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. PTall with yellow seeds = 9/(9 + 3 + 3 + 1) = 9/16 PTall with green seeds= 3/(9 + 3 + 3 + 1) = 3/16 PDwarf 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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