LS4_Week_3_Discussion

LS4_Week_3_Discussion - Life Sciences 4: Supplementary...

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Life Sciences 4: Supplementary Problems Extensions to Mendel 1. If a man of blood group AB marries a woman of blood group A whose father was of blood group O, what different blood groups can this man and woman expect their children to belong to? Solution : The woman must be AO, so the mating is AO X AB. Their children will be: Genotype Phenotype 1 AA A 1 AB AB 1 AO A 1 BO B 2. Erminette fowls have mostly light-colored feathers with an occasional black one, giving a flecked appearance. A cross of two ermines produced a total of 48 progeny, consisting of 22 Erminette, 14 blacks, and 12 pure whites. What genetic basis of the Erminette pattern is suggested? How would you test your hypotheses? Solution : You are told that the cross of two Erminette fowls results in 22 Erminette, 14 black, and 12 pure white. Two facts are important: (1) the parents consist of only one phenotype, yet the offspring have three phenotypes, and (2) the progeny appear in an approximate ration of 1:2:1. These facts should tell you immediately that you are dealing with a heterozygous X heterozygous cross involving one gene and that the Erminette phenotype must be the heterozygous phenotype. When the heterozygote shows a different phenotype from either of the two homozygotes, the heterozygous phenotype results from incomplete dominance or codominance. Because two of the three phenotypes contain black, either fully or in an occasional feather, you might classify the remitter as an instance of incomplete dominance because it is the Erminette has both black and white feathers, you might classify the phenotype as codominant. Your decision will rest on whether you look at the whole animal (incomplete dominance) or at individual feathers (codominance). This is yet another instance where what you conclude is determined by how you observe. To test the hypothesis that the Erminette phenotype is a heterozygous phenotype, you could cross an Erminette with either, or both, of the homozygotes. You should observe a 1:1 ratio in the progeny of both crosses.
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3. In a maternity ward, four babies become accidentally mixed up. The ABO types of the four babies are known to be O, A, B, and AB. The ABO types of the four sets of parents are determined. Indicate which baby belongs to each set of parents (a) AB x O, (b) A x O, (c) A x AB, (d) O x O. Solution : Both codominance (=) and classical dominance (>) are present in the multiple allelic series for blood type: A = B, A>O, B>O. Parents' Phenotypes Parents' Possible Genotypes Possible Offspring Genotypess a) AB X O AB X OO AO, BO b) A X O AA or AO X OO AO, OO c) A X AB AA or AO X AB AA, AB, AO, BO d) O X O OO X OO
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This note was uploaded on 03/28/2011 for the course LS 4 taught by Professor Ribaya during the Spring '08 term at UCLA.

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LS4_Week_3_Discussion - Life Sciences 4: Supplementary...

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