Need Help: Bacterial gene deletion- sacB sucrose resistance - (Aug/26/2009 )
Yeah, this occurred to me as I drove in today...
What fishdoc means is that perhaps you should have designed the experiment with the mutant's inability to grow in mind. If you put a copy of the gene on a replicating plasmid into the WT strain before you punched out the wild-type chromosomal copy of the gene, you could show that the mutation is complemented by an extrachromosomal copy of the gene.
If you used a locked-down expression vector that requires induction to express, you could show that under conditions of induction (plasmid-borne gene expressed in the mutant background), the mutant strain grows, but when not induced (thus the plasmid-borne copy is not expressed), the mutant strain fails to grow...
Since you'd need to re-create the mutant strain, perhaps this time you could do it using a sacB suicide vector, and speed the process of isolating double cross-out clones.
cloneboy on Aug 27 2009, 11:07 AM said:
fishdoc on Aug 27 2009, 09:39 PM said:
HomeBrew on Aug 27 2009, 07:19 AM said:
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
If you construct a complementation plasmid and insert it into PAO1, then you'll have the plasmid with the gene, and the native gene in the genome. Then you do 2 complementation experiments, one you conjugate the donor to the WT PAO1 and in the second you conjugate the donor to the PAO1 + complementation plasmid. Assuming recombination occurs with the genome copy (it may also occur with the plasmid-borne gene), you should end up with no mutants in the WT PAO1 conjugation, but genomic mutants in the complemented PAO1. They will grow because of the presence of the complementation plasmid expressing the mutated gene. If you can show the mutation can occur when the gene is expressed in trans, but cannot occur if the plasmid is not present, that may be evidence that your gene is absolutely required for growth.
As for the step you describe above, I'm not sure what it will tell you. The presence of the SM10 strain may confound things, I'm not sure.
What may work is to design the primers like you say, conduct the conjugation, and then plate the conjugation onto cent+kan+amp to select for the single crossovers that integrated the plasmid. In each colony, your PCR products should show a WT band and a mutant band, because both will be present in a plasmid integration. In the next step, you take the colonies that grew on cent+kan+amp and plate them on cent only. This will remove pressure for the integration, and some of your colonies will lose the plasmid, going to either WT or mutant. Patch those colonies then (from cent only) onto both cent only and cent+kan+amp (use the same inoculum for both plates... that is, if you use a toothpick to transfer the colonies, tranfer to cent only, and then immediately use the same toothpick to patch onto cent+kan+amp without getting more bacteria on the toothpick). After growth, compare the two patch plates, and see which ones lost their kan+amp resistance. Those will be your double crossovers. Do PCR on the patches that grew only on cent media and see if there is a WT or a mutant band. If it's only WT, then then it's likely the mutation is lethal, because you should expect that double crossover to yield roughly 50% WT and 50% mutant.
It's essentially the same thing you'd do with the sacB, except you have to do a couple extra plate growths to see which ones lose the ability to grow on kan+amp... the advantage of the sacB is that ONLY those that lose the integration will grow on the sucrose, and you can just use those colonies directly and immediately for PCR without having to check for antibiotic resistances.
fishdoc on Aug 28 2009, 12:41 AM said:
cloneboy on Aug 27 2009, 11:07 AM said:
fishdoc on Aug 27 2009, 09:39 PM said:
HomeBrew on Aug 27 2009, 07:19 AM said:
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
If you construct a complementation plasmid and insert it into PAO1, then you'll have the plasmid with the gene, and the native gene in the genome. Then you do 2 complementation experiments, one you conjugate the donor to the WT PAO1 and in the second you conjugate the donor to the PAO1 + complementation plasmid. Assuming recombination occurs with the genome copy (it may also occur with the plasmid-borne gene), you should end up with no mutants in the WT PAO1 conjugation, but genomic mutants in the complemented PAO1. They will grow because of the presence of the complementation plasmid expressing the mutated gene. If you can show the mutation can occur when the gene is expressed in trans, but cannot occur if the plasmid is not present, that may be evidence that your gene is absolutely required for growth.
As for the step you describe above, I'm not sure what it will tell you. The presence of the SM10 strain may confound things, I'm not sure.
What may work is to design the primers like you say, conduct the conjugation, and then plate the conjugation onto cent+kan+amp to select for the single crossovers that integrated the plasmid. In each colony, your PCR products should show a WT band and a mutant band, because both will be present in a plasmid integration. In the next step, you take the colonies that grew on cent+kan+amp and plate them on cent only. This will remove pressure for the integration, and some of your colonies will lose the plasmid, going to either WT or mutant. Patch those colonies then (from cent only) onto both cent only and cent+kan+amp (use the same inoculum for both plates... that is, if you use a toothpick to transfer the colonies, tranfer to cent only, and then immediately use the same toothpick to patch onto cent+kan+amp without getting more bacteria on the toothpick). After growth, compare the two patch plates, and see which ones lost their kan+amp resistance. Those will be your double crossovers. Do PCR on the patches that grew only on cent media and see if there is a WT or a mutant band. If it's only WT, then then it's likely the mutation is lethal, because you should expect that double crossover to yield roughly 50% WT and 50% mutant.
It's essentially the same thing you'd do with the sacB, except you have to do a couple extra plate growths to see which ones lose the ability to grow on kan+amp... the advantage of the sacB is that ONLY those that lose the integration will grow on the sucrose, and you can just use those colonies directly and immediately for PCR without having to check for antibiotic resistances.
I am not sure about the single crossover and double crossover events your talking about. Since when i plate my cells onto the cent and cent+kan+amp plates i get no growth. How can i then differentiate single crossovers and double crossovers cells?
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.
Ashley Madison
What may work is to design the primers like you say, conduct the conjugation, and then plate the conjugation onto cent+kan+amp to select for the single crossovers that integrated the plasmid. In each colony, your PCR products should show a WT band and a mutant band, because both will be present in a plasmid integration. In the next step, you take the colonies that grew on cent+kan+amp and plate them on cent only. This will remove pressure for the integration, and some of your colonies will lose the plasmid, going to either WT or mutant. Patch those colonies then (from cent only) onto both cent only and cent+kan+amp (use the same inoculum for both plates... that is, if you use a toothpick to transfer the colonies, tranfer to cent only, and then immediately use the same toothpick to patch onto cent+kan+amp without getting more bacteria on the toothpick). After growth, compare the two patch plates, and see which ones lost their kan+amp resistance. Those will be your double crossovers. Do PCR on the patches that grew only on cent media and see if there is a WT or a mutant band. If it's only WT, then then it's likely the mutation is lethal, because you should expect that double crossover to yield roughly 50% WT and 50% mutant.
Nature cleanse
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?
Acai Berry
cloneboy on Aug 28 2009, 04:18 AM said:
Sorry, I forgot that part.
That poses a problem then, I think.
When you conjugate, there are 3 things that can occur. First is nothing... that is, no recombination event occurs. The second option is that a complete recombination event occurs in which your WT allele is exchanged for the mutant allele. This requires a double crossover event meaning a region upstream and a region downstream of your plasmid mutation interact and "crossover" with the homologous regions in your genome, resulting in the exchange of the mutant sequence for the native sequence. The third option is that the crossover ONLY occurs once, either upstream or downstream. In this case, the entire plasmid integrates into the genome at the site of recombination. You then end up with both the native and mutant alleles in the genome, but you also end up with any antibiotic resistance cassettes carried by the plasmid in the genome.
So, if you do your conjugation by filter mating, and the first step after filter mating is to plate on cent+kan+amp, you're selecting for plasmid integration, for the ability to grow on kan and amp. By selecting for that phenotype, you're also selecting for there to be a WT and a mutant allele present in any bacteria that grow. In that instance, since you're selecting for the bacteria to still have the native gene of interest intact, there should be no lethality involved. You should get colonies. Only when you select for the double mutation should you lose the ability to grow on kan+amp.
If you're not getting any colonies immediately after plating your conjugation while selecting for single crossovers, then I think that you're not making any mutants, you've simply got a conjugation that is not resulting in any recombination, for whatever reason.
Let's use the terms "integration event" and "excision event".
If you conjugate a suicide plasmid containing a resistance marker into cells that are sensitive to the compound for which the marker provides resistance, and plate those cells on media containing the compound, only cells which have undergone the integration event will produce colonies, since the plasmid can not replicate in the cells. You have now what is called a meridiploid strain, which contains a wild-type copy of the gene, the mutant copy of the gene as carried by the plasmid, plus the plasmid backbone.
If you passage these cells without selection, the cells will at some frequency (usually low) undergo an excision event, wherein the plasmid excises from the chromosome and is lost. These cells are once again sensitive to the compound for which the plasmid's marker provides resistance. Since there are two copies of the gene in the meridiploid strain, there are two ways in which the plasmid can excise: one way leaves behind the wild-type copy of the gene, creating a revertant indistinguishable from the initial wild-type host strain, and the other possibility leaves behind the mutant copy of the gene originally carried by your plasmid.
Finding the cells that have undergone an excision event is difficult, as the event happens at low frequency. This is where a sacB vector comes in handy. Having a counter-selectable marker on your integrated vector allows you to isolate from the meridiploid cells only those that have undergone the excision event (i.e. lost the integrated plasmid), since all other cells (i.e. those that have not undergone the excision event) will die due to the presence of the counter-selectable marker.
If you were to clone a wild-type copy of your gene into a replicating plasmid (let's call this the "rescue vector") which contains a resistance marker different from the one used by your suicide plasmid, we could call these clones the "rescue clone" (rescue clone = rescue vector + your gene).
If you transform one population of your meridiploid cells with the rescue vector, and another population of your meridiploid cells with the rescue clone, and isolate from these transformants cells which have undergone an excision event while selecting for the marker carried by your rescue vector, you would expect to find only wild-type (revertant) cells arising from the population transformed with the rescue vector, but the population transformed with the rescue clone (which provides your gene in trans) should yield some isolates for which the excision event has left behind a mutant copy of your gene on the chromosome.
fishdoc on Aug 28 2009, 09:17 PM said:
cloneboy on Aug 28 2009, 04:18 AM said:
Sorry, I forgot that part.
That poses a problem then, I think.
When you conjugate, there are 3 things that can occur. First is nothing... that is, no recombination event occurs. The second option is that a complete recombination event occurs in which your WT allele is exchanged for the mutant allele. This requires a double crossover event meaning a region upstream and a region downstream of your plasmid mutation interact and "crossover" with the homologous regions in your genome, resulting in the exchange of the mutant sequence for the native sequence. The third option is that the crossover ONLY occurs once, either upstream or downstream. In this case, the entire plasmid integrates into the genome at the site of recombination. You then end up with both the native and mutant alleles in the genome, but you also end up with any antibiotic resistance cassettes carried by the plasmid in the genome.
So, if you do your conjugation by filter mating, and the first step after filter mating is to plate on cent+kan+amp, you're selecting for plasmid integration, for the ability to grow on kan and amp. By selecting for that phenotype, you're also selecting for there to be a WT and a mutant allele present in any bacteria that grow. In that instance, since you're selecting for the bacteria to still have the native gene of interest intact, there should be no lethality involved. You should get colonies. Only when you select for the double mutation should you lose the ability to grow on kan+amp.
If you're not getting any colonies immediately after plating your conjugation while selecting for single crossovers, then I think that you're not making any mutants, you've simply got a conjugation that is not resulting in any recombination, for whatever reason.
I think we still come back to the starting point of essentiality again with respect to your above suggestions. As said above 3 outcomes by conjugation and i suppose mine follows the third one " third option is that the crossover ONLY occurs once, either upstream or downstream. In this case, the entire plasmid integrates into the genome at the site of recombination".
For your comments "If you're not getting any colonies immediately after plating your conjugation while selecting for single crossovers, then I think that you're not making any mutants, you've simply got a conjugation that is not resulting in any recombination, for whatever reason".
I just want you to recap the table which i posted before (added below again). If you see my WT grows only in cent and not in cent+amp+kan. If the WT incorporates the antibiotic cassette from the plasmid obviously it might not grow even in cent alone. No growth in cent+amp+kan for sure. If no recombination then i should see growth at least in the cent right?? I think you got the catch now.
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
Coming back to mutant i would like to define MUTANT into category,
1. Mutant (non-essential) -In this case of non-essential genes, cells can survive the disruption and form kan+amp resistant colonies.
2. Mutant (essential) - no growth due to lack of recovery of antibiotic-resistant colonies.
My mutant is essential and no matter single or double crossover, my mutant will be lethal after wild type copy gene is disturbed .
This can be even verified from the NAR paper for which i gave the link in the previous posts. Why didn't the group try all the above steps which you mentioned above?
Does this mean NAR reviewers failed to ask the above questions?
@Homebrew:
I understand that you want me to try with another suicide vector with all the suggestions. I should have done that before deciding my vector but now its too late.
cloneboy on Aug 28 2009, 10:15 AM said:
For your comments "If you're not getting any colonies immediately after plating your conjugation while selecting for single crossovers, then I think that you're not making any mutants, you've simply got a conjugation that is not resulting in any recombination, for whatever reason".
I just want you to recap the table which i posted before (added below again). If you see my WT grows only in cent and not in cent+amp+kan. If the WT incorporates the antibiotic cassette from the plasmid obviously it might not grow even in cent alone. No growth in cent+amp+kan for sure. If no recombination then i should see growth at least in the cent right?? I think you got the catch now.
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
Coming back to mutant i would like to define MUTANT into category,
1. Mutant (non-essential) -In this case of non-essential genes, cells can survive the disruption and form kan+amp resistant colonies.
2. Mutant (essential) - no growth due to lack of recovery of antibiotic-resistant colonies.
My mutant is essential and no matter single or double crossover, my mutant will be lethal after wild type copy gene is disturbed .
This can be even verified from the NAR paper for which i gave the link in the previous posts. Why didn't the group try all the above steps which you mentioned above?
Does this mean NAR reviewers failed to ask the above questions?
I can't speak to what the NAR reviewers did or didn't ask. I'm simply going on the information you're telling me. If you have a vector that carries kan+amp resistance in its backbone, but not in the gene mutation, then the presence of the plasmid should result in growth on kan+amp, right? When a single crossover occurs, as Homebrew described, you get a merodiploid: one mutant allele and one WT allele in the genome of a single cell. The presence of that native allele *should* complement the activity of the mutated one, and thus you should get growth on kan+amp. Because you're selecting for the maintenance of kan+amp, you're selecting for a single crossover, which would result in the presence of the WT non-mutated gene in the genome. With that there, it should still grow, correct? Only once you remove the kan+amp pressure, thus not selecting for the plasmid integration, the plasmid can excise from the genome, removing the kan+amp resistance, and leaving either a WT or a mutant allele (but not both). This is when you'd see lethality.
If you were trying to insert kan or amp into the middle of your gene of interest, thus causing a mutation, then I think the inability to grow in the presence of the antibiotic might indicate lethality of a mutant. However, since the resistance cassettes are encoded by the plasmid backbone, and not inserted into the gene, you should definitely get growth of colonies on kan+amp following conjugation. They will be your single crossovers that, while they do carry a mutated allele, will also carry the WT allele. And since that WT allele is still intact, you should get growth if indeed you are getting that single crossover.
For your table, are those results you expect to happen or are those what you have actually observed?
If you do your conjugation and immediately plate onto cent, you should still get growth, because there will still be WT PAO1 present in the conjugation. Not all of the cells will have picked up the plasmid or have had a recombination event. Also, some that do have that event will revert back to the WT and lose the plasmid, thus growing on cent alone.
That is, unless I'm really misunderstanding something big.