Cumulative exam

Question 1

direction of DNA sequence

Answer 1

written/read 5’ to 3’

Question 2

nucleotide linkages

Answer 2

3’, 5’-phosphodiester bridges

Question 3

RNA vs DNA

Answer 3

• uracil instead of thymine in RNA
• 2’ OH missing in DNA: this can normally facilitate hydrolysis of phosphate backbone, therefore DNA is more stable

Question 4

division of common DNA/RNA bases

Answer 4

1) purines: two rings, includes adenine and guanine
2) pyrimidines: one ring, includes cytosine, uracil and thymine

Question 5

pyrimidine numbering system

Answer 5

1) assign priority to nitrogen adjacent to C=C
2) number in direction of second nitrogen

Question 6

purine numbering system

Answer 6

1) assign priority to the nitrogen in the larger ring, farthest away from the ring junction (C=c)
2) number in direction of second nitrogen in larger ring
3) move onto the smaller ring, number the double bonded nitrogen.
4) go clockwise for the rest of the atoms in the smaller ring

Question 7

ribose sugar numbering

Answer 7

• 1’ is anomeric carbon, continue clockwise
• ’ is associated with sugars

Question 8

nucleoside

Answer 8

• base attached to sugar with beta-glycosidic linkage (anomeric carbon to N, aka N-glycosidic linkage)
• purine: N9
• pyrimidine: N1

Question 9

nucleotide

Answer 9

• base + sugar + one or more phosphate
• phosphate attached to 5’ carbon through phosphoester link
• phosphates attached to each other through phosphoanhydride link

Question 10

phosphates in ATP

Answer 10

• closest to ribose sugar is alpha, then beta, gamma

Question 11

why does DNA contain thymine?

Answer 11

• cytosine can spontaneously convert to uracil (hydrolytic deamination)
• DNA repair enzymes recognize these “mutations” to replace Us with Cs
• thymine used in place of uracil in DNA to signal that T belongs in DNA

Question 12

purine derivatives

Answer 12

1) hypoxanthine: deaminate A
2) xanthine: deaminate G
3) inosine: hypoxanthine attached to ribose sugar

Question 13

what is special about inosine?

Answer 13

• wobble pairing: can base pair with C, A or U

Question 14

pyrimidine derivatives

Answer 14

1) thiouridine: uracil carbonyl replaced with thiol
2) ribothymidine: thymine + ribose sugar
3) pseudouridine: uracil linked to ribose sugar at C5 instead of N1
4) dihydrouridine: hydrogenation of uracil double bond

Question 15

where are pseudouridine and dihydrouridine found?

Answer 15

tRNAs

Question 16

timeline of DNA double helix discovery

Answer 16

1) nuclein
2) 4 distinct nuclein species identified (pentose + nitrogenous base)
3) carbohydrate or phosphate polymer
4) nucleosides crystallized, sugar-base connectivity defined
5) 1:1:1:1 base ratio incorrectly reported, led to tetrameric structure and cyclic tetramer theories
6) MW ~10^6 inconsistent with tetramer (heavier than expected), X-ray diffraction of extended structure found 3.4 angstrom spacing between nucleotides = penny stack
7) H-bonding discovered in enolic tautomers
8) 3-stranded structure, Chargaff reports
9) DNA double helix

Question 17

phosphate-based polymer (Steudel)

Answer 17

• linkages through phosphate
• problems: no negative charge, base bound to sugar attached to phosphates

Question 18

carbohydrate-based polymer (Levene/Jacobs)

Answer 18

• linkages through carbohydrates (ribose sugars)
• base/phosphate hang off that backbone

Question 19

cyclic tetramer (Takahashi’s tetranucleotide hypothesis)

Answer 19

• correct base-sugar-phosphate connectivity, but overall structure wrong

Question 20

tetranucleotide hypothesis (Levene/Tipson)

Answer 20

• linear, single-stranded tetramer units

Question 21

single-stranded DNA structure (Astbury/Bell)

Answer 21

• aka penny stacking model
• problems: single stranded DNA, planar sugars and bases stacked on top of each other

Question 22

Gullard’s discovery of H-bonding

Answer 22

• discovered in enolic tautomers
• H-bonds were in the wrong locations
• suggests possibility of multiple strands

Question 23

why do tautomers have different H-bonding paterns?

Answer 23

• enol form has OH (H-bond donor) instead of C=O (H-bond acceptor)

Question 24

Pauling’s inside-out DNA structure

Answer 24

• three strands
• backbone on inside, but this would have too many negative charges and would be unstable

Question 25

Chargaff reports

Answer 25

- observed 1:1 ratio of A:T, G:C, purines:pyrimidines - facilitated understanding of H-bond pattern (Watson-Crick base pairing)

Question 26

Watson-Crick base pairing

Answer 26

- 3 bonds between G-C - 2 bonds between A-T - purine uses larger ring (1, 2 for guanine only, 6) - pyrimidine uses face opposite to beta-glycosidic bond (4, 3, 2 for cytosine only)

Question 27

DNA double helix

Answer 27

- Watson and Crick, using x-ray crystallography by Franklin - also paved the way for understanding DNA replication

Question 28

forces that stabilize DNA structure

Answer 28

1) hydrogen bonding 2) dipole interactions 3) hydrophobic effect

Question 29

hydrogen bonding in DNA

Answer 29

- between base pairs - three types 1) canonical Watson-Crick base pairs: large distance between sugar C1s 2) Hoogsten base pair: any other variation, smaller C1 distance 3) wobble base pairs: important in anticodon/codon pairing and degeneracy of codons

Question 30

guanine H bonding in telomeres

Answer 30

- guanine rich strands can form quadruplexes using Hoogsteen hydrogen bonding - these can be found in telomeres

Question 31

measuring DNA melting temperature

Answer 31

- denaturation causes A260 to increase (exposes inner bases that have rings)

Question 32

DNA melting curve

Answer 32

- sigmoidal shape = cooperativity - melting temperature (Tm) is middle of curve, where there is 50% ds and 50% ss DNA - Tm is reflective of DNA stability

Question 33

G-C vs A-T melting temperature

Answer 33

- G-C has more H-bonds so higher Tm

Question 34

dipole interactions in DNA

Answer 34

- weak dipole-dipole interactions between adjacent bases (individually has minimal effect, but cumulative is stronger) - aka base-stacking - changes to atoms present can also affect Tm, ex. less polar = less dip-dip = lower Tm

Question 35

non-polar solvents and DNA

Answer 35

- destabilize duplex DNA - indicates that H-bonds are not the primary force in DNA stability

Question 36

hydrophobic effect in DNA

Answer 36

- most important driving force - aromatic hydrophobic base rings bury themselves inside DNA away from water, then base-stacking occurs - entropy driven: H-bond formation releases water molecules, also disruption of "iceberg" structure - more hydrophobic = more stable

Question 37

DNA intercalators

Answer 37

- hydrophobic, planar agents that use the hydrophobic effect to insert between base pairs - can fluoresce (visualization), but destabilize DNA so that it can't be effectively used for replication or transcription (carcinogenic)

Question 38

factors controlling DNA/RNA structure

Answer 38

1) base-sugar orientation 2) sugar pucker

Question 39

base-sugar orientation

Answer 39

- rotation around glycosidic bond can lead to syn (same side) or anti conformation

Question 40

syn vs anti base-sugar conformation

Answer 40

- syn less stable because of steric effects - syn more unfavourable for pyrimidines (basically does not exist) because of C=O

Question 41

sugar pucker

Answer 41

- C2 or C3 points out (C-2' or C-3' endo) - C-2' favoured for deoxynucleotides (DNA), leads to larger interphosphate distance (7.0 angstroms) - C-3' in ribonucleotides (RNA helices), smaller interphosphate distance (5.9A)

Question 42

types of helices

Answer 42

1) A-form (RNA) 2) B-form (DNA) 3) Z-form

Question 43

A-form helix

Answer 43

- short and broad - 2.3A per base pair - right-handed helix - narrow and deep major groove - broad and shallow minor groove (protein-binding here) - anti base-sugar orientation - C-3' endo

Question 44

B-form helix

Answer 44

- longer and thinner - 3.32 plus/minus 0.19A per base pair - right-handed helix - wide and intermediately deep major groove (protein-binding here) - narrow and intermediately deep minor groove - anti base-sugar conformation - C-2' endo

Question 45

Z-form helix

Answer 45

- elongated and slim - 3.8A per base pair - left-handed - flattened out major groove on helix surface - extremely narrow and deep minor groove - anti base-sugar at C, syn at G - C-2' endo

Question 46

protein-DNA interactions

Answer 46

- binding to major-groove with certain motifs (basic and polar AAs) - ex. Leucine zipper, helix-turn-helix (ex. Lac operon), zinc finger

Question 47

eukaryotic chromosome structure

Answer 47

- very compressed (2m to 5micrometer diameter nucleus) - DNA wrapped around nucleosomes, packed in helical filaments, arranged in loops associated with nuclear matrix

Question 48

nucleosomes

Answer 48

- fundamental unit of chromatin - 2x H2A, H2B, H3 and H4 create octameric core structures - (H1 is another histone that also exists, histones are positively charged) - DNA wound ~2 turns around core to create nucleosomes

Question 49

structures in increasing compression in DNA

Answer 49

1) DNA double helix 2) beads on a string chromatin 3) solenoid (6 nucleosomes per turn) 4) loops (50 turns per loop) 5) miniband (18 loops) 6) chromosome (~10^6 stacked minibands)

Question 50

RNA structures

Answer 50

- unlike DNA, complex secondary and tertiary structures are important for function - dsRNA structures: 1) transfer RNA (tRNA) 2) ribosomal RNA (rRNA) 3) catalytic RNAs (ribozymes) 4) riboswitches - ssRNA structures: 1) messenger RNA (mRNA) 2) microRNA (miRNA)

Question 51

how do RNA structures form?

Answer 51

- complementary sequences can join via intrastrand base pairing - when base pairing is incomplete, bulges and loops can form (ex. hairpin stem-loop structures) - bases that stick out in bulges or loops can facilitate noncovalent interactions in tertiary structures

Question 52

tRNA characteristics

Answer 52

- carry amino acids at 3' end to ribosomes for protein synthesis - each amino acid has at least one unique tRNA, sometimes more (because codon degeneracy) - small polynucleotide chains (73-94 residues) - contain modified bases and multiple interactions that create L-shaped molecule

Question 53

tRNA primary structure

Answer 53

- reflect different amino acid specificities ex. anticodon region - some highly conserved regions (sequences and post translational modifications) that are important for overall structure and general function

Question 54

tRNA secondary structure

Answer 54

- cloverleaf structure with four distinct base-paired regions + one variable loop: 1) acceptor stem: 5' and 3' end, tRNA nucleotidyltransferase attaches CCA to 3' end, terminal A attaches to AA 2) TψC loop: conserved T, pseudouridine, C 3) variable loop: lacks conserve sequence, varying length 4) anticodon loop: contains anticodon, which pairs with mRNA 5) D loop: conserved dihydrouridines

Question 55

tRNA tertiary structure

Answer 55

- TψC loop and D loop brought into close proximity - stabilized by internal noncovalent interactions: 1) Watson-Crick base pairing 2) Hoogsten pairing 3) triple base pairing - also stabilized by: 4) divalent metal ion binding 5) polyamine binding

Question 56

modified bases in tRNA

Answer 56

- function to facilitate specific interactions and prevent unwanted interactions - ex. methylation: 7-methylguanosine, 5-methylcytidine, 1-methylguanosine - also inosine and other derivatives

Question 57

Hoogsten basepairs in tRNA

Answer 57

- G18:ψ55 - two H-bonds (NH2 on C2, N1) from guanine to one C=O (C4) on pseudouridine

Question 58

how would pseudouridine synthase mutation affect tRNA structure or function?

Answer 58

- structure: different H-bonding pattern but minimal effect because only 2 H-bonds affected - function may be altered because ψ is conserved: could move H-bonding with ribosome (misaligned)

Question 59

triple base pairs in tRNA

Answer 59

- U12:A23-A9 - U12:A23 = normal WC base pair - A9 binds entering through major groove with 1 H-bond to phosphate backbone, 2-H bonds to A base

Question 60

divalent metal ions in tRNA

Answer 60

- Mg2+ interactions with RNA: 1) Mg2+/phosphate --> direct interaction 2) Mg2+/H2O/base --> indirect 3) Mg2+/H2O/2'-OH in sugar --> indirect

Question 61

polyamines in tRNA

Answer 61

- have positive charge at physiological pH because amines protonated (pH < pKa) - positive charges interact with negatively charged phosphate backbone

Question 62

rRNA characteristics

Answer 62

- about 2/3 of ribosome (other 1/3 is protein) - structural foundation for ribosomal proteins - contain certain modified nucleotides (ribothymidine, pseudouridine) - different species referred to by sedimentation coefficients

Question 63

sedimentation coefficient

Answer 63

- aka Svedberg units - larger, more dense settle faster - dependent on surface area, so not additive, ex. 16S + 16S does not make 32S (would be less)

Question 64

rRNA secondary and tertiary structures

Answer 64

- complex, vary between species - however, common function = homologous, similar structures

Question 65

ribozyme characteristics

Answer 65

- RNA folding creates active sites - play a role in catalysis via hydrolysis or transesterification reactions in the absence of proteins (but metal ions may be important) - ex. self-splicing (removal of introns, alternatively done by spliceosome) - ex. RNAse P is a type of ribonuclease that processes tRNA 5'-ends (phosphate backbone cleavage)

Question 66

riboswitch characteristics

Answer 66

- found on mRNA transcripts - binding of small molecules affects structure and can turn RNAs on/off

Question 67

riboswitch transcription regulation

Answer 67

1) ligand binding to aptamer region 2) terminator stem loop region formed, signals stop to RNA polymerase 3) no transcription

Question 68

riboswitch translation regulation

Answer 68

1) ligand binds to aptamer 2) ribosome-binding sequence hidden 3) ribosome cannot bind, no translation

Question 69

aptamers

Answer 69

- ssRNA protein-binding sequences - can be synthesized, ex. as complementary to a biomarker (analogous to and cheaper than antibodies)

Question 70

techniques for studying nucleic acids

Answer 70

- reading or locating sequences: 1) restriction analysis 2) synthetic DNA 3) blotting and hybridization 4) DNA sequencing - manipulating sequences (i.e. to study gene function) 1) PCR 2) site-directed mutagenesis 3) recombinant DNA technology 4) cloning

Question 71

restriction nucleases

Answer 71

- discovered in bacteria - enzymes that recognize 4, 6, or more base sequences with 2-fold axis of symmetry - cleave to create blunt or sticky (overhangs) ends - endonuclease: cleave phosphodiester bond within sequence - exonuclease: remove from ends

Question 72

restriction enzyme nomenclature

Answer 72

- 1st letter = genus - 2nd/3rd letters = species - 4th letter = strain (capital) - roman numeral for order of discovery

Question 73

restriction enzyme types

Answer 73

- type I cleaves randomly - type II/II cleave at selected sites - type II have no ATP requirement

Question 74

isoschizomers

Answer 74

restriction enzymes that recognize the same site

Question 75

restriction enzyme binding

Answer 75

- bind as dimers (noncovalent binding)

Question 76

gel electrophoresis

Answer 76

- samples loaded on to gel - negatively charged DNA migrates towards positive anode - larger fragments migrate slower - stain with intercalator to visualize - molecular weight ladder of known fragment sizes

Question 77

restriction enzyme digest analysis

Answer 77

- using restriction enzymes to analyze DNA sequences

Question 78

restriction enzyme analysis for disease dignostics

Answer 78

- can detect single base pair mutations that cause restriction cleavage sites to appear/disappear - can identify healthy, carrier or diseased individuals - ex. CTFR gene in cystic fibrosis

Question 79

why is synthetic DNA needed?

Answer 79

- probes and primers for blots or PCR

Question 80

solid phase synthesis nucleotide modifications

Answer 80

1) di-isopropyl group bound to the 3’ phosphate: good leaving group 2) β-cyanoethyl (βCE) group bound to the 3’ phosphate: protects backbone 3) dimethyoxytrityl (DMT) group bound to the 5’ end: protects 5'OH so that 3' phosphate reacts with another nucleotide (not itself) 4) base is also protected: preserve functional groups

Question 81

solid phase synthesis steps

Answer 81

1) coupling: growing chain (first base bound to resin) 5'OH attacks activated monomer 3' phosphate = phosphite triester intermediate 2) oxidation by I2 = phosphotriester intermediate (more stable) 3) deprotection with dichloroacetic acid: removes DMT to facilitate addition of next nucleotide = elongated chain 4) after all bases added, other protecting groups are removed

Question 82

solid phase vs natural DNA synthesis

Answer 82

- solid phase is reverse direction - leaving group in natural = pyrophosphate (PPi hydrolyzed to 2Pi drives reaction) - LG in solid phase is di-isopropyl group

Question 83

southern blotting steps

Answer 83

1) cleave DNA (ex. with restriction enzymes) 2) separate fragments (gel electrophoresis) 3) transfer to nitrocellulose in alkaline solution (capillary action), this also denatures 4) incubate with radiolabeled probe 5) visualize with autoradiogram

Question 84

southern blot probes

Answer 84

- generated from solid phase-synthesis - ideally want to bind with high specificity (good probe) - non-specific or imperfect pairing = bad-probe

Question 85

southern blot application example

Answer 85

forensics --> DNA match of suspects to crime scene evidence

Question 86

fluorescent in situ hybridization

Answer 86

- fluorescently tagged probes hybridize to chromosomes - visualized using fluorescence microscope - ex. used diagnostically to identify multiple copies of HER2 gene

Question 87

DNA sequencing methods

Answer 87

1) Maxam-Gilbert 2) Sanger chain-terminator 3) automated 4) single-molecule 5) proton (electronic) 6) pyrosequencing

Question 88

Maxam-Gilbert sequencing

Answer 88

1) radiolabel DNA sample for identification (5' P32 label) 2) react with different reagents - dimethyl sulfate (neutral) cleaves G - dimethyl sulfate (acidic) cleaves G and A - hydrazine + 1.5M NaCl cleaves C - hydrazine alone cleaves C+ T 3) each sample treated with piperidine: cleaves phosphate backbone 4) run samples on gel and visualize: 5' at bottom of gel

Question 89

Sanger chain-terminator sequencing

Answer 89

- use DNA polymerase - add low concentrations of dideoxyribonucleotides that stop polymerization (also missing 3'OH)

Question 90

Sanger sequencing steps

Answer 90

1) radiolabel primer 2) add ddNTPs, dNTPs and DNA polymerase: one tube for each; since ddNTPs are at low concentrations, they are randomly incorporated and can terminate synthesis at various points 3) electrophoresis and visualization: 5' at bottom, complementary to original template strand

Question 91

automated DNA sequencing

Answer 91

- also uses ddNTPs, but each has a different colour fluorescent tag - gel electrophoresis with capillary gel in sequencing machine = high resolution reading of sequence - emitted photons upon excitation identifies base

Question 92

DNA sequencing efficiency

Answer 92

- becoming more efficient and less costly - especially with next-generation sequencing techniques: single molecule, proton flow and pyrosequencing

Question 93

single molecule sequencing

Answer 93

- fluorescently labelled dNTPs used to determine sequence as it is being synthesized - bases flowed in through nanopore one at a time - incorporation of base leads to pulse of fluorescence

Question 94

proton (electronic) sequencing

Answer 94

- measures DNA sequencing byproducts (hydrogen and pyrophosphate) - flow bases one at a time, ion sensor detects byproduct release if base is incorporated (several in a row = multiple H+ released) - flow of H generates current

Question 95

pyrosequencing

Answer 95

- measures PPi release - after released, sulfurylase incorporates APS (adenosine 5' phosphosulfate) to make ATP - ATP and luciferin --> oxyluciferin via luciferase, this emits light that is detected (proportional to initial amount of PPi released)

Question 96

PCR reagents

Answer 96

1) dsDNA template 2) DNA polymerase, usually Taq because temperature resistant 3) Mg2+ 4) dNTPs 5) primers

Question 97

PCR steps

Answer 97

1) denaturation (heat) 2) annealing of primers 3) extension by polymerase using primers, dNTPs and Mg2+ (cofactor) 4) repeat cycle to amplify (exponentially)

Question 98

PCR application as a diagnostic tools

Answer 98

1) amplify regions of DNA that vary between individuals (ex. short tandem repeats or STRs vary in length aka variable number of tandem repeats/VNTR) 2) clone pieces of DNA using primers 3) introduce mutations (site-directed mutagenesis) 4) real-time PCR to detect and quantify presence of specific nucleotide sequences

Question 99

site-directed mutagenesis

Answer 99

- mismatched nucleotide in primer added to sequence - PCR to make many copies of plasmid with new mutation

Question 100

real-time PCR (qPCR)

Answer 100

- primer with reporter and quencher on different poles - quencher blocks reporter fluorescence because in close proximity - once synthesis done and primers are cleaved/released, reporter detected - measurements made during each cycle of polymerization: exponential increase in product can quantify the number of copies

Question 101

diagnosis of CML using qPCR example

Answer 101

- primers designed to flank bcr-abl sequence - positive = fluorescence - negative = no fluorescence

Question 102

recombinant DNA technology

Answer 102

- cleave DNA fragment and vector with restriction enzymes (flank sequence, best to have two different ones) - anneal DNA fragments and rejoin with DNA ligase - check for bacteria with vector (antibiotic resistance) - bacteria replicate DNA - fusion proteins can be expressed and purified

Question 103

green fluorescent protein

Answer 103

- acts as reporter protein, identifying location of fusion proteins

Question 104

RNA base nomenclature

Answer 104

- ribonucleoside: adenosine, guanosine, uridine, cytidine - ribonucleotide (nMP): adenlyate, guanylate, uridylate, cytidylate

Question 105

DNA base nomenclature

Answer 105

- deoxyribonucleoside: deoxyadenosine, deoxyguanosine, thymidine, deoxycytidine - deoxyribonucleotide: deoxyadenylate, deoxyguanylate, thymidylate, deoxycytidylate

Question 106

roles of nucleotides

Answer 106

1) precursors of DNA and RNA 2) universal energy (ATP, GTP) 3) components of cofactors (NAD+, FAD, CoA) 4) molecule activation (UDP-glucose, PRPP) 5) signal transduction (cAMP, cGMP)

Question 107

nucleotide metabolic pathways

Answer 107

1) de novo synthesis: base/sugar from simple precursors 2) salvage pathway: recycle free bases and nucleosides from polynucleotide breakdown 3) nucleotide degradation: modification of bases to generate products for excretion or reuse

Question 108

nucleotide metabolic pathway defects

Answer 108

1) de novo synthesis: orotic aciduria, causes anemia and growth retardation 2) salvage pathway: Lesch-Nyhan syndrome, causes neurological impairments and self-mutilation 3) nucleotide degradation: gout, causing uric acid crystal build up in joints and acute inflammatory arthritis

Question 109

nucleotide metabolism overview

Answer 109

1) 70% tissue de novo nucleotide synthesis (in liver) and 30% dietary purines and pyrimidines 2) contribute to cellular nucleotide use, leads to... 3) complete pyrimidine breakdown or purine breakdown to uric acid 4) salvage pathways from 3) recycle for more cellular use 5) uric acid: 70% excreted through kidneys, 30% excreted through large intestine

Question 110

common nucleotide metabolism intermediates

Answer 110

- pyrimidine: uridylate (UMP) - purine: inosinate (IMP) - xanthine, hypoxanthine

Question 111

pyrimidine de novo nucleotide synthesis

Answer 111

1) bicarbonate + NH4 --> carbamoyl phosphate using 2 ATP 2) carbomyl phosphate + aspartate = pyrimidine ring 3) add sugar (PRPP) to make UMP 4) UTP --> CTP (RNA) 5) CTP --> dCTP, TMP (DNA)

Question 112

purine de novo nucleotide synthesis

Answer 112

1) rings built off 5-phosphoriribosyl-1-amine (amine becomes N9, add glutamine, glycine, aspartate, N10-formyl-tetrahydrofolate) 2) IMP 3) ATP or GTP (RNA) 4) dATP or dGTP (DNA)

Question 113

substrate activation by phosphorylation

Answer 113

1) ATP molecule to phosphorylate C=O, creating a good leaving group 2) nucleophilic attach from amine group to form C-N bond

Question 114

displacement reactions and PRPP

Answer 114

1) ribose 5-phosphate --> phosphoriboyl diphosphate (PRPP) via PRPP synthetase (two Pi added) = pyrimidine pathway 2) PRPP --> 5-phosphoribosyl-1-amine: glutamine hydrolysis makes glutamic acid with amine group, amine attacks PRPP (PPi is good LG) = pyrimidine pathway, catalyzed by PRPP amidotransferase

Question 115

AMP regeneration to ATP

Answer 115

AMP + ATP --> 2 ADP --> ATP

Question 116

pyrimidine synthesis: production of carbamoyl phosphate

Answer 116

1) bicarbonate phosphorylated with ATP to make carboxyphosphate 2) NH3 from Gln displaces Pi from carboxyphosphate to make carbamic acid 3) carbamic acid phosphorylated with ATP to make carbamoyl phosphate

Question 117

carbamoyl phosphate synthetase

Answer 117

- 3 active sites: 1) glutamine hydrolysis: hydrolyze NH3 from glutamine, add to carboxyphosphate 2) bicarbonate phosphorylation 3) carbamic acid phosphorylation - substrates travel through with noncovalent interactions

Question 118

pyrimidine synthesis: ring formation

Answer 118

- rate-limiting step (regulated) 1) carbamoyl phosphate + Asp --> carbamoylaspartate + Pi via aspartate transcarbamolyase (ACTase) 2) ring formation: dehydration to dihydroorotate and redox reaction (CoQ or NAD+ reduced) to orotate (oxidized to make double bond)

Question 119

ACTase structure

Answer 119

tetramer with 2 regulatory and 2 active sites

Question 120

pyrimidine synthesis: UMP formation

Answer 120

1) orotate + PRPP --> orotidylate via orotate phosphoribosyltransferase 2) orotidylate decarboxylase: to UMP

Question 121

nucleoside monophosphates to triphosphates

Answer 121

1) UMP + ATP --> UDP + ADP (uridylate kinase) 2) UDP + NTP --> UTP + NDP (nonspecific kinase) *all with delta G = 0 (break and reform phosphate bond)

Question 122

pyrimidine synthesis: UTP to CTP

Answer 122

- cytidylate synthetase - substrate activation by phosphorylation (ATP) + ammonia from glutamine hydrolysis adds NH2 group

Question 123

feedback regulation of pyrimidine synthesis

Answer 123

- CTP negative feedback (inhibits) to ATCase - ATP activates ATCase

Question 124

pyrimidine de novo synthesis ATP cost

Answer 124

- 4 ATP: 2 for carbomoyl phosphate synthesis, 2 to make PRPP (ATP --> AMP functionally uses 2 ATP)

Question 125

deoxyribonucleotide synthesis

Answer 125

- enzyme = ribonucleotide reductase (RR), which is diphosphate specific - ribonucleoside diphosphate + NADPH + H+ --> deoxyribonucleoside diphosphate + NADP+ + H2O - diphosphokinases add 3rd phosphate

Question 126

ribonucleotide reductase structure

Answer 126

- R1 dimer: two alpha subunits, contains active site - R2 dimer: two beta subunits, contains tyrosyl-radical site (iron cluster to facilitate radical transfer to Tyr122)

Question 127

ribonucleotide reductase mechanism

Answer 127

1) radical transferred from R2 to Cys439 in R1 2) radical transferred to nDP sugar 3' carbon (H donated to Cys) 3) radical promotes release of water (2'OH + H from SH on Cys462) 4) Cys225 donates H to 2' carbon, disulfide bond formed between Cys225-Cys462, radical transferred back to Cys439 5) NADPH reduces disulfide bond (thioredoxin reductase)

Question 128

ribonucleotide reductase regulation

Answer 128

- R1 dimer contain two allosteric sites per subunit 1) specificity-determining site: - dATP/ATP: enhance pyrimidine (CTP, TTP) reduction - dTTP: inhibits pyrimidine reduction, promotes GDP reduction - dGTP: stimulates ADP reduction 2) activity-determining site: - ATP binding stimulates - dATP binding inhibits

Question 129

pyrimidine synthesis: thymine problem

Answer 129

- DNA polymerase cannot distinguish between TTP and dUTP - dUTP converted to dUMP by dUTP pyrophosphatase (hydrolyses PPi) - methyl group added using tetrahydrofolate derivative by thymidylate synthase to make (d)TMP - TMP phosphorylated to TTP

Question 130

TMP synthesis

Answer 130

1) enzyme Cys residue attacks C6 of dUMP 2) proton abstracted to enzyme Cys, hydride transferred to add methyl group, enzyme released

Question 131

drugs targeting TMP synthesis

Answer 131

- to inhibit DNA synthesis, since thymidylate synthase is the rate-limiting step in DNA synthesis 1) fluorouracil inhibits thymidylate synthase 2) analogs of dihydrofolate: aminopterin, methotrexate (amethopetrin) --> inhibit dihydrofolate reductase (which makes tetrahydrofolate --> N5,N10-methylene-THF with Ser to Gly) - many off target effects: healthy cells, also tetrahydrofolate needed in other pathways

Question 132

suicide inhibition of thymidylate synthase using fluorouracil

Answer 132

- fluorodeoxyuridylate has F instead of H: cannot be abducted to regenerate the enzyme - used for treatment of skin, colon, breast and stomach cancer

Question 133

purine synthesis: inosinate formation parts

Answer 133

1) 5-membered heterocycle installed on ribose 2) additional 6-membered heterocycle formed to complete purine

Question 134

inosinate formation part 1: 5-membered ring

Answer 134

1) glycine (N7, C4/5) attached to sugar using ATP 2) carbonyl (C8) added from 10-formyltetrahydrofolate 3) substrate activation by phosphorylation and glutamine hydrolysis to form C4=N 4) ring closure using ATP, C=N becomes amine group (N3) 5) bicarbonate (C6) activated by phosphorylation, attacked by C5) to start 2nd ring

Question 135

inosinate formation part 2: 6-membered ring

Answer 135

6) bicarbonate C=O activated by phosphorylation, attached by Asp amino group 7) fumarate removed, only nitrogen (N1) remains 8) 10-formyltetrahydrofolate donates C=O by forming C-N bond (C2) 9) dehydration reaction closes ring

Question 136

purine synthesis overall requirements

Answer 136

- 5 ATP - 1 Gly - 1 Gln - 1 Asp - 2 THF - 1 bicarbonate - produces 1 H2O

Question 137

functionalization of purines: IMP to AMP

Answer 137

1) inositate carbonyl activated by phosphorylation (using GTP!), attacked by Asp amino group 2) fumarate released to make AMP

Question 138

functionalization of purines: IMP to GMP

Answer 138

1) redox reaction (NAD+ reduced), inositate oxidized to form xanthylate (C2 carbonyl) 2) C2 carbonyl phosphorylated by ATP, glutamine hydrolysis adds amine to make GMP

Question 139

purine synthesis regulation

Answer 139

1) excess IMP, AMP, GMP: inhibit PRPP --> phosphoribosyl-amine 2) AMP: inhibits IMP --> adenylosuccinate 3) GMP: inhibits IMP --> xanthylate

Question 140

the DNA replication task (humans)

Answer 140

- 3x10^9 base pairs (haploid) - 10bp = 1 turn = 3.4nm/turn = 1.02m DNA per genome = 2.04m DNA per cell (diploid) - has to be replicated in S phase: 6-8 hrs

Question 141

the cell cycle

Answer 141

G0: resting G1: growth and metabolism (6-12hrs, DNA = 2n) S: DNA replication, (6-8hrs, DNA = 2n to 4n) G2: growth of structural elements (3-4hrs, DNA = 4n) M: mitosis (1hr, DNA = 4n to 2n)

Question 142

Lehman DNA polymerase assay components

Answer 142

1) E. coli: rapid growth cycle = constant DNA replication 2) ATP: energy for anabolic process 3) Mg2+: commonly associated with ATP 4) C14 labeled deoxythymidine: unique to DNA

Question 143

Lehman DNA polymerase assay steps

Answer 143

1) combine radiolabeled T and E.coli with other components 2) add acid and centrifuge: kills E.coli, separates by solubility --> acid soluble nucleosides in supernatant, acid insoluble polynucleotides and proteins in pellet 3) radioactivity detected in pellet (0.01%), but T could be bound to proteins, not necessarily DNA 4) resuspend pellet 5) add DNase 6) add acid and centrifuge --> acid soluble nucleosides in supernatant (denatured DNA), acid insoluble proteins only in pellet 7) radioactivity (0.01%) detected in supernatant!

Question 144

what was the issue with Lehman's initial assay?

Answer 144

- not enough DNA polymerization activity - needed to optimize

Question 145

Lehman DNA polymerase assay optimization

Answer 145

1) better substrates: - thymidine kinase: deoxythymidine + ATP --> dTMP + ADP - nucleotide kinase: ATP + dTMP --> ADP + dTDP - nucleoside diphosphate kinase: ATP + dTDP --> ADP + dTTP = strongest signal = best substrate 2) P32 as more sensitive radiolabel than C14: - label gamma phosphate of ATP: added as alpha phosphate on dTTP, which remains when incorporated into DNA - also a better substrate

Question 146

purifying DNA polymerase by size

Answer 146

1) fractionation column separated into simpler fractions 2) add radioactively labelled dTTP, Mg2+ and ATP 3) acid precipitation 4) check for which pellet has radioactivity: contains DNA polymerase and other necessary components for DNA polymerization

Question 147

purifying DNA polymerase by solubility

Answer 147

1) add streptomycin (+) to precipitate (-) proteins 2) Lehman DNA polymerization assay on supernatant and resuspended pellet = no activity in individual, but activity in control *conclusion: need components from both for DNA polymerization

Question 148

additional DNA polymerization components purified with fractionation

Answer 148

1) supernatant: - kinases: to convert dNMPs to dNTPs - dNTPs: all four needed 2) pellet: - DNA polymerase --> purified finally by ion exchange chromatography - DNA template

Question 149

revising the hypothesis of DNA polymerization

Answer 149

- original thought: existing DNA chains elongated by DNA polymerase - however... 1) other dNTPs needed to allow dTTP to be incorporated 2) new DNA had same AT:GC 3) composition of new DNA not affected by dNTP concentration - revised: DNA polymerization is directed by a DNA template

Question 150

other findings about DNA polymerase I

Answer 150

- can't start DNA chains de novo; needs a primer - synthesis in 5' to 3' direction only - not highly processive or rapid - contains both polymerase and exonuclease activity

Question 151

DNA polymerization reaction

Answer 151

dNTP + (dNMP)n <-> (dNPM)n+1 + PPi --> 2 Pi 1) base-pairing between new substrate and parental (template) strand: ensures correct fit and brings two into close proximity 2) nucleophilic attack: 3'OH attacks alpha phosphate, leading to release of PPi (hydrolysis into 2 Pi drives reaction forward through equilibrium)

Question 152

DNA polymerase fidelity factors

Answer 152

~1 error in 10^10 bases added 1) base-pairing: incorrect shape does not fit well (10^-2) 2) active site: shape = fit and amino acid interactions with P backbone (10^-3) 3) proofreading: exonuclease activity (10^-2) 4) mismatch repair: other enzymes recognize altered shape (10^-3)

Question 153

DNA proofreading vs synthesis

Answer 153

*not reverse reactions - polymerization active site: (dNMP)n + dNTP --> (dNMP)n+1 + PPi (normally faster rate of reaction) - exonuclease active site (dNMP)n --> (dNMP)n-1 + dNMP (faster if incorrect base is incorporated)

Question 154

DNA polymerase polymerization vs editing mode

Answer 154

1) polymerization: DNA 3'OH in active site (between fingers and thumb) 2) editing: distortion of DNApol due to incorrect nucleotide incorporation = 3'OH in active site of exonuclease domain (below palm)

Question 155

is DNA polymerase I enough?

Answer 155

- too slow for E. coli replication rate (every 20 mins) - therefore there must be faster polymerases

Question 156

searching for DNA polymerase mutants

Answer 156

1) random mutagenesis of E. coli strains 2) Lehman assay 3) detection of mutant with no DNA pol I activity --> proved DNA pol I is not essential for life, and therefore does not replicate the genome! 4) search mutant fraction for other polymerases: discovered DNA pol II/III

Question 157

DNA polymerase type functions

Answer 157

I) prototype enzymatically and structurally, involved in DNA repair, removing RNA primer, filling primer gaps II) DNA repair III) replication and synthesis of DNA

Question 158

types of DNA polymerases in eukaryotes

Answer 158

1) DNA-dependent DNA polymerases: DNA polymerases 2) RNA-dependent DNA polymerases: reverse transcriptase, telomerase 3) DNA-dependent RNA polymerase: RNA polymerases

Question 159

DNA polymerase alpha

Answer 159

- replication (complex with primase, aids in starting primer) - DNA repair - no exonuclease activity

Question 160

DNA polymerase beta

Answer 160

- DNA repair only - no exonuclease activity

Question 161

DNA polymerase gamma

Answer 161

- DNA replication in mitochondria - 3' to 5' exonuclease

Question 162

DNA polymerase delta

Answer 162

- replication (processive synthesis on lagging strand) - DNA repair - 3' to 5' exonuclease

Question 163

DNA polymerase epsilon

Answer 163

- replication (leading strand) - DNA repair - 3' to 5' exonuclease

Question 164

DNA polymerase kappa, eta, xi, iota

Answer 164

- DNA repair (bypass polymerase) - no exonuclease activity

Question 165

biotechnologies based of DNA polymerase understanding

Answer 165

PCR, DNA sequencing, CRISPR

Question 166

retrovirus replication

Answer 166

1) entry into cell 2) uncoating of capsid, release of RNA 3) reverse transcription 4) duplex DNA integrated into host chromosome 5) viral RNA synthesis 6) viral protein translation 7) virus assembly 8) release of new viruses

Question 167

AZT and HIV

Answer 167

- AZT has 3' N3 instead of OH - prevents next base from being added onto growing DNA chain, so prevents HIV replication - less impact in eukaryotes because exonuclease activity can remove altered base

Question 168

Meselson and Stahl experimental design

Answer 168

1) E.coli undergoing replication 2) grow in 15N for a few generations 3) switch to 14N 4) separate by CsCl gradient centrifuge: heavier closer to bottom of tube 5) visualize with intercalators (280nm UV absorption)

Question 169

Meselson and Stahl expected results

Answer 169

1) conservative: one heavy, one light --> heavy, more light 2) semiconservative: HL mix --> L, HL mix 3) random dispersive: HL mix --> lighter HL mix

Question 170

Meselson and Stahl actual results

Answer 170

- semiconservative

Question 171

possible ways for semi-conservative replication

Answer 171

1) continuous on both strands 2) semi-discontinuous 3) discontinuous on both strands 4) pieces

Question 172

Okazaki et al. experimental design

Answer 172

1) grow E. coli with 3H-thymidine to label newly synthesized DNA --> stop reactions with cyanide and ice (kills cells) before replication is completed 2) separate DNA by size: ultracentrifugation in alkaline (denatures) sucrose gradient = small at top 3) divide centrifuge tube into fractions, measure radioactivity

Question 173

why does alkaline solution denature DNA?

Answer 173

amino groups deprotonated on bases = inhibits H-bonding

Question 174

Okazaki et al. expected results

Answer 174

1) continuous = long fragments only, growing longer over time 2) discontinuous (both strands) = short fragments growing, long appears later 3) pieces = only short fragments 4) semi-discontinuous = short fragments roughly same size decrease by the end (ligation), long fragments getting longer and longer

Question 175

Okazaki et al. results

Answer 175

1) 2 peaks: short and long 2) both increase in number over time 3) long gets longer, short stays the same = semi-discontinuous model

Question 176

Huberman and Riggs experimental design

Answer 176

1) cells (Chinese hamster ovaries) grown with 5fdU = stops thymine synthesis, all cells primed and ready to go in G1 2) pulse: add 3H thymidine (hot) for a short time 3) chase: change to excess 1H-thymidine (cold) 4) lyse cells, spread DNA on slide 5) coat with photographic emulsion (silver halide crystals) 6) fibre autoradiography 7) develop and view under microscope

Question 177

Huberman and Riggs expected results

Answer 177

1) bidirectional: chase, pulse, chase 2) unidirection: chase, pulse --> only decreases in one direction

Question 178

Huberman and Riggs actual results

Answer 178

1) replication is bidirectional 2) mammalian cells have multiple origins of replication

Question 179

problems DNA polymerase cannot address on its own

Answer 179

1) DNA problem: can't unwind dsDNA, ssDNA template is unstable (reforms dsDNA too quickly or forms hairpins) 2) starting problem: can't start DNA chains de novo (need 3'OH to build off) 3) speed: too slow and distributive 4) lagging strand: no 3' to 5' polymerase activity

Question 180

solving the DNA problem

Answer 180

1) DNA helicase unwinds DNA 2) ssDNA binding proteins (SSBs) prevent re-annealing and protect DNA from damage 3) topoisomerase relieves supercoiling

Question 181

DNA helicase characteristics

Answer 181

- toroid hexamer - homo in prokaryotes, hetero in eukaryotes - binds dsDNA and ssDNA - ATPase - positively charged center

Question 182

DNA helicase mechanism

Answer 182

*not fully understood 1) two helicase bind to DNA, DNA fed 3' into the channel (unwinds 3' to 5') 2) DDK (kinase) phosphorylates both to start 3) ATP cycling causes rotation along the strands in opposite directions 4) leads to unwinding

Question 183

DNA helicase assay

Answer 183

- radiolabelled DNA under various conditions with ATP: 1) boil = positive control = ssDNA only 2) no boil = negative control = dsDNA only 3) increasing amounts of DNA helicase = increasing ssDNA, decreasing dsDNA

Question 184

ssDNA binding protein characteristics

Answer 184

- replication protein A (RPA) - heterotrimer: RPA70, 32, 14 (largest to smallest) - multiple DNA binding domains (3) - bend DNA around protein - prevent reannealing, protect from damage

Question 185

topoisomerase characteristics

Answer 185

- break phosphodiester bond downstream of replisome, relieving supercoiling (torsional strain) - type 1 = single-strand break - type 2 = double-strand break - reanneals later (has nuclease and ligase active sites

Question 186

replication termination

Answer 186

- helicases meet, replication near complete but DNA is tangled - Topo II (type 2 topoisomerase) cleaves to untangle

Question 187

resolving the starting problem

Answer 187

- RNA polymerases don't need 3'OH group (can hold two bases in active site) - two polymerases, four subunit complex 1) DNA-dependent RNA polymerase (primase): slow, adds 11-14 RNA bases 2) DNA-dependent DNA polymerase (Pol alpha): faster, extends by 10-20 bases, NO proofreading - conformational change to switch between the two

Question 188

why does the primase/pol alpha complex not need high-fidelity?

Answer 188

- primer is removed from final DNA product - prokaryotes: DNA pol I - eukaryotes: RNase H and Fen endonuclease

Question 189

solving the speed problem

Answer 189

- sliding clamp keeps DNA pol from sliding off DNA - called proliferating cell nuclear antigen (PCNA) in eukaryotes

Question 190

PCNA characteristics

Answer 190

- homodimer - toroid - no ATPase activity - binds DNA pol epsilon and delta

Question 191

PCNA prokaryote equivalent

Answer 191

- beta clamp

Question 192

PCNA affect on DNA polymerase processivity

Answer 192

- dramatic increase because can stay on DNA for longer and create longer strands 1) high conc. DNA pol with no pol delta auxillary protein or PCNA = small fragments only 2) low conc. DNA polymerase with poly delta aux or PCNA = large fragments

Question 193

loading sliding clamp onto DNA

Answer 193

1) RFC1/RFC5 (replication factor C) N-termini bind PCNA 2) ATP binds to open up PCNA 3) DNA attaches 4) ATP hydrolyzed, ADP released = PCNA separated and loaded onto primer-template junction (ss-dsDNA junction)

Question 194

solving the lagging strand problem

Answer 194

- replisome coordinates replication: 1) topoisomerase 2) primosome: DNA helicase + DNA primase 3) sliding clamps on both strands 4) ssDNA binding proteins - lagging strand wraps around to keep complex close together

Question 195

lagging strand synthesis

Answer 195

1) first round of synthesis 2) unwinding of parental strands, second round starts with new Okazaki fragment at replication fork 3) RNA primers removed 4) gaps filled with repair DNA polymerase 5) Okazaki fragments joined by DNA ligase

Question 196

DNA ligase

Answer 196

- forms phosphodiester bond to fill gaps

Question 197

DNA replication vs PCR

Answer 197

similarities: 1) function: copy DNA template 2) substrates: dNTPs, DNA template differences: 1) polymerase: epsilon/delta vs Taq 2) cofactors: Mg vs Mg + buffer 3) DNA problem: solved by helicase, SSBs, topoisomerase vs heat 4) starting problem: primase/DNA pol alpha vs primers 5) speed: PCNA vs E507K mutation 6) lagging strand: Okazaki fragments vs no lagging strand

Question 198

end replication problem

Answer 198

- no place for the final bases once the primer is removed on 5' end - leaves exposed ssDNA that is susceptible to nucleases

Question 199

solving the end replication problem

Answer 199

- telomerase (RNA-dependent DNA polymerase) adds repeat sequence (contains its own complementary template) - provides space for primase to add another primer

Question 200

norfloxacin mechanism

Answer 200

- quinolone antibiotic that inhibits ligation activity of bacterial topoisomerase (gyrase) - prevents DNA replication - nuclease activity remains = DNA damage = cell death - specific to bacteria (differences in AAs)

Question 201

is DNA very stable?

Answer 201

- wrong assumption - DNA is susceptible to damage: nicks, breaks, base alterations, chemical reactions - repair mechanisms prevent damage from becoming mutations

Question 202

endogenous vs exogenous DNA damage

Answer 202

- endogenous = spontaneous - exogenous = environmental

Question 203

DNA damage vs mutation

Answer 203

- damage = physical alteration to structure of DNA - mutation = change in sequence after replication = inherited by daughter cell, cannot be repaired

Question 204

types of DNA damage

Answer 204

endogenous: 1) depurination / depyrimidination 2) deamination 3) oxidative damage exogenous: 1) ionizing radiation 2) UV radiation 3) alkylation

Question 205

depurination / depyrimidination

Answer 205

- loss of nitrogenous base through hydrolysis of N-glycosidic bond - results in apurinic (more common) / apyrimidinic / abasic (AP) site = loss of coding info for DNA replication or transcription - also risk of DSB

Question 206

DSB from depurination / depyrimidination

Answer 206

1) free OH = potential for linearization of sugar 2) beta-elimination: base attack leads to loss of 3' connection to phosphate 3) no 3'OH to help with repair = DSB

Question 207

translesion (bypass) synthesis polymerase

Answer 207

1) replication or transcription stalls at site of damage (ex. AP site) 2) PCNA ubiquinated 3) recruitment and switch to bypass polymerase (eta), fills gap with random nucleotide 4) PCNA deubiquinated, switch back to replisome to continue process

Question 208

deamination

Answer 208

- exocyclic amine group replaced with C=O - can occur on A, G, C - loss of two H-bond donors, replaced with two H-bond acceptors - ex. C --> U changes base pairing (G-C to A-U)

Question 209

oxidative damage

Answer 209

- from ROS generated in cellular processes: ETC, drug metabolism, inflammation

Question 210

ROS reactions

Answer 210

O2 + electron = O2- (superoxide) O2- + e, 2H+ = H2O2 (hydrogen peroxide) H2O2 + e, H+ = H2O + OH radical OH radical + e, H+ = H2O

Question 211

protective mechanisms against ROS

Answer 211

- enzymes and anti-oxidants protect DNA, protein and lipids from ROS - ex. SOD, catalase

Question 212

oxidative damage example: guanine

Answer 212

- guanine + OH radical = 8-oxo-G - steric clash introduced = syn conformation preferred - base pairs with adenine instead of cytosine - 2nd round of replication = mutation

Question 213

ionizing radiation

Answer 213

- produces ROS (splitting water) - also directly damages DNA bonds

Question 214

UV radiation

Answer 214

- produces pyrimidine dimers - pi bond electrons excited by UV light, can cause stacked pyrimidines to bond

Question 215

pyrimidine stacking effects

Answer 215

1) pulls bases closer together (3.4 to 1.5A) 2) distortion (kink/bend) because of disrupted H-bonding 3) interferes with polymerase activity (binding)

Question 216

types of pyrimidine dimers

Answer 216

1) two bonds = cyclobutane pyrimidine dimer (CPD) 2) one bond = 6-4 photoproduct

Question 217

TLS in pyrimidine stacking

Answer 217

- fills gap to allow replication or transcription to continue - preference for adding A:A dimer = potential mutation

Question 218

alkylation

Answer 218

- transfer of methyl to base - ex. from SAM - ex. methylation of O6 on guanine alters H-bonding, can now base pair with T

Question 219

relative frequencies of DNA damage

Answer 219

1) UV radiation: 100 000 lesions/cell/day 2) depurination: 10 000 3) alkylation: 5 000 4) depyrimidination: 500 5) deamination: 100

Question 220

how does sunscreen protect skin from DNA damage?

Answer 220

- delocalized electrons in organic compounds absorb some energy from UV radiation - released photons are at lower energy

Question 221

DNA repair mechanisms

Answer 221

1) direct repair 2) base excision repair (BER) 3) nucleotide excision repair (NER) 4) mismatch repair 5) post-replication repair 6) DSB repair

Question 222

direct repair

Answer 222

- fixes damaged base with specific enzymes that reverse the damage

Question 223

O6-methylguanine-DNA methyltransferase (MGMT)

Answer 223

1) recognizes kink caused by G methylation 2) interdigitation: flips damaged base out into active site and inserts Arg into DNA helix to stabilize 3) Cys145 attacks methyl = suicide enzyme (degraded)

Question 224

general steps to replacing DNA damage

Answer 224

1) damaged DNA recognized by protein 2) enzymes recruited to break phosphodiester bond to remove damaged area 3) DNA polymerase rebuilds removed area 4) DNA ligase re-seals phosphodiester bonds

Question 225

base excision repair

Answer 225

remove and replace damaged nitrogenous base: 1) proteins recognize changes in DNA bases 2) enzymes recruited: glycosylase breaks N-glycosidic linkage to generate an AP site (specific for each type of DNA damage) 3) endonuclease nicks backbone 4) dRP lyase removes deoxyribose 4) DNA polymerase replaces nucleotide 5) DNA ligase seals phosphodiester bond

Question 226

nucleotide excision repair

Answer 226

remove and replace damaged nucleotide 1) XPC-HR23B recognizes changes in helix structure 2) TFIIH (XPB/XPD) is recruited to unwind DNA: XPC-HR23B interdigitation enhances kink severity to attract 3) RPA protects ssDNA (TFIIH bound to damaged strand only) 4) endonucleases (3' XPG/XPA and 5' XPF) cleave phosphodiester bond, removing larger stretch of nucleotides 5) replisome fills gap 6) DNA ligase seals phosphodiester backbone

Question 227

DNA damage and associated repair mechanism

Answer 227

1) direct repair = alkylation 2) BER = depurination / depyrimidination, deamination, oxidative damage 3) NER = UV radiation

Question 228

mismatch repair

Answer 228

- fixes DNA polymerases mistakes (mismatches and small insertions/deletions) - recognizes newly synthesized strand because of lack of methylation (epigenetic marks copied after complete replication) --> demonstrated in prokaryotic cells, only PREDICTED for eukaryotic cells

Question 229

mismatch repair steps

Answer 229

1) unwind DNA 2) endonucleases cleave phosphodiester bond 3) replisome replaces nucleotides 4) DNA ligase seals phosphodiester backbone

Question 230

xeroderma pigmentosum (XP)

Answer 230

- inherited condition (mutations in XPC and POLH (DNA pol eta)) - extreme sensitivity to UV radiation and risk of UV-induced cancers - XPC mutations = NER inhibited - POLH mutations = TLS inhibited = more damage, apoptosis

Question 231

single-strand breaks

Answer 231

- phosphodiester bond breaks on one side of the DNA - leads to DNA damage response

Question 232

causes of SSBs

Answer 232

1) ionizing radiation: ROS (sugar oxidation) and direct DNA damage 2) inhibition of DNA ligase = unligated Okazaki fragments 3) inhibition of topoisomerase 1 ligation activity 4) stochastic base damage, cytosine demethylation/alkylation, guanine deamination = BER (if fails, leads to SSB) 5) depurination = AP site = beta-elimination (or BER)

Question 233

DNA damage response for SSBs proteins involved

Answer 233

1) lesion identification: PARP1, XRCC1 2) repair of DNA ends: PNKP 3) replace DNA bases: a) short-patch = pol beta b) long-patch = pol beta, then epsilon/delta add more, then FEN1 endonuclease removes primers 4) ligation: DNA ligase

Question 234

indirect SSB (BER) DNA damage response

Answer 234

1) detection: APE1/lyase 2) end processing: XRCC1 scaffolds for PNKP and other proteins 3) short-patch filling 4) ligation

Question 235

direct SSB (ex. sugar damage) DNA damage response

Answer 235

1) detection: PARP1/PARG 2) end processing: XRCC1 scaffolds for PNKP and other proteins 3) long-patch filling with PCNA 4) ligation

Question 236

TOP1-SSB DNA damage response

Answer 236

1) detection: RNAP and other lesions 2) end processing: XRCC1 scaffolds for PNKP and other proteins 3) long-patch filling with PCNA 4) ligation

Question 237

double-stranded breaks

Answer 237

- break of phosphodiester bond on both sides of DNA (can be blunt or sticky end) - more serious than SSB, requiring immediate action - if not repaired, apoptosis initiated

Question 238

causes of DSBs

Answer 238

1) ionizing radiation 2) ROS 3) type II topoisomerases 4) meiosis: recombination events 5) SSBs during replication: if replication fork encounters SSB, lack of non-covalent interactions can cause replication arm to separate from the fork

Question 239

DSB repair pathways

Answer 239

1) if resection = homology-directed repair (HDR) 2) if no resection = non-homologous end-joining (NHEJ)

Question 240

which DSB repair pathway dominates?

Answer 240

- competitive binding of protein complexes - if MRN binds first, resection occurs = HDR - if Ku binds first, resection blocked = NHEJ

Question 241

NHEJ characteristics

Answer 241

- quick and messy - error-prone - all phases of cell cycle

Question 242

HDR characteristics

Answer 242

- finding match - fewer errors - S and G2 phases

Question 243

HDR types

Answer 243

1) homology between tails = single strand annealing 2) no homology between tails = homologous recombination

Question 244

resection

Answer 244

- removal of some bases around the break to create 3' overhang

Question 245

NHEJ vs HDR protein complexes

Answer 245

- NHEJ = Ku70/80 heterodimer - HDR = MRN complex

Question 246

NHEJ steps

Answer 246

1) Ku recognizes DSB, acts as scaffold 2) repair proteins recruited 3) ends of chromosome brought together 4) any homology used to reconnect DNA 5) IF NEEDED: gaps filled or excess DNA is removed (pol beta, gamma, mu) 6) backbone ligated together

Question 247

possible outcomes of NHEJ

Answer 247

1) perfect synapsis = no sequence change 2) incorrect microhomology = addition or deletion (indels)

Question 248

HDR (homologous recombination) steps

Answer 248

1) MRN (scaffold) recognizes DSB 2) CtIP recruited, activated by phosphorylation (ATM) and ubiquitylation (BRCA1) 3) MRN/CtIP begins resection: CtIP 5' to 3' endonuclease removes ~100nt 4) Exo1 continues: more efficient (~4kb/h), produces large ssDNA overhangs 5) RPA binds overhangs 6) BRCA2 replaces RPA with RAD51 and paralogs 7) strand invasion: RAD51 coated ssDNA invades dsDNA, searching for homologous region 8) D-loop formed 9) RAD54 removes RAD51, allows DNA polymerase to bind and replicate DNA from template 10) second strand invasion = Double Holiday Junction 11) nucleases resolve junction

Question 249

HDR template options

Answer 249

1) sister chromatid: ideal, because exact same sequence 2) homologous chromosome: could be different allele 3) retrotransposon (homologous sequences): detrimental effects ex. chromosomal translocation and rearrangement 4) artificially introduced repair templates

Question 250

BRCA2 evidence study components

Answer 250

1) long dsDNA with 3' overhang coated with RPA 2) radiolabeled short dsDNA repair template 3) RAD51: same conc. in all samples 4) BRCA2: various conc.

Question 251

BRCA2 evidence study methods

Answer 251

gel electrophoresis: 1) long fragment visible = transfer or repair template = strand invasion occurred 2) short fragment visible = no reaction

Question 252

BRCA2 evidence study results

Answer 252

1) negative control: RPA only = only template 2) positive control: RAD51 only = template and product 3) RPA + RAD51 +increasing BRCA2 conc. = increasing product

Question 253

why are BRCA1/2 mutation associated with increased cancer risk?

Answer 253

- HDR cannot be initiated - DSBs repaired using NHEJ, which is more error-prone

Question 254

sickle cell anemia

Answer 254

- single amino acid change (Glu6Val) in beta globin gene - hemoglobin cannot form tetramers, instead form rod-like aggregates that deform RBCs and occlude blood flow - can be treated with gene therapies

Question 255

genome editing

Answer 255

- alteration of genetic material of living organism by insertion, replacement or deletion of DNA sequence - typical aim: improve some characteristic of crop/farm animal or correct genetic disorder

Question 256

genome editing exampels

Answer 256

1) site-directed mutagenesis 2) restriction cloning 3) random mutagenesis 4) recombinant DNA cloning

Question 257

recombinant DNA cloning mechanism

Answer 257

1) artificially introduced recombinant DNA molecule triggers DSB repair pathway 2) molecule is homologous to gene of interest, leading to gene replacement

Question 258

recombinant DNA cloning efficiency

Answer 258

- low efficiency - worked best in organisms that prefer HDR over NHEJ, ex. yeast

Question 259

how to improve genome engineering efficiency

Answer 259

- introduce DSB in genome as well (in target gene) - end resection occurs on both genome and artificial DNA, allowing for single-strand anealing

Question 260

single-strand annealing

Answer 260

- short-cut for HDR - homologous tails base pair, flaps are cleaved, DNA ligated

Question 261

which proteins can be used to create DSBs?

Answer 261

1) topoisomerase II 2) restriction enzymes 3) nucleases 4) CRISPR-Cas

Question 262

limitations of topoisomerase, restriction enzymes and engineered nucleases in genome editing

Answer 262

- topoisomerases: cannot target specific sequences - REs: limited by specific target sequence - engineered nucleases: require extensive protein engineering

Question 263

types of engineered nucleases

Answer 263

1) zinc-finger nucleases 2) transcription activator-like effector nuclease (TALENs)

Question 264

ZFN vs TALEN

Answer 264

- both have nuclease attached to engineered proteins that recognize DNA sequences - ZFN: each domain recognizes 3nt - TALEN: each domain recognizes 1nt

Question 265

what does CRISPR do in bacteria?

Answer 265

1) invasive viral nucleic acids cleaved and incorporated into CRISPR site of bacterial genome 2) expressed as RNA 3) forms complex with endonuclease 4) complex searches for matching viral sequences 5) breaks them = protects against viral infection

Question 266

spCas9

Answer 266

Cas9 from streptococcus pyogenes

Question 267

spCas9 domains

Answer 267

- 2 nuclease domains: RuvC and HNH - C-terminal domain = PAM-recognition site - recognition (REC) lobe = sgRNA binding site - Arg-rich helix = DNA binding

Question 268

spCas9 conformational changes

Answer 268

- apo state: PAM site disordered, nuclease domains collapsed on each other - sgRNA bound: central channel opens, accommodating dsDNA

Question 269

Cas9 searching

Answer 269

- loaded with guide RNA - searches dsDNA for PAM site: protein-DNA interaction - if no site, Cas falls off

Question 270

spCas9 PAM site

Answer 270

5'-NGG-3'

Question 271

Cas9 mechanism

Answer 271

1) dsDNA unwound, stabilized by Arg-rich helix 2) interaction allowed with sgRNA: if enough complementary sequence is found, another conformational change occurs 3) nuclease domains brought into position 4) cleave on both sides to make blunt end: HNH on target strand, RuvC on non-target

Question 272

Cas9 off-target effects

Answer 272

- some mismatch permitted

Question 273

what part of sgRNA has greatest impact on binding?

Answer 273

sgRNA-DNA base-pairing nearest to PAM site

Question 274

what happens after CRISPR-Cas9 cleavage?

Answer 274

- DSB repair or cell death - NHEJ: can knock-out gene, silent mutation or no mutation - HDR: can provide repair template for targeted mutation SSA

Question 275

advantages of CRISPR/Cas9

Answer 275

1) targeted 2) sgRNA is cheap and highly efficient (compared to proteins) 3) Cas9 can be modified and/or be a transport protein (dCas9 = no nuclease activity)

Question 276

what proteins can Cas9 transport?

Answer 276

1) nickase: causes SSB (one Cas9 nuclease site mutated) 2) transcriptional regulators 3) chromatin modification: epigenetic modulators 4) tags: ex. GFP 5) base-modifiers

Question 277

applications of CRISPR

Answer 277

1) research: KO models, add tags/domains/modify proteins, change gene expression 2) biotechnology: agriculture and manufacturing 3) healthcare: correct gene mutations, treat disease

Question 278

gene-editing to treat sickle cell anemia

Answer 278

1) screen 2) collect patient stem cell (low risk of rejection) 3) gene-edit, test for function - turns off expression of BCL11A to allow gamma-globin expression to increase = fetal hemoglobin produced 4) infuse cells

Question 279

ethical concerns around genome editing

Answer 279

1) somatic (adult cells) vs germline (embryos) editing 2) preventing/treating disease vs enhancement 3) risk/benefit profile: availability of reasonable alternatives 4) supporting clinical or pre-clinical data 5) independent ethics reviews

Question 280

CCR5

Answer 280

receptor that HIV binds to get into T cells, may have other important functions that are not completely understood

Question 281

CCR5delta32

Answer 281

- leads to frameshift + premature stop codon - homozygous individuals for this mutation are resistant to some strains of HIV - HIV can still infect by interacting with other receptors

Question 282

He experiment (CRISPR babies)

Answer 282

1) CRISPR engineered mutations in CCR5 gene in embryos (mother HIV negative, father HIV positive) 2) no repair template = NHEJ 3) Nana: homozygote with different mutations (1bp insertion, 4bp deletion) 4) Lulu: WT + 15bp deletion (heterozygote) 5) both displayed mosaicism after birth because editing only occurred in some cells of embryo 6) no apparent off-target mutations

Question 283

ethical concerns of He experiment

Answer 283

1) other more effective ways to prevent HIV infection, ex. IVF: remove HIV from sperm first 2) no repair template = cannot guarantee mutation provides resistance 3) debatable informed consent

Question 284

central dogma of molecular biology

Answer 284

DNA --> RNA --> protein

Question 285

prokaryotic vs eukaryotic RNA polymerase similarieties

Answer 285

- overall structure quite similar - 3 stage synthesis: initiation, elongation, termination

Question 286

RNA polymerase functions

Answer 286

1) recognize initiation sites and promoters 2) helicase activity: unwinds dsDNA 3) correct ribonucleotide triphosphate selection: unidirectional/processive 4) termination 5) activation/repression: transcription factors 6) fundamental reaction: 3' OH attacks alpha P to create phosphodiester bond

Question 287

cis-acting elements

Answer 287

cis = on template strand being transcribed

Question 288

RNA synthesis direction

Answer 288

5' to 3', DNA is read 3' to 5'

Question 289

RNA polymerase vs DNA polymerase

Answer 289

1) no primer required 2) no proofreading capability (1 in 10^5 error rate vs 10^10): acceptable because of codon degeneracy 3) slower (50nt/s vs 800)

Question 290

E. Coli RNAP structure

Answer 290

- 400kDA - 4 subunits - holoenzyme is two alpha, one beta, one beta prime, one sigma (pentamer) - alpha2betabeta = core enzyme, contains catalytic site

Question 291

RNAP sigma subunit function

Answer 291

- decreases DNA binding affinity - released after 10nts synthesized, goes to bind another core enzyme

Question 292

RNAP active site

Answer 292

- similar to DNAP - Asp residues interact with DNA and Mg2+ - another Mg2+ cofactor interacts with ribonucleoside triphosphate gamma and beta phosphate = good LG

Question 293

footprinting

Answer 293

1) PCR amplification with radiolabeled DNA 2) add DNase I: cleaves randomly 3) denature dsDNA 4) gel electrophoresis 5) compare DNA/protein complex to naked DNA: gaps show where protein binds (DNase cannot access) 6) can use single base analysis to find sequence

Question 294

identification of prokaryotic promoter sites

Answer 294

- footprinting - comparisons between different genes reveals similar initiation sites

Question 295

prokaryotic promoter sites

Answer 295

core promoter = 40nts, contains: 1) -35 region 2) -10 TATA (pribnow) box

Question 296

consensus sequence

Answer 296

- average of different sequences of genes - number below shows percent frequency

Question 297

gene nucleotide numbering

Answer 297

+1 = initiation site negative = upstream positive = downstream

Question 298

RNAP interaction with promoter

Answer 298

- recognition helix in sigma subunit makes transient bonds with T and A - residues: Tyr, Trp, Thr, Gln, Arg

Question 299

non-template strand

Answer 299

coding or sense strand

Question 300

template strand

Answer 300

non-coding or antisense strend

Question 301

RNA transcript resembles...

Answer 301

non-template strand with U instead of T

Question 302

strong vs weak promoters

Answer 302

- strong = frequently transcribed (housekeeping genes), 17nt between -10/-35 is ideal - weak = multiple substitutions in -10/-35 regions, less frequently transcribed

Question 303

factors that impact promoter strength

Answer 303

1) regulatory proteins: bind near or to promoter regions 2) UP elements

Question 304

UP (upstream) elements

Answer 304

- bound by alpha subunit C-terminus - increases transcription efficiency - not highly conserved - typically A/T rich - 40-60 nts upstream

Question 305

alternative promoter sequences in E. coli

Answer 305

- bound by different variants of sigma subunit, ex: 1) standard = 70 2) heat-shock = 32 3) nitrogen-starvation = 54 - bacteria upregulates synthesis of different subunits in response to environmental changes

Question 306

prokaryotic transcription initiation mechanism

Answer 306

1) holoenzyme (sigma recognition helix) forms transient H-bonds with base pairs: rapid (10^10 M^-1 s^-1) and random search 2) recognition of promoter = closed promoter complex (STILL REVERSIBLE) 3) unwinding of DNA to form open promoter complex (IRREVERSIBLE) 4) RNA polymerase starts transcription, usually by adding a purine 5) sigma subunit falls off after 2-10 nts synthesized 6) core enzyme remains to continue to elongation

Question 307

open promoter complex characteristics

Answer 307

- 17bp segment = 1.6 DNA turns - lower G-C content is easier to unwind

Question 308

prokaryotic transcription elongation mechanism

Answer 308

1) formation of transcription bubble 2) nucleotide addition cycle: events required for addition of a nucleotide to the RNA product to form a cyclic process

Question 309

prokarytotic transcription bubble

Answer 309

- RNA polymerase unwinds DNA at the front, ssDNA enters active site - rewinds DNA at the back - RNA-DNA helix formed as RNA synthesis occurs with 3' end of RNA in active site - Phe in exit channel inserts into RNA-DNA helix to separate them, promoting exit of RNA strand

Question 310

RNA-DNA helix characteristics

Answer 310

- ~8bps long = 1 double-helix turn - moves 170 angstroms/sec = 50nt/sec - rotates because of re/unwinding

Question 311

nucleotide addition cycle important structures

Answer 311

- bridge helix - trigger loop - different conformations facilitate translocation and NAC

Question 312

pre-insertion site in RNAP

Answer 312

- inactive site - revealed by streptolydigin: inhibits transcription by sitting in insertion site, but RNAP could still bind NTPs

Question 313

NAC steps

Answer 313

catalysis: 1) open trigger loop accepts NTP into pre-insertion site 2) closure of trigger loop moves NTP into insertion site 3) catalytic incorporation releases pyrophosphate translocation equilibrium: 4) trigger loop opens 5) equilibrium formed between open with pre-translocated DNA, wedged with intermediate and open with post-translated DNA states 6) latter conformation allows another NTP to be accepted

Question 314

general steps of transcription termination

Answer 314

1) ceasing of phosphodiester bond formation 2) RNA/DNA hybrid dissociation 3) rewinding of dissociated DNA strands 4) release of DNA by RNAP

Question 315

two methods of termination in prokaryotes

Answer 315

1) intrinsic 2) protein-dependent

Question 316

intrinsic termination signal in prokaryotes

Answer 316

- palindromic GC region separated by ~4A (forms U loop) - followed by 4+ U (coded for by DNA) - forms stem loop structure = RNAP stalls - A/U rich area is weaker, leads to RNAP falling off

Question 317

protein-dependent termination signal in prokaryotes

Answer 317

- Rho factor recognizes Rho termination sites - used in rRNA synthesis: DNA template contains different Rho sites for 10S, 13S, 17S rRNA - experiment: depending on timing of Rho addition = different subunits

Question 318

how does Rho cause termination?

Answer 318

- binds to C-rich site: Rho utilization or rut site on synthesized RNA - catches up to RNAP, causes unwinding of RNA/DNA helix = stalling + dissociation

Question 319

Rho characteristics

Answer 319

- ATP-dependent helicase - hexamer

Question 320

drugs targeting transcription

Answer 320

1) rifampicin 2) actinomycin 3) amanitin

Question 321

rifampicin action

Answer 321

- binds RNA/DNA helix channel in prokaryotes (not eukaryotes) - stops initiation - no effect once elongation starts

Question 322

actinomycin action

Answer 322

- intercalates with DNA - DNA is no longer an effective template - also affects DNA polymerase (replication)

Question 323

steady-state level vs synthetic capacity of RNA

Answer 323

- synthetic capacity: cell's ability to synthesize - usually correlated, except in the case of mRNA: immediately processed and transcribed

Question 324

prokaryotic post-transcriptional modification of rRNA and tRNA

Answer 324

- usually on one transcript - undergoes cleaving, processing (ex. CCA to tRNA), and modification (to bases) - other mRNA typically undergoes little to no modification after synthesis

Question 325

rRNA/tRNA mRNA cleavage

Answer 325

1) RNase III: cuts out rRNA 2) M16, 5, 23 (specific varieties): trims rRNA 3) RNase P: 5' end of tRNA 4) RNase D: 3' end of tRNA

Question 326

mRNA base modification examples in prokaryotes

Answer 326

- ribothymidylate, pseudouridylate - methylation ex. 6-dimethyladenine

Question 327

prokaryotic vs eukaryotic transcription

Answer 327

1) spatial-temporal regulation: translation co-transcriptionally vs. must be exported out of nucleus (after fully processed) 2) processing: minimal vs extensive 3) number of RNAP: 1 vs 3 4) RNA subunits: 5 vs 7-10

Question 328

characteristics of eukaryotic RNAP

Answer 328

1) all large proteins with 8-15 subunits 2) RNAP II has unique CTD with repeats of YSPTSPS 3) phosphorylation of CTD (S/T) regulates activity 4) different responses to alpha amanitin

Question 329

eukaryotic RNAP I

Answer 329

1) location: nucleolus 2) cellular transcripts: 18S, 5.8S, 28S rRNA (one copy / primary transcript) 3) alpha amanitin effects: insensitive 4) promoters: A-rich upstream promoter element (UPE), ribosomal initiation element

Question 330

eukaryotic RNAP II

Answer 330

1) location: nucleoplasm 2) cellular transcripts: mRNA precursors and snRNA (splicing) 3) alpha amanitin effects: strong inhibition, Kd = 10nm 4) promoters: TATA box, initiator element, downstream promoter element

Question 331

eukaryotic RNAPIII

Answer 331

1) location: nucleoplasm 2) cellular transcripts: tRNA, 5S rRNA 3) alpha amanitin effects: inhibited by high conc., Kd = 1um 4) promoters: downstream promoters

Question 332

important RNAPII subunits

Answer 332

- RBP1 contains CTD, homolog to beta prime subunit - RPB4: promoter recognition

Question 333

alpha-amanitin mechanism

Answer 333

- inhibits open to closed trigger loop change: ribonucleotide cannot move to insertion site - inhibits wedged loop to open: RNAP cannot accept ribonucleotide

Question 334

RNAP I promoter

Answer 334

core promoter: 1) -200 to -150: UPE 2) +1 ribosomal initiation element

Question 335

RNAP II promoter

Answer 335

1) enhancer at least -1kB core promoter: 2) upstream TATA box -100 to -20 OR DPE around +30 3) ribosomal initiation element +1

Question 336

RNAP III promoter

Answer 336

type 1: 5srRNA 1) downstream A and C block type 2: tRNA 1) downstream A and B block *A/B/C all ~11bp

Question 337

TATA box mutations

Answer 337

- single base markedly impairs promoter activity

Question 338

upstream enhancers in RNAPII promoter

Answer 338

1) CAAT box 2) GC box - usually for housekeeping genes - increase transcriptional frequency - bound by RPE3

Question 339

DNA looping

Answer 339

- binding of transcription activators to enhancer or silencer - interaction with RNAP facilitated by DNA looping (to bring elements close together)

Question 340

eukaryotic transcription initiation

Answer 340

1) TFIID (multi-subunit complex) recognizes TATA box via TATA box binding protein (TBP) 2) TFIIA stabilizes complex 3) TFIIF recruits RNAPII 4) TFIIB, TFIIE, TFIIH also recruited = basal transcription apparatus (BTA) aka pre-initiation complex 5) TFIIH (helicase) unwinds DNA and phosphorylates CTD 6) all except TFIIF dissociate 7) elongation begins: NAC

Question 341

TFIID subunits

Answer 341

1) TBP 2) TAFs: recognize non-TATA elements, histone acetyltransferase activity

Question 342

TFIIB function

Answer 342

recruits RNAPII and TFIIF, helps start-site selection

Question 343

TFIIF function

Answer 343

- promoter targeting of RNAPII - destabilizes nonspecific RNAPII-DNA interactions

Question 344

TFIIE function

Answer 344

- recruited by RNAPII - recruits and modulates TFIIH helicase, ATPase, kinase activities

Question 345

TFIIH

Answer 345

- helicase activity - CTD phosphorylation via 2 cyclin:CDK pair subunits

Question 346

negatively regulated gene

Answer 346

- transcription prevented by a repressor - transcribed in absence of active repressor

Question 347

positively regulated gene

Answer 347

- transcription occurs in presence of activator - not transcribed in absence

Question 348

repressors and activators characteristics

Answer 348

- allosteric proteins - bind ligands

Question 349

positive transcriptional regulation methods

Answer 349

1) molecular signal causes binding of activator = transcription 2) molecular signal causes dissociation of activator = no transcription

Question 350

negative transcriptional regulation methods

Answer 350

1) molecular signal causes dissociation of regulatory protein = transcription 2) molecular signal causes binding of regular protein = no transcription

Question 351

operons

Answer 351

linearly organized: regulator gene: 1) promoter for regulator gene 2) regulator gene control sites: 3) promoter 4) operator (surrounding operon) structural genes: 5) ex. lactose operon: z, y, a

Question 352

polycistronic

Answer 352

mRNA that encodes multiple proteins

Question 353

lac operon components and functions

Answer 353

1) z = beta-galactosidase: lactose --> allolactose 2) y = permease: transports lactose in 3) a = transacetylase: acetylates lactose to be exported out

Question 354

lactose vs allolactose

Answer 354

1-4 beta-glycosidic linkage vs 1-6

Question 355

lac operon two control mechanisms

Answer 355

1) regulatory response to lactose 2) regulatory response to glucose

Question 356

lac operon lactose regulation

Answer 356

1) lac repressor prevents transcription 2) allolactose inactivates repressor and allows transcription

Question 357

lac repressor mechanism

Answer 357

1) binds DNA (operator regions) as dimer 2) dimers from both operator regions forms tetramer = DNA looping 3) blocks RNAP

Question 358

lac repressor transient binding

Answer 358

- transient binding ensures transcription occurs at least once = basal levels of lac operon genes

Question 359

lac repressor/DNA interactions

Answer 359

- lac repressor alpha helix binds major groove - H-bonding with AAs ex. Arg

Question 360

lac repressor bound vs unbound

Answer 360

- lactose = unbound by allolactose = bound to DNA = very ordered + lactose = bound by allolactose = unbound to DNA = more disordered (conformational change)

Question 361

results of lactose lac operon study

Answer 361

- adding lactose = beta-galactosidase and total bacterial protein (indicating growth) have direct relationship - removing lactose = plateau

Question 362

accessory proteins for promoters

Answer 362

some require additional accessory proteins to speed up transcription, ex, catabolite activator protein (CAP) for lac operon

Question 363

CAP characteristics

Answer 363

- dimer of 22.5 kD peptides - N-terminus binds cAMP - C-terminus binds DNA

Question 364

lac operon glucose regulation

Answer 364

1) increase in glucose = decrease in cAMP and vice versa 2) if glucose decreases, cAMP increase and binds CAP 3) CAP can bind to DNA: upstream of RNAP binding site (-41, -61 or -71 bp) 4) assists formation of closed promoter complex

Question 365

purpose of catabolite repression

Answer 365

- ensures that operons for metabolism of alternative energy sources are repressed until glucose exhausted

Question 366

components of eukaryotic transcription machinery

Answer 366

1) activators 2) repressors 3) coactivators 4) basal factors

Question 367

common DNA sequences bound by proteins

Answer 367

- two-fold axis of symmetry - palindromic - dimer binding - Glu/Asn often AT - Arg often CG

Question 368

DNA-binding motifs

Answer 368

80% DNA-binding proteins have one of: 1) helix-turn-helix motif 2) zinc-finger 3) leucine zipper-basic region (bZIP)

Question 369

helix-turn helix

Answer 369

- recognition helix binds major groove

Question 370

zinc fingers

Answer 370

- 10x small domains with Zn coordination to 4 Cys or 2Cys/2His - key helix that binds DNA

Question 371

leucine zippers

Answer 371

1) DNA binding region 2) 6-AA connector 3) leucine zipper: coiled coils held together by hydrophobic interactions between leucines (every 7th residue)

Question 372

estrogen: example of hormonal control of transcription in humans

Answer 372

- cholesterol-derived hormone required for development and ovarian cycle - diffuses across cell membrane - ligand for DNA-binding protein (estrogen receptor)

Question 373

estrogen receptor

Answer 373

- contain Zn fingers - bind at estrogen-receptor elements - three domains: transcription activation (variable), DNA binding (conserved, 66-68 residues), hormone binding (variable

Question 374

estrogen receptor conformational change

Answer 374

- binding of estrogen causes helix 12 to fold up into side of receptor - does NOT alter DNA binding affinity of receptor - increases affinity of COACTIVATOR: P160 family = chromatin remodelling = increases transcription

Question 375

drugs targeting estrogen receptor

Answer 375

- ex. tamoxifen - binds in ligand-binding pocket - prevents proper folding of helix 12, coactivator cannot bind - ER antagonist to treat ER+ breast cancers

Question 376

types of cellular RNA

Answer 376

1) mRNA: template for protein synthesis 2) tRNA: carries AAs to site of protein synthesis on ribosome 3) rRNA: part of ribosome 4) other: components of ribonucleoproteins with variety of functions

Question 377

rRNA splicing

Answer 377

1) RNAPI produces pre-RNA with 3 ribosomal RNAs: 18S, 5.8S, 28S, separated by spacer regions 2) nucleotide modification: methyl groups and pseudouridines 3) spacer regions cleaved out

Question 378

rRNA splicing machinery

Answer 378

- rDNA with many branches of pre-rRNA - SSU (small subunit) processome (aka large ribonucleoprotein): at the ends for processing premRNA

Question 379

tRNA splicing

Answer 379

1) RNAPIII produces tRNAs 2) 5' cleavage by RNAase P 3) 3' cleavage by RNase D 4) nucleotide modification 5) tRNA nucleotidyl transferase adds CCA to 3' end, releasing 3 PPi = AA attachment site 6) eukaryotes only: intron splicing near anticodon

Question 380

eukaryotic mRNA delivery to ribosomes

Answer 380

- must be processed into mature mRNAs - must be transported from nucleus into cytosol

Question 381

eukaryotic vs prokaryotic mRNAs

Answer 381

- eukaryotic are monocistronic

Question 382

mRNA capping

Answer 382

- 7-methyl-G capped added to increase transcript stability and define start site for protein synthesis - occurs co-transcriptionally

Question 383

mRNA capping steps

Answer 383

1) RNAP II produces mRNA 2) GTP added by guanylyl transferase: a) hydrolysis of terminal phosphate on 5' end of transcript b) diphosphate attacks alpha phosphate of GTP, PPi released = 5'-5' triphosphate linkage 3) methylation by RNA methyltransferase using S-adenosyl methionine: a) cap O: guanine methylated at position 7 b) cap 1: 2'OH of sugar c) cap 2: 2'OH of sugar

Question 384

mRNA 3' polyadenylation steps

Answer 384

1) RNAP transcribes past consensus AAUAAA sequence (polyA addition site) 2) 10-35 nts past, RNAP stalls, resulting in cleavage and polyadenylation specificity factor (CPSF) recruitment 3) CPSF binds consensus sequence and invariant GU in mRNA = looping 4) cleavage factors (CFs) cleave downstream of consensus sequence 5) CFs and 3' fragment dissociate 6) poly(A)-adenylation protein (PAP) recruited to add 100-200 As 7) CPSF dissociates

Question 385

timing of polyadenylation

Answer 385

closely linked to transcription termination

Question 386

RNA editing

Answer 386

- change in nt sequence not as a result of splicing - deamination of C to U or A to I, changing coding possibilities in transcript - introduces diversity from same gene transcript

Question 387

apolipoprotein

Answer 387

fatty acid and steroid transport protein

Question 388

apolipoprotein RNA editing

Answer 388

1) in liver: unedited mRNA translated to ApoB-100 = LDL, VLDL surface 2) in small intestine: enzymatic deamination of CAA to UAA = early stop codon = no LDL receptor binding part = chylomicron surface

Question 389

methods of increasing protein diversity by RNA editing

Answer 389

1) altering amino acid coding possibilities 2) introducing premature stop 3) changing splice site in transcript

Question 390

organization of split eukaryotic genes

Answer 390

DNA: 1) promoter/enhancers 2) start of mRNA 3) exons and introns 4) poly-A addition signal RNA: 1) 5' UTR 2) exons and introns 3) 3' UTR mature mRNA 1) 7-mG cap 2) exons: as little as ~1/3 of DNA coding region! 3) 3'UTR with poly-A tail

Question 391

alternative splicing

Answer 391

- more diversity of EXONs - 2^n possibilities, where n is the number of exons that can be alternatively spliced

Question 392

RNA sequences in intron for splicing

Answer 392

1) 5' splice site: invariant GU 2) branch site: key A residue 3) pyrimidine tract: ~10nts 4) 3' splice site: invariant AG

Question 393

transesterification reaction

Answer 393

- exchange of ester groups - requires no energy

Question 394

two transesterification reactions in splicing

Answer 394

1) 2'OH on key A in branch site attacks phosphate of 5' splice site = precursor to lariat intermediate 2) 3'OH of exon attacks phosphate of exon 2 = spliced product + lariat form of intron

Question 395

splicing branch point: lariat form of intron

Answer 395

adenine has 3 phosphodiester linkages: 1) 5' to 3' 2) 3' to 5' 4) 2' to 5' of guanine

Question 396

snRNP

Answer 396

- small nuclear ribonucleoprotein particles - required for splicing - small RNA (100-200 bases) + ~10 proteins (some general and specific) - regions that bind RNA are protein-deficient to facilitate base-pairing - major snRNP species are abundant, >100 000 per nucleus

Question 397

key snRNP in pre-mRNA splicing

Answer 397

1) U1: 5' splice site 2) U2: branch site 3) preformed complex of U4-6: 5' splice, recruitment of branch point to 5'splice site

Question 398

splicing steps

Answer 398

1) U1 recognizes 5' splice site 2) ATP hydrolysis for U2 binding to branch site 3) ATP hydrolysis for U4-6 complex formation + ATP for binding 4) once U5 aligned at 5' splice site, ATP for U4 dissociation unmasks U6 activity = U2/U6 catalytic site forms across pre mRNA 5) U5 uses ATP to align 2’OH of the A branch site and 5’ intron splice site for first transesterification reaction 6) U5 uses ATP to align3’OH of exon 1 with the 3’ intron splice site for second transesterification 7) ATP releases U5/6/2, another to dissociate U2/6

Question 399

spliceosome catalytic center

Answer 399

- U6/2 base pair - U2 base pairs with branch site = key A residue protrudes - dissociated by ATP dependent helicase

Question 400

CTD recruits which proteins to pre-mRNA

Answer 400

1) capping enzymes 2) splicing components 3) endonuclease

Question 401

how does CTD recruit proteins?

Answer 401

phosphorylation

Question 402

CTD coupling transcription to premRNA processing mechanism

Answer 402

1) phosphorylated CTD has capping enzymes, splicing factors and polyadenylation factors (starting from end inwards) 2) capping 3) spliceosome recruited, splicing factors splice 4) poly(A) added

Question 403

issue of complex RNA processing

Answer 403

difficult to trace mature mRNA back to original gene sequence

Question 404

result of mutations that affect pre-mRNA splicing

Answer 404

50% normal RNA 50% abnormal RNA --> degraded = disease ex. beta globin subunit in thalassemia

Question 405

thalassemia splicing

Answer 405

1) A-->G mutation in an intron, creating new 5' splice site 2) 50/50 chance to assemble on mutated or original splice site 3) if mutated: in-frame stop codon retained, leads to truncated protein

Question 406

tissue-specific splicing and processing

Answer 406

- same gene with multiple poly(A) sites - recognition of site and alternative splicing due to different regulatory factors in tissues - ex. calcitonin/CGRP

Question 407

calcitonin vs CGRP processing

Answer 407

in thyroid: - 1st poly(A) site recognized - CGRP removed - splicing: calcitonin exon kept in brain: - 2nd site recognized - CGRP gene kept - splicing: calcitonin removed

Question 408

self-splicing introns vs spliceosome

Answer 408

- spliceosome more intricately regulated: influenced by snRNP availability, regulated by CTD of RNAP - self-splicing introns do not require proteins

Question 409

types of self-splicing introns

Answer 409

Group 1: 1) GMP/DP/TP cofactor required 2) 3'OH attacks exon 1 5' splice site 3) 3'OH of exon 1 attacks exon 2 (5' end) 4) NO lariat formed Group 2: 1) key A (part of transcript) has 2'OH that attacks exon 1 5' splice site 2) 3'OH of exon 1 attacks exon 2 (5' end) 3) lariat formed

Question 410

ribozyme classes

Answer 410

1) Group 1 introns: eukaryotic genomes (rRNA, tRNA) 2) Group 2 introns: mitochondrial genomes, lower eukaryotes 3) viral RNAs: a) hammerhead ribozyme: plant viruses b) hairpin ribozyme: tobacco plant virus c) hepatitis delta virus ribozyme 4) RNase P 5) spliceosome 6) ribosome

Question 411

Group 1 intron mechanism

Answer 411

1) internal guide sequence (IGS) base pairs with exon 1 (3' region of upstream exon) 2) folding allows binding of G cofactor, aligned so that 3'OH attacks for first TE 3) 2nd TE, spliced RNA and LINEAR intron released

Question 412

IGS characteristics

Answer 412

rich in G

Question 413

self-splicing introns in tetrahymena

Answer 413

1) release of 414nt linear intron 2) intron self-splices 2x: -15, -4 = final L19 catalytic molecule

Question 414

self-splicing RNA 3D structure

Answer 414

- complex, necessary for function

Question 415

self-splicing intron experiments

Answer 415

1) magnesium/protein dependence 2) guanosine cofactor binding site 3) role of magnesium

Question 416

magnesium/protein dependence experiment

Answer 416

1) 26S rRNA requires intervening sequence (IVS) removal for mature produce 2) conditions: a) negative control: unspliced version b) positive control: circular 414, linear L19 c) +/- Mg2+ with no treatment, pronase or proteinase K 3) results: Mg2+ required for splicing, occurred in presence of proteases = Mg-dependent, protein-independent

Question 417

metal ions and self-splicing RNA

Answer 417

stabilizes structure

Question 418

guanosine binding with group 1 intron

Answer 418

guanosine stabilized through triple-base pair formation with G264, C3111

Question 419

guanosine binding site experiments

Answer 419

1) WT RNA a) + guanosine = most efficient b) + 2-aminopurine = most activity lost 2) mutant RNA: A264-C311 or A264-U311 a) + guanosine = limited activity b) + 2-aminopurine = some activity restored for double mutant

Question 420

2-aminopurine as cofactor with mutated guanosine binding site

Answer 420

creates isosteric triple base pair = same # of electrons in same arrangement, but different atoms

Question 421

L19 function

Answer 421

catalyzes nucleotidyl transfer reactions

Question 422

L19 mechanism

Answer 422

1) (C)5 oligomer binds to IGS 2) TE reaction incorporates 1 C at 3' end, releasing (C)4 oligomer 3) another (C)5 oligomer binds IGS 4) TE reaction transfers C to make (C)6 5) L19 restored

Question 423

Mg structural role in group 1 intron

Answer 423

1) coordinates with cofactor 2) stabilizes transition state of TE reaction (positive charge with 2 negative O's)

Question 424

Mg functional role experiment

Answer 424

1) normal or mutated (S-link instead of O to phosphate) 2) Mg2+ (binds oxygen only) or Mn2+ (binds sulfur and oxygen = positive control) cofactor 3) results: a) normal + Mg2+ = splicing b) mutated + Mg2+ = no splicing c) normal + Mn2+ = splicing d) mutated + Mn2+ = splicing 4) conclusion: Mg at the active site is required for splicing

Question 425

characteristics of viral ribosymes

Answer 425

- use metals or base functionalities to assist catalysis

Question 426

translation definition

Answer 426

genomic sequences to protein sequences

Question 427

ribosome subunit functions

Answer 427

large = catalyzes reaction + has exit channel for polypeptide, small = interacts with mRNA

Question 428

active sites in ribosome

Answer 428

Exit: uncharged tRNAs exit Peptidyl: peptide bonds formed Aminoacyl: charged tRNAs enter *in both small and large subunit

Question 429

protein synthesis direction

Answer 429

N to C terminal

Question 430

tRNA role in translation

Answer 430

- deliver coded amino acids to ribosome - requires different sequences to transmit coding info - but similar structure to be used with equivalent efficiencies

Question 431

genetic code determination

Answer 431

- examined which polypeptides were produced in response to synthetic RNA polymers with repeating sequences of three or four bases

Question 432

genetic code features

Answer 432

1) used by all living organisms 2) codon = triplets of nts in mRNA 3) nt sequence written 5' to 3', codons translated in same direction 4) 3nts = one AA 5) synonyms = codons for same AA 6) most synonyms differ in only the last base 7) code is non-overlapping

Question 433

genetic code patterns

Answer 433

- similar codons = similar properties - ex. U first base = hydrophobic

Question 434

non-overlapping meaning

Answer 434

- every three nts = one AA - no overlap of reading frame = this would mean one mutation affects multiple codons vs just one

Question 435

genetic code translation features

Answer 435

1) unambiguous 2) multiple codons for most AAs = degeneracy, helpful against mutations 3) first two nts specify AA, third is variable 4) similar codon = chemically similar AAs 5) only 61/64 are AAs, 3 are stop 6) no punctuation

Question 436

polypeptide chain growth reaction (protein synthesis reaction aka transpeptidation)

Answer 436

1) N-terminal of AA in A site attacks C=O of peptide in P site 2) transition state 3) peptide bond formed, protons released --> H2O = dehydration reaction

Question 437

probability of protein with no errors formula

Answer 437

p = (1-e)^n e = frequency of inserting n = number of AA residues

Question 438

max error rate for large proteins

Answer 438

cannot exceed 10^-4, otherwise excessive production of potentially mutant proteins

Question 439

codon-anticodon pairing

Answer 439

anticodon 3 pairs with codon 1, 2 with 2, 1 with 3

Question 440

tRNA and tRNA synthetase interaction regions

Answer 440

1) position 73 2) acceptor stem 3) variable pocket 4) variable loop 5) variable stem loop 6) anticodon

Question 441

tRNA sequence conservation

Answer 441

conserved = AA attachment site, TΨC loop variable = variable loop, anticodon, other nucleotides near anticodon

Question 442

tRNA structure

Answer 442

1) L-shaped 2) 2 double helix segments from 4 base paired regions 3) bases in non-helical regions = H-bonding participants 4) CCA from 3' end 5) anticodon at other end of L-shape

Question 443

how are tRNAs charged?

Answer 443

specific aminoacyl-tRNA synthetases covalently link AAs with tRNAs

Question 444

tRNA charging mechanism (Class 1)

Answer 444

1) activated by adenylation: AMP added (requires 1 more ATP to regenerate ATP!) 2) 2'OH nucleophilic attack 3) transesterification to 3'-O aminoacyl tRNA

Question 445

tRNA charging mechanism (Class 2)

Answer 445

1) adenylation 2) 3'OH nucleophilic attack *NO transesterification

Question 446

adenylation of AA is analogous to...

Answer 446

charging of fatty acyl coA

Question 447

tRNA nomenclature

Answer 447

1) aminoacyl-tRNA--amino acid--anticodon, ex. alanyl-tRNA = tRNA^Ala a) if wrong: Y-tRNA^Ala 2) aminoacyl-tRNA synthetase, ex. threonyl-tRNA synthetase

Question 448

amino acid specificity for aminoacyl-tRNA synthetases

Answer 448

required for correct addition of AA

Question 449

Thr-tRNA synthetase key residues

Answer 449

- Asp, 2xHis, Cys in active site interact with Thr with Zn2+ coordination

Question 450

importance of OH for Thr-tRNA synthetase

Answer 450

- Zn2+ coordinates with OH and NH3 - Asp H-bonds to OH

Question 451

mischarging of Thr-tRNA synthetase

Answer 451

- Ser common (also has OH) - Val has no OH, not mischarged

Question 452

Thr-tRNA synthetase proofreading site

Answer 452

- incorporation of Ser flips CCA into editing site - hydrolysis to remove - analogous to DNAP proofreading

Question 453

size discrimination proofreading by tRNA synthetases

Answer 453

- larger amino acids rejected from activation site - smaller mischarged amino acids accepted by editing site - ex. Ile-tRNA and Thr-tRNA both use this mechanism

Question 454

discovery of aminoacylation specificity

Answer 454

- mutational studies in yeast, E. coli - different regions important in different tRNAs

Question 455

alanyl-tRNA recognition by alanyl-tRNA synthetase

Answer 455

- relies on key G3:U70 microhelix in acceptor stem - mutation of G3:C70 to G3:U70 in Cys-tRNA leads to recognition by Ala-tRNA = mutant proteins

Question 456

class I vs II tRNA synthetases interaction with tRNA

Answer 456

class 1: interacts with hairpin CCA on face of tRNA class 2: interacts with helical CCA on side of tRNA

Question 457

class I vs II tRNA synthetase amino acids

Answer 457

- each 10/20

Question 458

key differences between class I vs II tRNA synthetases

Answer 458

1) 2'OH vs 3'OH 2) different ATP binding (conformation) 3) monomeric vs dimeric

Question 459

tRNA synthetases and isoacceptor tRNAs

Answer 459

- specificity but can recognize isoacceptors: different tRNAs that for the AA

Question 460

prokaryotic vs eukaryotic ribosomes

Answer 460

1) entire ribosome: 70S (50/30) vs 80S (60/40) 2) large subunit: 23S rRNA, 5sRNA, 31 proteins vs. 5.8S, 28S, 5S, 40 proteins 3) small subunit: 21 proteins, 16S vs 30 proteins, 18S

Question 461

ribosome protein composition

Answer 461

~1/3 protein, areas between subunits have no proteins for RNA binding

Question 462

polysomes

Answer 462

- translation and transcription coupled in prokaryotes - many ribosomes on each mRNA transcript

Question 463

what is NOT the start signal for translation?

Answer 463

first 3 nts of each mRNA

Question 464

where is the start signal in translation?

Answer 464

- approximately 25nts away from 5' end - all mRNAs contain beginning and end signals

Question 465

mRNA unique feature in prokaryotes

Answer 465

all are polycistronic/polygenic

Question 466

shine-dalgarno sequences

Answer 466

- recognized by prokaryotic ribosomes - ~10 nts upstream - purine-rich - base pairs with 3' end of 16SrRNA in small subunit to establish correct reading frame

Question 467

initiator tRNA in prokaryotes

Answer 467

- adds formylated Met (fMet) - different from internal Met tRNA

Question 468

fMet tRNA formation

Answer 468

1) charged with unmodified Met by Met-tRNA synthetase 2) transformylated by Met-tRNAi^fMet formyl transferase: additional formyl group mimics peptide bond, allowing binding to P site instead of A

Question 469

specificity of Met-tRNAi^fMet formyl transferase

Answer 469

does not interact with free Met or Met-tRNA^Met

Question 470

general protein synthesis mechanism

Answer 470

1) tRNAi in P site 2) aminoacyl-tRNA binds in A site 3) peptide bond formation, leading to positional shift in acceptor arms (large subunit) 4) translocation to position next codon 5) uncharged tRNA dissociation from E site

Question 471

translocation in prokaryotes requires...

Answer 471

elongation factor G, which uses GTP hydrolysis

Question 472

protein sequence is coded by...

Answer 472

mRNA sequence only: if AA modified after adenylation and charging, the wrong AA will be incorporated

Question 473

wobble hypothesis

Answer 473

- first two bases of codon and last two of anticodon = canonical WC pairing - last pair less stringent

Question 474

inosine in wobble hypothesis

Answer 474

- appears in first position of many anticodons - can pair with U, C, A = steric freedom - ex. yeast tRNA^Ala anticodon = IGC, recognizes GCU, GCC, GCA

Question 475

U base pairs with...

Answer 475

A or G

Question 476

C base pairs with...

Answer 476

only G

Question 477

A base pairs with...

Answer 477

only U

Question 478

G base pairs with...

Answer 478

U or C

Question 479

I base pairs with...

Answer 479

U, C or A

Question 480

generalizations in codon reading

Answer 480

1) codons that differ in first bases must be recognizes by different tRNA 2) first base of anticodon determines whether tRNA reads 1, 2, or 3 codons

Question 481

monitoring of codon-anticodon base pairing

Answer 481

- 16S rRNA - A1493, A1492 hydrogen bond with with first two codon spots, ensuring WC - G530 also involved - wobble site not proofread

Question 482

initiation factors (IFs)

Answer 482

- aid assembly of 30S initiation complex, then 70S initiation complex - P-loop NTPases (G protein family) that use GTP

Question 483

prokaryote translation initiation steps

Answer 483

1) IF-1/3 bind, IF3 prevents premature joining of 50S 2) mRNA-IF2-tRNAi^fMet bind with GTP in correct reading frame = 30S initiation complex 3) GTP hydrolysis = conformational change, releasing IF1/2/3 and allowing 50S association to form 70S initiation complex

Question 484

elongation factors (EFs)

Answer 484

- deliver aminoacyl tRNAs to A site of ribosome - prokaryotes: Ef-Tu

Question 485

purpose of Ef-Tu binding to aminoacyl tRNA

Answer 485

1) protects ester linkage from hydrolysis 2) GTP hydrolysis only occurs with appropriate codon-anticodon base pairing = fidelity, no hydrolysis means tRNA leaves A site 3) does not interact with fMet-tRNAf (whereas IF2 only recognizes this)

Question 486

Ef-Tu cycling mechanism

Answer 486

1) EF-Tu-GTP binds aminoacyl tRNA 2) GTP hydrolysis if correct pairing, allowing tRNA to bind A site firmly 3) EF-Tu-GDP dissociates 4) Ef-Ts (GEF) regenerates EF-Tu-GTP

Question 487

prokaryote translocation mechanism

Answer 487

1) EF-G binds 50S 2) GTP hydrolysis = conformational change 3) EF-G dissociates

Question 488

EF-G structure

Answer 488

mimics EF-Tu bound amino-acyl tRNA complex, allowing it to fit easily near the A site

Question 489

stop codon recognition

Answer 489

1) aminoacyl tRNA will not bind to A site 2) recognized by release factors (RFs)

Question 490

translation termination reaction

Answer 490

termination factors contain H2O for hydrolysis of polypeptide from tRNA

Question 491

termination factor association with H2O

Answer 491

- Gln, Gly, Gly interact with H2O

Question 492

prokaryote termination mechanism

Answer 492

1) RF1 or RF2 recognizes stop codon, binds with RF3-GTP 2) hydrolysis of polypeptide = release from P site of ribosome 3) RF3 GTP hydrolysis = RF dissociation 4) RRF GTP hydrolysis releases entire translocation complex

Question 493

RF-1 recognizes...

Answer 493

UAA, UAG

Question 494

RF-2 recognizes...

Answer 494

UAA, UGA

Question 495

total required ATP/GTP for protein synthesis

Answer 495

ex. n AA in length 1) charging tRNA: 2 ATP x n 2) initiation (IF2): 1 GTP 3) elongation + translocation (EF-Tu, EF-G): 2 GTP x n 4) termination (RF3): 1 GTP 5) ribosome dissociation (RRF): 1 GTP

Question 496

eukaryotic vs prokaryotic translation initiation

Answer 496

- mostly differ in translation initiation - more IFs (eIFs) - no SD sequence - initiator tRNA is Met - precomplex assembled first to search for start - starts at first AUG near 5' end - circularization for scanning process

Question 497

eukaryotic translation mechanism

Answer 497

1) eIF4F complex and 43S ribosomal (initiator) complex assembled 2) bind mRNA at 5' cap (eIF4F first through eIF4E recognition of m2G) 3) mRNA circularization 4) mRNA scanning with eIF4H (helicase) ATP hydrolysis 5) start codon recognition = GTP hydrolysis = 60S binding, eIFs dissociate

Question 498

eIF4F complex

Answer 498

- heterotrimer: eIF4G bound to eIF4E and eIF4A

Question 499

43S ribosomal complex

Answer 499

- 40S small subunit - many eIFs, including eIF2 - eIF2 bound to GTP, brings in Met-tRNAi^Met

Question 500

Met-tRNAi^Met

Answer 500

different from internal Met, but not formylated!

Question 501

eukaryotic mRNA circularization in translation

Answer 501

eIF4G serves as bridge: - N-terminal binds eIF4E and polyA binding proteins - eIF4E also bound to methyl-G cap

Question 502

eukaryotic vs prokaryotic translocation

Answer 502

1) EF1alpha = EF-Tu 2) EF1betagamma = EF-Ts 3) EF2 mediates GTP driven translocation = EF-G

Question 503

eukaryotic vs prokaryotic termination

Answer 503

- eRF1 recognizes all stop codons - no equivalent of RRF or RF3

Question 504

antibiotics interfering with translation

Answer 504

- can be prokaryote or eukaryote specific or both - can inhibit different steps of translation

Question 505

puromycin mechanism

Answer 505

- analog of terminal aminoacyl A residue of aminoacyl tRNA, allows fit into A site - has amide linkage instead of ester = no more ester to be nucleophilically attacked once incorporated - unstable in P site, leading to dissociation

Question 506

streptomycin mechanism

Answer 506

- highly basic - interferes with initiator tRNA binding in prokaryotes

Question 507

Diptheria toxin mechanism

Answer 507

1) cleaved into A and B fragments in cells 2) A fragment catalyzes covalent modification of diphthamide (modified His in eEF2) using NAD+ = ADP-ribose added 3) EF2 function blocked (no translocation)

Question 508

ricin mechanism

Answer 508

1) heterodimeric catalytic A chain and B chain by S-S bond: released in cell cytoplasm 2) A chain (N-glycosidase) removes A4324 in 28S rRNA = abasic site 3) prevents elongation factor binding