A cross is made between an Hfr that is met+ thi+ pur+ and an F that is met thi p

Question

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 thi+ and pur+ alleles. The following numbers of individuals are found for each genotype: met+ thi+ pur+ 280 met+ thi+ pur- 0 met+ thi- pur+ 6 met+ thi- pur- 52

What is the gene order? It may help to draw the possible gene orders and recombination events necessary to produce each set of genotypes.

What are the map distances between each locus?

Explanation / Answer

Answer:

a. We know that met enters the recipient 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 requires two additional crossovers. Each possible order predicts a different class that arises by four crossovers rather than two. If the order were met, thi, pur, then met+thipur+ recombinants would be very rare.

On the other hand, if the order were met, pur, thi, then the four-crossover class would be met+purthi+. From the information given in the table, it is clear that the met+purthi+ class is the four-crossover class.

So, the gene order is: met, pur, thi

b. To compute the distance between met and pur, we compute the percentage of met+purthi:

52/388=15.4 m.u.

Similarly, the distance between pur and thi is: 6/338=1.8 m.u.

Gene map:

15.4 m.u. 1.8 m.u.

met+---------------------------pur+--------------------------thi+

met----------------------------pur----------------------------thi-

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