BIOLOGY 2C03 Lecture Notes - Lecture 8: Heritability, Plant Pathology, Necrosis

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Problem 1: how many cm does each allele contribute to the height di erence of 90 cm di = 90 cm alleles = 10 di /allele = 90cm/10 = 9cm, plant 1 genotype: aabbccddee. How? parent 1: aabbccddee parent 2: aabbccddee o spring: aabbccddee and aabbccddee --> 7 additive alleles --> 193 cm. Aabbccddee --> 8 additive alleles --> 202 cm: all genes are unlinked. Determine the probability of an abcde gamete from plant 1 and abcde from: what will the genotype and phenotype of a plant from these gametes be and what is the probability? parent 1: aabbccddee --> abcde. = 1(a) x 1/2(b) x 1(c) x 1/2(d) x 1(e) = 1/4 parent 2: aabbccddee --> abcde. = 1(a) x 1/2(b) x 1(c) x 1/2(d) x 1(e) = 1/4 o springs genotype/phenotype. Aabbccddee: 4 alleles x 9cm/allele + 130cm = 166cm probability = 1/4 (p. gamete 1) x 1/4 (p. gamete 2) = 1/16.

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