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.

1) How many viable layers can you expect from these eggs?

Answer 1

On mating heterozygotic parents for the W and C gene, we should have a dihybrid ratio of :

9:3:3:1 as shown below:

WwCc x WwCc =

	WC	Wc	wC	wc
WC	WWCC	WWCc	WwCC	WwCc
Wc	WWCc	WWcc	WwCc	Wwcc
wC	WwCC	WwCc	wwCC	wwCc
wc	WwCc	Wwcc	wwCc	wwcc

We get the progeny with genotype and phenotypes as follows:

1. large wing /long claws: 1 WWCC, 2 WWCc, 2 WwCC, 4 WwCc

2. Large wing/short claws: 1 WWcc, 2 Wwcc,

3. Small wing/ Long claws: 2 wwCC, wwCc

4. Small wing/ short claws: 1 wwcc

Out of total 360 eggs if we divide it into the dihybrid ratio we would have :

1. large wing /long claws (202.5) :  22.49 of WWCC, 44.99 of WWCc, 44.99 of WwCC & 89.99 of WwCc

2. Large wing/short claws (67.5): 22.5 of WWcc and 45 of Wwcc

3. Small wing/ Long claws (67.5): 45 of wwCC and 22.5 of wwCc

4. Small wing/ short claws (22.5):   22.5 of wwcc

Since, W has a lethal penetrance of 18% and C has a lethal penetrance of 38% therefore, among those progenies having W or C or both, 18% or 38% or 56% of the eggs respectively would be non-viable  .

Hence, finally we have total viable eggs = 9.9 WWCC + 19.79 WWCc + 19.79 WwCC + 39.59 WwCc + 18.45 WWcc + 39.9 Wwcc + 39.9 wwCC + 18.45 wwCc + 22.5 wwcc = 228.27 eggs which amounts to approximately 228 eggs. Hence after cross of heterozygous hawks we would expect to get 228 viable layers.