Question

In dihybrid cross, when \(F_1\) plants (Rr Yy) are self hybridized, the ratio of segregation of yellow and green in \(F_2\) is

• A. 1:2:1
• B. 3:1
• C. 9:3:3:1
• D. 1:1:1:1

Hint:

Don't be distracted by the term "dihybrid cross." If a question asks for the segregation of a \textit{single} trait within that cross, the answer will always be the monohybrid ratio (3:1 for phenotype or 1:2:1 for genotype).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In a dihybrid cross, we track two pairs of contrasting traits. However, according to Mendel's Law of Independent Assortment, the inheritance of one pair of characters is independent of the other. This means we can analyze each trait (seed color and seed shape) as a separate monohybrid cross.
Step 2: Detailed Explanation:
1. The $F_1$ genotype is $RrYy$. Selfing this gives $RrYy \times RrYy$. 2. To find the ratio for yellow vs. green seeds, we only look at the $Yy \times Yy$ part of the cross. 3. A cross of $Yy \times Yy$ results in: - $YY$ (Yellow) - $Yy$ (Yellow) - $Yy$ (Yellow) - $yy$ (Green) 4. Counting the phenotypes, we see 3 Yellow seeds for every 1 Green seed. 5. Even though the overall dihybrid phenotypic ratio is 9:3:3:1, if we isolate just one trait (seed color), it still follows the monohybrid segregation ratio of 3:1.
Step 3: Final Answer
The ratio of segregation of yellow and green in \(F_2\) is 3:1.