microbial genetics

Question 1

Discuss the classical (forward) approach for studying bacterial genetics

Answer 1

Phenotype (biological function) to Gene.

• random gene wide mutagenesis
• phenotypic screen for desired mutant
• biochemical/physiological characterization of mutant
• gene analysis
• gene isolation
• gene sequence determination

Question 2

Why are bacteria good model organisms for genetic studies?

Answer 2

• haploid organisms. phenotype of mutant can be seen immediately.
• fairly small genome
• genetic manipulation is straight forward
• relatively easy to make mutants with desired combination of mutations

Question 3

Compare and contrast forward and reverse genetics

Answer 3

Forward = phenotype to gene. advantageous as there is emphasis on derived phenotype so can find mutants in essential genes. Disadvantaged as might be impossible to identify all genes mutated in given phenotype.
Reverse = gene to phenotype. focus on a gene of interest (eg that has a similar homology to a known sequence) mutate this gene and complement with wild type allele. Determine the phenotype for the given gene.

Question 4

Discuss the use of conditional lethal mutants in bacterial genetics.

Answer 4

Can grow temperature sensitive mutants - Permissive temp will grow both wild type and mutant strains. Restrictive only wild type will grow. This can allow selection of mutant phenotypes.

Question 5

What ways can mutants be used in bacterial genetics?

Answer 5

can define genes used in particular functions, or inform about metabolic pathways.
Can help match a protein to its biological function.
Can help clone a gene by complementation

Question 6

What is slip strand mispairing?

Answer 6

This occurs during DNA replication, often in areas of long repeats. Completes synthesis with extra codons, forming a loop. Is used in some pathogenic bacteria as a switch on or off for expression of surface exposed proteins.

Question 7

What is methyl mismatch repair?

Answer 7

Mismatched pair eg GT AC, cut out of nascent dna strand. Newly synthesized strand is recognized by unmethylated nature (if methylated likely to be parent strand)

Question 8

Discuss Thymine dimer repair

Answer 8

T-T can become covalently linked in the same strand (error). Caused by UV radiation. Repaired by UVrAB complex - ATP dependent.

Question 9

What enzyme catalyzes recombination repair

Answer 9

RecA recombinase

Question 10

Explain SOS repair in bacteria

Answer 10

induced by extensive DNA damage. Inactives repressor LecA which normally represses transcription of SOS genes. RecA interacts with ssDNA after extensive DNA damage and breaks down LecA. Causes rapid polymerization of DNA.

Question 11

Why is Error prone repair benefitial?

Answer 11

SOS repair is error prone but can cause some mutations which can give survival advantages to the cell.

Question 12

What is phenotypic selection of bacteria?

Answer 12

Direct selection procedure. Use media that only allows growth of mutant colonies (eg ONPG is lactose analogue, if metabolized by wild type E.coli with be toxic if metabolized by lacZ or lacY mutant)

Question 13

What is phenotypic screening of bacteria?

Answer 13

Have to test all bacterium, either by an indicator medium or pH changes. Replica plating can be used to detect conditional mutants (temperature sensitive)

Question 14

Discuss penicillin enrichment

Answer 14

Incubate cell in optimal growth conditions for wild type mutant. Include penicillin in growth culture. This will kill growing cells. (roughly 99% wt cells killed) Was the cells and transfer to fresh media. Cell growth = enriched mutant cells.

Question 15

How does the RecBCD complex act in homologous recombination?

Answer 15

RecBCD enters and scan DNA fragment (damaged) by unwinding. Nicks DNA at Chi site. continues to unwind.

Question 16

How does RecA act in homologous recombination?

Answer 16

RecA molecules assemble on newly created ssDNA. Catalyse strand invasion and D loop formation on homologous section found in dsDNA

Question 17

How does branch migration occur in homologous recombination?

Answer 17

RuvAB assembles at the crossover point of single strand crossover formed by RecA. Pulls donor and recipient strands in opposite direction.

Question 18

How is the holliday junction resolved in homologous recombination?

Answer 18

Resolved by RuvC, horizontally. If donor and recipient are both circular, forms a cointegrate molecule. If donor is linear a second crossover is required to resole the viability of the DNA.

Question 19

what features are required on a plasmid designed for delivery of lamda red genes?

Answer 19

Carry lamda red genes, have selectability marker such as antiobiotic resistance gene and are temperature sensitive. Control by araC promoter to induce growth on arabinose medium.

Question 20

What are the problems associated with lamda red gene manipulation, how can this be overcome?

Answer 20

If lamda red genes are left and continue to be expressed in the cell after the desired recombination has occurred, illegitimate recombination might start to happen. This can be overcome by growing the cells at restrictive temperature after successful recombination, replate these colonies on fresh media

Question 21

What features are required for lamda red primer design?

Answer 21

Must design two primers, containing the promoters for the resistance gene in the plasmid vector and flanked by regions from the target gene to allow for homologous recombination.

Question 22

Briefly discuss lamda mediated recombination

Answer 22

Transform PCR product into (containing resistance gene and hybrid primers) into E.coli cells by electroporation. add arabinose to initiate lamda genes, allows homologous recombination. Select for successful mutations on antibiotic plates.

Question 23

How does using 1-sec1 meganuclease benefit a lamda red mediated recombination?

Answer 23

If a plasmid codes for 1-sec1, it will cut the target plasmid containing resistance gene and SacB gene (death in presence of sucrose) giving two linear fragments available for recombination. 1-sec1 will also cut its own plasmid from which it was expressed blocking further expression of this plasmid. This means lamda red gene activity is repressed after initial use.