcription. Bulky base adducts can also change structure, results in GC to TA.

Question 67

Describe the type of machinery used to repair the damage.

Answer 67

• Repair can be a direct reversal of damaged bases (ex: MGMT), base excision/nucleotide excision/mismatch repair, tolerance/bypass of the damage, or strand break repair of damage to the backbone.

Question 68

Explain the basic steps of mismatch repair, the type of damage repaired by this pathway, and how eukaryotic cells distinguish between old and new strand of DNA.

Answer 68

• Mismatch repair removes mis-incorporated nt’s during DNA replication; distinguishes between template strand and new strand. Fixes errors in nucleotide incorporation using MutS and MutL (or in humans, MSH and MLH) proteins that recognize incorrectly incorporated bases on the new strand (which is not yet methylated). Endonuclease chews the backbone of the DNA, helicase separates and an exonuclease chews away that chunk of the strand, and DNA pol repairs the gap (ligase seals).
• nascent lagging strand marked by 5’ ends of Okazaki fragments; nascent leading strand marked by presence of ribonucleotides.
• Lynch syndrome if not working.

Question 69

Describe the basic mechanisms of base excision repair by homologous recombination and NHEJ, and the types of lesions corrected by these DNA repair pathways.

Answer 69

Base excision repair = Glycosylase recognizes wrong base and hydrolyzes the bond to yield an AP site (which contains just the sugar and phospho- bond, no nitrogenous base). An AP endonuclease cleaves the backbone on the 5’ end and another endonuclease cuts on the 3’ end of the AP site, removing the group so that DNA polymerase can add a new base and ligase can seal. BER repairs base damages that do not distort DNA. Removes ONE base.

Question 70

Describe the basic mechanisms of nucleotide excision repair by homologous recombination and NHEJ, and the types of lesions corrected by these DNA repair pathways.

Answer 70

Nucleotide excision repair = repair to base damages that do distort the DNA structure and block polymerase function, and it does so by removing an oligonucleotide that contains the damaged base. Multi-protein complex recognizes the damaged side, helicase unwinds that area, the damaged strand is cut out by endonucleases (removes an actual chunk of the strand), and DNA polymerase fills it, ligase seals it. Aided by TFIIH (helps to unwind DNA). Global genome NER recognizes damage anywhere in the genome (defects in this will cause cancer, like xeroderma pigmentosum). Transcription-coupled NER recognizes damage within a transcribed region (so defects can cause CNS disorders, like Cockayne syndrome).

Question 71

Describe the basic mechanisms of double-strand break repair by homologous recombination and NHEJ, and the types of lesions corrected by these DNA repair pathways.

Answer 71

Double strand breaks can be fixed by nonhomologous end-joining or homologous recombination. In NHEJ, it is a ds break between non-homologous chromosomes, so no homology is needed to fix, but it’s often inaccurate and can lead to insertion or deletion of nt’s at the breakpoint. In homologous recombination, there is extended sequence homology between the two broken ends needed and it’s more accurate. PARP fixes single strand breaks by amplifying damage signal, enriching repair proteins in the area, and changing the local chromatin structure.

Question 72

Describe the mechanism that enables replication to continue in the face of DNA lesions that other repair pathways fail to remove, and know the unfortunate consequence of this process for the cell.

Answer 72

If DNA is too damaged for normal repair mechanisms (generally happens as a result of thymine dimers, which block DNA replication), cells will use a “bypass” mechanism in which they use DNA polymerases that are not so tightly specific to the template strand and continue through damaged regions. These polymerases lack 3’-5’ proofreading ability and have much higher error rates = higher risk of mutation and disease as a consequence.

Question 73

Explain the concept of DNA damage checkpoint and its role in maintaining genome stability.

Answer 73

Checkpoints exist between stages of the cell cycle that prevent cell from moving on to next stage of replication if there is something wrong with the DNA. Signals such as cellular metabolism, UV light exposure, radiation, chemical exposure, and replication areas trigger sensors that pick up on DNA damage. Those activate transducers/checkpoint kinases (like ATM, ATR, p53), which turn on effectors that cause delay/arrest of the cycle, activation of transcriptional programs, DNA repair, or apoptosis.