In sweet pea, Lathyrus odoratus, flower color is controlled by two independently

Question

In sweet pea, Lathyrus odoratus, flower color is controlled by two independently assorting gene loci (A, a and B, b). In order to produce purple-colored flowers a plant must have at least one capital allele at each locus (i.e., must be A_B- to have purple-colored flowers). Homozygous recessive genotypes for either aa or bb or both result in white-flowered plants. Suppose the following cross is made: AaBb times AaBb. What is the probability that the first offspring produced has purple flowers? What is the probability that the first two plants produced are white-flowered? Given that the first plant is purple-flowered, what is the probability that it has the same genotype as its parents?

Explanation / Answer

We can have the below matrix for the cross AaBb x AaBb

(A)

Purple flower is possible for AaBb, AABb, AaBB, AABB

Hence the probability that the first offspring produced has purple flowers is 4/16 = 1/4

(B)

Probability that a plant produced are white flowered = 1-1/4 = 3/4

Hence the probability that the first two plants produced are white flowered = 3/4 * 3/4 = 9/16

(C)

There are four possible outcomes for the genotype same as of parents

Hence probability that the plant produced has the same genotype as of parent is 4/16 = 1/4

BB Bb bb Bb AA AABB AABb Aabb AABb Aa AaBB AaBb Aabb AaBb aa aaBB aaBb aabb aaBb Aa AaBB AaBb Aabb AaBb

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