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The diagram below illustrates the normal expression of three different groups of genes (A, B, and C) within four different whorls of a developing flower. Normally, whorl 1 produces sepals, whorl 2 produces petals, whorl 3 produces stamens, and whorl 4 produces carpels. Furthermore, expression of class A genes inhibits class C genes, and expression of class C inhibits class A.
-What kind of mutation would produce carpels in whorls 1 and 4 and stamens in whorls 2 and 3?
A) defective in A
B) defective in B
C) defective in C
D) defective in A and B
E) defective in B and C

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The diagram below illustrates the normal expression of three different groups of genes (A, B, and C) within four different whorls of a developing flower. Normally, whorl 1 produces sepals, whorl 2 produces petals, whorl 3 produces stamens, and whorl 4 produces carpels. Furthermore, expression of class A genes inhibits class C genes, and expression of class C inhibits class A. 
-What kind of mutation would produce carpels in whorls 1 and 4 and stamens in whorls 2 and 3?
A) defective in A 
B) defective in B 
C) defective in C 
D) defective in A and B 
E) defective in B and C

Quizplus · Accepted Answer

A mutation defective in A would lead to the transformation of sepals (whorl 1) into carpels and petals (whorl 2) into stamens, while whorls 3 and 4 remain as stamens and carpels, respectively.

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