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n Drosophila melanogaster, black body (b) is recessive to gray body (b+), purple eyes (pr) are recessive to red eyes (pr+), and vestigial wings (vg) are recessive to normal wings (vg+). The loci encoding these traits are linked, with the following map distances:b and pr:          6 mupr and vg:        13 mupr lies in the middle of b and vgThe interference among these genes is 0.5.A fly with a black body, purple eyes and vestigial wings is crossed with a fly homozygous for a gray body, red eyes and normal wings. The female progeny are then crossed with males that have a black body, purple eyes and vestigial wings. If 1000 progeny are produced from this test cross, what will be the phenotypes and proportions of the progeny?

Question

n Drosophila melanogaster, black body (b) is recessive to gray body (b+), purple eyes (pr) are recessive to red eyes (pr+), and vestigial wings (vg) are recessive to normal wings (vg+). The loci encoding these traits are linked, with the following map distances:b and pr:          6 mupr and vg:        13 mupr lies in the middle of b and vgThe interference among these genes is 0.5.A fly with a black body, purple eyes and vestigial wings is crossed with a fly homozygous for a gray body, red eyes and normal wings. The female progeny are then crossed with males that have a black body, purple eyes and vestigial wings. If 1000 progeny are produced from this test cross, what will be the phenotypes and proportions of the progeny?

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Solution

This is a three-point test cross problem in genetics. The first step is to determine the parental, single crossover, and double crossover phenotypes.

Parental phenotypes are the easiest to determine because they are the same as the original parents. In this case, the parental phenotypes are:

  1. Gray body, red eyes, normal wings (b+ pr+ vg+)
  2. Black body, purple eyes, vestigial wings (b pr vg)

Next, we need to determine the single crossover phenotypes. These are the phenotypes that result from one crossover event. Since pr is in the middle, there are two possibilities for single crossovers:

  1. Crossover between b and pr: Gray body, purple eyes, normal wings (b+ pr vg+) and Black body, red eyes, vestigial wings (b pr+ vg)
  2. Crossover between pr and vg: Gray body, red eyes, vestigial wings (b+ pr+ vg) and Black body, purple eyes, normal wings (b pr vg+)

Finally, we need to determine the double crossover phenotypes. These are the phenotypes that result from two crossover events. In this case, the double crossover phenotypes are:

  1. Gray body, purple eyes, vestigial wings (b+ pr vg)
  2. Black body, red eyes, normal wings (b pr+ vg+)

Now, we need to calculate the expected proportions of these phenotypes. The map distances given in the problem are actually percentages, so we can use them to calculate the expected proportions.

The expected proportion of parental phenotypes is 1 - (0.06 + 0.13) = 0.81 or 810 out of 1000.

The expected proportion of single crossover phenotypes is the sum of the map distances, so 0.06 + 0.13 = 0.19 or 190 out of 1000. However, since there are two types of single crossovers, we divide this number by 2 to get 95 out of 1000 for each type.

The expected proportion of double crossover phenotypes is the product of the map distances, so 0.06 * 0.13 = 0.0078. However, since the interference is 0.5, we multiply this number by (1 - 0.5) to get 0.0039 or 3.9 out of 1000. Again, since there are two types of double crossovers, we divide this number by 2 to get 1.95 out of 1000 for each type.

So, the expected proportions of the progeny are:

  1. 810 gray body, red eyes, normal wings
  2. 810 black body, purple eyes, vestigial wings
  3. 95 gray body, purple eyes, normal wings
  4. 95 black body, red eyes, vestigial wings
  5. 1.95 gray body, purple eyes, vestigial wings
  6. 1.95 black body, red eyes, normal wings

Please note that these are expected proportions, the actual proportions may vary due to random chance.

This problem has been solved

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