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barry barry
wrote...
Posts: 11630
12 years ago
A man and a woman with normal phenotypes have several children, one of whom has albinism.
a. What can you conclude about the genotype of the mother?
b. What is the probability that the children who do not display albinism are heterozygous?
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wrote...
Staff Member
12 years ago
(a) Both parents must be heterozygous (a carrier) in order to have an offspring with albinism.

(b) With parents of genotype Aa × Aa, the genotypes possible for offspring not displaying the traits of albinism are AA, Aa, and aA. Two of these three are carriers, so there is a   (66%) chance that a child who does not display albinism is a carrier.
- Master of Science in Biology
- Bachelor of Science
barry Author
wrote...
12 years ago
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