Biology and Penetrance Question

Question

Assume that you have identified a unique gene in hawks which controls wing span in these birds. You have determined that the dominant allele (W) has a non-additive lethal penetrance of 18%. You have also identified a second gene which controls claw length and you determined that the dominant allele (C) of this gene has a non-additive lethal penetrance of 38%. After mating birds with a heterozygous genotype you obtain 360 eggs.   

How many viable layers can you expect from these eggs?

Answer 1

Hi Asked,

There's just too much left to assumption in this problem! but, let's assume that heterozygous means crosses of WwCc and WwCc, then WW:Ww:ww = 1:2:1, and CC:Cc:cc = 1:2:1 . Since these are dominant alleles (W and C), both of them, that will be 3/4 x 0.18 for lethal by W, and 3/4 x 0.38 for lethal by C. That runs up as a mortality fraction 0.135 for W and 0.285 for C, minus(0.135 x 0.285) [you only die once, contrary to the late Ian Fleming's conjecture]. Oh yes, and of those survivors, only the females lay!

you finish up the math,

--- Cheers, --- Mr. d.