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F2 plants segregate 9/16 colored: 7/16 colorless. If one plant among the colorless.

a) The probability is 1/16 for a colored plant.
b) The probability is 3/4 for a colored plant.
c) The probability is 7/16 for a colored plant.
d) The probability is 9/16 for a colored plant.

User Goamn
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1 Answer

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Final answer:

In the F2 generation of a dihybrid cross considering only the trait of seed color, the ratio is 9 colored to 7 colorless. Thus, for one plant that is colorless, the probability that another plant is colored is 9/16.

Step-by-step explanation:

When considering the F2 generation in Mendelian genetics, particularly the segregation of plant seed color and texture, a dihybrid cross of two heterozygous parents gives a phenotypic ratio of 9:3:3:1. This ratio represents the offspring's possible combinations of two characteristics, such as seed color and seed texture. If we look at the color only, assuming a dominant and recessive pattern, three-quarters of the F2 plants will be colored (dominant trait) and one-quarter will be colorless (recessive trait).

To answer the student's question, given that F2 plants segregate 9/16 colored to 7/16 colorless, if one plant is among the colorless, the probability that another plant is colored is the proportion of colored plants in the F2 generation, which is 9/16. Therefore, the correct answer is:

d) The probability is 9/16 for a colored plant.

User Masinde Muliro
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