TION S ebody color (e) is recessive to wildtype body color...

TION S ebody color (e) is recessive to wildtype body color (e*) and curled wings (cu) is recessive to wildtype straight wings (cu*). The ebony and curled genes are both located on chromosome 3 and the distance between the two genes is 18 m.u. The recessive allele (bw) leads to flies with brown colored eyes and the dominant wildtype allele (bw*) leads to flies with red eyes. Fill in the expected proportions of progeny for each phenotypic group when a female heterozygous fly of the genotype given below is tested crossed to a homozygous recessive male fly. \frac{e* \quad cu* \quad bw*}{e \quad cu \quad bw} % of Progany Phenotypes brown body, curled wings, red eyes (e*,cu,bw*) brown body, curled wings, brown eyes (e*,cu,bw) ebon body, straight wings, red eyes (e,cu*,bw*) ebon body, straight wings, brown eyes (e,cu*,bw) brown body, straight wings, red eyes (e*,cu*,bw*) brown body, straight wings, brown eyes (e*,cu*,bw) ebon body, curled wings, red eyes (e,cu,bw*) ebon body, curled wings, brown eyes (e,cu,bw)

Transcript

00:01 So in fruit flies, black body b is recessive to gray b+.

00:05 Purple pr are recessive to red pr+.

00:09 Curved wings c are recessive to normal wing c+.

00:12 A female fly that is heterozygous for all three traits is test crossed with a male fly that is black with purple on curved wings.

00:23 So they are homozygous recessive for all three genes.

00:26 And the following 1 ,000 offspring are observed.

00:30 So you got 409 gray body red eye normal wing, 61 gray body red eye curved wing, and two gray body purple eyes normal wing.

00:42 On 29 gray body purple eyes curved wing, 27 black body red eye normal wing, and one black eye red eyes curved wing.

00:54 60 black body purple eye normal wings.

00:57 And then 411 black body purple eyes curved wing.

01:01 If you were to do the chi -square analysis on the data above the test for linkage, what would be the expected number of offspring in this category? so this is a test cross.

01:17 So if the three genes are not linked and they assort independently, then let's take a look at this cross.

01:48 So you have a heterozygous b plus b, pr plus pr, and c plus c crossed with bb, pr, pr, c, and c.

02:15 So the homozygous recessive for all three genes are very simple.

02:19 It produces only one type of gametes.

02:20 But you will have to find out all the combination of the heterozygous parent.

02:27 So the heterozygous parents produce eight kind of different gametes because for each, you have three genes.

02:35 Each gene has two alleles.

02:36 So the combination is 2 to the third power, 8.

02:40 So let's write out all the gametes produced from this heterozygous parents.

02:46 So you have b plus, pr plus, c plus.

02:56 And then you have b plus, pr plus.

03:05 Then you have b plus, pr, c.

03:12 And then you have b plus, pr, c plus.

03:18 And these are four of them.

03:20 And then continue.

03:20 B, pr plus, c plus.

03:26 B, pr plus, c.

03:35 Then you have b, pr, c plus.

03:41 And b, pr, c.

03:48 So you have eight different gametes from this particular parent.

03:52 And the other parent, since it is homozygous recessive for all three genes, it only produces one type of gametes.

03:59 B, pr, c.

04:01 That's it.

04:02 So when you cross these together and you can see that the first one is b plus, pr plus, c plus pair up with b, pr, and c.

04:16 So this is the one type of, the first category of the offspring.

04:23 So i'm going to use red color to represent the test cross homozygous recessive.

04:32 And the second, obviously same.

04:34 B, pr, c...

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