Question at position 18 The self-fertilization of an F1 pea plant produced from a parent plant homozygous for yellow and wrinkled seeds and a parent homozygous for green and round seeds resulted in F2peas (yellow and round are dominant). What is the probability that a given pea is yellow and wrinkled?3/161/161/43/42/3

Question

BlackTom AI · Accepted Answer

To tackle this genetics problem, I’ll map out the inheritance patterns involved and then assess each answer option.
Option 1: 3/16. This matches the expected product of two independent probabilities: winner yellow (dominant) from the color gene and wrinkled (recessive) from the shape gene, calculated as 3/4 × 1/4 = 3/16.
Option 2: 1/16. This would require both the color to be yellow with probability 1/4 and the shape to be wrinkled with probability 1/4, which forgets that yellow is dominant (the color trait in F2 from Yy × Yy is yellow 3/4, not 1/4).
Option 3: 1/4. This could arise if someone only counted the wrinkled phenotype frequency (rr) as 1/4 and ignored the need for the color to be yellow as well, effectively multiplying by 1 instead of 3/4 for color.
Option 4: 3/4. This would be the probability of yellow color alone (dominant phenotype) but does not specify the seed shape; it omits the requirement that the seeds be wrinkled, which is essential for the combined phenotype.
Option 5: 2/3. This fraction doesn’t align with classic Mendelian dihybrid ratios for the specific combined phenotype of yellow and wrinkled; it lacks a basis in the independent 3/4 (yellow) and 1/4 (wrinkled) probabilities.

Now, let’s confirm the underlying genetics to justify why 3/16 is the correct value:
- The first cross (parent1 YY rr) × (parent2 yy RR) yields F1 progeny with genotype Yy Rr, displaying yellow and round seeds (since Y is dominant for color and R is dominant for shape).
- When F1 plants self-fertilize (YyRr × YyRr), the color gene (Yy × Yy) segregates to yield yellow phenotype 3/4 of the time and green phenotype 1/4 of the time.
- The shape gene (Rr × Rr) segregates to yield wrinkled phenotype 1/4 of the time (rr) and round phenotype 3/4 of the time.
- Since color and shape genes assort independently, the probability of yellow (3/4) and wrinkled (1/4) together is (3/4) × (1/4) = 3/16.

Putting this together, option 1 (3/16) correctly represents the probability that a randomly chosen F2 pea is both yellow and wrinkled, while the other options misstate either the color probability, the shape probability, or ignore the need for both traits simultaneously.

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