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Need Help: Bacterial gene deletion- sacB sucrose resistance - (Aug/25/2009 )

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Hi All,
I have a problem regarding selecting allelic exchange mutants. I know that counterselectable markers are often instrumental for the construction of mutants when under appropriate growth conditions, a counterselectable gene promotes the death of the microorganisms harboring it. Hence, transformants which have integrated a suicide vector containing a counterselectable marker, either by a single event of homologous or illegitimate recombination, retain a copy of the counterselectable marker in the chromosome and are therefore eliminated in the presence of the counterselective compound.

Most commonly used is a "sacB gene conferring sensitivity to sucrose in Gram-negative bacteria". The problem is i am using a suicide vector pFS100 a derivative of pGP704 carrying a "pir" dependent origin of replication from plasmid R6K and gene encoding resistance to ampicillin and kanamycin but no sacB gene marker in it.

I am trying to knockout a gene which is essential for growth of the pathogen Pseudomonas aeruginosa PAO1 (has no sacB gene in it). I check for the gene knockout in "centrimide+kan+amp" plate with PAO1 using the gene inserted suicide vector pFS100 and get no growth (showing the gene is essential) whereas the control plate without antibiotic markers show growth.

By using the sacB gene onto the suicide plasmid will convert sucrose to levans which are toxic to the gram-ve bacteria. As a result we get few sucrose resistance clones. About 50% of these clones will have replaced the original wild type gene with a mutated one and at the same time got rid of the suicide plasmid backbone from the chromosme. Rest of the 50% will have a wild type gene.

Now both my vector (pFS100) as well as PAO1 has no sacB gene. Am i wrong in choosing the suicide vector without sacB gene? Is there any other method for this problem?

Please advise me!!!

Thanks in advance


Regards,
cloneboy

-cloneboy-

cloneboy on Aug 26 2009, 01:06 AM said:

Hi All,
I have a problem regarding selecting allelic exchange mutants. I know that counterselectable markers are often instrumental for the construction of mutants when under appropriate growth conditions, a counterselectable gene promotes the death of the microorganisms harboring it. Hence, transformants which have integrated a suicide vector containing a counterselectable marker, either by a single event of homologous or illegitimate recombination, retain a copy of the counterselectable marker in the chromosome and are therefore eliminated in the presence of the counterselective compound.

Most commonly used is a "sacB gene conferring sensitivity to sucrose in Gram-negative bacteria". The problem is i am using a suicide vector pFS100 a derivative of pGP704 carrying a "pir" dependent origin of replication from plasmid R6K and gene encoding resistance to ampicillin and kanamycin but no sacB gene marker in it.

I am trying to knockout a gene which is essential for growth of the pathogen Pseudomonas aeruginosa PAO1 (has no sacB gene in it). I check for the gene knockout in "centrimide+kan+amp" plate with PAO1 using the gene inserted suicide vector pFS100 and get no growth (showing the gene is essential) whereas the control plate without antibiotic markers show growth.

By using the sacB gene onto the suicide plasmid will convert sucrose to levans which are toxic to the gram-ve bacteria. As a result we get few sucrose resistance clones. About 50% of these clones will have replaced the original wild type gene with a mutated one and at the same time got rid of the suicide plasmid backbone from the chromosme. Rest of the 50% will have a wild type gene.

Now both my vector (pFS100) as well as PAO1 has no sacB gene. Am i wrong in choosing the suicide vector without sacB gene? Is there any other method for this problem?

Please advise me!!!

Thanks in advance


Regards,
cloneboy



I'm currently using a pGP704 derivative containing sacB to do some bacterial mutagenesis. Thus far, it works very well and is quite efficient. I would be happy to try to help you resolve your issues if I can. I'm not quite sure I understand your question, though.

You certainly can use a vector without sacB in it. Depending on what kind of mutation you're making, sacB may not be needed. If you're inserting a resistance cassette into your gene of interest, you can do positive selection of clones based on the presence of that selectable marker. However, if you're trying to make markerless mutations (e.g., deletions) then sacB is very useful for negative selection.

-fishdoc-

fishdoc on Aug 26 2009, 08:14 PM said:

cloneboy on Aug 26 2009, 01:06 AM said:

Hi All,
I have a problem regarding selecting allelic exchange mutants. I know that counterselectable markers are often instrumental for the construction of mutants when under appropriate growth conditions, a counterselectable gene promotes the death of the microorganisms harboring it. Hence, transformants which have integrated a suicide vector containing a counterselectable marker, either by a single event of homologous or illegitimate recombination, retain a copy of the counterselectable marker in the chromosome and are therefore eliminated in the presence of the counterselective compound.

Most commonly used is a "sacB gene conferring sensitivity to sucrose in Gram-negative bacteria". The problem is i am using a suicide vector pFS100 a derivative of pGP704 carrying a "pir" dependent origin of replication from plasmid R6K and gene encoding resistance to ampicillin and kanamycin but no sacB gene marker in it.

I am trying to knockout a gene which is essential for growth of the pathogen Pseudomonas aeruginosa PAO1 (has no sacB gene in it). I check for the gene knockout in "centrimide+kan+amp" plate with PAO1 using the gene inserted suicide vector pFS100 and get no growth (showing the gene is essential) whereas the control plate without antibiotic markers show growth.

By using the sacB gene onto the suicide plasmid will convert sucrose to levans which are toxic to the gram-ve bacteria. As a result we get few sucrose resistance clones. About 50% of these clones will have replaced the original wild type gene with a mutated one and at the same time got rid of the suicide plasmid backbone from the chromosme. Rest of the 50% will have a wild type gene.

Now both my vector (pFS100) as well as PAO1 has no sacB gene. Am i wrong in choosing the suicide vector without sacB gene? Is there any other method for this problem?

Please advise me!!!

Thanks in advance


Regards,
cloneboy



I'm currently using a pGP704 derivative containing sacB to do some bacterial mutagenesis. Thus far, it works very well and is quite efficient. I would be happy to try to help you resolve your issues if I can. I'm not quite sure I understand your question, though.

You certainly can use a vector without sacB in it. Depending on what kind of mutation you're making, sacB may not be needed. If you're inserting a resistance cassette into your gene of interest, you can do positive selection of clones based on the presence of that selectable marker. However, if you're trying to make markerless mutations (e.g., deletions) then sacB is very useful for negative selection.


I want markerless mutations ie. delete the gene of interest and see if it is essential for growth of the bacterium. In my case the gene is essential as seen from no growth in the antibiotic selective plates. I have basically "knocked out" the gene but how will i show that the gene has been knocked out? Negative selection??
For which i must have used a vector with sacB gene but i missed that.

Are there anyother counter selectable markers very specific for my problem so that i can confirm i have knocked out the gene??

Cloneboy

-cloneboy-

Ultimately, if you make a mutant, you need to sequence the DNA to confirm it. You have to be able to grow the mutant, PCR the mutated region, and then sequence the PCR products.

To help me understand better, could you lay out the procedures you went through to construct your mutant including the vectors you used and what antibiotic resistance genes are encoded by them?

By what you describe above, you get growth on media, but in the presence of kan+amp, there is no growth. Couldn't that simply be WT P. aeruginosa growing?

I assume you're transforming the P. aer by conjugation. Following conjugation, what if any antibiotics are you using in your plates? If you simply plate the conjugation products onto no-antibiotic-media, you're likely to just have a ton of WT growing. Interspersed in that is likely some single and double crossovers.

However, if your first plating following conjugation is onto kan-amp, you are selecting in the first round for single crossover events. At least that way, the plasmid and construct are in the genome. After that, without sacB, things get hairy. You then need to select for the second crossover event, which sacB is used for. In your case, there is no selection, so it amounts to a whole bunch of screening on your part. From the colonies you'd get on the kan amp plates, you'd want to patch those onto no-antibiotic-media. That will take away the pressure to maintain the kan-amp resistance provided by the integrated plasmid. That plasmid then could loop out (the 2nd recombination). Your results then are the 50% WT 50% mutant. Of those colonies that grow on the plate, you'd have to patch to kan-amp media to verify the plasmid is gone, and then of the colonies that don't grow on kan-amp, screen them to determine which reverted to WT and which maintained the mutation.

It ends up being a lot of extra work to find those mutants. Throw in that it sounds like you're trying to make a mutation in a required gene, then at every step you also need to provide the substrate to allow the bacteria to grow.

What the sacB does is prevent those kan-amp colonies from growing after you try to induce the 2nd crossover event. That way, what you're left with are the final double crossovers not contaminated with a bunch of single crossovers.

-fishdoc-

fishdoc on Aug 26 2009, 09:38 PM said:

Ultimately, if you make a mutant, you need to sequence the DNA to confirm it. You have to be able to grow the mutant, PCR the mutated region, and then sequence the PCR products.

To help me understand better, could you lay out the procedures you went through to construct your mutant including the vectors you used and what antibiotic resistance genes are encoded by them?

By what you describe above, you get growth on media, but in the presence of kan+amp, there is no growth. Couldn't that simply be WT P. aeruginosa growing?

I assume you're transforming the P. aer by conjugation. Following conjugation, what if any antibiotics are you using in your plates? If you simply plate the conjugation products onto no-antibiotic-media, you're likely to just have a ton of WT growing. Interspersed in that is likely some single and double crossovers.

However, if your first plating following conjugation is onto kan-amp, you are selecting in the first round for single crossover events. At least that way, the plasmid and construct are in the genome. After that, without sacB, things get hairy. You then need to select for the second crossover event, which sacB is used for. In your case, there is no selection, so it amounts to a whole bunch of screening on your part. From the colonies you'd get on the kan amp plates, you'd want to patch those onto no-antibiotic-media. That will take away the pressure to maintain the kan-amp resistance provided by the integrated plasmid. That plasmid then could loop out (the 2nd recombination). Your results then are the 50% WT 50% mutant. Of those colonies that grow on the plate, you'd have to patch to kan-amp media to verify the plasmid is gone, and then of the colonies that don't grow on kan-amp, screen them to determine which reverted to WT and which maintained the mutation.

It ends up being a lot of extra work to find those mutants. Throw in that it sounds like you're trying to make a mutation in a required gene, then at every step you also need to provide the substrate to allow the bacteria to grow.

What the sacB does is prevent those kan-amp colonies from growing after you try to induce the 2nd crossover event. That way, what you're left with are the final double crossovers not contaminated with a bunch of single crossovers.


I am checking gene essentiality. For eg. a virulent gene is responsible for pathogenicity of the bacterium which in turn explains that the virulent gene is responsible for survival of the pathogen. I have a gene which is found to be "always" essential for PAO1, so knocking out the gene should result in killing the bacteria.

I designed my experiment with pFS100 suicide vector kanR,AmpR (a derivative of pGP704). Other strains used for transformation E.coli MC1061 and SM10lambda phir. After conjugation i should see the following outcome which i have summarized in the table below,

Growth on Centrimide ||| Growth on Cent+kan+amp

Wild Type (WT) - G (Growth) ||| NG (No Growth)

WT recombine with
recomb plasmid but
gene NON ESSENTIAL - G ||| G


WT recombine with
recomb plasmid and
gene ESSENTIAL - NG ||| NG

No recombination - G ||| NG



For my gene i found it to be essential and hence i find no growth in both the above combination(I get the 3rd combo as outcome) suggesting that i have knocked out the virulent gene that is responsible for pathogen survival. These are the only controls i used to show that i have knocked out the gene.

There is a paper in NAR link given below which also checks the essentiality of the genes and they say "A total of 347 candidate reading frames were subjected to disruption analysis, with 113 presumed to be essential due to lack of recovery of antibiotic-resistant colonies" (in abstract) meaning they have knocked out the essential gene whereas non essential genes will keep growing in antibiotics.


http://www.pubmedcentral.nih.gov/articlere...bmedid=12136097

Also read the section "Streptococcus pneumoniae competent cell preparation and gene disruption" where they explain " Plates were then examined and colony numbers compared to controls. The lytA non-essential controls typically gave 150–200 colonies per transformation and the ftsZ positive controls gave 0 colonies per transformation. Experimentally, most non-essential genes gave 200–300 colonies per transformation (lytA was found to be on the low end of the spectrum) and essential genes gave results similar to ftsZ. An occasional single chloramphenicol-resistant colony was seen for some experiments, presumably due to contamination or non-homologous integration of the construct. For verification, all essential genes were subjected to a second independent gene disruption experiment"


Do the above control expts enough for me to prove that i have knocked out the gene or do i know need to find a negative selection marker??


Cloneboy

-cloneboy-

OK, I think I understand. The gene you're working with is a gene required for the bacterium to live under any condition, not simply a virulence gene required only under certain conditions in the host. Correct?


If that's the case, I understand what you're doing, and if there is previous work done using those data as evidence of gene deletion, then I think you're on the right track. I can't say for sure that what you've done is sufficient to prove your statement, because I'm not familiar with that type of mutagenesis (knocking out an absolutely essential gene).

One thing I can think of that may strengthen your argument is to clone into a sacB plasmid, and then conjugate. Following conjugation, select for a single crossover event by plating on the cent+amp+kan. In that case, the plasmid will have integrated, and you'll have a mutant and native gene copy in the genome. Take any colonies that grow on cent+amp+kan and plate them onto media containing cent. This will remove the pressure to maintain the plasmid integration. Take the colonies that grow on cent only, and then plate them onto media containing 5% sucrose. This will select for the 2nd crossover by eliminating any bacteria that have maintained the plasmid integration. Basically, as you stated in your original post, theoretically 50% of your final double crossovers SHOULD be mutants and the other 50% should be WT revertants. Select a bunch of colonies and do colony PCR using primers that flank the deleted region. If you get 100% WT PCR products, that would also indicate that the mutants would not survive, because you would expect around 50% to be mutants. Now, that percentage will likely vary a little bit depending on the flanking regions used for homologous recombination, but as long as the flanking regions are somewhere near the same length, you still expect 50%.

Of the two mutagenesis projects I've done with the sacB vector, one of them had 22/48 mutants after the 2nd crossover, and the other had 6 of 24. In that second one, however, there were 11 that went back to WT and 7 that were inconclusive, because after colony PCR they had neither the WT or the mutant band. I haven't gone back yet to check those 7 again to see if any are actually mutants (and I may never do that).

This type of thing may not be needed, but I think it would certainly strengthen your case of a required gene if you can have a system set up where you expect X number of colonies to be mutants, but end up with zero.

-fishdoc-

Thanks for your suggestion but right now i am short of time and hence not in a position to clone into a sacB vector.


Anyother suggestion from other forum members?


Cloneboy

-cloneboy-

SacB is just confusing the issue. It is useful in selecting for double cross-over events, where allelic replacement has occurred, but you already have those.

You need two additional pieces of data to prove your hypothesis that the gene you've knocked out is required for viability:

1. You need to show that the colonies that won't grow on centrimide agar have lost the gene you suspect is required. You can do this by selecting two primers that anneal outside of the gene presumed to be mutated, amplify the product from WT and your mutant, and sequence the two products. You might be able to show this just on the basis of the difference in size of the PCR products on an agarose gel, or by Southern blot using the PCR product as a probe, but sequencing the products is more definitive.

2. You need to show that it's just the loss of the gene that's causing the lack of growth. You can do this by cloning a WT copy of the gene presumed to be required into a non-suicide expression vector, and transforming the mutant cells with it. If the mutant cells grow when complemented by an extrachromosomal copy of the mutated gene, then that gene is necessary and sufficient to restore growth. To be a bit more robust, you should also transform the mutant cells with the expression vector alone, and show that the presence of the vector alone does not restore growth.

-HomeBrew-

HomeBrew on Aug 27 2009, 07:19 AM said:

SacB is just confusing the issue. It is useful in selecting for double cross-over events, where allelic replacement has occurred, but you already have those.

You need two additional pieces of data to prove your hypothesis that the gene you've knocked out is required for viability:

1. You need to show that the colonies that won't grow on centrimide agar have lost the gene you suspect is required. You can do this by selecting two primers that anneal outside of the gene presumed to be mutated, amplify the product from WT and your mutant, and sequence the two products. You might be able to show this just on the basis of the difference in size of the PCR products on an agarose gel, or by Southern blot using the PCR product as a probe, but sequencing the products is more definitive.

2. You need to show that it's just the loss of the gene that's causing the lack of growth. You can do this by cloning a WT copy of the gene presumed to be required into a non-suicide expression vector, and transforming the mutant cells with it. If the mutant cells grow when complemented by an extrachromosomal copy of the mutated gene, then that gene is necessary and sufficient to restore growth. To be a bit more robust, you should also transform the mutant cells with the expression vector alone, and show that the presence of the vector alone does not restore growth.



I agree that would be useful to accomplish, but how can he do any sort of amplification or complementation if the mutant doesn't grow at all to begin with? Perhaps he could insert the complementation plasmid prior to mutagenesis?

Can the mutation be supplemented by a nutrient in the media, allowing it to grow?


As for the sacB step taking too long, the actual work would go pretty fast. All you'd need is to clone the construct into the sacB vector, and conjugate (or electroporate, however you do it). The time involved would be securing a copy of the plasmid, and if that takes too long, that's understandable.

-fishdoc-

fishdoc on Aug 27 2009, 09:39 PM said:

HomeBrew on Aug 27 2009, 07:19 AM said:

SacB is just confusing the issue. It is useful in selecting for double cross-over events, where allelic replacement has occurred, but you already have those.

You need two additional pieces of data to prove your hypothesis that the gene you've knocked out is required for viability:

1. You need to show that the colonies that won't grow on centrimide agar have lost the gene you suspect is required. You can do this by selecting two primers that anneal outside of the gene presumed to be mutated, amplify the product from WT and your mutant, and sequence the two products. You might be able to show this just on the basis of the difference in size of the PCR products on an agarose gel, or by Southern blot using the PCR product as a probe, but sequencing the products is more definitive.

2. You need to show that it's just the loss of the gene that's causing the lack of growth. You can do this by cloning a WT copy of the gene presumed to be required into a non-suicide expression vector, and transforming the mutant cells with it. If the mutant cells grow when complemented by an extrachromosomal copy of the mutated gene, then that gene is necessary and sufficient to restore growth. To be a bit more robust, you should also transform the mutant cells with the expression vector alone, and show that the presence of the vector alone does not restore growth.



I agree that would be useful to accomplish, but how can he do any sort of amplification or complementation if the mutant doesn't grow at all to begin with? Perhaps he could insert the complementation plasmid prior to mutagenesis?

Can the mutation be supplemented by a nutrient in the media, allowing it to grow?


As for the sacB step taking too long, the actual work would go pretty fast. All you'd need is to clone the construct into the sacB vector, and conjugate (or electroporate, however you do it). The time involved would be securing a copy of the plasmid, and if that takes too long, that's understandable.


@Homebrew:
I can't do PCR or the complementation as i don't get my mutant to grow on the media(explained in the above posts about the gene essentiality).

I have an idea to workout. I do conjugation by filter mating. At the end with cell lysate before adding to the selective media to check for growth can i try PCR as you said above. In that way the cell lysate would have 3 combinations: 1. ecoli SM10 (donor with construct) 2. PAO1 wildtype (if no recombination) 3. PAO1 mutant(Recombined one). I will design the primers as suggested and my PCR products will either be of the wildtype or mutant. Gel cut the bands and then send for sequencing. Am i right in doing this step?

@Fishdoc:

Perhaps he could insert the complementation plasmid prior to mutagenesis? What do you mean by this?

I will try with a sacB vector if nothing conclusive happens.

cloneboy

-cloneboy-
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