Brainscape's Knowledge GenomeTM

Browse over 1 million classes created by top students, professors, publishers, and experts.


See full index

MBB267 > Microbial Genetics - Green > Flashcards

Microbial Genetics - Green Flashcards

1
Q

name 6 advantages of using microbes for genetics

A
  • 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
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

what is forward genetics?

A

know the phenotype - want to find responsible gene

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

how is forward genetics studied?

A 
Random genome wide mutagenesis
Phenotypic screening for desired mutants
Genetic analysis (eg complementation test or genetic mapping)
Gene isolation
Gene sequence determination
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

name an advantage and a disadvantage of forward genetics

A
  • 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
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

what is reverse genetics?

A

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?)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

how is reverse genetics studied now?

A

•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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

name 5 potential uses of mutants

A
  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
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

name 2 types of point mutation and briefly explain them

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

* Transversion: purine ↔ pyrimidine

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

name 3 types of chromosomal mutation

A

insertion
deletion
inversion

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

name 2 types of physical mutagens/mutations

A

electromagnetic radiation

spontaneous tautomers during replication

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

name 3 chemical mutagens

A

• 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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

name a biological mutagen

A

transposon

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

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

A

slip strand mispairing

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

briefly describe slip strand mispairing

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

how can slip strand mispairing be utilised by pathogenic bacteria?

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

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

A

• 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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

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

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

explain methyl mismatch repair

A

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

19
Q

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

A

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

20
Q

explain base excision repair

A

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

21
Q

explain recombinational repair and state why/when it occurs

A

• 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

22
Q

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

A

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

23
Q

what is the difference between mutation rate and mutation frequency?

A

Mutation rate: no of mutations per cell division

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

24
Q

describe the differences in phenotypic selection, phenotypic screening and enrichment

A

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

25
Q

describe the mechanism of homologous recombination

A
  • 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
26
Q

in homologous recombination, what structure(s) form when:

1) donor and recipient dna are circular
2) donor is linear, recipient is circular

A

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

27
Q

is recombineering with lambda red recA independent or dependent?

A

recA independent - site specific recombination doesn’t use RecA

28
Q

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

A
  • 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
29
Q

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

A

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

30
Q

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

A
  • 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)
31
Q

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

A

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).

32
Q

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)

A

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.

33
Q

describe the difference between insertion sequence and transposon

A

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

34
Q

describe nonreplicative transposition

A
  • 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
35
Q

describe replicative transposition

A
  • 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
36
Q

briefly describe class I and II transposons

A 
class I - retrotransposons (composite)
class II - DNA transposons (non-composite/complex)
37
Q

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

A

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

38
Q

state the structure of Tn10 and Tn3

A

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)

39
Q

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

A

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

40
Q

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

A

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

41
Q

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

A

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

42
Q

what is a null mutant?

A

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

43
Q

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

A

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.

MBB267 (3 decks)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2024 Bold Learning Solutions. Terms and Conditions