The testcross is between an eyeless female and a shaven bristle male where the W

Question

The testcross is between an eyeless female and a shaven bristle male where the Wild female-588 and Wild male-615 are the first progeny generation. Therefore, the test cross is sv+/+ey x svey/svey.

Then the Wild female from the F1 generation was mated with a eyeless and shaven bristles male. Ignoring the sex, I got 626 ey, 602 sv, 9 svey, 6 +

Draw a map that shows the map distance (in map units or centimorgans) between the locus for the shaven bristle allele and the locus for the eyeless allele.

Could you explain what the map distance is and what the drawing should look like, including the calculations?

Explanation / Answer

Hi, the question is about the genetic linkage. The no. Of progeny is used to calculate the gene order and gene distance. Since the question deals with 2 genes we only have to determine the distance between them.

Let us begin with the genotype of parents. It is already given that they are homozygotes and hence the F1 progeny are hetero zygote. With svey+/sv+ey genotype.

The cross between F1 Female and male gave progeny with 626ey, 602 sv, 9 svey and 6 +. The parental combination always occur in highest number. So the 626, 602 are parental combination. 9 and 6 are mutang types which have occurred due to recombination. To determine the genetic distance we need to find the recombination frequency first.

RF = mutant type / total no of progeny

= 15/ 1243 (adding 626+602+9+6)

= 0.012

Gene distance is given by RF * 100

= 0.012 × 100

= 12cM

So the genes are located 12 cM apart on the chromosome.

SV________12______EY

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