BIOLOGY 2C03 Lecture Notes - Sickle-Cell Disease, James V. Neel, Vernon Ingram

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Let"s review what we learned last class with an example. P generation: perform a cross between two black-furred rabbits. In the f1 generation you get 6 black-furred offspring and two white-furred offspring. 1 b/b:2b/b genotypic ration among the black f1 progeny. F1 generation: perform a cross between two black-furred rabbits from the f1 progeny. To get a white rabbit, both rabbits must be b/b in genotype and a b/b offspring must be produced by these parents. We need to calculate two probabilities: probability the parents are heterozygous. There will be a 1/3 chance that the f1 black rabbit is b/b and a. 2/3 chance that the black rabbit is b/b: probability heterozygous parents will produce b/b offspring. The chance of two b/b parents giving offspring is . P (white offspring) = p(both f1 rabbits are b/b) x p(b/b offspring) (2/3 x 2/3) x (1/4) = 1/9. Sickle cell disease is an autosomal recessive disorder.

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