Molecular Biology Flashcards

1
Q

Microscopy: explain dispersion

A
  • The phase velocity of a wave depends on its frequency.
  • This causes focused white light going into a lens to disperse into different colors (because wavelengths of light vary by the color) when exiting the lens.
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2
Q

Microscopy: explain chromatic aberration

A
  • White light focused through a lens disperses into the different wavelengths (colors) due to their different phase velocities.
  • Chromatic aberration is the failure of the lens to focus all the colors to the same point.
  • This can be offset using achromatic lenses.
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3
Q

Microscopy: explain diffraction

A

Interference of waves at obstacles results in contructive and destructive interference patterns due to the wave nature of light.

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4
Q

Microscopy: explain refraction

A

Refraction is a change of the angle of light (or other waves) when passing through a boundary between 2 mediums of a different refractive index.

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5
Q

What is the minimum distance resolvable by optical microscopy?

A
  • λ/2NA (or λ/2nsinα, or about 1/2 the wavelength)
    • where λ = wavelength of light, n = refractive index, α = opening angle of lens, NA = numerical aperture of lens
    • ~ 200 nm (0.2 um) = diffraction barrier
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6
Q

Explain constructive and destructive interference.

A
  • Constructive interference:
    • If two coherent waves with phase shift 0 or 2π interfere, a wave with doubled amplitude results
  • Destructive interference:
    • If two coherent waves with phase shift π interfere, both waves are annihilated

(π = one part of the phase, like half the wavelength)

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7
Q

What is Fourier Analysis and Synthesis?

A
  • Fourier Synthesis:
    • Waveforms (signals) can be generated through superposition of different harmonic waves with wavelength λ and frequency ν and different amplitudes
  • Fourier Analysis:
    • Signals can be disassembled into harmonic waves with different amplitude to describe the signal (waveform) in the frequency domain
    • This simplifies computation of waves
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8
Q

What does the Numerical Aperture (NA) of an objective describe?

A
  • NA describes the relationship between the opening angle of that objective’s lens and the refractive index of the medium (air, water, oil) between the objective and the sample being imaged (or its coverslip)
  • NA = nsinα, where n = refractive index, α = opening angle of lens (theta can also be used as the symbol instead of alpha)
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9
Q

What are phase objects?

A

Unstained objects that do not absorb light but slightly alter the phase of the light diffracted by the specimen, usually by retarding such light approximately 1/4 wavelength as compared to the undeviated direct light passing through or around the specimen which is unaffected.

The light is slowed by the specimen because of its thickness, refractive index, or both.

  • Amplitude objects absorb light as it passes through them, resulting in differences in amplitude of the wave.
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10
Q

What is fluorescence?

A

Fluorophores that emit light at another wavelength after irradiation with light

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11
Q

Explain the basics of Time-Correlated Single-.Photon Counting (TCSPC).

A

A way to measure fluorescence lifetime

  • Fluorescence is excited repeatedly by short laser pulses. When a laser pulse hits the sample, time is started. When fluorescence is emitted (and detected by a PMT), time is stopped.
  • Time delay between excitation and emission is measured, and the resulting data is a histogram of times to emission, with each line representing one time bin, and the intensity of each line representing a count of how many events took that amount of time (the count of events at each given time t).
  • This cycle is repeated many times to generate the histogram.To avoid pile-up (overlap between the cycles/artefacts), <5% of the cycles should result in fluorescence.
  • Fluorescence lifetime Tau is then calculated by fitting the histogram to an exponential decay function: N(t) = e-t/T N number of counts as a function of t time = e to the -t/Tau
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12
Q

What is the difference between direct and indirect immunolabeling?

A
  • Direct: using fluorescently labeled primary antibodies to bind directly to your sample of interest
  • Indirect: using fluorescently labeled secondary antibodies - these bind to the primary antibody (labeled or unlabeled) bound to the sample of interest
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13
Q

What is photobleaching?

A

The destruction of a fluorophore/fluorescent protein through irradiation/exposure to light.

Irreversible loss of fluorescence of a fluorophore during irradiation.

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14
Q

What are chemical tags, give examples and explain their mode of action.

A

The interaction of genetically encoded tags with small molecules allows for fluorescent labeling of proteins in live cells

Covalent

  1. Substrate is conjugated to the fluorescent probe -> substrate-probe conjugate
  2. Tag (self-modifying protein/enzyme) is bound to the POI
  3. Enzyme tag irreversibly transfers the substrate alkyl group to one of its amino acid residues creating a covalent bond of the substrate-probe conjugate to the tag
  4. Result: The probe tags enzyme and its fusion protein irreversibly
  • Example: SNAP-Tag
    • Enzyme = mutant hAGT
    • Substrate = modified benzylguanine, an O6-benzylguanine (BG) derivative
    • Amino acid residue = cysteine
  • Example: CLIP-Tag
    • Enzyme = a different mutant hAGT
    • substrate = benzylcytosine
    • Amino acid residue = cysteine
  • Example: Halo-Tag
    • Enzyme = Mutant dehalogenase enzyme
    • substrate = alkylhalide derivatives
    • amino acid residue = asparagine, and mutated His289 prevents hydrolysis of this covalent bond

Non-covalent

  • TMP-tag
  • Tag = eDHFR (E. coli dihydrofolate reductase)
  • Substrate = trimethoprim derivatives
    • Trimethoprim binds to eDHFR with high affinity as a receptor-ligand pair
  • TMP is conjugated to the fluorophore and then binds to eDHFR
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15
Q

What are Quantum Dots? Advantages and limitations of QDs.

A
  • CdSe, for example. These are semi-conducting particles which exhibit stable fluorescence. Wavelength of absorption and emission depends on their size.
  • Advantages – they do not photobleach, and they can be labeled with antibodies
  • Disadvantages – large size of the QD can affect the molecule of interest
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16
Q

Describe confocal fluorescence microscopy.

Why does confocal fluorescence microscopy achieve improved axial sectioning?

A
  • Confocal fluorescence microscopy uses a pinhole to exclude out of focus light from the image
    • One is after the light source
    • One is before the detector
  • Only light from the plane in focus is imaged at one time, without the additional light scatter from the other planes. Then you only need to change the plane of focus axially in order to image the next portion of the sample in the z direction, acquiring a series of images which can be reconstructed.
  • The size of the pinhole determines the thickness of the optical section
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17
Q

Describe two-photon microcopy.

Does two-photon microscopy achieve a higher spatial resolution?

A
  • Longer wavelength light is used at high intensity to excite the fluorophores, so that 2 photons are hitting the fluorophore at the same time. The highest probability of this happening is at the laser focal point, so only fluorophores directly at in the focal point are absorbing 2 photons and getting excited. This eliminates any light coming from the rest of the sample/gives darker background. This reduces phototoxicity and improves light detection.
  • No, it does not actually achieve higher spatial resolution because the wavelength being used is longer (spatial resolution = ~ ½ the wavelength), even though fluorescing volume is reduced.
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18
Q

Why does two-photon microscopy achieve a higher imaging depth?

A
  • Because the longer wavelength excitation light can penetrate deeper into the sample due to minimized scattering, and the background signal is strongly suppressed.
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19
Q

Describe how a confocal fluorescence microscope works. Why do you get a better depth of field?

A
  • Confocal fluorescence microscopy uses a pinhole to exclude out of focus light from the image
    • One is after the light source
    • One is before the detector
  • Only light from the plane in focus is imaged at one time, without the additional light scatter from the other planes. Then you only need to change the plane of focus axially in order to image the next portion of the sample in the z direction, acquiring a series of images which can be reconstructed.
  • The size of the pinhole determines the thickness of the optical section
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20
Q

Schematically draw the beam path (from the sample to the detector) of a confocal fluorescence microscope.

A
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21
Q

Explain why electron microscopy has a higher resolution than light microscopy.

What limits the resolution of electron microscopy?

A
  • Because the wavelength of the electrons used in EM are so short, like around 0.004 nm. This is achieved by increasing the velocity of the electrons. The faster the velocity, the shorter the wavelength.
  • Current resolutions are ~ 0.1 nm for EM
  • Aberrations of optical components limits the resolution of EM
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22
Q

How does EM work?

A
  • A heated cathode/tungsten wire produces electrons. These electrons are accelerated in a vacuum. Magnetic lenses focus and collect the electrons.
  • In EM, you see electron density.
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23
Q

Explain FRET.

Which distances can be determined experimentally with FRET?

A

*

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24
Q

What different functional principles of ion channels do you know?

How to determine the activity of ion channels?

A
  • xxx
  • You can use the patch clamp technique, where single ion channels are pulled up into the tip of a micropipette. Voltages can be applied and voltages measured to determine when the ion channel is in the on or off state.
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25
Q

Explain the functional principle of multiphoton fluorescence microscopy.

Why do you achieve a higher depth of field?

A
  • Longer wavelength light is used at high intensity to excite the fluorophores, so that 2 photons are hitting the fluorophore at the same time. The highest probability of this happening is at the laser focal point, so only fluorophores directly at in the focal point are absorbing 2 photons and getting excited. This eliminates any light coming from the rest of the sample/gives darker background. This reduces phototoxicity and improves light detection.
  • Because the longer wavelength excitation light can penetrate deeper into the sample due to minimized scattering, and the background signal is strongly suppressed.
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26
Q

How to determine interactions of proteins in the range of 2-10nm in cells?

Briefly explain the mechanism.

What methods do you know to mark the proteins involved with a dye?

A

*

27
Q

What determines the resolution of an optical microscope?

A
  • Wavelength of light, refractive index of the medium between the lens and the sample, opening angle of the lens (its NA)
  • Resolution (minimal resolvable distance) = λ/2NA (or λ/2nsinα, or about 1/2 the wavelength)
    • where λ = wavelength of light, n = refractive index, α = opening angle of lens, NA = numerical aperture of lens
    • ~ 200 nm (0.2 um) = diffraction barrier
28
Q

What are protein tags? Name four tags you know.

A
29
Q

How does pyrosequencing work?

A
  • Sequencing by synthesis/2nd generation sequencing/massively parallel sequencing
  • DNA strand to be sequenced is attached by the 3’ end to a solid surface
  • Incorporation of the next dNTP by polymerase causes cleavage and release of a pyrophosphate (PPi)
  • Next, the PPi is converted to ATP which interacts with luciferase to release light (this happens when luciferase converts luciferin to oxyluciferin and at the same time this oxidation causes ATP to release a flash of light)
  • This means an incorporation of that nucleotide causes a flash of light which is then detected by a CCD
    • The intensity of light released is proportional to the number of bases incorporated in each location on the chip (2 dNTPs incorporated into the strand -> 2x the light in that spot)
  • Only one dNTP type is added to the reaction at a time, then apyrase is added to degrade all unincorporated dNTPs and ATP, it is washed, and the next dNTP is added to the reaction, so you only get incorporation of that one nucleotide type at a time as determined by the DNA sequence.
  • Modified ATP (dATP-S) is used as the base in the dATP reaction b/c addition of dATP would result in light production even in absence of incorporation
30
Q

How does Acyclovir (Acyclo-guanosine) work?

A
  • Aciclovir is a modified nucleotide/nucleotide analogue/guanosine analogue which can be converted by viral thymidine kinase to acyclo-guanosine monophosphate with 3000x more affinity than human cellular thymidine kinase.
  • Then it is converted to acyclo-GTP. This has 100x more affinity for viral polymerase than for human cellular polymerase.
  • Once incorporated, it inhibits DNA polymerase and DNA synthesis stops.
  • Viral enzymes cannot remove acyclo-GTP from the DNA chain, so the virus particle replication ends.
31
Q

Which methods of single-molecule detection are you familiar with?

A
  • STORM – stochastic optical reconstruction microscopy
  • dSTORM
  • PALM – photoactivated localization microscopy
  • Patch clamp
  • Force spectroscopy (optical tweezers, ATM)
32
Q

Explain the functional principle of atomic force microscopy.

A
  • A tiny tip attached to the end of a cantilever moves over the surface of a sample. Displacement of the tip/cantilever can be detected using lasers. This can provide an image of whatever is on the surface.
  • Attraction forces such as Van der Waals interactions or electrostatic attraction can also be measured as a molecule attached to the tip is brought close to a molecule attached to the slide surface, and they interact. This pulls the tip toward the surface and the displacement is measured.
  • Molecules such as proteins can be attached to the tip and to the surface of the slide, and the forces applied while pulling the molecule apart or stretching it can be measured. This provides information about the binding strength of bonds or protein folds, etc.
33
Q

What is PNA?

Why is the binding strength of PNA greater than DNA?

A
  • Peptide nucleic acid - It is a synthetic nucleic acid polymer / nucleotide analogue which uses glycine (an amino acid) instead of a deoxyribose or ribose + phosphate backbone
  • It has higher binding strength because there are no negative charges
  • Less specific b/c of the higher binding strength, but also has higher hybridization efficiency
34
Q

How does localization microscopy (PALM/dSTORM/STORM) work?

What defines the resolution?

A
  • Photoswitchable fluorophores are excited at low power so that only a few at a time turn “on” and are imaged. Those turn “off” and others turn on, and imaging continues like this. Because the fluorophores stochastically switch on and off, and having only a few switched on at one time reduces light scatter from neighboring fluorophores (background noise), individual molecules can be localized with high precision.
  • This is basically temporal control (rather than positional) but is stochastic in these examples.
  • For STORM, different buffers are used to induce the blinking fluorophores.
  • In PALM, low power is used to excite only a few at a time, they photobleach, then you change something to excite the next set.
  • The localization of the molecule is calculated from intensity of the signal and the gaussian function of the point spread function. The image is reconstructed.
  • Resolution is ~10 nm
35
Q

What is a dideoxinucleotide triphosphate (ddNTP) and what is its use?

A
  • It is a dNTP which is missing the 3’ OH group.
  • When it is incorporated into a growing DNA strand, DNA elongation is terminated.
  • This is the basis of Sanger sequencing.
36
Q

What are the differences and similarities between bacterial polymerase I and III?

A
  • Pol I
    • 5’ to 3’ polymerization
    • slower polymerization speed
    • low processivity
    • proofreading exonuclease activity = 3’ to 5’ direction
    • also has 5’ to 3’ exonuclease activity - can replace already synthesized strand in 5’ to 3’ direction when proofreading
    • only 1 subunit
  • Pol III
    • fast polymerization speed
    • very high processivity
    • main replication enzyme
    • >10 subunits
    • proofreading exonuclease activity = 3’ to 5’ direction
37
Q

What are the 3 main typesof bacterial Pol III subunits and their functions?

A
  • clamp-loading
    • ATPase: recognizes a primed DNA with 3’ OH end available for sliding clamp to load and bind
  • core
    • 5’ to 3’ polymerase
    • 3’ to 5’ exonuclease
  • sliding clamp
    • binding to DNA/processivity
38
Q

What is the function of the clamp-loading complex?

A

Bacterial DNA replication

  • Clamp loader binds ATP which opens the associated sliding clamp which is bound to the Pol
  • Sliding clamp surrounds the primed DNA (3’ OH end available)
  • Clamp loading complex binds to primed DNA and pulls it through the open sliding clamp
  • ATP hydrolysis -> dissociation of clamp loaders, leaving only sliding clamp and polymerase
39
Q

What is the Trombone model of replication?

A
  1. DNA helicase unwinds/separates ds DNA strands
  2. ssDNA binding proteins keep the strands open
  3. Pol III synthesizes the leading strand in 5’ to 3’ direction continuously
    • sliding clamp keeps the Pol III associated with the DNA strand
  4. DNA primase synthesizes RNA primer on the other strand so Pol III can synthesize Okazaki fragments (the lagging strand) in the 5’ to 3’ direction. This happens in discontinuous sections.
  5. clamp loading protein coordinates synthesis from replication fork b/c it is bound to polymerase, helicase, and loads the sliding clamp
  6. DNA Pol I replaces RNA primers with DNA
  7. DNA ligase links the Okazaki fragments
40
Q

What are the parts of the replisome?

A
  • Helicase
  • Clamp-loading complex
  • Polymerase III (2 or 3 of them)
  • Sliding clamp (beta clamp)
  • Primase
  • single strand DNA binding proteins
  • DNA gyrase to remove supercoils
41
Q

What is methyl directed mis-match repair (MMR)?

A
  • DAM methylase transfers a methyl group to the A of a 5’-GATC-3’ sequence on both strands, but directly after DNA synthesis, the newly replicated strand is non-methylated while the template strand is methylated
  • If a mismatch is detected between the template and the new strand,
    • MutS binds to the mismatch
    • MutL binds to MutS
    • MutH (an endonuclease) binds to next GATC sequence (for DAM methylation) and the 5’ end of the G is cleaved
    • 3’ exonuclease removes the daughter strand from there to the mismatch
    • Gap is filled in by Pol III and ligated
42
Q

Which proteins are involved in the initiation of bacterial replication?

A
  • DnaA: initiator protein/origin binding protein - absolutely essential for survival
    • binds oriC to open double helix
  • DnaB: helicase
    • undwinds double helix at replication fork
  • DnaC: Helicase-loader
    • loads helicase at origin
  • DnaG: primase
    • primes both strands of DNA with a short stretch of RNA
43
Q

What is oriC?

A
  • Origin of replication, where replication is initiated.
  • In bacteria, there is only 1 origin of replication, so replication of the circular genome goes bidirectionally until it terminates at the terminus of replication (Ter sites) ~180 degrees from the origin.
44
Q

How is the bacterial nucleoid organized?

A
  • histone-like proteins organize supercoiled loops of DNA
  • the circular dsDNA has been broken by topoisomerase (ds breaks), then twisted to introduce negatively coiled loops, then rejoined
  • there are loops within the loops to produce supercoiled DNA - 1000x condensed in E. coli
45
Q

How does chromosome partitioning occur in bacteria?

A

localization of sister-chromosomes to the opposite cell-poles

  • Replication machinery is located in the middle of cell
  • oriC regions of each strand being replicated move to the cell poles
  • Condensins are located in the ¼ and ¾ positions and condense newly-synthesized DNA
  • Cleaner proteins transport chromosomes out of the region; septum is formed
  • Integral membrane proteins and the actin-like protein MreB are also involved in chromosome “Partitioning“
46
Q

What are the 3 phases of transcription?

A
  • initiation
  • elongation
  • termination

all done by RNA pol

47
Q

What are the functions of RNA pol?

A

1) Detects promoters
2) Unwinds dsDNA to yield a ssDNA template
3) Catalyses formation of phosphodiester bonds between ribonucleotide triphosphates
4) Detects termination signals
5) Interacts with transcription factors

48
Q

What are the basic requirements for transcription to occur?

A

1) Template, double-stranded DNA
2) Activated ribonucleotide triphosphates
3) Metal ion (Mg2+ or Mn2+)
4) RNA polymerase

49
Q

Compare and contrast prokaryotic RNA pol vs eukaryotic RNA pol II.

A

Bacterial RNA pol

  • Core enzyme: α2ββ’ω (5 subunits); 400 kDa (transcription elongation)
  • Holoenzyme: core + sigma (σ) (transcription initiation)
    • sigma recognizes promotor/initiation site and then dissociates once transcription begins

Eukaryotes use many different polymerases

  • e.g. RNA pol I for rRNA, RNA pol II for mRNA, RNA pol III for regulatory RNAs

Eukaryotic RNA pol II

  • Core enzyme : 10 polypeptides catalytic core
  • Holoenzyme: 2 subunits (Rpb4/7) (transcription initiation)

They all share certain subunits with high homology, esp the active sites

50
Q

What are the -10 and -35 boxes?

A

Conserved sequences which act as promotors upstream of a transcription start site, 10 bp upstream (often TATAAT) and 35 bp upstream (often TTGACA).

Promotor sequences lie on the untranscribed coding/sense/+ 5’ to 3’ strand.

Promotors are assymetric and provide a ‘direction’ so that the polymerase knows in which direction to transcribe / which strand is the codeing strand.

The template/antisense/- strand is the 3’ to 5’ strand and transcription (RNA synthesis) takes place 5’ to 3’ for the newly formed RNA molecule.

51
Q

In bacteria, how is gene family transcription controlled?

A
  • The various different promotors provide variable transcription initiation efficiencies, with the weakest ones even requiring regulator proteins to help.
  • By changing the rate of either synthesis or degradation of a particular sigma factor
52
Q

Define the following:

  • polycistronic mRNA
  • transcriptional units
  • cotranscribed genes
  • operon
  • open reading frame
  • mRNA
  • tRNA
  • rRNA
  • promoter
  • UP sequences
  • transcription terminator
  • primary transcript
  • mature transcript
A
  • polycistronic mRNA – mRNA which contains multiple open reading frames
  • transcriptional units – segments of DNA that are transcribed into a single RNA molecule bounded by their initiation and termination sites
    • the resulting primary transcript is subsequently “processed” by proteins that cut them to form mature transcripts
  • cotranscribed genes – two or more genes are transcribed together to form the RNA
  • operon – multiple genes under the control of a single promoter which are transcribed together
    • genes that encode several enzymes of a particular metabolic pathway in prokaryotic cells are often clustered together in an operon, allowing their expression to be coordinated
  • open reading frame - portions of mRNA that actually encode amino acids
  • mRNA – messenger RNA; this is then translated to obtain the amino acid code -> to form protein
  • tRNA – transfer RNA; this brings amino acids to the growing aa chain during translation
  • rRNA – ribosomal RNA; specific subunits of rRNA come together to form the translational machinery
  • promoter – upstream initiation site which needs to be recognized in order for transcription to begin
  • UP sequences - these are recognized by the alpha subunit of bacterial RNA Pol to help properly position it on the DNA strand prior to transcription starting
  • transcription terminator - specific sequence in the transcribed RNA which signal the end of transcription
  • primary transcript – initial product of transcription which termination signals, and may include multiple RNA molecules with spacers in between
  • mature transcript – final RNA products with spacers and termination sequence removed
53
Q

What are 2 ways transcription can be terminated in bacteria?

A

Rho independent

  • GC rich palindrome followed by AT rich sequence (inverted repeat) in the DNA
  • After being transcribed, RNA-Pol stops at the resulting stable hairpin structure formed in the RNA strand
  • An adenine run directly following results in uridine on the RNA strand, which bonds weakly to the DNA strand
  • Most likely, the RNA dissociates first from the DNA template and then from the RNA-Pol

Rho dependent

  • RNA pol can also pause on a Rho-dependent transcription termination sequence
  • Transcription terminator protein Rho binds and then moves along the nascent RNA and catches up to RNA Pol
  • Causes RNA and RNA pol to be released from the DNA strand
  • If the RNA is not being actively translated, Rho can also stop transcription
54
Q

How is the initiation of replication regulated in E. coli?

A
  • Hemi-methylation indicates that a chromosome has recently undergone replication.
  • Inhibitor protein SeqA binds with high affinity to the 11 hemi-methylated DAM methylation sites (GATC) at the origin of replication, preventing binding of DnaA (initiator protein), thereby preventing replication initiation.
55
Q

study the following transcription table for bacteria vs eukaryotes

A

compare and contrast

56
Q

What is the first amino acid in a new polypeptide chain in bacteria?

A

f-Met ( N-formyl-methionine)

57
Q

How are amino acids brought to the growing peptide chain during bacterial translation?

A
  1. Amino acid synthetases specifically load tRNAs with the respective amino acid
    • example: tryptophanyl-tRNA-synthetase loads a matching tRNA with tryptophan
    • a tRNA specificity is based on its complimentary sequence (anticodon) to that amino acid’s codon on the mRNA being translated
    • the aa is loaded onto the 3’ aa attachment site of the tRNA, always CCA
  2. Elongation factor Tu guides loaded tRNA to the ribosome
  3. Codon-anticodon interactions guarantee incorporation of the amino acid into the growing polypeptide
58
Q

How can some bacteria have fewer tRNAs than possibilities with codons?

Example: 30 tRNAs for 60 codons

A
  • tRNAs can be used for multiple codons b/c in the 3rd position, there are modified nts in the tRNA that can read 2 or 3 different amino acids
59
Q

What is the Shine-Dalgarno sequence?

A
  • A sequence on the bacterial mRNA which is recognized by the 16S rRNA.
  • Recognition of the AUG start codon depends on recognition of the SD sequence.
  • The SD sequence is the ribosome binding site, and the start sequence will be 5-10 nucleotides downstream from that.
  • The ideal SD sequence in E. coli has the sequence: AAGGAGG
60
Q

How is bacterial translation initiated?

A
  • Formation of initiation complex: 30S subunit of ribosome bound to initiation factor 1 and IF3 binds to RBS in mRNA
  • Charged fMet-tRNA activated by IF2, GTP binds to initiation codon
  • IF1, IF2, IF3 dissociate and GTP is hydrolyzed, then 50S subunit binds to 30S subunit to form 70S ribosome
  • fMet-tRNA is now positioned in the peptide site which leaves the acceptor and exit site empty
61
Q

What are the steps in bacterial translation elongation?

A
  • Elongation factor (EF) Tu binds to loaded tRNA which is then positioned in the A site of the ribosome by GTP hydrolysis
  • The α- amino group of the amino acid forms a peptide bond with the carboxyl group of the amino acid in the P (peptide) site
  • EF-G (elongation factor Tu + GTP) mediates translocation of the complex along the mRNA. Translocation of mRNA requires hydrolysis of GTP. As a result, the uncharged tRNA is released and the A (acceptor) site becomes free
  • This requires much energy and many GTPases
62
Q

How is bacterial translation terminated?

A
  • ribosome moves to stop position/stop codon
  • A (acceptor) site remains empty because there is no appropriate tRNA for this sequence
  • release factors RF1 or RF2, plus RF3 bind to the A site to form termination complex
    • Note: release factors can only bind if there is a stop codon
  • polypeptide is released
  • mRNA, release factors, tRNA and the inactive 70 S ribosome dissociate
  • the inactive 70S rib is available for another event
63
Q

What is trans-translation?

A

In bacteria, this is essential and cannot be deleted.

  • If a ribosome gets stalled, it sits on an mRNA and doesn’t move
  • A tmRNA (transfer-messenger RNA) which is charged with an alanine binds to A site and helps to release the mRNA from the ribosome
  • the tmRNA is then the template, and a tag sequence is translated
  • once the stop codon is released, the peptide is released
  • the tagged, released unfinished protein is then degraded by a protease
64
Q

How does tmRNA rescue bacterial translation?

A
  • Through trans-translation
  • Without the tmRNA, the ribosomes remain sequestered and further translation of this mRNA is prevented
  • With tmRNA allowing release of the mRNA from the ribosome, the mRNA can continue to be translated by multiple ribosomes