Wings rare dominant to black body and vestigial wings in...

Wings are rare dominant to black body and vestigial wings in wild type fruit flies. A gray body (G) and regular homozygous wild type fly is crossed with a homozygous recessive (g) fly. Show this cross using the letter symbols: G x g. If the resulting heterozygous offspring were test crossed with flies that were not linked, what would be the expected phenotypic ratio? However, the actual results from the above test cross yielded the following phenotypes: gray-regular, black-vestigial, gray-vestigial, black-regular. Which of the above is the recombinant phenotype and which is the parental phenotype? Has linkage occurred? How do you know? Calculate how many map units the body color locus is from the wing length locus?

Transcript

00:01 We're looking at an f1 and an f2 generation.

00:02 So for this f1 generation, we know that all of them have the same genotype where they're all heterozygous.

00:09 They have one dominant v, one reccessive, one dominant v, and one recessive.

00:14 And we're told that we're going to cross someone from the f1 generation with an organism that is homozygous recessive for both traits.

00:26 Now, if we were to do this, we can see what we would.

00:32 Expect given a punnet square because we're looking at two different genes it's going to be a four by four punnet square and so you can take the first parent we're going to see all of the comedic combinations so i'm going to do first to be with first v and see it's all dominant first b with second v second b second b first v or second of both and we're going to do the same for the other parent first b second v, v, second v, second v, first, v, second of both.

01:22 And when we do that, you can actually notice that in this punnet square, you know, on the top, there's all these different options, but along the side from our second pair, they can only give the same gammy.

01:34 Which means that when we complete this punnet square, it actually gets easier.

01:38 Because, like, i've done this first square, i figured that one out.

01:42 All the ones below it in the same column are going to be the exact same.

01:46 Because we're going to get the exact same type of yami from the second parent.

01:54 So that means that solving a punnet square just becomes a lot faster.

02:10 And analyzing our punnets square is also easier because now we see that it's broken up into fourths.

02:20 So like these four squares in the first column are all the same.

02:25 So i would expect one -fourth of the offspring of this cross to have both dominant traits and then i would expect another fourth to have dominant b but recessive v traits another fourth recessive b but dominant v and then the last fourth to have only recessive traits so that's what we would expect from this kind of cross we can see it just with our little pun it square but we can also compare that to what we're told about the f2 generation.

03:09 So for the f2 generation, we have all of those four different phenotypes, right? two dominant, only dominant b, only dominant v, no dominant.

03:21 And here we have, i'm going to list them in order from the most common to the least common...

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