Which one of the following relationships correctly represents the narrow sense heritability? ($$V_{A}-$$ additive variance, $$V_{D}-$$ dominance variance, Vi - interaction variation, $$V_{GE}-$$ genotype x environmental variance, $$V_{T}-$$ total variance)​

Correct option is A

Correct Answer:
​Option (a) — VA / VT
Explanation:
Narrow-sense heritability (h²) represents the proportion of total phenotypic variance that is due to additive genetic variance (VA).

Mathematically,
h² = VA / VT

Where,

VA = Additive genetic variance (effects of individual alleles that are transmitted from parents to offspring)

VD = Dominance variance (effects due to interaction between alleles at the same locus)

VI = Interaction or epistatic variance (interactions between alleles at different loci)

VGE = Genotype × Environment interaction variance

VT = Total phenotypic variance = VA + VD + VI + VGE + VE (where VE is environmental variance)

Thus, narrow-sense heritability isolates the additive component because it determines the resemblance between parents and offspring and is used in selection and breeding programs.

Information Booster:
Broad-sense heritability (H²) = (VA + VD + VI) / VT
→ Includes all genetic components (additive + dominance + interaction).

Narrow-sense heritability (h²) = VA / VT
→ Includes only the additive component (transmissible variance).

Importance:
-h² helps predict response to selection in breeding programs.
-Higher h² means that selection for the trait will be more effective because most phenotypic variation is genetically controlled.

Additional Knowledge (Incorrect Options Explained):

Option (b) VA / VGE: Incorrect — genotype × environment variance (VGE) is only one component of environmental variation, not the total variance.

Option (c) (VA + VD + VI) / VT: This represents broad-sense heritability (H²), not narrow-sense.

Option (d) VD / VT: Dominance variance contributes to genetic variance but not directly to resemblance between parents and offspring; hence not narrow-sense heritability.