Molecular Genetic Screening in Bacteria

Question 1

What is a genetic screen?
What does a genetic screen tell you about the gene(s) identified?

Answer 1

Series of experiments designed to discover the function of specific genes/sets of genes

Genes identified are important for surviving the selection pressure applied

Question 2

Process of a genetic screen? (3 steps)
What selection pressure will we use?
- How will this help identify cell division genes?

Answer 2

Make a mutagenised library
Apply selection pressure
Further study

Temperature selection pressure of 42°C causes bacteria to elongate as they cannot divide; Eventually elongate so much they kill themselves
- Strain with mutations in cell division genes will elongate if they have a temperature sensitive mutation

Question 3

Process of genetic screen –> Identifying genotype? (hint - 5 steps)

Answer 3

1. Hypothesis – Molecular genetic screening will identify new genes/proteins involved in these pathways
2. Mutagenesis – Chemical mutagenesis
3. Screen selection – Temperature 30°C - 42°C; Searching for filament temperature sensitive mutants
4. Hit validation – Complementation and gene mapping
5. Identify genotype

Question 4

When using nitrosoguanadine to induce chemical mutation, we look for streptomycin resistance, which is caused by a known point mutation in ribosome of E. coli
What are we looking for to get 1 mutation in bacteria?

Answer 4

Looking for a sweet spot between getting enough cells with 1 mutation to study and not killing them

Question 5

If you get 80 temperature sensitive (TS) mutants for every 10^8 cells, how many colonies would need to be screened for 10 TS mutants?
How long would this take?

Answer 5

Screen 12.5 million colonies

Would take years of time; Harsh reality of these screens

Question 6

Go through Selection section of (5)

Question 7

When do we stop screening?

Answer 7

When you find something new; Reality is you don’t have infinite time and funding

Question 8

How do we know our 2/3 strains are not just ftsA ‘re-discoveries’?

Answer 8

Use complementation to determine if the phenotype is associated with ftsA or a new gene
- Remove gene, observe phenotype, put gene back in and see if that phenotype disappears

Question 9

How does λ phage complementation work?
What is the B region?

Answer 9

Enters E. coli genome through recombination of attP and attB regions

B region is not essential for phage growth and can be deleted and replaced with DNA/gene insert of interest

Question 10

Go through phage complementation in (5)

Question 11

What does graph in (5) - identify genotype show?
What does this confirm?

Answer 11

At 42°C, the cfu is static but the absorbance is still increasing; Cells are growing and not dividing

Confirms elongation phenotype

Question 12

Why are old bacterial genomes marked with minutes?

Answer 12

By putting conjugation plasmid in donor, they would transfer entire genome to recipient through pilus
Time at which gene was transferred to recipient was the time it was assigned

Question 13

Using conjugation data, where is the new fts gene found on the E. coli genome? ((5) - Gene mapping)

Answer 13

Between LeuC and NadC marker

Question 14

How do we identify the new genes primary DNA sequence using λ16-2 phage? (hint - restriction enzymes)

Answer 14

λ16-2 phage can be used to ‘cure’ TS phenotype in PAT84; Gene of interest is somewhere in λ16-2 – Specifically the b region
Sequence the phage by fragmenting with restriction enzymes and ligating it and then seeing if it still complements our strain

Question 15

Using the digestion data and complementation information available, identify the order of the genes in the λ16-2 phage (and therefore the E. coli genome); + indicates that it complements ((5) - last page)

Answer 15

MurC and dd1 (can’t tell order)
ftsA
New gene (PAT84)
envA

Question 16

Summary of data analysis 1:
Hypothesis
Mutagenesis
Screen selection
Hit validation
Identify genotype

Answer 16

Hypothesis - Hypothesised there was more about bacterial cell division to discover

Mutagenesis - Used Nitrosoguanadine to generate temperature sensitive mutants

Screen ‘Selection’ - Selected filamentous temperature sensitive mutants out of these

Hit Validation - Used the A16-2 phage to prove the gene was not ftsA and therefore a new gene

Identify Genotype/Gene mapping - Used bacterial conjugation to map this new gene to the 2 min region of the E.coli genome
- Mapped the gene to a region downstream of ftsA and named it ftsZ

Question 17

In the modern era, what can we identify now rather than just genotype?

Answer 17

Gene function

Question 18

There are currently unidentified Class A PBP enzymes in S. pneumoniae (gram +ve)
In gram -ve bacteria, these are controlled by regulatory proteins in outer membrane, LpoA and LpoB; These 2 protein complexes are vital for cell growth and division
Why doesn’t this apply to gram +ve?

Answer 18

They have no outer membrane; What is controlling these enzymes?

Question 19

What is meant when we say GeneA and geneB have synthetic lethal relationship?

Answer 19

Need at least one to encode for essential function and survive – Without both the cell dies

Question 20

If 2 genes (geneA and geneB) have a synthetic lethal relationship, what genes will become essential if geneA is inactivated?

Answer 20

GeneB + Any genes required for function of GeneB

Question 21

How does transposon mutagenesis work?

Answer 21

Transposon mutagenesis works by gene disruption which is predicted to inactivate the gene it jumps into through homologous recombination into the gene
- Inactivates through gene truncation

Question 22

How do we pair transposon mutagenesis and next generation sequencing?
What is this process called?

Answer 22

Called Tn-seq

Map large numbers of transposons

Question 23

How does Tn-seq work?

Answer 23

Saturate the genome with many transposons and have every possible insertion represented
To interpret ‘gaps’ in this profile, the transposon insertions must have happened and then been lost from the population because they were selected against (killed) - These genes are essential for what we are looking for

Question 24

If transposon sequencing works by transposons being lost from a complex population, how big do you need your Tn-seq transposon library to be?
- S. pneumoniae genome size – 2,046,115 bp
- S. pneumoniae gene number – 2069

Answer 24

2,000,000 members (~1 per 1 bp) is best to ensure enough saturation of genome with transposons; 200,000 (~1 per 10 bp) is more realistic and you still get good saturation

Question 25

How do we screen/select for synthetic lethal genes?
How do we measure the difference between these 2 profiles? (2 ways)

Answer 25

Looking for a profile that tolerates gene of interest insertions in WT, but can’t tolerate gene of interest insertions in the knockout
- Look for gaps

Read count ratio; (No. of Insertions in A)/(No. of Insertions in B) – Bigger is better (>1)
Statistical measure; Mann Whitney U test

Question 26

What is a p-value in statistical analysis?

Answer 26

Probability that the null hypothesis is true (essentially what you’re seeing is due to random chance)

Question 27

If the number of genes in the S. pneumoniae genome is 2069, how many genes do we reasonably know are false positives using a p-value of <0.05?; What does this mean?
What should we do instead?

Answer 27

103 (2069 x 0.05); We’d have to screen 103 genes to see if just 1 is correct - Too much time

Instead use a p value of <0.005
- 10.3 genes need screening

Question 28

What is a divide by 0 error and what causes it?
What are it implications?

Answer 28

We are looking for scenarios with 0 insertions in a particular gene; Causes errors in read count calculation and algorithm doesn’t include it in dataset

Question 29

Summarise the genetic screen steps in the modern day PBP1a and PBP2a screen we did (5 steps):
Hypothesis
Mutagenesis
Screen Selection
Hit Validation
Gene Function

Answer 29

Hypothesis - We hypothesised that there are unidentified protein required for PBP1a and PBP2a function
Mutagenesis - We used a transposon-based Tn-seq screen to identify genes which are synthetically lethal with pbp1a (and pbp2a)
Screen Selection - The screen identified 10 genes synthetic lethal with pbp1a and more data analysis identified 2 more giving 12 in total
Hit Validation - Used a conditional expression systems to confirm SPD_0876 + SPD_0878 were synthetic lethal pbp1a
Gene Function - Used microscopy data to show SPD_0876 (MacP) was required for pbp2a function in vivo

Question 30

Explain the 2 diagrams in (6) - Hit validation
You have identified 3 genes of interest and tested each in a pbp1a conditional expression strain with zinc (lets us turn it on and off at will)

Question 31

How can we work out how the cells die in this synthetically lethal scenario with zinc expressed pbp1a? (look at (6) - Gene Function if you need diagram)
How do we get quantification of microscopy data?

Answer 31

Deny the cells zinc and they’ll slowly die
In non-WT they shrink without zinc

Analyse image using software (e.g. MicrobeTracker)
Measure the area of 250 cells/condition

Question 32

Given the microscopy quantification data in (6) - Gene Function, what is the genetic relationship between SPD_0876, pbp1a, pbp2a?
Why is cell area measured rather than volume?

Answer 32

SPD_0876 is not required for pbp1a function, but is required for pbp2a function
Use a t-test to prove changes in cell area

Cell area was measured instead of volume as we have a 2D data set, so we can’t infer volume