25. In fruit flies, the yellow-bodied (y), echinus (ec) and white-eyed (w) mutat

Question

25. In fruit flies, the yellow-bodied (y), echinus (ec) and white-eyed (w) mutations are all recessive to their corresponding normal alleles. A homozygous recessive yellow-bodied, echinus, white-eyed female (y ec w / y ec w, the gene order is arbitrary) was mated to a normal male, and the resulting F1 females were mated to homozygous recessive males. After collecting 1000 F2 progeny, the following numbers were observed with each phenotype:
+ + + 475
y ec w 469
y + + 8
+ ec w 7
y + w 18
+ ec + 23
+ + w 0
y ec + 0
What is the genetic map distance from y to w?
The Answer is 1.5 m.u... My question is in regards to how we go about acquiring these numbers. Please show your work..

Explanation / Answer

The total number of progeny = 1000 The number of recombinants between y and w are = 8+ 7 = 15 Map distance is calculated by using the number of recombinants. percentage of recombinants betwen y and w = 15 / 1000 * 100 = 1.5m.u So, the answer 1.5 mu is the distance between the genes y and w.

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