BVB313 Lecture Notes - Lecture 2: Allele Frequency, Phenylketonuria, Genotype
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Question 1: in order to establish a breeding flock of chickens, you purchased 20 birds with normal feathers and 30 birds with frizzled feathers. The alleles controlling feather type are f and f. after allowing these chickens to interbreed, you count the number of normal, intermediate, and frizzled individuals among their offspring. F 40 = 0. 4 f 60 = 0. 6. Ff 2(0. 4 x 0. 6) = 0. 48 ff 0. 62 = 0. 36. Question 2: phenylketonuria is inherited as an autosomal recessive. Ff 2(0. 9999 x 0. 0001) = 0. 0001998 ff 0. 00012 = 0. 0000001. Ff 2(0. 99 x 0. 01) = 0. 0198 (carriers) about 2% ff 0. 012 = 0. 0001 (disease) Question 3: the allele for curly hair is dominant over the allele for straight hair. In a population of 2,500 individuals, 20% show the recessive phenotype. Ff 2(0. 80 x 0. 20) = 0. 32 ff 0. 202 = 0. 04. Ff 2(0. 55 x 0. 45) = 0. 495 ff 0. 452 = 0. 2025.
