Note that the two parental gamete types (+ + and bl pu) are the most abundant, as expected. Use the data to calculate the recombination frequency and the genetic map distance between the two genes. Record the map distance in your notes.

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Linkage Mapping Using Dihybrid Testcrosses in Fruit Flies
In fruit flies, the normal honey-gray body-color (bl*) is dominant to black (bl),
having normal red eyes (pu+) is dominant to purple (pu), and having normal
wings (vg+) is dominant to vestigial (vg). The three genes are located close
together on the same chromosome.
Your task is to use data from three dihybrid testcrosses to (1) calculate
recombination frequencies, (2) convert recombination frequencies into map
distances, and (3) build a map of the chromosome interval covered by the
three genes, indicating their order and distances between them.
BLACK PURPLE DIHYBRID CROSS
In the parental generation, you mate a pure-breeding wild-type female
(bl+/blt;put/pu+) with a pure-breeding black, purple male (bl/bl;pu/pu) to
produce an F1 generation that is all wild-type (bl+/bl;pu*/pu). Note that the F1
flies are all dihybrid.
Next, you mate several F1 dihybrid females (bl+/bl;pu*/pu) with tester males,
which are black, purple (bl/bl;pu/pu). The offspring of this dihybrid testcross
are:
Phenotype Genotype Tester Gamete Dihybrid Gamete Number
+/bl;+/pu
444
bl/blipu/pu
443
Wild-type
Black, purple
Purple
Black
34
+/bl;pu/pu
bl/bl;+/pu
30
bl pu
bl pu
bl pu
bl pul
++
bl pu
+ pu
bl +
The columns in blue (phenotypes and numbers of offspring) are what you can
see and count. Generally, only the mutant phenotypes are listed, so a fly with
normal body color and purple eyes is listed as purple in the table.
The genotypes of the testcross offspring (orange) must be deduced from the
phenotypes and knowing that the tester contributed bl pu gametes. Finally, you
can deduce the dihybrid gametes (green) by subtracting the tester gamete
contribution from the offspring genotypes.
Note that the two parental gamete types (+ + and bl pu) are the most
abundant, as expected. Use the data to calculate the recombination frequency
and the genetic map distance between the two genes. Record the map
distance in your notes.
Transcribed Image Text:Linkage Mapping Using Dihybrid Testcrosses in Fruit Flies In fruit flies, the normal honey-gray body-color (bl*) is dominant to black (bl), having normal red eyes (pu+) is dominant to purple (pu), and having normal wings (vg+) is dominant to vestigial (vg). The three genes are located close together on the same chromosome. Your task is to use data from three dihybrid testcrosses to (1) calculate recombination frequencies, (2) convert recombination frequencies into map distances, and (3) build a map of the chromosome interval covered by the three genes, indicating their order and distances between them. BLACK PURPLE DIHYBRID CROSS In the parental generation, you mate a pure-breeding wild-type female (bl+/blt;put/pu+) with a pure-breeding black, purple male (bl/bl;pu/pu) to produce an F1 generation that is all wild-type (bl+/bl;pu*/pu). Note that the F1 flies are all dihybrid. Next, you mate several F1 dihybrid females (bl+/bl;pu*/pu) with tester males, which are black, purple (bl/bl;pu/pu). The offspring of this dihybrid testcross are: Phenotype Genotype Tester Gamete Dihybrid Gamete Number +/bl;+/pu 444 bl/blipu/pu 443 Wild-type Black, purple Purple Black 34 +/bl;pu/pu bl/bl;+/pu 30 bl pu bl pu bl pu bl pul ++ bl pu + pu bl + The columns in blue (phenotypes and numbers of offspring) are what you can see and count. Generally, only the mutant phenotypes are listed, so a fly with normal body color and purple eyes is listed as purple in the table. The genotypes of the testcross offspring (orange) must be deduced from the phenotypes and knowing that the tester contributed bl pu gametes. Finally, you can deduce the dihybrid gametes (green) by subtracting the tester gamete contribution from the offspring genotypes. Note that the two parental gamete types (+ + and bl pu) are the most abundant, as expected. Use the data to calculate the recombination frequency and the genetic map distance between the two genes. Record the map distance in your notes.
Expert Solution
Step 1

Given information

  • The parental genotypes bi+bi+/pu+pu+ and bi bi pu pu.
  • The F1 generation bi+bi pu+pu is test crossed with bi bi pu pu.

From the data obtained

  • The maximum number of offspring 444 and 443 represents the offspring having parental genotype.
  • While the lowest number of offspring 34 and 40 represent offspring having recombinant genotypes.
  • Total offspring are 444 + 443 + 34+ 40 = 961
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