Answer:
C. Luria Agar + kanamycin + tetracycline
Explanation:
Given that the recipient strain is tetracycline resistance and the donor strain is tetracycline sensitive.
Recipient strain is kanamycin sensitive and the donor strain is kanamycin resistance sensitive.
If we grow the recipient cell in Luria agar containing tetracycline then only recipient cells will grow and all the donor cells will die.
If we grow the recipient cell in Luria agar containing kanamycin then only recipient cells that have undergone transposon mutagenesis will grow and along with donor cells.
If the Luria agar has both kanamycin and tetracycline then only the recipient cells that have undergone transposon mutagenesis will grow.
Since recipient cells will acquire pir gene. So it will not require pir in the media.
So correct answer is option C. Luria Agar + kanamycin + tetracycline
Answer:
C. Luria Agar + kanamycin + tetracycline
Explanation:
As you can see in the question above, the recipient strain is resistant to tetracycline and not resistant to kanamycin, while the donor strain is not resistant to tetracycline and is resistant to canycin.
Based on this, we can visualize how the growth of these cells would be in relation to the composition of the culture media. If the culture medium has agar + tetracycline, only resistant strains would grow, in this case, donor strains would die. However, if the culture medium has agar + kanamycin in its composition, the recipient cells can grow, in the same way as donor strains, if a mutation caused by transposons has occurred.
A third situation can happen. There may be a culture medium containing luria agar + kanamycin + tetracycline. In this case, cells that are neither resistant to kanamycin nor tetracycline will be able to grow, however, recipient cells that have been mutated by transposons will be able to grow
