BACTERIAL GENETICS Flashcards

1
Q

what did they use to study gene function before whole genome sequencing existed

A

used streptomycin resistance in bacteria

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2
Q

by what mechanism are bacteria resistant to STR

A

either they mutate their S12 ribosomal protein or their 16S rRNA or use efflux to pump the antibiotic out of the cell so that STR cannot recognise these targets and thus wont be effective in the bacterium

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3
Q

what experiment did they do to study STR resistance

A

they used mycobacterium fortuitum FC1 strain to test antibiotic resistance in this strain

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4
Q

why did they use Mbt. fortuitum in the experiment for STR resistance

A

because it is naturally resistant to STR and so they only had to use a minimum inhibitory concentration of 50ug/ml which is the MIC used to inhibit growth of the bacteria

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5
Q

step 1 of the STR resistance experiment: generating a chromosomal library

A

they first had extract the genome and then fragment the FC1 genome.
then they placed these fragments into pSUM36 plasmids. the plasmids were then transformed into Mbt. smegmatis plated on STR

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6
Q

What were the results of the STR resistance experiment

A

they found that 2 colonies grew on the STR plate

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7
Q

what did they do with the 2 Mbt. smegmatis colonies that grew on the STR plate

A

they extracted the plasmids from these colonies and called them pAC5 and pAC6. they then took these plasmids and did sanger sequencing on them

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8
Q

what did they find from the sanger sequencing of pAC5 and pAC6

A

they found this common 2.5kb region that is conferring the STR resistance in the 2 plasmids.

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9
Q

what did they do with the 2.5kb fragment

A

they further fragmented a 1kb fragment from it; containing only full orfc and placed these fragments into 2.5= pSAN19 and 1=pSAN26. they placed these plasmids into more Mbt. smegmatis on STR plates

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10
Q

what did they conclude about the genes responsible for STR resistance in FC1 strain

A

that because both pSAN19 and pSAN26 transformed colonies grew and so orfc must be the gene responsible for the resistance

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11
Q

what does sanger sequencing rely on

A

relies on a chain terminating dideoxynucleotide that is irreversible in its function

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12
Q

explain how sanger sequencing works

A

first they add the FC1 common 2.5kb fragment(DNA) with free nucleotides, primers and polymerase. then they add these terminator bases that are fluorescently labelled. then they heat everything to 96 degrees so the ds DNA denatures and then cool down to 50 so that the DNA primers can bind to the template strands and then heat to 60 so DNA polymerase can bind and start making complimentary strand. polymerase stops when terminator base is added. heat to 96 degrees again so that new and template strands can seperate. this is repeated over and over so that mutiple new strands of different lengths are made. then we use electrophoresis to seperate the new strands by length and we run through a capillary tube which at the end has a light that lights up the fluorescently labelled terminator bases. the strands are read from shortest to longest. so we end up with the sequence of bases for our 2.5kb fragment

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13
Q

what are the limitations of sanger sequencing

A
  1. it is expensive
  2. can only sequence 300-1000bp
  3. poor quality of the first 15-40 reads
  4. quality degrades for the last 700-900bp
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14
Q

what do we call NGS/illumina sequencing

A

massively parallel sequencing because it can simultaneously sequence 1000s of fragments

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15
Q

what are the 4 steps of illumina sequencing

A
  1. library preparation
  2. cluster generation
  3. sequencing
  4. data analysis
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16
Q

step 1 for NGS: library preparation

A

to fragment the DNA and add adapters/linkers to the ends of the fragments. these are complimentary to the oligos on the flow cell

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17
Q

step 2 of NGS: cluster generation

A

then hybridization of the adapter that is complimentary to the first type of oligo on the flow cell occurs. DNA polymerase will generate a complementary strand to the hybridized strand. the ds molecule is denatured and then original template strand is washed away. then the strands left are amplified through bridge amplification where the strands fold over and hybridizes to the other type of oligo on the flow cell. DNA polymerase will make complimentary strand. the ds bridge is denatured but both strands stick to flow cell. this process is repeated over and over and then all the reverse strands are washed away and only forward strands are left stuck to the oligo. the 3 prime ends are blocked to prevent unwanted priming. note this process is occuring simultaneously for 1000s of DNA fragments.

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18
Q

step 3: sequencing

A

extension of the first sequencing primer occurs to generate the first read. then fluorescently labelled nucleotides are added one by one and a light excites the flow cell so that a signal is given off after each base is added. call this sequencing by synthesis. then the read products are washed away and then 3 prime end is unblocked so that more reads can be generated so that we are left with millions of reads for all the DNA fragments.

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19
Q

step 4 for NGS: data analysis

A

then the reads are overlayed and compared to a reference genome so the whole genome sequence can be generated

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20
Q

limitations of NGS

A

there is a high error rate

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21
Q

how do they ensure a big enough signal is produced in NGS

A

use bridge amplification to clonally amplify the fragments so a larger signal is given off

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22
Q

What is genome annotation

A

determining the structural and functional properties of the genome once its sequence has been determined

23
Q

what is manual curation

A

annotation a genome using a persons prior knowledge. it is accurate but extremely slow process. so instead we use automated computational methods

24
Q

structural annotation: extrinsic method

A

where a sequence is compared to an annotated sequence in a database. more accurate for more closely related organisms

25
Q

structural annotation: intrinsic method

A

using the properties of DNA to determine the coding and non coding regions

26
Q

sequence alignment in extrinsic method

A
  1. local alignment: BLAST is an example of this where try find best matching region of the query and annotated sequences. dont need the sequences to be the same length
  2. global alignment: try to match as many positions across the entire length of the query and annotated sequences. must be same length
27
Q

functional annotation

A

there is an underlying assumption that similiar sequences share the same function because they are related by ancestry.

28
Q

functional annotation: homologs

A

genes related by ancestry ie; share the same ancestry

29
Q

functional annotation: paralogs

A

homologs arising from duplication events

30
Q

functional annotation: orthologs

A

homologs found in different species

31
Q

what are shuttle plasmids

A

plasmids that can replicate in more than one organisms

32
Q

requirement for shuttle plasmids

A

need origin of replication for both organisms

33
Q

shuttle phagemids

A

these are plasmids that have been transformed with elements of mycobacterial viruses ie; viruses that only infect mycobacteria.

34
Q

how do we generate phagemids

A

the plasmids must be transformed in E.coli and replicate here where they wont produce viral particles. this way we can generate large amounts of phagemid DNA

35
Q

what does it mean that mycobacteriophages are conditionally replicating

A

means that they only replicate to produce viral particles at certain temperatures. ie; if a phagemid is transformed into Mbt. smegmatis at 42 degrees the mycobacteriophage wont replicate but at 30 degrees it will to form plaque.

36
Q

transposons

A

these are additional MGEs that can be added to phagemids

37
Q

what is the most common transposon

A

Himar1- it has a transposase flanked by inverted repeats. the transposase then binds these inverted repeats and introduces double stranded breaks in the transposon, allowing it to excise. following excision the transposon can enter into another piece of DNA.

38
Q

in what scenario would the transposon excise

A

at 42 degrees the mycobacteriophage wouldnt replicate and produce viral particles and so the transposon would excise and jump into chromosomal DNA of the host instead

39
Q

where does the transposon jump into

A

jumps into TA sites in the genome of the bacterial host. and remember that there are many TA sites it can jump into and this is happening in many cells so we generate a transposon mutant library. ie; many different Mbt. smegmatis mutants.

40
Q

how did they study phage transposon mutagenesis

A

used Mbt. Tuberculosis

41
Q

the tuberculosis experiment

A
  1. they incubated phagemids with Mbt. Tuberculosis at 42 degrees.
  2. this generated a transposon mutant library
  3. grew the transposon mutant library on cholesterol and found only some colonies survived
    and some died
42
Q

how did the tuberculosis experiment help us understand gene function

A

well they found that the colonies that died on cholesterol must have had the transposon jump into genes essential for survival on cholesterol. so the transposon disrupts this essential gene so colony no longer survives. whereas the colonies that survived had the transposon jump into TA sites of non essential genes.

43
Q

results from the tuberculosis experiment

A

found 10^5 TA insertion sites.
they realised that to try and map each missing colony would take too much effort so instead they used illumina sequencing to sequences the input and output pools to find which colonies were missing in the output pool.
-the found around 40000 insertion sites randomly distributed throughout the genome and gaps would be essential genes. so absence of reads predicts essentiality.

44
Q

advantageous or disadvantageous transposon insertion

A

found that the insertion of a transposon into some genes lead to rapid colonial growth ie; advantageous insertion

45
Q

what mycobacterium did they study in cystic fibrosis patients

A

Mbt. abscessus
- it is non tuberculosis causing
-easily transmissable
-infects the lungs, soft bone and tissue

46
Q

2 methods used prior to NGS to study Mbt. abscessus transmission and strains

A
  1. PFGE- this method uses electric current in the electrophoresis gel to as the fragments are too large. relies on base changes occuring in the restriction endonuclease cut sites. this causes changes in the cutting pattern by the restriction endonuclease of the strain. see different banding pattern on the gel as a result.
    - important that these changes are only recognised if they occur in the restriction cut sites.
    - low resolution method.
  2. multi locus sequence typing- this method detects more changes between the strains than PFGE does but still not all the changes in the genomes of the strains
47
Q

what method is preferably used to study Mbt. abscessus strains and transmission

A

NGS- whole genome sequencing as it is the highest resolution method.

48
Q

the experiment done on CF patients

A

-they took 168 isolates from 31 patients over 5 years
-they genotyped all the strains, generated a phylogenetic tree and did antibiotic susceptibility tests.

49
Q

results from the phylogenetic tree

A
  1. 1 persons strains were all clustered together which indicates aquistion of these diverse strains from the environment
  2. several patients were found in the same cluster; although individuals strains still more closely related to each other; clustering indicates some relationship between the patients strains= strains diverged from a common ancestor.
    3.found some strains from patients 1 and 2 were more closely related to each other than to own respective strains. indicates transmission
50
Q

results from the antibiotic resistance tests

A
  • found 3 patients who had never been exposed to antibiotic treatment by clarithromycin were resistant
    -found 5 patients who had never been exposed to amikacin treatment were resistant
    -indicates transmission of these resistant strains between patients
51
Q

mechanisms of resistance to clarithromycin

A
  1. inducible mechanism
  2. change in base pair from A to C/G
52
Q

opportunities for transmission between the CF patients

A

-mapped the hospital admissions and found several patients who were in the same cluster ie; had similar strains diverging from a common ancestor but they had no chance of transmission between them.
-evidence for dominant strains circulating in that area that arose from a common ancestor and independent aquisition of these dominant strains

53
Q

conclusions from the CF study on 3 modes of infection

A
  1. independent aquisition of genetically diverse strains
  2. transmission
  3. independent aquisition of dominant strains