Assume that a bacterial plasmid carries the gene for resistance to the antibioti

Question

Assume that a bacterial plasmid carries the gene for resistance to the antibiotic kanamycin.
Using restriction enzyme A, you open the plasmid and insert a segment of a biologically
important gene isolated from a mouse. The gene was excised from the chromosome as part of
a fragment cut from whole DNA by using the same restriction enzyme A. After conducting
the appropriate steps in a typical bacterial transformation, you plate the transformed cells (+)
and control cells (-) on LB agar containing kanamycin and on LB agar alone.

a. What do you expect to see?

b. What you actually observe is no growth on either LB/Kan plate, but growth on both LB
plates. You try the experiment again using a different restriction enzyme B. This time you get
growth of transformed (+) cells on LB/Kan but no growth of control (-) cells on LB/Kan.
You get growth of both transformed (+) and control (-) cells on LB plates. How might you
explain these observations? Propose a map for the bacterial plasmid and the restriction sites
for restriction endonucleases A and B.

Explanation / Answer

a. Since kanamycin resistance gene is inserted therefore, we can assume that kanamycin resistance is needed for cells to grow. In such a case we expect to see growth on both LB agar alone plates as well as on LB/Kan plate plated with transformed cells and no growth of non transformed control cells on LB/Kan plate.

b. The observations can be explained as follows:

No growth on LB/Kan plate and growth only on LB agar only plate signifies that kanamycin resistance was actually not necessary for cell growth and they can grow on LB only.

But after restriction enzyme B has been used for the experiment the results come as were expected because now kanamycin resistance is necessary for the cells to grow and survive.

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