Week 5: Chapter 6: Homework
#44: (see image)
-- answer is on phone
#62: (see image)
-- answer is on phone
#13: In humans, PKU (phenylketonuria) is a recessive disease caused by an enzyme inefficiency at step A in the following simplified reaction sequence, and AKU (alkaptonuria) is another recessive disease due to an enzyme inefficiency in one of the steps summarized as step B here: phenylalanine --A--> tyrosine --B--> CO2 H2O A person with PKU marries a person with AKU. What phenotypes do you expect for their children? All normal, all having PKU only, all having AKU only, all having both PKU and AKU, or some having AKU and some having PKU?
-- assuming homozygosity for the normal gene, the mating A/A - b/b - a/a - B/B would only result in A/a - B/b heterozygous offspring. All their children would be normal. (the dash above indicates a dot, which the book uses to denote that it is unknown whether the genes are on two different chromosomes or linked)
#20: Radishes may be long, round, or oval, and they may be red, white, or purple. You cross a long, white variety with a round, red one and obtain an oval, purple F1. The F2 shows nine phenotypic classes as follows: 9 long, red; 15 long, purple; 19 oval, red; 32 oval, purple; 8 long, white; 16 round, purple; 8 round, white; 16 oval, white; and 9 round, red. a. Provide a genetic explanation of these results. Be sure to define the genotypes and show the constitution of the parents, the F1, and the F2. b. Predict the genotypic and phenotypic proportions in the progeny of a cross between a long, purple radish and an oval, purple one.
-- rest of the answer is in the answer book
#27: Two normal-looking fruit flies were crossed, and, in the progeny, there were 202 females and 98 males. a. What is unusual about this result? b. Provide a genetic explanation for this anomaly. c. Provide a test of your hypothesis.
a) The expected sex ratio is 1:1, but in this case it is 2:1 b) The female parent was heterozygous for an X-linked recessive lethal allele. This would result in 50 percent fewer males than females. c) Half of the female progeny should be heterozygous for the lethal allele and half should be homozygous for the nonlethal allele. Individually mate the F1 females and determine the sex ratio of their progeny. Of these F1 females, mated by normal males, half should produce a normal male: female ratio (1:1), while the other half should produce a ratio 2:1.
#32abc: In roses, the synthesis of red pigment is by two steps in a pathway, as follows: colorless intermediate ---gene P---> magenta intermediate ---gene Q---> red pigment a. What would the phenotype be of a plant homozygous for a null mutation of gene P? b. What would the phenotype be of a plant homozygous for a null mutation of gene Q? c. What would the phenotype be of a plant homozygous for null mutations of genes P and Q?
a) colorless b) magenta c) colorless