Microbial Genetics - Green

Question 1

name 6 advantages of using microbes for genetics

Answer 1

• Reproduce rapidly
• Simple to maintain and culture
• Large colony numbers, means populations of spontaneous mutants (freq of which can be increased by mutagens) -This is useful when doing mutant screens
• Bacteria are haploid – phenotype seen immediately
• Small genome
• Genetic manipulation straightforward
• Make strains with combinations of mutations easily

Question 2

what is forward genetics?

Answer 2

know the phenotype - want to find responsible gene

Question 3

how is forward genetics studied?

Answer 3

Random genome wide mutagenesis
Phenotypic screening for desired mutants
Genetic analysis (eg complementation test or genetic mapping)
Gene isolation
Gene sequence determination

Question 4

name an advantage and a disadvantage of forward genetics

Answer 4

• Advantage – can find mutants with defects in essential genes (conditional lethal) as you pick the phenotype you’re interested in
• Disadvantage – slow and may be impossible to find all the genes in a species for a given phenotype

Question 5

what is reverse genetics?

Answer 5

knowing the important gene and want to find out the phenotype (ie if this bacteria has this gene, will my bacteria’s version do something similar?)

Question 6

how is reverse genetics studied now?

Answer 6

•Now this is done by comparing readily available microbial genome sequences:
Mutate your gene in vitro
Substitute the mutated allele for the WT allele in the genome
Determine the phenotype of the resulting mutant strain

Question 7

name 5 potential uses of mutants

Answer 7

1. Mutants define genes involved in a particular function
2. Mutant phenotypes can be informative
   - If an intermediate accumulates we know a pathway has been blocked and we may be able to interpret what the next step of the pathway is
   - If a TF is a mutated and 4 genes are affected, we know this TF regulates 4 genes
3. Permit matching a protein to its biological function
4. Conditional lethal mutants eg temperature sensitive mutants
5. Having a mutant can help to clone a gene
   - If the WT phenotype is selectable we can transform a mutant cell with a gene library by fragmenting non-mutant chromosomal DNA and adding each fragment to a mutant cell plasmid. Mutant cell + WT gene on plasmid –> complementation and the WT phenotype is restored

Question 8

name 2 types of point mutation and briefly explain them

Answer 8

• Transition: purine –> purine or pyrimidine –> pyrimidine

* Transversion: purine ↔ pyrimidine

Question 9

name 3 types of chromosomal mutation

Answer 9

insertion
deletion
inversion

Question 10

name 2 types of physical mutagens/mutations

Answer 10

electromagnetic radiation

spontaneous tautomers during replication

Question 11

name 3 chemical mutagens

Answer 11

• Analogs of bases (chemicals which resemble a purine/pyrimidine base)
• Base-modifying chemicals
- Nitrosoguanidine, nitrous acid
• Intercalators (chemicals that insert between bases)
- Cause frameshift mutations

Question 12

name a biological mutagen

Answer 12

transposon

Question 13

name the type of dna replication mistake that can occur when tandem repeats are being replicated

Answer 13

slip strand mispairing

Question 14

briefly describe slip strand mispairing

Answer 14

the formation of a small single strand loop of dna due to the mispairing of bases in dna replication

Question 15

how can slip strand mispairing be utilised by pathogenic bacteria?

Answer 15

•Used in some pathogenic bacteria to switch expression of surface exposed proteins on or off for immune evasion (phase variation)

Question 16

name 4 ways in which pieces of dna can be inserted or deleted into regions of dna

Answer 16

• Homologous recombination
- Between homologous chromosomes (different alleles are exchanged)
• Non-homologous/illegitimate recombination
- Between non-homologous chromosomes (different genes can be exchanged)
• Site-specific recombination
- Mobile genetic elements are moved to non-homologous regions eg the use of an att site by a bacteriophage (recA independent)
• Replicative recombination/transposition
- The movement of a transposable element to other regions of DNA

Question 17

name 4 methods of error-proof dna repair (in bacteria)

Answer 17

methyl mismatch repair
nucleotide excision repair -thymine dimer
base excision repair - damaged bases
recombinational repair

Question 18

explain methyl mismatch repair

Answer 18

DNA pol makes a mistake and inserts a normal, but incorrect base
Uses the GATC methylation site to determine which is old and which is new DNA
MutS binds to the mismatch and recruits MutL and MutH
MutL recognises the old, methylated strand (GATC position) and loops the DNA to meet MutS and MutH
MutH cleaves the new, unmethylated strand containing the mutation, near the GATC site
UvrD unwinds the cleaved strand, exonucleases remove the damaged strand, DNA pol synthesises a new strand

Question 19

explain nucleotide excision repair and state the type of mutagen that causes this repair mechanism

Answer 19

Induced by UV damage
UvrA and UvrB form a complex which binds to the damaged (thymine dimer) DNA
UvrA bends the DNA and is then ejected
UvrB lures the cleaving enzyme UvrC to the site
UvrC cleaves the phosphodiester backbone at 2 places (on the same strand)
UvrD helicase removes the ss fragment containing the mutations from the DNA
DNA pol fills the gap

Question 20

explain base excision repair

Answer 20

DNA glycosylase binds to and excises the damaged base
An endonuclease cleaves the phosphodiester backbone
DNA polymerase synthesizes a replacement strand
DNA ligase seals the DNA strand

Question 21

explain recombinational repair and state why/when it occurs

Answer 21

• Occurs when DNA replication takes place before a UV-induced thymine dimer can be excised by nucleotide excision repair
Replication fork approaches thymine dimer
DNA pol skips damaged region (forms gap in strand)
The RecA protein binds to the sister double helices at the single stranded segment
RecA-dependent recombination replaces the damaged-strand gap with a section of the homologous undamaged strand
Gap in undamaged strand is repaired by DNA polymerase
The thymine dimer can now be repaired by nucleotide excision repair

Question 22

describe error-prone repair and state an advantage of this method of repair

Answer 22

Extensive DNA damage inactivates LexA repressor protein
Activation of many repair genes occurs
Rapid polymerisation of DNA
Error prone but its better than no repair
Promotes mutations, some of which could be advantageous to survival

Question 23

what is the difference between mutation rate and mutation frequency?

Answer 23

Mutation rate: no of mutations per cell division

Mutation frequency: ratio of the number of mutant cells to total cells in the population

Question 24

describe the differences in phenotypic selection, phenotypic screening and enrichment

Answer 24

phenotypic selection - you are applying a selection pressure that only allows growth of mutants
phenotypic screening - the testing of every bacterium to work out which colonies are mutants - mutants don’t survive (use replica plating)
enrichment - the enrichment for the mutant ie the killing of the WT to grow large amounts of a mutant

Question 25

describe the mechanism of homologous recombination

Answer 25

* RecBCD enters end of DNA fragment and unwinds it * When it reaches the chi site (8bp) it nicks DNA and continues unwinding * A recA filament assembles on the ssDNA and scans the dsDNA for homology * RecA catalyses strand invasion and D-loop formation (ss crossover) * RuvAB assembles at the crossover point and pulls the donor and recipient strands in opposite directions (branch migration) * Endonuclease cleaves one end of the D-loop * Displaced ends are ligated to opposite strands * Holliday structure that is generated is resolved by RuvC cleaving the DNA across the junction * Ligation of broken ends completes a single crossover

Question 26

in homologous recombination, what structure(s) form when: 1) donor and recipient dna are circular 2) donor is linear, recipient is circular

Answer 26

1) when both are circular a cointegrate forms (ie a large single circular molecule containing the donor dna) 2) when donor is linear and recipient is circular 1 circular and 1 linear molecule are regenerated

Question 27

is recombineering with lambda red recA independent or dependent?

Answer 27

recA independent - site specific recombination doesn't use RecA

Question 28

name the 3 genes encoded by the lambda red system and explain what they do

Answer 28

* Exo: 5’-3’ exonuclease that degrades 5’ ends of linear DNA (creates 3’ overhang) * Beta: binds to ss 3’ ends generated by Exo and promotes annealing to complementary DNA (ie. bacterial chromosome) * Gam: Gam binds to the host RecBCD complex to inhibit exonuclease activity

Question 29

at what bacterial DNA structure/region does lambda red induce homologous recombination?

Answer 29

• The bacteriophage λ encodes a system (red) that promotes homologous recombination at the att site

Question 30

state what gene structures are present in pKD46 that make it a useful tool for inducing homologous recombination and inserting genes into donor DNA

Answer 30

* pKD46 contains pBAD promoter used in the araC operon (araC used as TF – can switch β/γ/exo on/off) - creates inducible lambda red region * Origin of replication that is dependent of RepA protein, RepA allele that is present on plasmid is temp sensitive (at 42oC there will be no folding of the protein and the plasmid doesn’t replicate)

Question 31

describe how lambda red genes in pKD46 can be used in the ins/del method of gene knockout

Answer 31

gene knockout can be achieved by replacing (knocking out) target genes by inserting a selectable marker gene eg drug resistance (this is the ins/del method) this achieved by generating primers complementary to the flanking regions of the cassette and of the target gene. This results in PCR copies of the cassette, which contain regions of homology to the flanking regions of the target gene. Homologous recombination can then occur between the cassette and the target gene (double crossover event).

Question 32

what are the problems of leaving lambda red genes in the cell for a long time? how are they solved? (this involves the pKD46 plasmid)

Answer 32

problem: leaving lambda red genes in a cell for too long can result in off-target illegitimate recombination solution: remove the red genes by culturing at the non-permissive temperature - RepA protein needed for replication cannot fold. Move the mutated dna into a WT cell by transduction.

Question 33

describe the difference between insertion sequence and transposon

Answer 33

insertion sequence: ONLY contains genes required for transposition eg transposase as well as flanking inverted repeats transposon: an insertion sequence with additional genes eg antibiotic resistance genes

Question 34

describe nonreplicative transposition

Answer 34

* Transposase aligns inverted repeats and flanking DNA forming a transpososome * One phosphodiester bond is cleaved on each strand at opposite ends of the IS element * 3’OH attack of the remaining phosphodiester bond on the complementary strand produces a hairpin structure and the host carrier DNA is ejected and repaired * Hairpins are re-nicked with the 3’ OH attacking recipient DNA (with help from transposase) * The IS element has moved from one DNA site to another

Question 35

describe replicative transposition

Answer 35

* Transposon (Tn) is copied and simultaneously ligated at the target site, creating a transient cointegrate molecule (as circular DNA and circular DNA) * The Tn is duplicated. This is because the nicks are made at the 3' end of the transposon on both strands of the DNA. The end result of this is that a cointegrate forms with the Tn attaching the plasmids at both the 3' and 5' of the DNA molecule

Question 36

briefly describe class I and II transposons

Answer 36

``` class I - retrotransposons (composite) class II - DNA transposons (non-composite/complex) ```

Question 37

what is the difference between composite and non-composite/complex transposons?

Answer 37

composite transposons: more complex than IS elements (eg contain drug resistance genes); contain 2 IS elements that flank the gene noncomposite: no IS elements (but contain inverted repeats), contain other genes as well as genes for transposition eg transposase

Question 38

state the structure of Tn10 and Tn3

Answer 38

tn10 - Tet resistance gene flanked by 2 IS elements, 4 sets of inverted repeats (either side of IS's) but only outer 2 are used Tn3 - no IS elements, contains flanking inverted repeats, amp resistance gene, transposase gene, resolvase gene (unlinks cointegrate at res site)

Question 39

what is a conjugative transposon? are the hosts homologous or heterologous?

Answer 39

large, complex transposons that contain tra genes (for conjugative transfer). can insert into hosts from a different species - heterologous

Question 40

name a conjugative transposon and describe its key gene and its role in antibiotic resistance

Answer 40

Tn916 encodes tet resistance low levels of tetracycline dont kill bacteria containing this Tn - increases Tn916 frequency in population

Question 41

how can we be sure Tn916 has inserted into a bacterial chromosome?

Answer 41

use a temp sensitive plasmid (contains Tn916) at 42o there will be no plasmid replication at 30o there will be plasmid replication and integration plate on tet media at 30o then shift up to 42o - if plasmid hasnt integrated Tn916 will be degraded and the cell will die on the tet media, transformants will survive

Question 42

what is a null mutant?

Answer 42

a mutant in which a coding sequence has been interrupted - no protein encoded

Question 43

explain how Transposon mediated differential hybridisation (TMDH) works and why we use it

Answer 43

used to locate essential genes within bacterial chromosome random mutagenesis by transposons used to generate a mutant library - one Tn per chromosome genome of each cell fragmented to isolate individual genes on single fragments use modified Tn with phage promoters (T7 and SP6) to visualise on array T7 creates green transcripts and SP6 creates red transcripts when in vitro transc. of DNA fragments carried out RNA hybridised to microarray - creates unique pattern of red and green this acts as control - mutant library then transferred to mouse - if an essential gene for colonisation contains a Tn then the cell will die and in the microarray there will be no signal.