3) A cross is made between an Hfr that is met thi pur* and an F that is met thi
ID: 213461 • Letter: 3
Question
3) A cross is made between an Hfr that is met thi pur* and an F that is met thi pur Interrupted-mating studies show that met enters the recipient last, so met recombinants are selected on a medium containing supplements that satisfy only the pur and thi requirements These recombinants are tested for the presence of the thit and pur alteles. The following numbers of individuals are found for each genotype: met thi pur 280 met thi pur met thi pur6 met" thi pur 52 a) Why was methionine (Met) left out of the selection medium? b) What is the gene order? c What are the map distances in recombination units? F6 8Explanation / Answer
a. Methionine was left out of the selection medium because we were interested in isolating met+ bacteria. Met+ signifies that these bacteria have the ability to synthesize methionine. While, met- bacteria cannot synthesize methionine. So, they can only grow if methionine is provided from outside that is, added in the medium). If the selection medium contained methionine, then both bacteria met+ and met- will be able to grow and therefore we would not be able to differentiate met+ from the met- bacteria.
b. Since met enters the recipient at the last, there are only two possible gene orders if the first marker enters on the right: met, thi, pur or met, pur, thi.
One of the four possible classes of recombinants needs two additional crossovers. Each possible order predicts a different class that arises by four crossovers rather than two. For example, if the order were met, thi, pur, then met+thipur+ recombinants would be very exceptional. On the other hand, if the order was met, pur, thi, then the four-crossover class would be met+purthi+. From the information given in the table above, it is clear that the met+purthi+ class is the four-crossover class and therefore that the gene order would be met, pur, thi.
c) Distance between met and pur = 52/338 = 15.4 m.u.
Distance between pur and thi = 6/338 = 1.8 m.u
Gene map: met pur thi (as per above conclusion)
15.4 m.u 1.8 m.u
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