Another cross in Drosophila involved the recessive, X-linked genes yellow (y), white (w), and cut (ct). A yellow-bodied, white-eyed female with normal wings was crossed to a male whose eyes and body were normal but whose wings were cut. The F₁ females were wild type for all three traits, while the F₁ males expressed the yellow-body and white-eye traits. The cross was carried to an F₂ progeny, and only male offspring were tallied. On the basis of the data shown here, a genetic map was constructed. Phenotype Male Offspring y + ct 9 + w + 6 y w ct 90 + + + 95 + + ct 424 y w + 376 y + + 0 + w ct 0 Could the F₂ female offspring be used to construct the map? Why or why not?

Video transcript

Hi everyone. Let's look at our next problem. It says 850 out of 1000 individuals. In a cross between individuals. Homos, I guess for X. Y. and the wild type were of the parental type. The distance that separates X from Y is then, well, we can recall our formula for calculating map units. Since we see all four of our answer choices are in map units. For distance Is the number of recombinant offspring divided by the total number of offspring times 100 is the recombination frequency in which equals the map units. So of 850 are the parental type than to get the recombinant offspring. We subtract 850 from the total number of offsprings. 1000 total -850 parental types Equals 150 recombination off string. So now we plug that into our equation. We have 150 re competent offspring over 1000 total times 100. And that's going to equal 15 map units apart for those two genes. So when we look at our answer choices, we see that choice D. Is indeed 15 map units. That's going to be our answer for the distance that separates X from Y. See you in the next video