In the above cross, the probability of producing $F_{2}$ homozygous plants with red fruits, heterozygous plants with red fruits, and plants with yellow colored fruits respectively is

1/4 , 1/4 , and 2/3

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1/4, 1/2 and ¼

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3/4 , 1/2 and 1/4

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1/4, 3/4 and 1/4

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Solution

The correct option is B
1/4, 1/2 and ¼

The $F_{2}$ genotypic ratio of this monohybrid cross is 1:2:1. As the Punnett Square explains, out of the 4 combinations, the probability of homozygous dominant character is $\frac{1}{4}$ , that of heterozygous dominant character is $\frac{2}{4}$ = $\frac{1}{2}$ , and homozygous recessive character is $\frac{1}{4}$ .

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