Another recessive mutation in Drosophila, ebony (e), is on an aut...

Question

Another recessive mutation in Drosophila, ebony (e), is on an autosome (chromosome 3) and causes darkening of the body compared with wild-type flies. What phenotypic F₁ and F₂ male and female ratios will result if a scalloped-winged female with normal body color is crossed with a normal-winged ebony male? Work out this problem by both the Punnett square method and the forked-line method.

Accepted Answer

Hey everyone, let's take a look at this question together, hemophilia is an X linked recessive disorder because the abnormal gene that causes it is located on the X chromosome. The eye color gene on the other hand, is located on chromosome 15 determine the expected ratios of hemophiliacs and brown eyed female and male offspring of the F. Two generation. If a normal female with Hamas, I guess blue eyes marries a hemophiliac man with homesickness, Brown eyes, remember that brown eye color is dominant over blue eye color. Let's first star with the jena types of both parents. Here we can see both genotype of the mom and the dad. So with the mom we get Big X. Little B. So for the dad we can have the X. That is linked with the recessive disorder and Big B. As well as Y. And big B. Our F one generation looks something like this. And so now that we have the F one generation, Jenna types, go ahead and cross those in a punnett square to see what our outcomes are. And so here's how our opponent square looks like. So let's go ahead and do the crosses to get the different genotype that we would end up with when we make that male and female from the F. One generation. And so here we have all 16 Jenna types of the F. Two generation. And so let's go ahead and write what the phenotype is for all, all of them. So that way we can get that ratio of Emma. Filic and brown eyed female to male offspring. So here we have all of the different types for the F. Two generation. And so we want to know the ratio of male to female or the Hema filic and brown eyes. So looking at our different types, we can see that here we have him a Philip brown. We have it here as well and we have it here now. All three of these are all Hema filic brown eyed males. We have three and then we have zero hem a Philip brown eyed females in the F. Two generation. And we have a total of 16 different types. And so our ratio of male to female is zero female and three out of 16 male. Which looking at our answer choices is answer choice C. Because answer choice C. Says that we have that zero out of 16 females with brown eyes and hemophilia. And we have three out of 16 males with hemophilia and brown eyes. So answer choice C. Is the only correct answer here. I hope you found this video to be helpful. Thank you and goodbye

Key Concepts

Autosomal Inheritance

Punnett Square

Forked-Line Method

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