The autosomal recessive alleles for waltzing wz hooded hd an

The autosomal recessive alleles for waltzing (wz), hooded (hd) and tailless (ta) are all on different chromosomes in mice. A waltzing mouse is crossed to a hooded, tails mouse, and the F1 females and F1 males are inbred to produce 1000 progeny. How many waltzing, tailless (but not hooded) mice are expected amount the progeny?

The answer is 46.9. I need help figuring out how to get this answer…

Solution

Assume that the gene coding for normal phenotype are W, H and T (tail) and are dominant over the recessive characters waltzing (w), hooded (h) and tailless (t).

The genotype of waltzing mouse is, wwHHTT and the genotype of hooded, tails mouse is, WWhhtt. Cross between them will have the progeny with the following genotypes:

wwHHTT* WWhhtt --à Ww Hh Tt (all wild type) -à F1

Cross between F1 offspring will have the progeny with the following genotypes:

WwHhTt* WwHhTt -à

Ww*Ww -à W_ (3/4, wild type), ww (1/4, waltzing)

Hh*Hh -à H_ (3/4, wild type), hh (1/4, hooded)

Tt*Tt à T_ (3/4, wild type), tt (1/4, tailless)

The probability of waltzing, tailless (but not hooded) mice is = ¼*3/4*1/4 = 3/64

Thus, in 1000 offspring, the number of waltzing, tailless (but not hooded) mice is = 3/64 * 1000 = 46. 87 or 46.9