Question

The population mean is proportional to gene frequency if there is

• A. no dominance
• B. complete dominance
• C. over dominance
• D. partial dominance

Hint:

When there is no dominance, the heterozygote phenotype is exactly intermediate between the two homozygotes, making gene effects purely additive.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In quantitative genetics, the mean value of a quantitative trait in a population depends on the allele (gene) frequencies and the genotypic values of the locus.

Step 2: Detailed Explanation:
Let us define the genotypic values for a single biallelic locus with alleles \(A_1\) and \(A_2\):
- \(A_1 A_1\): value is \(+a\)
- \(A_1 A_2\): value is \(d\) (dominance deviation)
- \(A_2 A_2\): value is \(-a\)
Let the frequency of \(A_1\) be \(p\) and \(A_2\) be \(q\).
The population mean (\(M\)) is defined by the formula: \[ M = a(p - q) + 2pqd \]
Let us analyze how the presence of dominance (\(d\)) affects this relationship:
If there is dominance (complete, partial, or overdominance), the term \(2pqd\) is non-zero.
Because \(2pq = 2p(1-p)\), the population mean becomes a quadratic function of the allele frequency, and is not directly proportional.
If there is no dominance (codominance/additive gene action), the dominance deviation \(d\) is zero.
Substituting \(d = 0\) into the formula: \[ M = a(p - q) \] \[ M = a(p - (1 - p)) = a(2p - 1) \]
In this case, the population mean is a linear function of the gene frequency \(p\), making it directly proportional.

Step 3: Final Answer:
The population mean is proportional to gene frequency if there is no dominance.