S2 - S1.A heterozygous pea plant that is tall with yellow seeds TtYy is allowed to self-fertilize What is the probability that an offspring will be

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S1.A heterozygous pea plant that is tall with yellow seeds, TtYy, is allowed to self-fertilize. What is the probability that an offspring will be either tall with yellow seeds, tall with green seeds, or dwarf with yellow seeds? Answer: This problem involves three mutually exclusive events, and so we use the sum rule to solve it. First, we must calculate the individual probabilities for the three phenotypes. The outcome of the cross can be determined using a Punnett square. P Tall with yellow seeds = 9/(9 + 3 + 3 + 1) = 9/16 P Tall with green seeds = 3/(9 + 3 + 3 + 1) = 3/16 P Dwarf with yellow seeds = 3/(9 + 3 + 3 + 1) = 3/16 Sum rule: 9/16 + 3/16 + 3/16 = 15/16 = 0.94 = 94% We expect to get one of these three phenotypes 15/16, or 94%, of the time. S2. As described in chapter 2, a human disease known as cystic fibrosis is inherited as a recessive trait. A normal couple’s first child has the disease. What is the probability that their next two children will not have the disease? Answer: A phenotypically normal couple has already produced an affected child. To be affected, the child must be homozygous for the disease allele and, thus, has inherited one copy from each parent. Therefore, since the parents are unaffected with the disease, we know that both of them must be heterozygous carriers for the recessive disease-causing allele. With this information, we can calculate the probability that they will produce an unaffected offspring. Using a Punnett square, this couple should produce a ratio of 3 unaffected : 1 affected offspring.
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The probability of a single unaffected offspring is P Unaffected = 3/(3 + 1) = 3/4 To obtain the probability of getting two unaffected offspring in a row (i.e., in a specified order), we must apply the product rule. 3/4 × 3/4 = 9/16 = 0.56 = 56% There is a 56% chance that their next two children will be unaffected. S3. A pea plant is heterozygous for three genes ( Tt Rr Yy ), where T = tall, t = dwarf, R = round seeds, r = wrinkled seeds, Y = yellow seeds, and y = green seeds. If this plant is self-fertilized, what are the predicted phenotypes of the offspring and what fraction of the offspring will occur in each category? Answer: One could solve this problem by constructing a large Punnett square and filling in the boxes. However, in this case, there are eight possible male gametes and eight possible female gametes: TRY, TRy, TrY, tRY, trY, Try, tRy, and try. It would become rather tiresome to construct and fill in this Punnett square, which would contain 64 boxes. As an alternative, we can consider each gene separately and then algebraically combine them by multiplying together the expected phenotypic outcomes for each gene. In the cross
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This note was uploaded on 04/07/2009 for the course BIO 325 taught by Professor Saxena during the Spring '08 term at University of Texas at Austin.

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S2 - S1.A heterozygous pea plant that is tall with yellow seeds TtYy is allowed to self-fertilize What is the probability that an offspring will be

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