Answer:
Part B:-
Case 1 -
parents-- genotype -- PP (Hazel) x PP (Hank)
gametes-- P P
cross -- PP - all PP as no 'p' allele, so % of carriers = 0
Case 2 -
parents-- genotype-- PP (Hazel) x Pp(Hank)
gametes-- P P, p
cross--
P | |
P | PP - normal |
p | Pp - carrier |
Hence % of carriers = 50%
Case 3 -
parents-- genotype-- PP(Hazel) x pp (Hank)
gametes-- P p
cross-- Pp - 100% carriers.
Part C:
So, chance of girl child to be carrier = chance of Pp / total chances possible = 1/2
= 50%
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