Dr. Beatty went over in class the reason for mRNA to be 2.5kb while the DNA was only about 1kb (topic 10, P5). I just want to clarify that, is it because that that the crt gene must be the first gene in the operon and the extra kb came from other genes in the operon? It must be the first gene because it is very close to the promoter region?
Also, for topic 9, B2, how can we be 100% sure that it is trans-complementation by doing a second conjugation? Won't there be still be occurance of homologous recombination and mutation in the plasmid recipient?
I assume you say 1kb because we have a 1.7 kb fragment that contains the whole gene (so, fair enough, some of it is the promoter and the rest must be the coding region). You are right, considering that we are in bacteria, the best explanation for the RNA being so much longer than the DNA is that the crt gene is part of an operon. And again, it makes perfect sense that the crt gene is the first one in the operon since it's adjacent to the promoter.
For 9 B2, you need to think about frequencies and probabilities. Trans-complementation (assuming that we have the whole gene in the plasmid) works very efficiently, homologous recombination is rare, and mutation in the recipient is ultra-rare! in reality, we would do an experiment in parallel where we'd conjugate in a plasmid that does not contain the DNA fragment of interest. If, with the plasmid containing the sequence of interest, we get a red colony every 10, while in the control conjugation we get a red one every million colonies, we can pretty much be sure that our fragment of interest does indeed contain the crt gene!
This question is regarding topic 11, page 2, II.C.
When we use a reporter gene (i.e. lacz) to find location of promoter and regulatory sequence, how can we screen for recepient cells that has plasmid inserted with the 3rd PCR product? Wouldn't it have the same phenotype as a recepient cell that has the plasmid but without insertion? Since both are unable to yeild blue colonies and both are Kan. rsistant
Excellent question, and excellent thinking on your part. You are perfectly correct: in the case described you can select for colonies that have received the plasmid, but whether they have the insertion or not you can't tell just by looking at the colonies.
This is when restriction mapping comes in really handy: it is very easy to take a colony (or many of them, seperately), grow it up and isolate and purify the plasmid that it contains. That plasmid can then be subjected to restriction mapping, and comparing its restriction digestion patterns with that of the original plasmid (without the insertion) will tell you whether what you have isolated is just the plasmid or the plasmid with an insertion.
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6 comments:
hi Pam, could you post this week's summary regarding problem set 3?
thanx
yes, they are now available!
Cheers
Pam
Hi Pam,
Dr. Beatty went over in class the reason for mRNA to be 2.5kb while the DNA was only about 1kb (topic 10, P5). I just want to clarify that, is it because that that the crt gene must be the first gene in the operon and the extra kb came from other genes in the operon? It must be the first gene because it is very close to the promoter region?
Also, for topic 9, B2, how can we be 100% sure that it is trans-complementation by doing a second conjugation? Won't there be still be occurance of homologous recombination and mutation in the plasmid recipient?
Thanks Pam!!! :)
Thank you Pam!
I assume you say 1kb because we have a 1.7 kb fragment that contains the whole gene (so, fair enough, some of it is the promoter and the rest must be the coding region).
You are right, considering that we are in bacteria, the best explanation for the RNA being so much longer than the DNA is that the crt gene is part of an operon. And again, it makes perfect sense that the crt gene is the first one in the operon since it's adjacent to the promoter.
For 9 B2, you need to think about frequencies and probabilities. Trans-complementation (assuming that we have the whole gene in the plasmid) works very efficiently, homologous recombination is rare, and mutation in the recipient is ultra-rare!
in reality, we would do an experiment in parallel where we'd conjugate in a plasmid that does not contain the DNA fragment of interest. If, with the plasmid containing the sequence of interest, we get a red colony every 10, while in the control conjugation we get a red one every million colonies, we can pretty much be sure that our fragment of interest does indeed contain the crt gene!
Hope this helps.
Pam
Hello Pam,
This question is regarding topic 11, page 2, II.C.
When we use a reporter gene (i.e. lacz) to find location of promoter and regulatory sequence, how can we screen for recepient cells that has plasmid inserted with the 3rd PCR product? Wouldn't it have the same phenotype as a recepient cell that has the plasmid but without insertion? Since both are unable to yeild blue colonies and both are Kan. rsistant
Thanks in advance!
Excellent question, and excellent thinking on your part.
You are perfectly correct: in the case described you can select for colonies that have received the plasmid, but whether they have the insertion or not you can't tell just by looking at the colonies.
This is when restriction mapping comes in really handy: it is very easy to take a colony (or many of them, seperately), grow it up and isolate and purify the plasmid that it contains. That plasmid can then be subjected to restriction mapping, and comparing its restriction digestion patterns with that of the original plasmid (without the insertion) will tell you whether what you have isolated is just the plasmid or the plasmid with an insertion.
Does that make sense?
Cheers
Pam
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