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CHEE370 > Gene Expression and Replication > Flashcards

Gene Expression and Replication Flashcards

1
Q

Semiconservative model of DNA replication

A

Each strand of template DNA used to synthesize new strand + each template remains annealed w/new strand

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2
Q

DNA polymerases

A

Enzymes that synthesize new complementary DNA strands - most can only extend pre-existing nucleic acid strand at 3’ end of molecule by adding nucleotides

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3
Q

Replication bubbles

A

Suggest presence of replication forks on each side of bubble where enzymes operate to replicate new DNA

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4
Q

Okazaki fragments

A

Small new single-stranded DNA molecules that suggest that DNA synthesis at each replication fork is continuous on the leading strand while it’s discontinuous on the lagging strand

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5
Q

What did Okazaki conclude?

A

That DNA replication is semicontinuous

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6
Q

RNA primers

A

Fragments at 5’ end consisting of RNA

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7
Q

Primase

A

Synthesizes RNA primers

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8
Q

DNA replication in 3 quick steps

A
  1. Initiation
  2. Elongation
  3. Termination
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9
Q

Enzymes involved in DNA replication

A
  • Helicase
  • Single strand binding protein
  • Primase
  • DNA polymerase
  • Topoisomerase
  • Exonuclease
  • Tus protein
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10
Q

Helicase

A

Separates + unwinds DNA strands

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11
Q

Single strand binding protein

A

Prevents separated parental strands from re-annealing after being separated into 2 single strands by helicase

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12
Q

Topoisomerase

A

Cleaves 1/both of parental DNA strands to relieve tension caused by supercoiling of strands due to movement of replication forks - changes DNA topography (degree of unwinding)

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13
Q

Exonuclease

A
  • Removes RNA primers from Okazaki fragments - when necessary it removes improperly incorporated nucleotide before resuming DNA synthesis (proofreading)
  • Also removes RNA from lagging strand + replaces w/DNA as elongation continues
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14
Q

Tus protein

A

In prokaryotes only + binds to terminator site - prevents DNA unwinding by helicase which arrests movement of replication fork as it reaches Ter site

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15
Q

Terminator site

A

Sequence on bacterial chromosome that leads to arrest of helices + ends DNA replication

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16
Q

Replication origin/ori sequence

A
  • Specific sequence at prokaryotic/eukaryotic DNA - generally bound by proteins + when cell division occurs additional proteins are recruited to ori site
  • DNA replication starts once essential proteins have been recruited to the ori site
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17
Q

Prepriming complex

A

oriC + DnaA + DnaB + DnaC protein complex - once formed, DnaC released

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18
Q

Stringent control

A

Plasmids that divide only once per cell generation are under this

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19
Q

Relaxed control

A

Plasmids that replicate independently from cell division are under this - plasmids used in biotech are under this which allows for presence of several plasmids/engineered cell

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20
Q

2 steps of formation of protein complex which initiates DNA replication

A
  1. Formation of pre-replicative complex
  2. Activation of pre-replicative complex (occurs during S phase) - activated pre-replicative complex known as initiation complex
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21
Q

Pre-replicative complex

A

Forms during G1 phase of cell cycle

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22
Q

Origin recognition complex

A

In eukaryotes the ori site is 1st recognized by this group of proteins which recruits other proteins including MCM complex (acts as helicase) - once all protein in pre-replicative complex have bound ori site the pre-replicative complex is “licensed”

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23
Q

Kinases

A
  • Required to phosphorylate several proteins during activation of pre-replicative complex - they are enzymes that phosphorylate other proteins
  • Phosphorylation events in activation of pre-replicative complex leads to binding of primase, polymerase + replication protein A (single stranded binding protein in eukaryotes)
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24
Q

Why do primers contain RNA instead of DNA?

A

Higher error rate occurs during primer synthesis in comparison to error rate during elongation process - so RNA use allows for subsequent recognition + removal of primers from newly synthesized DNA strands + reduces error rate compared to expected error rate if DNA primers used

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25
Q

How does DNA polymerase catalyze the extension of the new DNA strand once loaded onto DNA?

A

It catalyzes the addition of nucleotides at 3’ end of new strand

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26
Q

How can DNA polymerase catalyze 4 dif specific reactions?

A

It recognizes substrates that form acceptable Watson-Crick base pairs w/template DNA strand bound to enzyme so it allows for rapid entry + exit of 4 possible dNTP molecules

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27
Q

Sliding clamp mechanism

A
  • DNA polymerase is a clamp that remains open until dNTP enters + forms Watson-Crick base paired structure w/parental strand - clamp then closes thus catalyzing addition of this nucleotide to growing DNA strand
  • Clamp slides to next position along parental DNA+ opens and is ready to catalyze next rxn
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28
Q

Nicks

A

Missing phosphodiester bonds between Okazaki fragments

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29
Q

Replisome

A

Proteins involved in DNA replication form this complex - it moves continuously w/out detaching from leading strand but periodically detaches from lagging strand to allow discontinuous synthesis of Okazaki fragments on lagging strand

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30
Q

Telomeres

A

Nucleotides at 5’ extremity of lagging strand which can’t be replicated - shorten as a function of # of cell doublings in adult animal cells

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31
Q

Telomerases

A

Can extend length of telomeres to counteract shortening that occurs upon each replication cycle

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32
Q

Fts proteins

A

Forms ring in centre of elongating cell that aids in chromosome segregation between 2 daughter cells

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33
Q

Mitosis

A

Occurs during M phase - asexual cell division in eukaryotes happens via this process + includes prophase (DNA condensation + mitotic spindle formation) + metaphase (chromosomes align) + anaphase + telophase (mitotic spindle disassembles nuclear membrane closes around segregated chromosomes + DNA condensation decreases)

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34
Q

Excision repair

A

Nucleotides (nucleotide excision repair) or nitrogenous bases (base excision repair) are removed + replaced by Watson-Crick base paired nucleotides

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35
Q

How does DNA glycosylase alter the nitrogenous bases of nucleotides?

A

It removes glycosidic bond between nitrogenous base + deoxyribose on DNA backbone - endonuclease then cleaves 1 side of deoxyribose residue + adjacent nucleotides removed by exonuclease (gap is filled by polymerase + nicks sealed by ligase)

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36
Q

RNA polymerase

A

Carries out RNA synthesis

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37
Q

Coding strand

A

Primary RNA transcript has same sequence as DNA strand complementary to this strand - sequence provided in 5’ to 3’ direction

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38
Q

Post-transcriptional modifications

A

Series of modifications that primary RNA transcript undergoes prior to obtaining mature mRNA - increase diversity of mature RNA molecules obtained from single primary transcript

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39
Q

Initiation in transcription

A

RNA polymerase binds to promoter sequences on DNA - RNA polymerase recruited to promoter indirectly via proteins that recognize specific promoter sequences

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40
Q

Promoters

A

DNA sequences containing sequence bound by RNA polymerase - define location of transcription initiation site + control rate of gene transcription

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41
Q

What are the proteins that recognize promoters?

A

Sigma factors (subunits of RNA polymerase that detach from other subunits when transcription initiated) in prokaryotes + transcription factors in eukaryotes

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42
Q

Elongation in transcription

A

RNA polymerase synthesizes RNA starting at initiation site - RNA elongation occurs via addition of ribonucleotides at free 3’ OH group of growing RNA strand

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43
Q

Termination in transcription

A

RNA polymerase removed from DNA + RNA synthesis stops - occurs due to presence of termination sequences on RNA chain

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44
Q

Holoenzyme

A

RNA polymerase is this in a prokaryote - it’s an enzyme composed of several protein subunits

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45
Q

Conserved sequences

A

Have high similarity between species

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46
Q

Consensus sequences

A

Have high similarity between genes

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47
Q

Pribnow box

A

Consensus sequence that’s highly conserved between species in prokaryotic cells

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48
Q

Upstream recognition sequence

A

Consensus sequence which determines type of sigma factor that associates w/the promoter

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49
Q

Efficient promoters

A

Lead to higher transcription rates

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50
Q

What happens once the RNA polymerase holoenzyme has bound the promoter?

A
  • Promoter melts - 2 strands of promoter separate + RNA polymerase core protein binds to template strand
  • Also sigma factor dissociates from RNA polymerase core enzyme
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51
Q

What is the first base present on template strand/5’ end of RNA chain?

A
  • T residue / A residue
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52
Q

Regulatory sequences

A

Regulate transcription factor binding to promoter - eukaryotic promoters may include these upstream or downstream from transcription initiation site

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53
Q

What 3 types of RNA polymerase do eukaryotes possess?

A
  • RNA polymerase I which synthesizes most rRNAs
  • RNA polymerase II which synthesizes primary transcripts of mRNA
  • RNA polymerase III which synthesizes tRNAs + 5s subunit of rRNA + other short RNA molecules
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54
Q

Housekeeping genes

A

Expressed constitutively + contain sequence rich in GC (GC box)

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55
Q

TATA box

A

AT-rich sequence in genes that are selectively expressed - plays role in initial binding of RNA polymerase to DNA + in formation of transcription bubble

56
Q

Types of transcription factors involved in regulation of promoters that bind RNA polymerase II

A
  • General transcription factors
  • Upstream transcription factors
  • Inducible transcription factors
57
Q

General transcription factors

A

Required for synthesis of all mRNA - include 6 transcription factors that form pre-initiation complex

58
Q

Upstream transcription factors

A

Bind upstream from transcription site + repress/activate general transcription factor/RNA polymerase complex

59
Q

Inducible transcription factors

A

Must 1st be activated/inhibited by post-translational modifications before binding DNA + acting in similar fashion to upstream transcription factors - play important role in tissue specification

60
Q

Enhancer

A

DNA sequences that aren’t transcribed but are required for full activation of promoter - contribute to tissue specific expression of genes in eukaryotes

61
Q

Silencers

A

DNA sequences that overturn transcription if they’re bound to repressor proteins

62
Q

Does RNA polymerase require the concerted action of helicase or topoisomerase?

A

No

63
Q

Can transcription on a given gene be initiated before the prior RNA can reach the termination site?

A

Yuh so several RNA polymerases can transcribe a gene at a given time

64
Q

Why does transcription termination in prokaryotes occur?

A

Due to intrinsic terminators or with assistance of the Rho factor

65
Q

Intrinsic terminators

A
  • Termination occurs via presence of GC rich RNA sequence w/high sequence homology that forms hairpin structure - sequence followed by U residues
  • This leads to termination b/c RNA-DNA hybrid pulls out from RNA polymerase
66
Q

Rho factor dependent terminators

A
  • Rho factor is protein that binds to specific sequences on rho utilization sites - it then travels along RNA strand towards polymerase
  • If it can overtake RNA polymerase then it unwinds DNA/RNA complex + removes RNA polymerase from transcription bubble
  • Requires presence of DNA sequence that causes pause for RNA polymerase (rho-dependent pause site)
67
Q

In prokaryotes can one prokaryotic gene code for several genes (cistrons)?

A

Yes so prokaryotic mRNA can be polycistronic

68
Q

Monocistronic

A

mRNA molecules in eukaryotes are this - each mRNA codes for single gene/protein

69
Q

What post-transcriptional modifications lead to conversion of primary RNA transcript into mRNA?

A
  • 5’ capping
  • Splicing
  • Polyadenylation
70
Q

5’ capping

A
  • 1 of 3 P groups at 5’ end is removed followed by addition of G residue which is then methylated - methyl group then transferred to 7 position on G residue + final cap obtained is m7G residue at 5’ end of RNA
  • m7G cap determines location of translation start site - its presence is required for translocation of mRNA from nucleus to cytosol + increases mRNA stability
71
Q

Introns

A

Regions of RNA that aren’t translated

72
Q

Exons

A

Coding sequences that are translated to proteins

73
Q

Splicing

A

Process in which introns are removed from primary transcript to form mRNA - can occur via self-splicing or spliceosome catalyzed splicing

74
Q

Self-splicing

A

1st is cleavage of intron at its 5’ end + then intron catalyzes joining of 2 axons flanking intron as well as cleavage of intron at its 3’ end - intron also reacts w/itself to form a partial cyclical structure (intron lariat) or a cyclized intron

75
Q

Ribozymes

A

RNA molecules that have catalytic activity + act as enzymes

76
Q

Spliceosomes

A
  • RNA-protein complexes that catalyze reactions like RNA splicing - include small nuclear RNAs, proteins that bind to mRNA + primary RNA transcript to be spliced
  • Recognize introns via snRNAs + intron excision proceeds in process that requires splicing factors
77
Q

Alternative splicing

A

When splicing of primary RNA transcript doesn’t occur in same location - some introns can become exons + vice versa

78
Q

Evolutionary advantage of intron presence

A

Increases probability of rearrangement of genes allowing insertion of exon into new gene w/out disrupting other exons + increases diversity of proteins produced from given RNA transcript + contain regulatory sequences that alter level of gene expression they’re associated with

79
Q

Polyadenylation

A
  • Primary transcripts in eukaryotes are located at a site prior to a GU rich sequence recognized by cleavage stimulation factor protein
  • Then poly(A) chain is added by poly(A) polymerase + poly(A) binding protein II binds to poly(A) tail
80
Q

Is the half-life of an mRNA molecule correlated w/length of the poly(A) tail?

A

Yuh

81
Q

Nuclear pore complex

A
  • Channels that span nuclear membrane + allow free passage of small molecules + selective transport of larger molecules like mRNA which passes through starting at 5’ cap end
  • Once in cytosol nuclear receptor protein dissociates from mRNA + re-enters nucleus
82
Q

Inducers + co-repressors

A
  • Environmental triggers that lead to changes in gene expression - small molecules involved in cell metabolism + they increase/decrease expression of certain proteins
  • In prokaryotes they often turn on/off entire set of genes
83
Q

Operon

A

Unit formed by structural genes + regulatory proteins that control expression of genes present on same mRNA molecule after transcription

84
Q

Activators

A

Proteins that bind to regulatory DNA regions + increase level of gene expression - exert positive control on downstream genes + act on weak promoters that don’t effectively bind to RNA polymerase

85
Q

Repressors

A

Proteins that bind regulatory DNA regions + decrease level of gene expression - exert negative control on gene expression but the effect depends on its binding affinity for regulatory DNA regions + its expression level

86
Q

Activator-binding sites

A

Regulatory DNA regions bound by activators

87
Q

Operators

A

Regulatory DNA regions bound by repressors in prokaryotes

88
Q

What is control of lactose operon facilitated by?

A
  • A repressor + an activator - activator is CAP which binds to binding site in presence of cAMP
  • In absence of glucose the cAMP intracellular concentration increases leading to CAP binding to its binding site
  • Repressor is lacI which binds to operator
89
Q

When does the maximal expression of genes in the lactose operon occur?

A

In the presence of lactose + absence of glucose

90
Q

Enhancer + silencer sequence in eukaryotes

A

Similar to activator-binding sites + operators in prokaryotes but they can be found far from the promoter - their effect on promoter activity doesn’t depend on direction/ exact location compared to promoter

91
Q

In eukaryotes what does binding of transcription factors to regulatory DNA regions depend on?

A

Nuclear concentration of transcription factors + their activation (modulated by their tertiary/quaternary structure which is regulated by presence/absence of ligands/post-translational modifications)

92
Q

What is a common post-translational modification required for transcription?

A

Phosphorylation (performed by kinases)/dephosphorylation

93
Q

Translocate

A

Cross nuclear membrane

94
Q

Protein motifs

A

Local tertiary structures - can recognize changes in DNA strand shape resulting from diff nucleotide sequences

95
Q

Epigenetic modifications

A

Methylation of certain nucleotides + methylation/acetylation of histones - play important role in silencing of certain genes

96
Q

Open reading frame

A
  • In sequence of nucleotides that code for polypeptide chain - spans from start codon to codon just before stop codon
  • Prokaryotic mRNA contain several while eukaryotic mRNA contain one
97
Q

Difference in fates of secreted + membrane proteins in prokaryotes vs eukaryotes

A
  • Prokaryotes: proteins translocated to cytoplasmic membrane (reducing)
  • Eukaryotes: translocated to ER (oxidizing) + several post-translational modifications applied to polypeptide chain here
  • So disulphide bonds don’t form in cytosol of prokaryotes/eukaryotes but do in ER of eukaryotes
98
Q

Is the genetic code triplet?

A

Yuh - each sequence of 3 nucleotides codes for 1 amino acid + sequence of 3 nucleotides on mRNA is called codon

99
Q

Is genetic code degenerate?

A

Yuh - several codons code for same amino acid

100
Q

Is genetic code non-overlapping?

A

Yuh - each nucleotide part of single codon + is read once

101
Q

Is genetic code comma free?

A

Yuh - all nucleotides in open reading frame are part of a codon

102
Q

Main start codon in prokaryotes + eukaryotes

A

AUG

103
Q

2 untranslated regions on mRNA molecule

A
  • 5’ untranslated region (between transcription initiation site + start codon) + 3’ untranslated region (between stop codon + transcription termination site or poly(A) tail on eukaryotes)
104
Q

Adaptor hypothesis

A

Adaptor molecules that recognize specific nucleotide sequences lead to addition of specific amino acids to polypeptide chain - adaptors consist of tRNA thus allowing recognition of mRNA sequences via Watson-Crick base pairing

105
Q

2 levels of specificity required to explain how genetic code is translated into a specific polypeptide chain in the adaptor hypothesis

A
  1. Specific pairing of codons on mRNA w/adaptor molecule
  2. Specific pairing of adaptor molecules w/amino acids
106
Q

tRNA molecules

A

Act as adaptor between mRNA + amino acids via Watson-Crick base pairing between codons on mRNA + anticodons on tRNA - interacting codons + anticodons are antiparallel

107
Q

Why are there only 20 dif amino acids present in polypeptide chains if there are 61 codon sequences that don’t code for a stop codon?

A

Lack of specificity at 3rd position of codon (wobble position) + certain tRNA molecules w/dif anticodons can be linked to same amino acid

108
Q

Isoaccepting tRNAs

A

Different tRNAs that carry same amino acid

109
Q

Aminoacyl-tRNA synthetases

A
  • Enzymes that attach amino acids to 3’ end end of tRNA molecules - they recognize tRNAs via interactions between protein + anticodon region of tRNA
  • Can bind >1 tRNA so each enzyme can bind a set of isoaccepting tRNAs
110
Q

What explains why the genetic code is degenerate?

A

Wobble in codon/anticodon base pairing + capacity of aminoacyl-tRNA synthetases to add same amino acids to dif tRNAs

111
Q

Codon usage bias

A

Bias towards using a given codon for 1 amino acid - often necessary to modify genetic sequence to fit the bias of the host organism

112
Q

Transpeptidation

A

Leads to peptide bond formation between amino acids - catalyzed by rRNA molecule + occurs at C-terminus of growing polypeptide chain which is why mRNA read in 5’ to 3’ direction

113
Q

What occurs as ribosome moves along mRNA strands in 5’ to 3’ direction?

A

Growing polypeptide chain emerges from ribosome - N-terminus of chain emerges 1st + chain is elongated at C-terminus until ribosome encounters stop codon

114
Q

Can several ribosomes travel along open reading frame at once?

A

Yuh

115
Q

Summarized translation initiation

A

Leads to binding of small ribosomal subunit + of 1st tRNA to start codon followed by binding of large ribosomal subunit - process requires hydrolysis of GTP molecule that’s bound to an initiation factor

116
Q

Peptidyl site (P site)

A

Elongation process able to proceed as long as aminoacyl-tRNA is bound to this

117
Q

Aminoacyl site (A site)

A

Aminoacyl-tRNA can enter this after P site - A site located on 3’ end side of P site + the aminoacyl-tRNAs that enter this site are bound to an elongation factor

118
Q

Elongation factor

A

Bound to GTP molecule - when proper codon-anticodon base pairing is established the GTP is hydrolyzed + elongation factor released from aminoacyl-tRNA bound to A site so that transpeptidation rxn can occur

119
Q

Translocation

A

Once transpeptidation done the ribosome moves by 1 codon towards 3’ end of mRNA - this requires energy + uncharged tRNA that was present in P site now enters exit site (E site)

120
Q

E site

A

Uncharged tRNA released from ribosome so aminoacyl-tRNA that was in A site is now in P site + N-terminus of amino acid on the tRNA is attached to growing polypeptide chain while its C-terminus is available for next round of elongation

121
Q

Release factor

A

Enters A site instead of aminoacyl-tRNA when ribosome reaches stop codon - so instead of new amino acid being transferred to polypeptide chain an OH group from water molecule is transferred to new polypeptide chain causing its hydrolysis instead of transfer of chain to aminoacyl-tRNA
- So polypeptide chain leaves ribosome

122
Q

Molecular chaperones aka heat-shock proteins

A

Proteins that aid in folding of other proteins - their expression levels increase when cells briefly exposed to heat which increases amount of improperly folded proteins which have exposed hydrophobic surfaces that are bound by molecular chaperones to create conditions conducive to proper folding

123
Q

Inclusion bodies

A

Protein aggregates that result from collisions + aggregation if they’re not folded properly - must be purified, denatured + re-folded to obtain proper protein structure

124
Q

Signal peptide

A
  • Sequence at N-terminus of polypeptide chain generates this - targets polypeptide chain to membrane + bound by proteins that transport polypeptide chain to translocator proteins bound to membranes
  • Translocators allow passage of polypeptide chain across membrane (found on cytoplasmic membrane in prokaryotes + ER membrane in eukaryotes)
125
Q

Signal recognition particle

A
  • Signal peptide is recognized by this for proteins that are translocated across membranes at same time as translation - binds to signal peptide + carries growing polypeptide chain along polypeptide chain along w/ribosome to signal receptor particle receptor on membrane
  • Signal peptide then binds to translocator + translocation occurs to other membrane side w/ribosome attached to membrane
126
Q

Signal peptidase

A

Recognizes peptide sequence following signal peptide in secreted proteins - located on other side of membrane + in its absence the polypeptide chain will remain bounded to membrane

127
Q

Mutations

A

Permanent changes in DNA sequences that can be passed onto daughter cells

128
Q

Recombination

A

When DNA sequences removed/inserted into DNA by viruses/enzymes that lead to DNA segment movement between plasmids + chromosomes

129
Q

Mobile genetic elements

A

DNA segments prone to recombination

130
Q

Wild-type

A

Describes organisms that don’t carry the mutation

131
Q

Missense mutations

A

Change a codon leading to change in amino acid coded by that codon

132
Q

Nonsense mutations

A

Introduce stop codon leading to premature termination of translation - polypeptide chain obtained is incomplete

133
Q

Silent mutations

A

Result from sub of single nucleotide - new codon obtained codes for same amino acid as original codon + polypeptide chain obtained after translation is same

134
Q

Frameshift mutations

A

Leads to insertion/deletion of 1/2 nucleotides so that reading frame downstream of mutation changes + all codons downstream from mutation are in dif reading frame compared to wild-type sequence

135
Q
A

CHEE370 (6 decks)

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