Discussion 2-1

Questions to think about and/or work on in discussion 2 1. Give the genotype and phenotype ratios that may appear among the offspring of simple crosses that involve two genes and the genotypes of the parents that may give rise to each ratio. 2. In a family of seven children, what is the probability of obtaining the following number of boys and girls? You may need the binomial equation: (p + q)7 = p 7 + 7p 6 q + 21p 5 q 2 + 35p 4 q 3 + 35p 3 q 4 + 21p 2 q 5 + 7pq 6 + q 7 a. All boys b. All children of the same sex c. Six girls and one boy d. Four boys and three girls e. Four girls and three boys 3. Height in the marsh mallow plant is controlled by three genes: A, B, and C. If you cross a homozygous dominant plant (AA BB CC) with a homozygous recessive plant (aa bb cc) you produce a trihybrid (Aa Bb Cc). If the F1 are allowed to interbreed, what is the probability that the F2 offspring has the following genotypes: 9 :3 :3 :| ( 1/2 ) > = I / 128 21128 7p6q → 71112141 ' / 2) = > 1128 351 ' / 2) * ( ' 1213=351128 351128

a. AA bb CC b. Aa Bb Cc c. AA Bb cc 4. Phenylketonuria (PKU) is an autosomal recessive disease. Suppose that two unaffected parents produce a child with PKU. a. What is the probability that a sperm from the father will contain the PKU allele? b. What is the probability that an egg from the mother will contain the PKU allele? c. What is the probability that their next child will have PKU? d. What is the probability that their next child will be heterozygous for the PKU gene? 5. In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. The F1 are then intercrossed to produce the F2. a. What are the phenotypic ratios in the F2 (i.e., ratio and phenotype)? b. If an F1 plant is backcrossed with the bitter, yellow-spotted parent, what phenotypes and proportions are expected in the offspring? ' 14 × 114 × 114 = ' 144 112 × ' 12 × ' 12 = 4/64 114 × ' 12 ✗ ' 14=2/64 Aa ✗ Aa 112 112 114 112

b. Give the most likely genotypes of the parent in each cross (use B for the burnsi allele and B + for the pipiens allele). c. Use a chi-square test to evaluate the fit of the observed numbers of progeny to the number expected on the basis of your proposed genotypes. (a separate test for each cross) 8. Pink-eye and albino are two recessive traits found in the deer mouse Peromyscus maniculatus . In pink-eyed mice, the eye is devoid of colour and appears pink because of the blood vessels within it. Albino mice are completely lacking in colour, both in their fur and in their eyes. F.H. Clark crossed pink-eyed mice with albino mice; the resulting F1 had normal coloration in their fur and eyes. He then crossed the F1 mice with mice that were pink-eyed and albino and obtained the following progeny. It is hard to distinguish between mice that are albino and mice that are both pink-eyed and albino, so he combined these two phenotypes (Clark 1936 Journal of Heredity 27). Phenotype Number of Progeny Wild-type fur, wild-type eye colour 12 Wild-type fur, pink-eye 62 Albino Albino, pink-eye 78 Total 152 a. Give the expected numbers of progeny with each phenotype if the genes for pink-eye and albino assort independently. b. Use chi-square to determine if the observed numbers of progeny fit the number expected with independent assortment. Phenotype Observed Expected (Obs-Exp)2/Exp Wild-type Wild-type, pink eyes

Albino, wild-type, and albino, pink eyes Total