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 Construct a map, assuming that white is at locus 1.5 on the X chromosome.

Video transcript

Hi everyone. Welcome back. Let's look at our next problem. It says which of the following is true when genes are located close to each other on chromosomes and cannot be separated even by crossing over. Well, we can sort of draw a little diagram to help us envision this scenario and think through our answer choices. So we imagine genes called B. And E. That are located so close to each other that no crossing over occurs to separate them. And imagine an individual hetero cygnus for Big B. Little B Big E. Little E. With these two genes on the same chromosome. So we imagine there are two chromosomes. We know that one of them. We're going to say that this individual is the result of a cross between Hamas, dominant individual and homogeneous recessive individual. So we imagine their chromosomes. We see they inherited from one parent. This Big B. Biggie um from the homicide is dominant parent and little B. Little lee from the homesickness recessive parent, one chromosome from each parent. Because these genes cannot be separated. When this hetero sickos individual produces gametes. Only two types of gametes will be produced. Big B. Little Biggie, little B. And little E. There will be no crossing over, No recombination. You won't see any mixing of the dominant and hetero salio. I'm sorry the dominant recessive values. So only gametes similar to the parental types will be there. There's no recombination. So with this in mind let's look at our answer choices. Choice A says there are more recombination. Parent types. Well this is not correct. As we see the gametes that can be produced can only be parental types. Choice B says there are equal numbers of recombinant and para types. Again, not correct. They're only parental types. Choice C says there are fewer recombination, parental types. Again, not correct. We have the only correct answer here. In D only parental types will exist, and this is correct. If they cannot be separated at all, there's no crossing over. There will be no recombination of those genes, and only parental types of gametes will exist. See you in the next video.