The data for the trihybrid cross below is given. By convention, only the alleles

Question

The data for the trihybrid cross below is given. By convention, only the alleles contributed by the beterozygous female parent is given since that will determine the phenotype of the progeny Use the data to calculate and draw a genetic map for the three gene loci, yellow body(),white eyes (wh), and cut wings (ct. The genes occur in the order given and all of the mutations are recessive Trihybrid Test Cross: Progeny Phenotypes 803 795 18 y wh et y+16 +et185 Total 2000 Draw a genetie map for the lociy, wh, and cr based on your calculations of recombination frequencies between yowh and whet Show your work above and draw the map below for full points Calculate the interference below

Explanation / Answer

Here + + + and y wh ct is the parental type

Now number of recombinant progeny for y wh = 18+16+3+1= 38

the frequency of recombination between y and wh = 38/2000 = 0.019 = 1.9%

again number of recombinant progeny for wh ct = 185+179+3+1 = 368

frequency of recombination between wh and ct = 367/2000 = 0.184 = 18.4 %

so the genetic map for given gene is

y-----1.9m.u.-----wh-------18.35m.u.--------ct

now expected double cross over frequency = 0.019 x 0.1835 = 0.003496

so expected number of double cross over progeny = 0.003496 x 2000 = 6.992 = 7 (approx)

however actual number of double crossover progeny = 3+1 =4

Hence coefficient of coincidence =

actual number of double crossover/ expected number of double crossover = 4/7= 0.5714

Hence, interference = 1- coefficient of coincidence = 1 - 0.5714 = 0.4286 = 42.86%

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