Question

A tall greengram plant was crossed with a dwarf plant. All the $F_1$ plants were of intermediate height. The $F_2$ plants produced different phenotypic classes (plant height) in the ratio of 1:4:6:4:1. How many genes are involved in controlling the plant height?

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

To quickly find the number of genes in polygenic ratios, count the number of phenotypes ($N$) and use $n = (N-1)/2$.
Common ratios: 1:2:1 ($n=1$), 1:4:6:4:1 ($n=2$), 1:6:15:20:15:6:1 ($n=3$).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question describes a cross involving plant height where the $F_1$ generation shows an intermediate phenotype, and the $F_2$ generation exhibits a specific distribution ratio of 1:4:6:4:1.
This pattern is characteristic of polygenic inheritance (also known as quantitative inheritance), where multiple genes contribute cumulatively to a single trait.

Step 2: Key Formula or Approach:
In polygenic inheritance, the number of phenotypic classes in the $F_2$ generation is given by the formula $2n + 1$, where $n$ is the number of genes involved.
The ratio of phenotypes follows the coefficients of the binomial expansion $(a+b)^{2n}$.
Detailed Explanation:

• Identification of the Ratio: The provided ratio is 1:4:6:4:1.

• Sum of the Ratio: Adding the components of the ratio gives $1 + 4 + 6 + 4 + 1 = 16$.

• Relating to Genotypes: In a dihybrid cross involving two genes (4 alleles), the total number of zygotic combinations is $4^n$.

• Calculation: Since the total frequency is 16, we set $4^n = 16$.

• Solving for $n$: $4^2 = 16$, thus $n = 2$.

• Verification via Phenotypic Classes: The number of phenotypic classes is 5 (1, 4, 6, 4, 1).

• Using the formula $2n + 1 = 5$, we get $2n = 4$, so $n = 2$.
Final Answer:
Based on the $F_2$ phenotypic ratio of 1:4:6:4:1 and the existence of 5 phenotypic classes, it is concluded that 2 genes are involved in controlling the plant height.