GenCh6answers

GenCh6answers - 2. 4. Assuming homozygosity for the normal...

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2. Assuming homozygosity for the normal gene, the mating is A/A · b/b × a/a · B/B. The children would be normal, A/a · B/b (see Problem 12). 4. Growth will be supported by a particular compound if it is later in the pathway than the enzymatic step blocked in the mutant. Restated, the more mutants a compound supports, the later in the pathway it must be. In this example, compound G supports growth of all mutants and can be considered the end product of the pathway. Alternatively, compound E does not support the growth of any mutant and can be considered the starting substrate for the pathway. The data indicate the following: a. and b. A E C B D G 5 4 2 1 3 vertical lines indicate the step where each mutant is blocked c. A heterokaryon of double mutants 1, 3 and 2, 4 would grow as the first would supply functional 2 and 4, and the second would supply functional 1 and 3. A heterokaryon of the double mutants 1, 3 and 3, 4 would not grow as both are mutant for 3. A heterokaryon of the double mutants 1, 2 and 2, 4 and 1, 4 would grow as the first would supply functional 4, the second would supply functional 1, and the last would supply functional 2. 5. a. If enzyme A was defective or missing ( m 2 /m 2 ) , red pigment would still be made and the petals would be red. b. Purple, because it has a wild-type allele for each gene, and you are told that the mutations are recessive. c. 9 M 1 /– ; M 2 /– purple 3 m 1 /m 1 ; M 2 /– blue 3 M 1 /– ; m 2 /m 2 red 1 m 1 /m 1 ; m 2 /m 2 white d. The mutant alleles do not produce functional enzyme. However, enough functional enzyme must be produced by the single wild-type allele of each gene to synthesize normal levels of pigment. 6. a. If enzyme B is missing, a white intermediate will accumulate and the petals will be white. b. If enzyme D is missing, a blue intermediate will accumulate and the petals will be blue. c. P b/b ; D/D × B/B ; d/d
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Chapter Six 108 F 1 B/b ; D/d purple d. P b/b ; D/D × B/B ; d/d F 1 B/b ; D/d × B/b ; D/d F 2 9 B/– ; D/– purple 3 b/b ; D/– white 3 B/– ; d/d blue 1 b/b ; d/d white The ratio of purple : blue : white would be 9:3:4. 7. The woman must be A/O , so the mating is A/O × A/B . Their children will be Genotype Phenotype 1 / 4 A/A A 1 / 4 A/B AB 1 / 4 A/O A 1 / 4 B/O B 9. a. The original cross and results were P long, white × round, red F 1 oval, purple F 2 9 long, red 19 oval, red 8 round, white 15 long, purple32 oval, purple 16 round, purple 8 long, white 16 ova l, white 9 round, red 32 long 67 oval 33 round The data show that, when the results are rearranged by shape, a 1:2:1 ratio is observed for color within each shape category. Likewise, when the data are rearranged by color, a 1:2:1 ratio is observed for shape within each color category. 9 long, red
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GenCh6answers - 2. 4. Assuming homozygosity for the normal...

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