IUBio

Antibiotic resistance and growth rates

Emir KHATIPOV khatipov at REMOVE-THISnibh.go.jp
Wed May 20 08:30:19 EST 1998


-----Original Message-----
From: Lisa R. Williams <lrwill01 at homer.louisville.edu>
To: Emir KHATIPOV <khatipov at nibh.go.jp>
Date: Wednesday, May 20, 1998 9:44 PM
Subject: Re: Antibiotic resistance and growth rates


>Emir,
>  Have you tried transforming the kanamycin plasmid into the WT and
>measuring B levels?   Perhaps this is the better control so all of your
>samples would be grown under the same conditions (presence of antibiotic).
>Then you wouldn't have to worry about comparing "apples and oranges".
>Good luck with your experiments.
>
>

Sorry if I did not make it clear enough, but this is exactly what I did -
transformation of WT with blank Km plasmid.
That means that I compared B levels in WT, and in WT transformed with the Km
plasmid. B levels in the latter are 3 times lower than in WT. That's the
problem.
Just to remind, I have:
(1) WT having normal A and B levels
(2) mutant of WT with higher B (Bx) and altered A (Ax - changed pigment
composition)
(3) tansformant of WT transformed by an expression Km plasmid carrying
normal A gene
(4) control transformant of WT transformed by the same expression Km
plasmid, but without A gene.

What we do to prove that increased B levels in (2) are due to chromosomal
alterations? We do the following. When we produce the transformant (3) we
want to restore the normal A gene in case that in the mutant (2) the point
of the mutation is somewhere in the A gene, i.e. we do complementation. In
fact, we should get the wild phenotype then (normal B level), but we cannot,
because (3) is growing in the presence of Km, wereas WT is not.
    The only correct thing we can do in this case is to compare B levels in
(3) and (4). And we found that difference in B levels between (3) and (4) is
the same as between (1) and (2).

To help you to anderstand more here is a table:

Strain                A level                B level            Km
(1)                          2                        20                  -
(2)                          1                        30                  -
(3)                          1                        10                 +
(4)                         0.5                       6                  +

Can we say then from these data that increased B levels in (2) are due to
mutation in A? We believe we can, but our referees may not agree. To argue
we have to present the published evidence that (3) and (4) are proper
controls, despite their B levels are lower. Did you come across any
publications showing similar problems? Your comments welcome.

Emir







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