2) A series of two-point crosses among fruit flies is...

2) A series of two-point crosses among fruit flies is carried out between genes for brown eyes (bw), arc wings (a), vestigial wings (vg), ebony body color (e), and curved wings (cv). The following number of nonrecombinant and recombinant progeny were obtained from each cross. Genes in cross Progeny (NR) Progeny (R) a, bw 2224 117 a, cv 2609 823 a, e 3200 3200 a, vg 5172 2379 bw, cv 4614 1706 bw, e 4150 4150 bw, vg 2796 1434 cv, e 3116 3116 cv, vg 2102 305 vg, e 4559 4559 Using these data from two-point crosses, create a genetic map for these genes, providing the order of the genes and the best estimates of the distances separating them. Recombination frequencies: a, bw a, cv a, e a, vg bw, cv bw, e bw, vg cv, e cv, vg vg, e The genetic map:

Transcript

00:02 So a series of two -point crosses among fruit flies is carried out between the genes for several genes such as brongi, vw, arkwing, a, vestral wing, vg, abony body color, e, and curved wing c .v.

00:17 The following number of non -recombinant and recombinant progeny were obtained from each crosses.

00:22 And as you can see on a board, i put the genes in the first row column.

00:29 The progeny from non -recombinant stands for nr in the same.

00:32 The second column and the progeny for recombinant from the third column.

00:37 So now use these data from the two point crosses created genetic map for these genes, providing order of the genes and best estimate of the distance separating them.

00:51 So first of all, let's take a look at each cross so we can figure out the distance between each two genes.

00:57 So let's see, this is a distance.

00:58 So calculate the distance between two genes.

01:03 All you have to do is take the recombinant and then divide by the total, which is 117 plus 2 -224.

01:15 So this is a total progeny and any times 100 to get the total recombinant rate.

01:24 So in this case, you have a recombinant rate of 5%, so that the distance is basically 5 cm.

01:33 So don't forget to times 100%.

01:38 Okay.

01:39 So you can tell that between a and bw, it's about 5 % recombinant rate, and this also tells you the distance between g and a and bw is 5 centimorgan.

01:51 So use the same method we can calculate the distance between a and cv.

01:56 So 8 ,23 divide 823 plus 2 ,609.

02:04 Again times 100, this gives you around 24 % recombinant rate.

02:11 So we know between c and v, again, has 24 centi morgan in distance.

02:20 Now, between a and e, you can see the non -recombinant and recombinant.

02:25 They're equal.

02:27 So it's 50 -50.

02:28 So this means that a and e not linked because you have the same amount of non -recombinant and recombinant offspring.

02:40 This means that they are assorted independently.

02:44 So they are not linked.

02:45 They are on separate chromosome.

02:48 The next one, a and vg, same idea.

02:52 You can see there is more non -recombinant than the recombinant.

02:58 So that means these two genes are somewhat linked.

03:01 So same thing, two, three, seven, nine, divide by the total 5172 plus 2379 times 1 ,000 100%, so between a and vg is about 31 .5%.

03:25 So the distance between a and vg is about 31 .5 centi morgan.

03:29 The next one, again, more non -recombinant than recombinant offspring.

03:37 Bw and vg are linked.

03:40 So 1706 divide by 1706 plus 461414 and times 100.

03:58 So between bw and cv, 27 % recombinant rate.

04:05 So the distance between bw and cv is about 27 centmorgan...

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