BACTERIAL GENETICS

Question 1

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

Answer 1

used streptomycin resistance in bacteria

Question 2

by what mechanism are bacteria resistant to STR

Answer 2

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

Question 3

what experiment did they do to study STR resistance

Answer 3

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

Question 4

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

Answer 4

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

Question 5

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

Answer 5

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

Question 6

What were the results of the STR resistance experiment

Answer 6

they found that 2 colonies grew on the STR plate

Question 7

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

Answer 7

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

Question 8

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

Answer 8

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

Question 9

what did they do with the 2.5kb fragment

Answer 9

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

Question 10

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

Answer 10

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

Question 11

what does sanger sequencing rely on

Answer 11

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

Question 12

explain how sanger sequencing works

Answer 12

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

Question 13

what are the limitations of sanger sequencing

Answer 13

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

Question 14

what do we call NGS/illumina sequencing

Answer 14

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

Question 15

what are the 4 steps of illumina sequencing

Answer 15

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

Question 16

step 1 for NGS: library preparation

Answer 16

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

Question 17

step 2 of NGS: cluster generation

Answer 17

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.

Question 18

step 3: sequencing

Answer 18

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.

Question 19

step 4 for NGS: data analysis

Answer 19

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

Question 20

limitations of NGS

Answer 20

there is a high error rate

Question 21

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

Answer 21

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

Question 22

What is genome annotation

Answer 22

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

Question 23

what is manual curation

Answer 23

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

Question 24

structural annotation: extrinsic method

Answer 24

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

Question 25

structural annotation: intrinsic method

Answer 25

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

Question 26

sequence alignment in extrinsic method

Answer 26

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

Question 27

functional annotation

Answer 27

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

Question 28

functional annotation: homologs

Answer 28

genes related by ancestry ie; share the same ancestry

Question 29

functional annotation: paralogs

Answer 29

homologs arising from duplication events

Question 30

functional annotation: orthologs

Answer 30

homologs found in different species

Question 31

what are shuttle plasmids

Answer 31

plasmids that can replicate in more than one organisms

Question 32

requirement for shuttle plasmids

Answer 32

need origin of replication for both organisms

Question 33

shuttle phagemids

Answer 33

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

Question 34

how do we generate phagemids

Answer 34

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

Question 35

what does it mean that mycobacteriophages are conditionally replicating

Answer 35

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.

Question 36

transposons

Answer 36

these are additional MGEs that can be added to phagemids

Question 37

what is the most common transposon

Answer 37

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.

Question 38

in what scenario would the transposon excise

Answer 38

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

Question 39

where does the transposon jump into

Answer 39

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.

Question 40

how did they study phage transposon mutagenesis

Answer 40

used Mbt. Tuberculosis

Question 41

the tuberculosis experiment

Answer 41

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

Question 42

how did the tuberculosis experiment help us understand gene function

Answer 42

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.

Question 43

results from the tuberculosis experiment

Answer 43

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.

Question 44

advantageous or disadvantageous transposon insertion

Answer 44

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

Question 45

what mycobacterium did they study in cystic fibrosis patients

Answer 45

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

Question 46

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

Answer 46

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

Question 47

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

Answer 47

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

Question 48

the experiment done on CF patients

Answer 48

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

Question 49

results from the phylogenetic tree

Answer 49

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

Question 50

results from the antibiotic resistance tests

Answer 50

- 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

Question 51

mechanisms of resistance to clarithromycin

Answer 51

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

Question 52

opportunities for transmission between the CF patients

Answer 52

-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

Question 53

conclusions from the CF study on 3 modes of infection

Answer 53

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