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Why can the principles of probability be used to predict the outcomes of genetic crosses?
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13. Mendelian Genetics
Punnett Square Probability
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A blue-eyed female that is homozygous recessive and a brown-eyed male that is heterozygous mate, producing two offspring. What is the probability that one child will have blue eyes AND one will have brown eyes? (Eye color is controlled by a single gene).
A
½.
B
¼.
C
¾.
D
1.
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Identify the genotypes of the parents. The blue-eyed female is homozygous recessive, so her genotype is 'bb'. The brown-eyed male is heterozygous, so his genotype is 'Bb'.
Set up a Punnett square to determine the possible genotypes of the offspring. Place the alleles of one parent on the top and the alleles of the other parent on the side. For the female, use 'b' and 'b', and for the male, use 'B' and 'b'.
Fill in the Punnett square by combining the alleles from each parent. The possible genotypes for the offspring are 'Bb', 'Bb', 'bb', and 'bb'.
Determine the phenotypic ratio from the Punnett square. 'Bb' results in brown eyes, and 'bb' results in blue eyes. Therefore, the ratio is 2 brown-eyed: 2 blue-eyed, or simplified, 1:1.
Calculate the probability of one child having blue eyes and one having brown eyes. Since each child is an independent event, multiply the probability of one child having blue eyes (1/2) by the probability of the other child having brown eyes (1/2), and then multiply by 2 to account for the two possible orders (blue-brown or brown-blue).
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