In the DetBrisk plant the synthesis of aqua pigment in the p

In the DetBrisk plant, the synthesis of aqua pigment in the petals is controlled by two genes, B and Q; the WT allele is dominant for both genes. (Unless told otherwise, you can always assume that all other genes in a pathway are homozygous WT in these individuals.) The pathway is... a) If a pure breeding plant was unable to catalyze the first reaction, what color of petals would you expect to see? b) If a pure breeding plant was unable to catalyze the second reaction, what color of petals would you expect to see? c) If the plants from parts a and b were crossed, what color petals will the F1 plants have? d) What would be the expected phenotypic ratios In a cross between F1 plants? Make sure you include phenotypic class information with your ratios. Use this complementation chart to answer the questions about these mutant true breeders, with regards to petal pattern a) If you mated mut #1 and mut #3, would the offspring display the wildtype or mutant phenotype? b) Are mut #2 and mut #5 allelic? Explain your answer. (1-2 sentences) c) Let\'s say that you\'ve mated mut #1 and mut #2. Then, you decided to perform a self cross with those F1 offspring... what would be your expected phenotypic ratios?

Solution

a, If a pure breeding plant was unable to catalyze the first reaction, we would get blue colour pink.

Because, the first reaction is covertion of pink colour in to blue colour, if the plant was unable to catalyze the first reaction it wont turn in to blue colour and it remains pink.

b, If a pure breeding plant was unable to catalyze the second reaction, we would get blue colour blue.

Because, the first reaction occurs which is covertion of pink colour in to blue colour, but the second reaction which is conversion of blue to aqua pigment will be failed. Eventully, it appears blue.

c, Since, the pathway is controlled by dominant genes the recessive state may produce colourless petals.

d, It would be 9(aqua):3 (pink):3 (blue):1(colourless)

2, Please consider left column as female and top row as male and the first mentioned one is the female and the second mentioned one in the question is the male. Eventually we can find as follows.

a, Mutant type.

b, yes, it is allelic. Because these two mutations do not complement in trans position eventually mutant phenotype occur.

c, If you self the F1 which is a wild type, could segregate in the ratio 9:3:3:1.