Question

If the values of phenotypic and environmental variances are 6.8 and 1.2, respectively, the heritability (broad sense) of the trait would be-

• A. 8.23%
• B. 82.3%
• C. 823%
• D. 0.823%

Hint:

To find heritability, subtract the environmental variance from the phenotypic variance to get the genotypic variance: \(6.8 - 1.2 = 5.6\).
Then divide by the phenotypic variance: \(5.6 / 6.8 \approx 0.823\), or 82.3%.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Heritability in the broad sense (\(H^2\)) is the proportion of total phenotypic variance in a population that is attributable to genetic differences among individuals.
Key Formula or Approach:
The formula for broad-sense heritability is: \[ H^2 = \frac{V_G}{V_P} \] where \(V_G\) is the genotypic variance and \(V_P\) is the phenotypic variance.
The total phenotypic variance is the sum of genotypic and environmental variances: \[ V_P = V_G + V_E \] Rearranging this equation gives: \[ V_G = V_P - V_E \]

Step 2: Detailed Explanation:
We are given the following values: - Phenotypic variance (\(V_P\)) = 6.8
- Environmental variance (\(V_E\)) = 1.2
First, calculate the genotypic variance (\(V_G\)): \[ V_G = V_P - V_E \] \[ V_G = 6.8 - 1.2 = 5.6 \] Next, calculate the heritability in the broad sense (\(H^2\)): \[ H^2 = \frac{V_G}{V_P} \] \[ H^2 = \frac{5.6}{6.8} \] \[ H^2 \approx 0.8235 \] Converting this decimal value to a percentage: \[ \text{Percentage heritability} = 0.8235 \times 100 \approx 82.35% \] Rounding to the nearest decimal place gives 82.3%.

Step 3: Final Answer:
Therefore, the broad-sense heritability of the trait is 82.3%, which corresponds to option (B).