Midterm 2 Flashcards

1
Q

Describe denaturing DNA molecules

A
  • denaturation refers to the separation of DNA strands
  • can be done using heat, decreasing salt concentrations, or increasing pH
  • they can be renatured, which makes hybrid molecules possible (strands from different sources)
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2
Q

How do we measure Tm? What makes Tm higher?

A
  • melting temperature refers to the temperature at which half of the DNA denatures
  • higher TM means that the DNA molecule is more stable
  • dependant on ionic strength (salt, temperature), length, and GC %
  • TM measures absorbency: becomes more absorbent as they denature
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3
Q

What is GC content? How does it impact melting temperature?

A

GC is the amount of GC bonds (%) present in the molecule: makes for a higher melting temperature/ more stable due to the presence of 3 Hydrogen bonds

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4
Q

Why does salt change the stability of DNA?

A
  • increased salt concentrations can increase stability
  • negative phosphorus in the backbone of DNA repel each other and cause instability, however ions in salt (Na+) shields the negative charge and DNA stabilizes!
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5
Q

Why is melting temperature useful?

A
  • for classifying organisms: GC% is specific in species
  • identifying rare mutations: mutated melt at different temperatures
  • for molecular biology: denaturing DNA used in western blotting and qPCR
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6
Q

Are longer DNA molecules more or less stable?

A

More!

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7
Q

What are the 3 proposed models for DNA synthesis?

A
  • Conservative, semi-conservative, dispersive
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8
Q

Explain who and what experiment was performed to determine the correct model for DNA replication

A
  • Meselson and stahl produced generations of E.Coli bacteria on N15 and N14 medium
  • used equilibrium density gradient centrifugation to determine isotope composition of DNA
  • found initial heavy N15 –> replicated to form lighter (50% heavy, 50% light)–> replicated again to see two lines: lightest and 50:50
    = semi conservative!
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9
Q

What are the 4 requirements for DNA synthesis?

A
  1. Single stranded DNA strand
  2. all 4 deoxyribonucleoside 5’ triphosphate
  3. requires DNA polymerase and other enzymes and proteins
  4. free 3’ OH group : absolute necessity
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10
Q

What are some important rules regarding DNA polymerase?

A
  • only synthesizes in a 5’ to 3’ direction (but template is READ in a 3’ to 5’ direction)
  • cannot create new DNA, only elongated chains
  • free 3’ OH is required
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11
Q

Describe synthesis at the replication fork

A
  • replication begins at a specific nucleotide sequence: the origin of replication
  • synthesis occurs in a replication bubble!
  • both DNA strands transcribed simultaneously in the 5’ to 3’ direction (two new strands being made)
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12
Q

What does it mean for replication to be semi-discontinuous?

A
  • leading strand synthesizes normally, unbroken
  • lagging strand must go in other way and keep moving back towards the replication fork: forms Okazaki fragments
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13
Q

What are the modes of replication?

A

Circular genome and linear genome
- circular genome: theta replication and rolling replication
- linear replication: eukaryotes

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14
Q

Describe theta replication

A
  • single replicon (for bacteria = entire chromosome)
  • bidirectional at replication fork
  • semi-discontinuous
  • product: two circular genomes!
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15
Q

Describe rolling replication

A
  • no replication bubble
  • continuous
  • uncoupling of replication of the two strands of DNA
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16
Q

Describe linear replication in eukaryotes

A
  • multiple: replicons, origins of replication, replication forks
  • semi-discontinuous
  • bidirectional: travels in two directions, two forks
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17
Q

What are the four stages of replication in eukaryotic DNA replication in prokaryotes ?

A
  • initiation, unwinding, elongation, termination
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18
Q

Describe initiation in prokaryotes

A
  • initiation proteins bind to the origin of replication
  • a section of DNA unwinds and proteins bind to it
  • single stranded binding proteins keep the strands separated
  • helicase binds to lagging strand template and breaks H bonds
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19
Q

Describe unwinding in prokaryotes

A
  • helices breaks hydrogen bonds
  • DNA gyrase travels ahead of the fork to break and reform bonds to prevent supercoiling
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20
Q

Describe elongation in prokaryotes

A

A short stretch of RNA nucleotides (RNA primer) is synthesized by primase
- RNA primer provides a free 3’OH end for DNA polymerase to use
- RNA primer later removed, replaced with nucleotides

–> Why RNA primer? Because production of RNA does not need a 3’ end

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21
Q

Describe E.Coli polymerase

A
  • E.coli has 5 polymerase: Pol I– > Pol V
  • all 5 have 5’ to 3’ polymerase activity
  • some polymerase have exonuclease activity : so they can remove incorporated nucleotides that do not match the template strand

–> Pol III: primary replication enzyme
–> Pol I: removes and replaces RNA primers with DNA

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22
Q

How does Pol I remove and replace RNA in prokaryotes?

A
  • DNA pol I exonuclease activity starts at 5’ end to remove RNA primers
    –> DNA pol I 5’ to 3’ polymerase activity fills in the gap with nucleotides
    –> DNA ligase used to heal the nicks in sugar phosphate backbone
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23
Q

What conditions initiate termination in prokaryotes?

A

termination occurs when two replication forks meet or a special sequence is met

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24
Q

Does eukaryotic DNA have primers?

A

nope, only polymerase!

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25
Q

What are the 3 polymerases in eukaryotic DNA replication we cover?

A

alpha : 5’–>3’ activity
delta : 5’–>3’ activity –> 3’ - 5’ activity
epsilon : 5’–> 3’ activity –> 3’-5’ activity

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26
Q

What are the roles of the eukaryotic DNA polymerases?

A

alpha: initiation of DNA function/repair, has primate activity

delta: guides lagging strand synthesis of nuclear DNA, DNA repair, transslesion, DNA synthesis

epsilon: leading strand synthesis

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27
Q

Describe the initiator of replication in eukaryotic DNA, and describe the steps

A
  • origins of replication clustered (20-80 at a time): called replication units
  • Controlled initiation:
    1) must be licensed (to prevent more than one copy)
    2) origin then activated and replication begins
    3) once activated/replicated origin is deactivated
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28
Q

Describe the rate of DNA synthesis in prokaryotes and eukaryotes. Why does it differ in eukaryotes?

A
  • prokaryotes: 4.6 million in genome, I origin
  • eukaryotes: >3 billion genome, 8 hours (slower), >10,000 origins of replication
  • S phase would take ~ 1 month with only one origin of replication: multiple is the solution to a larger, slower
    –> but much more costly when mistakes are made: slower = less cost
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29
Q

Describe nuclear dissasambnlry and reassmabley

A

DNA is packaged into chromatin: must be disassembled, produce more histones and reassemble nucleosomes

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30
Q

What is one problem with the replication of 3’ ends?

A
  • when primase is removed, there is a large gap left behind
  • telomeres fix this: a long nucleotide section with many Gs that stabilizes the ends
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31
Q

What is the solution to the problem of 3’ synthesis ends?

A
  • telomerase! A specialized reverse transcriptase, extends end of parental DNA by RNA templated DNA synthesis
  • erspondibel for replciatopn of chromosome ends
  • telomerase extends DNA, filling in the gap!
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32
Q

Describe what transcription, translation, and protein really does

A
  • transcription: information is transferred from DNA to RNA
  • translation: Information from RNA is transferred into a protein through a code that specifies amino acid sequences
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33
Q

What is the ‘equation’ for gene expression? Is gene expression simpler or more complex in prokaryotes?

A
  • transcription + translation = gene expression
  • simpler in prokaryotes!
  • eukaryotes need RNA processing (mRNA) –> translation
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34
Q

What are the different classes of RNA for prokaryotes and eukaryotes?

A
  • both: mRNA, tRNA, rRNA
  • eukaryotes: pre-mRNA, snRNA, snoRNA, miRNA, siRNA, pi-RNA
  • prokaryotes: criRNA (crispr)
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35
Q

Describe the structure of the RNA? Is it single stranded or not?

A
  • RNA has uracil instead of thymine
  • has an OH group on the sugar carbon, instead of DNA which has a hydrogen (RNA is more reactive!)
  • generally single stranded but can fold into a complex secondary structure (hairpin loops or stem loops)
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36
Q

What are hair pin loops/stem loops?

A
  • RNA is normally single stranded but sometimes forms a hair pin as a result of no-RHO termination (two inverse copies)
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37
Q

What are the 3 steps for transcription?

A
  • initiation does not require a primer
  • ribonucleotides are added to the 3’ OH group of growing RNA chain
  • DNA unwinds at front of transcription bubble and then rewinds
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38
Q

What are the 3 requirements for transcription?

A

1) DNA template
2) RNA nucleotides
3) RNA polymerase and other proteins

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39
Q

What is the substrate for RNA synthesis?

A
  • ribonucleoside 5’ triphosphate
  • triphosphate + sugar + base
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40
Q

which strand is used as the template strand for RNA synthesis? How is it read?

A
  • either can be used as the template strand!
  • Template is read in a 3’ to 5’ direction, and RNA is synthesized in a 5’ to 3’ direction
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41
Q

What is the transcription unit?

A
  • region of DNA that codes for an RNA molecule and the sequences necessary for transcription
  • 3 critical regions: promoter, RNA coding region, termination site
  • Only the RNA coding region is transcribed
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42
Q

What are the 3 parts of the transcription unit ?

A
  • the promoter
  • the RNA coding region
  • the termination site
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43
Q

What does the promoter do?

A
  • the DNA sequence recognized and bound by the transcription apparatus (RNA polymerase and other proteins)
  • binding of RNA polymerase to promoter region orients the enzyme towards start site
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44
Q

Describe the 3 stages of prokaryotic transcription

A

1) initiation: assembly of transcription apparatus on the promoter and begins synthesis of RNA
2) elongation: DNA threaded through RNA polymerase, unwinds DNA, adds nucleotides to 3’ end of growing RNA strand
3) termination:recognition of the end of transcription

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45
Q

describe prokaryotic RNA polymerase

A
  • many bacteria have multiple layers of sigma factors, which help in recognition of multiple classes of promoters
  • without the sigma, the core enzyme initiates transcription randomly
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46
Q

What is the holoenzyme?

A

-the complete enzyme complex composed of core RNA polymerase and the sigma factor

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47
Q

what is a consensus sequence?

A
  • promoters contain short stretches of DNA conserves among promoters of different genes
  • most encountered sequences at at -10 (pribnow box) and -35 upstream of start stie
  • binding orients polymerase towards start site (+1)
  • variation effects strength of the promoter (it’s the most common sequence) - up regulation and downregulation
  • reCA is a strong promoter
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48
Q

How is RNA transcription initiated in prokaryotes?

A
  • when core RNA polymerase binds to the promoter with the help of sigma
  • RNA backbone appears too (ribonucleoisde 5’ triphosphate)
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49
Q

Describe termination of transcription in prokaryotes

A
  • transcription ends after a terminator sequence is transcribed
  • two major types of terminators: rho dependant (requires rho protein) and rho independent (intrinsic terminator)
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50
Q

What is rho-dependant termination?

A

1) rho binds to RNA upstream of terminator
2) RNA polymerase pauses when it reaches the terminator sequence and Rho catches up
3) rho unwinds the DNA RNA hybrid using helices activity

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51
Q

What is rho independent termination?

A
  • inverted repeated, polymerase pauses at the Us
  • hairpin formation destabilizes RNA DNa hybrid
    (A-U BP relatively weak)
  • transcription terminates when inverted tepreats from a hairpin follows by a string of uracils
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52
Q

What makes up the 3 parts of the eukaryote promoter?

A
  • core promoter + regulatory promoter
  • enhancer sequence: transcriptional activator protein, enhance
  • regulatory: transcriptional activator protein
  • core promoter:
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53
Q

What is the core promoter?

A

-extends upstream and downstream of transcription start site
- minimal sequence reuqired for accrut transcription initiation
- includes a number of consensus sequences (eg; TFIIB, TATA, initiator, DCE) for transcription factor binding

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54
Q

What is the regulatory promoter in eukaryotes?

A
  • located upstream of core promoter, exact location can vary
  • transcriptional activator proteins bind to consensus sequence and affect rate of transcription
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55
Q

Look at basal transcription apparatus diagram

A
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56
Q

Describe the termination of RNA polymerase II transcription

A
  • no specific sequence for termination, requires cleavage of mRNA at a specific site
  • exonuclease activity degrades remaining mRNA terminating transcription
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57
Q

What is post transcriptional gene silencing?

A
  • destroying RNA before it is turned into a protein
  • or, inhibiting translation so protein is not made
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58
Q

What is transcriptional gene silencing?

A
  • condenses chromatin to suppress transcription, mRNA is not made
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59
Q

Describe the origins of miRNA

A
  • the precursor of miRNAs, pri-mRNA, are encoded by the genome (RNA polymerase II)
  • in nucleus, primRNA is cleaved by Drosha (primase III enzyme) into pre-mRNA (hairpin structure)
  • in cytosol: dicer cleaves pre-miRNA into 19-25 nucleotide miRNA: miRNA duplex with no stem loop
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60
Q

What are the three components required to make pre-miRNA?

A
  • RNA Pol II, Drosha, and dicer
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61
Q

Describe the RNA induced silencing complex?

A
  • two forms: siRNA and miRNA
  • siRNA is perfectly complementary, promotes cleavage
  • miRNA is not exactly complementary, promotes degradation, repression, and cleavage if enough complementarity
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62
Q

What are the fundamental steps of RISC?

A
  • double stranded, loses the tag along strand so only leading strand is left (this is RISC) - it binds to a section of mRNA and silences it
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63
Q

What is the state prior to dicer processing, structure, complementaritity, mRNA target and mechanism of gene regulation of siRNA?

A

state prior: double stranded RNA with up to 100 nucleotides

structure: 21-23 nucleotide RNA duplex with 2 nucleotides 3’ overhang

Complementarity: fully complementary = one mRNA target

Mechanism of gene regulation: endonucleolytic cleavage of mRNA

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64
Q

What is the state prior to dicer processing, structure, complementaritity, mRNA target and mechanism of gene regulation of miRNA?

A
  • prior: 70-100 nuvrlotides with mismatches and a hairpin
  • structure: 19-25 nuceltodies RNA duplex with 2 nucleotides 3’ overhang
  • partially complementary, van target 100+ at the same time (targets 3’ untranslated region)
  • translation repression, degradation, endonucleolytic cleavage (if high complementary)
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65
Q

What does small interfering RNAs do to transcription? What about miRNA? What about siRNA which are transcriptional ?

A

siRNA post transcription: cause cleavage of mRNA

miRNA: inhibits translation and/or mRNA cleavage and degradation

Some siRNA: inhibits transcription through methylating enzymes

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66
Q

Describe transcriptional gene silencing. What are RITS?

A
  • RNAi also works to form localized repressed chromatin formation
  • RNAi duplexes loaded onto a form of RISC called RITS (RNA induced transcriptional silencing complex)
  • similar to RISC in that they bind to homologous base pairs interactions involving the guide strand of small RNA
  • RITS mediates genetic silencing via heterochromatin formation
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67
Q

How does RNA induced transcriptional silencing complexes mediate genetic silencing?

A
  • mediates gene silencing via heterochromatin formation
68
Q

Where is RITS and RISC found?

A
  • RISC: cytoplasm (RISC used for post-transcriptional silencing)
  • RITS: nucleus (RITS is used for transcriptional silencing)
69
Q

Describe chromatin structure

A
  • Double stranded helical DNA –> nucleosomes (8 proteins that DNA is wrapped around) –> (chromosome (nucleosome + H1 histone)) –> nucleosomes fold to form fibres –> loops –> finer –> chromatid
70
Q

What is heterochromatin? What is euchromatin?

A
  • heterochromatin: inactive, non coding form of chromatin
  • euchromatin: open, active form of chromatin: allows for transcription
71
Q

What are the 3 forms of histone modifications?

A
  • occurs at multiple sites: methylation, acetylation, phosphorylation
72
Q

How does RITS work?

A
  • other siRNAs bind to complementary sites on RNA and attract methylating enzymes
  • methylation of histones inhibits transcription
73
Q

Why does RNAi matter?

A
  • 30% of human genes thought to be regulated by RnAi: thousands of genes code for miRNA
  • important for regulating gene expression during embryo development
  • important research tools for ‘knowckingout’ particular genes
  • lead to major advances in diseases treament (eg; chemo and siRNA inhibits translation)
74
Q

Describe the collinear model

A
  • collinearity between genes and protein (crick in 1958)
  • continuous sequence of nucleotides results in continuous sequence of amino acids
  • number of nucleotides is proportional to the number of amino acids

eg; 24 nucleotides = 8 codons = 8 aa

75
Q

Is the collinear model more accurate for prokaryotes or eukaryotes?

A
  • prokaryotes (eukaryotes have more introns)
76
Q

Describe the structure of mRNA in prokaryotes

A
  • right before the start codon it has a shine-dalgamo sequence (found in prokaryotes only)
77
Q

Describe the steps of mRNA production in eukaryotic cells

A
  • RNA polymerase on DNA
  • forms pre-mRNA
  • forms RNA
  • forms mRNA
  • to ribosome for translation
78
Q

Describe the steps of mRNA production in prokaryotic cells

A
  • RNA polymerase on DNA
  • transcription feeds high into ribosome for:
  • translation
79
Q

What is the difference between pre-mRNA and mRNA in eukaryotes?

A
  • pre-mRNA has introns and exons which need to be spliced out
  • mRNA has no introns, it also has a 5’ cap and a polyA tail
80
Q

What is the major type of intron?

A
  • nuclear pre-mRNA: protein encoding genes in the nucleus of euakrtoties
  • function: spliceosomal splicing mechanism
81
Q

Describe the prokaryotic protein coding genes

A
  • usually found in a continuous array in DNA called an operon
  • operons operate as a unit utilizing a single transcription start site for multiple genes
  • contains few introns: DNA translated into mRNA which is translated into a protein as it is being produced
82
Q

Describe the eukaryotic protein coding genes

A
  • each gene transcribed from its own start site to yield a pre-mRNA that is processed into a functional mRNA encoding single protein
83
Q

What are the post transcriptional modifications to eukaryotic mRNA?

A
  • 5’ cap: facilitates binding of ribosome to 5’ end of mRNA: promotes stability, RNA splicing
  • 3’ cleavage+ Poly A tail: increases mRNA stability, aids in export of mRNA from nucleus, facilitates binding of ribosome to mRNA

RNA splicing: moves interns, multiple proteins can be formed, facilities export of mRNA to cytoplasm

84
Q

what are the 3 modifications to eukaryotic mRNA?

A
  1. 5’ cap
  2. 3’ cleavage and polyadenylation
  3. splicing introns
85
Q

Describe 5’ capping

A
  • in eukaryotes
  • a methylated (CH3) guanine nucleotide jointed to the 5’ end of pre-mRNA by 5’ to 5’ linkage involving phosphate groups
  • necessary for initiation of translation, transportation of mRNA out of nucleus, protection from degradation , enhances RNA splicing
86
Q

Describe 3’ polyadenylation

A
  • in eukaryotes
  • 50-250 adenine nucleotides added to the 3’ en of pre-mRNA
  • per-mRNA is cleaved 11-30 nucleotides downstream of consensus sequence in the 3’ UTR
  • the polyadenylation occurs adding Adenine tail
87
Q

How is the poly a tail added? Why is it important?

A
  • through cleavage (of 3’ end) and polyadenylation
  • necessary for efficient translation, protects RNA from degradation
88
Q

What are the 3 consensus zones of splicing?

A
  • 5’ splice site, 3’ splice site, branch point
  • consensus sequences used by spliceosome to recognize/remove introns
89
Q

What is the process of splicing?

A
  • introns are removed in the form of a lariat and exons are spliced together by two successive reactions
  • spliceosome snRNP proteins U1 and U2 bind to the 5’ splice site and branch point respectively
  • form the lariat, which is then removed first, and then exons are spliced together and translated!
90
Q

What is a spliceosome?

A
  • splicing takes place on the spliceosome
  • a ribonucleoprotein complex (300 proteins and 5 snRNAs)
  • 5 snRNPS (snRNA + proteins): U1, U2, U4, U5, U6
  • U1 and U2 bind onto the intron
  • snRNPS are central to activity of the spliceosome
91
Q

What component is essential to the activity of the spliceosome?

A

snRNPs: protein + snRNA

92
Q

Describe the spliceosome unit

A
  • exon (5’ splice site: U1) + intron (branch point (U2)) + exon w/ 3’ splice site
93
Q

What is the relationship between mRNA transription and processing? What is responsible for this relationship and how does it work?

A
  • they are coupled!
  • functional coupling of mRNA transcription and processing by mRNA polymerase II
  • coupling of events are mediated by tail or CTD (c-train repeat domain) of the largest subunit of Pol II
  • processing enzymes are recruited to the CTD during transcription of mRNA
94
Q

Describe 5’ capping in mRNA transcription

A
  • CAP is added as soon as the 5’ end of the pre-mRNA emerges from the polymerase ( capping followed by methylation)
  • capping enzymes are recruited to the CTD of pol II during early stages of transcription
95
Q

Describe the assembly of the spliceosome

A
  • assembly of the spliceosome occur co-transcriptionally, while RNA Pol II is still actively transcribing the template
  • components of the splicosome recruited to RNA while trnscirptin is happening
    = the CTD interacts directly with the splicing proteins to recruit them to RNA
96
Q

Describe the termination/polyadenylation of mRNA transcription using CTD

A
  • polyadenlyation factors recruited to the CTD
  • RNA is cleaved at the poly (a) cleavage site,
  • degradation of the remaining RNA by rat 1 terminates transcription
97
Q

What are the three processes CTD coordinates?

A
  • coupled transcription and processing
  • 5’ capping
  • assembly of the spliceosome
  • polyadenlyation termination
98
Q

What is alternative processing?

A
  • increase protein diversity
  • the human genome sequence: ~20,000 genes, proteonome complicated as a single gene can give rise to many proteins
  • 100000+ proteins from 20,000 genes = performed by alternative splicing or alternative poly a sites
99
Q

What are the ways pre-mRNA can be processed?

A
  • alternative splicing : pre-mRNA can be spliced in multiple different ways
    or
    alternative poly(A) sites:” 3 different cleavage sites
100
Q

Describe alternative splicing

A

each mRNA has a different combination of exons
- each mRNA when translated produces a different protein (isoforms)

101
Q

What are different forms of proteins from the same gene sequence called?

A
  • isoforms of protein
102
Q

What are the different kinds of alternative splicing patterns

A
  • intron retention
  • exon skipped
  • alternative 5’ or 3’ splice site
  • mutually exclusive exons (one can not be with the other)
103
Q

What are alternative poly(A) cleavage sites?

A
  • a form of alternative processing
  • pre-mRNA contains multiple 3’ cleavage sites which determines the length of the mRNA transcript
104
Q

Why are regulations of alternative processing important?

A
  • can result in a functional or non functional protein
    -different forms of a protein may be produced in different cells
105
Q

Many genetics diseases arise from mutations that impact pre-mRNA splicing. What are some of the potential impacts of these mutations?

A
  • affects use of splice sites, splicing machinery, regulators of alternative splicing
    = ~15% of all point mutations that cause disease alter pre-mRNA splicing
106
Q

What are some affects of a mutation on splice sites?

A
  • make 3’ or 5’ splice site unrecognizable
  • initiate use of cryptic splice site
  • exons skipping or intron retention (partial or complete) = gene loss of function by generating non functional protein or altering mRNA stability (degraded and no protein)
107
Q

True or false: splicing requires consensus sequences

A

TRUE
- there are three consensus points used in pre-mRNA: 3’ site, 5’ site, and branch point
- used by spliceosome to recognize/remove introns

108
Q

What is beta thalassemia?

A
  • caused by a mutation in the B-globin gene (in hemoglobin): many of which lead to incorrect splicing
  • one of the most common genetic disorders
  • results in excessive breakdown of red blood cells leading to anemia
109
Q

What are the two ways a protein may be modified?

A
  • alternative processing or RNA editing
110
Q

Describe RNA editing. What are the two methods to do this?

A
  • occasionally a gene is found with a. sequence that does not match the seance of its RNA product
  • editing the RNA alters the coding information of the mRNA transcripts
  • done through substation editing or insertion editing
111
Q

What is substitution editing?

A
  • chemical alteration of individual nucleotides by specific enzymes (eg; U converted to U)
112
Q

What is insertion editing?

A
  • addition of U nucleotides by cleavage of the mRNA, insertion of the U nucleotide, and ligation of the ends
  • reactions catalyzed by gRNA (guide RNA)
113
Q

What is gRNA?

A
  • guideRNA: binds as best it can to the mRNA and serves as a template for the addition of nucleotides
  • gRNA adds nucleotides to the mRNA strand that are not encoded by the DNA
114
Q

What is the process by which the two exons are fused together after splicing?

A
  • two subsequent reactions of transesterification to fuse them after the removal of the lariat
115
Q

What are the two forms of alternative processing?

A

-alternative splicing and alternative poly A sites

116
Q

What are the two forms of RNA editing?

A
  • substitution editing
  • insertion editing
117
Q

How many amino acids are there? how many codons are there? What are sense codons and stop codons?

A
  • 20 amino acids
  • 64 codons, 61 sense codons (codons that aren’t stop codons), 3 stop codons
118
Q

Describe the amino acid format

A
  • central carbon connected to hydrogen, amino group (N) and carboxyl group (C) and R group (side branch specific to that amino acid)
119
Q

How are amino acids joined together?

A

by peptide bonds

120
Q

Describe the protein structure

A
  • primary: chain of amino acids
    -secondary: interactions between amino acids form helix
    -tertiary: secondary folds further, hydrophobic in hydrophilic out
    -quaternary : 2+ polypeptide chains
121
Q

What is a triplet code? What is a codon? Why are they in triplets?

A
  • triplet code are a sequence of 3 nucleotides in DNA
  • codon: 3 nucleotides in mRNA
  • 20 amino acids, 4 nucleic acid bases
    4^3 = 64 codons is enough for 20 Amino acids!
122
Q

What are tRNAs?

A
  • all tRNAs from all organisms have a similar structure
  • amino acid attachment site is the same for all tRNA molecules
  • sequence for attachment is always 5’ -CCA-3’
123
Q

What are the parts of the tRNA molecule?

A
  • acceptor arm: holds out - CCA
  • anticodon arm and anitocodn arm
124
Q

Do stop codons code for any proteins?

A
  • No, stop codons simply terminate translation
125
Q

What does it mean for amino acids to have degeneracy?

A
  • some amino acids are specified by more than one codon
126
Q

What does it mean for codons to be synonymous?

A
  • different codons that specify the same amino acid
127
Q

What does it mean for a code to be degenerate but not ambiguous?

A
  • a codon never specifies more than one amino acid
128
Q

Which positions of the codon are most important for distinguishing the amino acid?

A
  • 1st and 2nd most important, 3rd position rarely changes amino acid specification
129
Q

What are the two types of degeneracy?

A
  • partial: changing 3rd based from purine –> purine or pyrimidine –>pyrimidine
  • complete: purine –> pyrimidine or pyrimidine–> purine
130
Q

How is the degeneracy of the genetic code accommodated?

A
  • isoaccepting tRNA: tRNAS bind same amino acid but recognize different codons (by using different anticodons)
  • wobble effect: allows some amino acid tRNAs to pair with more than one codon
131
Q

What is the wobble effect?

A
  • only 30 anticodons
  • all 61 sense codons are used so the same anticodons bind to different codons
  • the wobble effect allows some anticodons to base pair with more than one codon
  • wobble positions always on the 5’ side of antiocodn, 3’ side of codon
  • wobble from normal position to forma non-watson and crick base pair with the codon
132
Q

What is the wobble position called?

A
  • non-watson and crick base pair
133
Q

What is an insonene?

A
  • an intermediate in the metabolism of pure.
  • essential for translation of genetic code in wobble base pairs
134
Q

What are reading frames (RFs?)

A
  • refers to the protein coding region of the mRNA
  • specifies a single protein starting and ending at internal sites within the mRNA
  • All RFs being with AUG (methionine)
135
Q

What are the 3 rules of the reading frames?

A
  • codons consisting of all 3 nucleotides read 5’ to 3’
  • codons are not overlapping
  • no gaps, each base is part of a codon
  • message translated set by AUG initiator
136
Q

What are the forms of ORFs?

A

1 RF: first nucleotide
2nd RF: 2nd nucleotide
3rd RF: 3rd nucleotide

137
Q

What does it mean when reading frame has a stop codon?

A
  • it is NOT a reading frame, a reading frame must have no stop codons
138
Q

How many possible RFs are there in a DNA molecule?

A

6: 3 on top, 3 on bottom 3 on top

139
Q

What are point mutations or base substitutions?

A
  • mutations
    base substitutions: nonsense, missense, silent/synonymous
140
Q

What are frameshift or non frameshift mutations?

A

frameshift: insertion and deletion
non frameshift: doesn’t move other codons, frameshift resets them all

141
Q

Where does translation occur? How is the mRNA read? What end do the amino acids attach?

A
  • within a ribosome
  • mRNA is read 5’ to 3’
  • amino acids attaches on the carboxyl (C) terminus of the polypeptide chain (N terminus is the first out of the ribosome)
142
Q

What is the term for when multiple ribosome simultaneously translate the same mRNA?

A
  • polyribosome can occur in prokaryotes and eukaryotes
143
Q

What are the requirements for translation in prokaryotes? What is unique about mRNA in prokaryotes?

A
  • mRNA template, tRNA, amino acids, ribosomes, many accessory proteins, energy provided by GTP hydrolysis
  • polycystrone: multiple ORFs
  • also has shine-delgarno sequence
144
Q

Describe the ribosome unit in prokaryotes?

A
  • RNA-protein complex ribo- nucleoprotein
  • prokaryote version has 3RNA molecules and 2 sub-units
  • 50S and 30S combine to form 70S
145
Q

Describe mRNA synthesis of proteins

A
  • synthesis takes place in the cavity between subunits
  • small subunits hold mRNA
  • growing polypeptide exits through tunnel in the large subunit
  • adds 20 amino acids / second
146
Q

What is the rate of protein synthesis?

A
  • 20 amino acids added/second
147
Q

What are the 3 tRNA binding sites?

A

A: Aminoacyl binding site
P: peptidyl binding site
E: Exit site

148
Q

What is the role of the small subunit? Why is the tRNA binding site across both the small and large subunits? What is the catalytic region of the large subunit?

A
  • small subunit holds mRNA in place
  • tRNA binding site all over so that the anticodon loop and make contact with mRNA
  • catalytic region of the large subunit joins amino acids together to form peptide bonds
149
Q

Describe the position of the accepting arm of tRNA?

A
  • place in a position that allows the polypeptide chain toe sit though a tunnel in the back of the ribosome
150
Q

True or False: the growing polypeptide chain is always connected to the mRNA

A

False: the growing polypeptide chain is always attached to a tRNA

151
Q

How do antibiotics attack the ribosome?

A
  • attack translational apparatus (ribosome)
    eg; binding to tRNA sites, blocking exit tunnel
152
Q

what are the 4 steps to translation?

A
  • tRNA charging
  • initiation
  • elongation
  • termination
153
Q

Descrive tRNA charging? Where are the amino acids?

A
  • attachment of an amino acid to the tRNA is referred to as tRNA charging
  • all amino acids are attached to the adenine nucleotide in the 3’ end acceptor stem
  • the carboxyl group of an amino acid is linked to the 3; OH group the A nuceltidie of the tRNA
154
Q

Where is the energy required to bind amino acids to tRNA found?

A
  • provided by ATP
155
Q

What is responsible for binding tRNA to amino acids?

A
  • aminoacyl-tRNA
156
Q

how many different aminoacyl-tRNA synthetase are there? What do they do?

A
  • 20
  • each recognizes ONE amino acid and attaches it to the correct set of tRNAs
157
Q

Describe the key steps and requirements for initiation of translation

A
  • assmbley of ribosome subunits at the translation start site
  • base pairing of initiator-tRNA anticodon with start side code in the mRNA
  • requirements mRNA, small and large ribosome subunit, initiator-tRNA, initiation factors (Fs), guanosine triphosphate (GTP)
158
Q

Describe translation initiation in prokaryotes?

A
  • IF-3 binds to the small subunit preventing the large subunit from binding
  • small ribosome subunit binds to the mRNA
  • 16S rRNA of small ribosome subunit complementary base pairs the shinedalgorna sequence
  • initiator-tRNA anticodon UAC base pairs with mRNA start codon *UAG)
  • initiator tRNA is charged with fMET amino acid
  • IF-3 protein keeps large and small ribosome separate
  • IF1 and 2 facilitates the intiator-tRNA binding to the correct site
  • forms the 30-s initiation complex
159
Q

What occurs after the formation of the 30-S initiation complex?

A
  • imitation factor proteins dissociate and the large ribosome subunit bids
  • forms 70-S initiation complex (fully formed ribosome)
160
Q

Describe the key steps in elongation

A

1) entry of aa-tRNA into a site of the ribosome
2) peptide bond formation
3) ribosome translocation
4) tRNA exit from E site

161
Q

What are the requirements of elongation?

A
  • aa trnas, 70s initiation complex, elongation factors, GTP
162
Q

Describe the role of the 3 sites used to carry tRNAs?

A

A site: aminocyl - accepts aa-tRNA carrying next aa
P site: peptidyl site: holds tRNA carrying polypeptide
E site: discharge tRNAs leave ribosome from this site

163
Q

Describe the first step of elongation

A

aa-tRNA is delivered to the A site
- incoming aa-tRNA binds to the A-site
- elongation factors guides incoming aa-tRNA to the correct site
- aa-tRNA anticodon binds with mRNA codon

164
Q

Describe the second step of elongation

A

peptide bond formation
- peptide transferase catalyses peptide bond formation
- enzyme activity performed by rRNA in the large ribosome subunit (ribozyme)

  • peptide between aa on p-site tRNA and A site tRNA
  • peptide bond formation release aa from tRNA at p site
  • growing polypeptide chain not attached to A site tRNA = (dipeptide)
165
Q

Describe the 3rd step of elongation

A
  • ribosome translocated in the 5; to 3’ direction on mRNA
  • translocation facilitated by elongation factors (EFG+GTP in, GDP, Pi out)
  • P-site tRNA now located at E site, this tRNA exits
  • A site now at P site and A site is free
166
Q

Describe termination of elongation

A
  • protein synthesis terminates when ribosome translocates to a stop codon
  • no aa-tRNA enters of A site that has a stop codon
  • release factors (RF) protein binds to A site and triggers release of polypeptide from P site tRNA