Topic G - Transformation

Transformation

Process by which bacteria take up naked DNA from their environment

-Recipient Bacterium - Acquires free DNA from the environment

-Donor Bacterium - Source of free DNA

Types of Transformation

Natural Transformation - Bacteria takes up DNA using gene-encoded proteins (on its own)

Artificial Transformation - Forcing a bacteria to take up DNA (does not have natural means)

Transformant

Bacterial cell that has received DNA by transformation

Fates of DNA in the Recipient

1. Degraded → No transformation observed

2. Donor DNA recombines with recipient chromosome → yields recombinant transformants

-Replaces recipient DNA with new alleles from donor

-Requires sequence similarity (occurs via strand displacement, not homologous recombination)

3. Exists in the cytoplasm as a plasmid

-DNA must have origin of replication

-DNA must be an intact circle

Commonly seen in artificial transformation

Detecting Transformants

-Requires a difference in phenotype between donor/recipients

-Easiest if phenotype can be selected for

Griffith’s Experiment on Transformation

-“S” cells (smooth), make a capsule, is pathogenic

-”R” cells (rough), don’t make capsule, 

S cells injected in mice kill mice, capsule helps to evade immune system

R cells injected in mice don’t kill mice, they lack capsule so they’re eradicated

-Heat killed S cells, then injected into mice, mice did not die

-But combining heat killed S cells with live R cells killed mice

R cells were transformed into S cells

General Steps of Natural Transformation

1. Recipient produces competence pilus, binds to free DNA (from dead donor)

2. Pilus retracts, bringing DNA to cell surface

3. One strand of DNA is graded, the other is internalized

4. DNA recombines in the recipient’s chromosome

Competence

A state in which bacteria can take up naked DNA from their environment

Allows cells to be transformed

-Only become competent under conditions where free DNA is likely in the environment

Artificial and Natural Competence

-Artifical: Cells modified in the lab, allow DNA to cross cell-membrane

-Natural: Bacterium produces proteins for the binding and uptake of DNA

Naturally Competent Bacteria

Gram Positive - Bacillus subtilis (isn’t pathogenic), Streptococcus pneumoniae (Griffith used ts)

Gram Negative - Haemophilus influenzae, Neisseria gonnorrhoeae, Helicobacter pylori, Vibro cholerae

com Genes

Competence is determined by the com genes

-Mutants in these genes produce a competence-defective phenotype, encode for:

1. Regulatory proteins

2. Competence pilus

3. DNA transport proteins

4. Nuclease enzymes

Requires a lot of protein + energy to make a bacterial cell competent

Regulation of Competence

Competence genes only expressed under certain conditions:

1. Starvation

2. High cell density (likely some cells have lysed thus releasing their DNA)

-Controlled by Quorum Sensing - process whereby bacteria communicate with each other

Detection of High Cell Density

Ex; Competence in Bacillus spp. 

1. Cells produce and secrete peptides (ComX, always expressed)

2. High peptide concentration = high cell density

3. Peptides bind to receptor proteins on cell surface

4. Activates a signal transduction cascade, turns on expression of com genes

-Activates transcription factor → activates expression of com genes

DNase Sensitivity Experiment

Uptake of DNA can be demonstrated by a DNase sensitivity experiment

1. Treat cells + DNA (labelled fluorescently/radioactively) mixture with DNase

If DNA is outside the cell, they’re degraded

-Bacteria will not be labelled

If DNA inside the cell = they’re protected

-Bacteria will be labelled

Letting bacteria + DNA mixture incubate for a while BEFORE addition of DNase allows DNA to be taken up

Gram Positive DNA Uptake - dsDNA Brought to Surface

dsDNA binds to extended pilus (ComG is the pilin that makes the pilus)

-Pilus retracts, bring dsDNA near cell surface

-dsDNA binds to DNA-binding protein (ComEA), NOT sequence specific

Gram Positive DNA Uptake - Breaking down of dsDNA

Membrane-bound extracellular nucleases break down dsDNA

-dsDNA cleaved into 8.5-12kbp fragments by an Endonuclease (Endo = Cut Within)

-One strand of dsDNA is broken down into single nucleotides (by Exonuclease, cut at the end of the DNA molecule)

-Selection of strand is random

Gram Positive DNA Uptake - Import of ssDNA

ssDNA is imported by the membrane transport complex

-ssDNA transported across cytoplasmic membrane through ComEC channel (powered by ComFA ATPase)

-ssDNA is coated with SSBs (Single-stranded DNA binding proteins)

-Protects ssDNA from degradation

DNA is then incorporated into chromosome

Gram Negative DNA Uptake (N.gonn)

Mechanism of import similar to Gram-positive bacteria

-Additional channel needed to cross the outer membrane (PilQ, Secretin)

Gram Negative DNA Uptake - dsDNA Brought to Surface

PilE forms Pilus to grab DNA, retracts

-PilQ (Secretin) is the outer membrane channel DNA travels through

-ComE: dsDNA binding protein

Gram Negative DNA Uptake - dsDNA broken down

dsDNA broken down by nucleases

Gram Negative DNA Uptake - Import of DNA

Imported into the cytoplasm:

-Crosses through ComA (Cytoplasmic membrane channel)

-PilT (ATPase) powers the ComA channel

SSBs are added to ssDNA

Uptake Sequences

Some Gram-negative bacteria take up DNA ONLY from their own species

Chromosomal DNA contains “Uptake Sequences"

-Only performed by Haemophilus influenzae and Neisseria gonorrhoeae

Uptake sequences are long enough so tehy won’t occur randomly in other species DNA

-All gram positive/some gram-negative bacteria DON’T have uptake sequences (take up any dNA)

Experimental Evidence for dsDNA Binding to Pilus

dsDNA containing an Arg+ (protrophic) marker, mixed it with an arg- recipient cell

-Plated on medium lacking arginine

When DNA double-stranded, they were able to Arg+ transformants 

When DNA was denatured (single-stranded) and mixed with arg- recipients/plated, no Arg+ transformants formed

Experimental Evidence for ssDNA Uptake

1. Combine dsDNA continuing arg+ marker (donor) and competent arg- bacterial cells (recipient)

2. Incubate for various times (time 1 = no incubation, time 2 = 10 minutes, time 3 = 15 minutes), allows transformation to initiate

3. Treat with DNase enzyme → destroys dsDNA outside of cells

4. Extract DNA from the recipient cells → only DNA from inside the cell will be present

5. Use DNA extracted to transform new arg- recipient cells 

-Only dsDNA is capable of transformation

6. Plate on medium lacking arginine to select for arg+ transformant

ssDNA Uptake Experiment - Times

Time 1 = Not long enough for DNA to enter, DNA is degraded

Time 2 = Long enough for DNA to enter, but not long enough for recombination (exists in the cytoplasm as ssDNA)

-Bacteria can’t take up ssDNA, no new arg+ transformants are created

Time 3 = Long enough for DNA to enter and to recombine (exists in chromosome as dsDNA)

-Bacteria can take up dsDNA, arg+ transformants are created

Results and Conclusion of ssDNA Uptake

Results: Once the DNA has entered the cell

-Non-transformable form at early time points

-Transformable form at later time points

Conclusion:

-DNA enters cells in single-tranded form

Eclipse Period

Time immediately following entry of donor DNA until the time it recombines in the recipient chromosome

-DNA enters as ssDNA, cannot be re-isolated and used to transform another bacterium UNTIL recombination

Formation of Recombinants

1. RecA protein binds to donor ssDNA, coating it and displacing SSB proteins

2. RecA allows Strand Invasion at a site of sequence homology

3. Nickase enzyme cuts recipient chromosome in the displaced strand of the recipient chromosome

4. Displaced recipient strand is degraded by nuclease enzyme

• New donor DNA is base-paired to remainin strand of recipient chromosome (forming a Heteroduplex)

5. Ligase seals nicks to attach donor DNA into recipient chromosome

6. Sequence mismatch in the heteroduplex are corrected by mismatch repair (MMR)

• Mismatch in heteroduplex causes distortion, recognized by MMR

Mismatch Repair (MMR)

Removes either the top (donor) or bottom (recipient) strand

-The remaining strand is used as a template

If it uses the Bottom Strand → Sequence will be same as the recipient

If it uses the Top Strand —> Sequence will be different (donor strand)

Requirement of Strand Displacement

Sequence similarity is REQUIRED for Donor DNA to be incorporated stably in recipient chromosome

-ssDNA not replicated in the cytoplasm (difference between conjugation)

-Therefore can only integrate via strand-displacement

After Strand Displacement

1. DNA is double-stranded again

2. DNA can be re-isolated and used to transform another bacterium

-Eclipse period is over

Theories of Natural Transformation

1. DNA used as nutrition during starvation

2. Repair of DNA Damage

3. Increases Genetic Diversity

Theories - DNA as Nutrition Source

-Nucleotides used as carbon/nitrogen source

Problem: More energy required to uptake DNA, than the energy the DNA provides

Theories - Repair of DNA Damage

UV radiation damages DNA at random positions

-Some cells die/release DNA, others take up free DNA

-Recombination between similar sequences can be used to repair damage

Problem: Competence genes are not induced by DNA damage

Theories - Increasing Genetic Diversity

Most likely theory

Example: Antigenic variation in N. gonorrhoeae

-Pili (competence pilus) are recognized by immune system

-Pilin protein encoded by pilE gene; has a conserved region and a variable region

-Also have pilS, that are not expressed (silent loci)

During infection, recombination in pilE with the pilsS changes sequence of pilus proteins

-DNA sequence changes in pilE, thus amino-acid sequence changes, preventing recognition 

But is this through intragenic recombination or transformation?

Evidence for Transformation in Antigenic Variation

1. Variation induced by addition of pil DNA to cultures

2. Variation is reduced by addition of DNase (demonstrates free-floating DNA is involved)

3. Variation requires ssDNA recombination mechanism to occur (strand displacement)

-N.gonn only take its own DNA to allow for antigenic variation

Antibiotic Resistance - Campylobacter Jejuni

Natural transformation may be involved in the spread of antibiotic resistance between naturally-competent pathogens

Example: Campylobacter jejuni

-Naturally competent, food-borne pathogen that causes bloody diarrhea

-Forms biofilms on abiotic/biotic surfaces in the environment

Formation of Biofilm - Campylobacter Jejuni

Bacteria form a community, releasing sugars and DNA in the environment (eDNA-extracellular DNA)

-eDNA functions as an ssential structural component in biofilm

Can C. Jejuni transfer resistance DNA via Transformation?

Used 2 C. jejuni strains: one was kanamycin resistance (kanR) one is chloramphenicol resistant (CmR)

-Mixed strains and allowed to form biofilms on plastic surface (added/didn’t add DNase)

Biofilm formation was measured by straining

-Tested both free-floating bacteria and bacteria in the biofilm for KanR and CmR (would indicate transformation)

Found a greater number of transformants living in biofilms (with no DNase added)

-When DNase added, no transformants (indicates free-floating DNA is required)

Is eDNA Required in Biofilm?

Measured eDNA in the biofilm (structural component) following treatment with DNase

-The addition of DNase reduced Biofilm formation

Conclusions of C. Jejuni

1. Coculture of KanR and CmR C. Jejuni formed double-resistant recombinants by natural transformation

-Addition of DNase stopped recombinants from forming

2. Double resistant recombinants formed at a higher frequency in biofilms → biofilm formation facilitates the natural transformation process

3. DNase treatment decreased the level of biofilm formation

-Supports existing theory that extracellular DNA (eDNA) helps to stabilize biofilm formation in C. jejuni

Methods of Artificial Transformation

1. Calcium Ion Induction

2. Electroporation

Calcium Ion Induction

-Cells treated with calcium ions 

-DNA added and forms calcium-DNA complexes that bind to cell surface

Heat shock causes DNA to enter the cell

-No step where dsDNA is degraded to ssDNA

-Cells can take up any DNA (ss/ds, etc)

Electroporation

-Cells + DNA briefly exposed to an electric field

Small pores temporarily form in the membrane, allowing DNA to enter the cells

-Much more efficient than calcium transformation

-Cells can take up any DNA

More efficient, but more expensive

Requirements of Artificial Transformation

Works best with circular, double-stranded plasmid DNA

ssDNA and linear DNA will enter the cell, but will tend to be degraded

-if circular dsDNA has an oriV, it still stay in the cytoplasm as a plasmid

-if circular dsDNA has homology to recipient chromosome, it can recombine with the chromosome and form recombinants

Mapping with Transformation

Principle: Gene will only transform together if they are close together on the same piece of DNA

Ex: Donor has genotype X+/Y+, both located far from each other

-Recipient has X-/Y-

Donor DNA is broken into pieces, X/Y will be located on different pieces of DNA

-Donor DNA mixed with recipient

Some bacteria will be X+/Y-, some will be X-/Y+, very few will be X+/Y+ 

-If they were close together, most transformants will be X+/Y+

Cotransformation Frequency

Cotransformants - taking up both genes of interest during transformation

Cotransformation Frequency - % of total transformants selected for one marker that are also recombinant for another marker

Formula: Cotransformation Frequency = # of cotransformants (x+/y+) /# total transformants for one gene (x+ OR y+) x 100%

Ex; % of x+ transformants that are also y+

-HIGHER cotransformation frequency = closer the genes are

Determining CoF - Example Problem

Transformation experiment with donor a+/b-/c+ and recipient a-/b+/c-

-What is the order of the three genes?

Determining CoF - Start with a+ and b-

-Look at # that receive a+ and b-

CF = 12 + 3 / 12 + 3 + 100 + 135 x 100%

=6%

6% of bacteria that received a+ also received b- from the donor

Determining CoF - a+ and c+

-Look at # that received a+ and b+

Next start with a+ and c+

CF = 12 + 100 / 12 + 3 + 100 + 135 x 100% = 45%

45% of bacteria that received a+ also received c+ from the donor

Determining CoF - c+ and b-

-Look at # that receive c+ and b-

CF = 17 + 13 / 17 + 85 + 13 + 142 x 100%

=12%

Determining CoF - Final Order

a—b—c or c—b—a