Answer:
The correct answer is 0.75.
Explanation:
The phenotypic variation (VP) in quantitative genetics refers to the combined influence of the genotype (VG) and the environment (VE). The equation is VP = VG + VE
Heritability predicts that how much of the phenotypic variation can be illustrated by genetic-environmental or genetic effects. The broad-sense heritability (H2) refers to the inclusion of all the probable sources of genetic variation (dominance, additive, paternal, and maternal effects).
H2 + VG / VP --------- (1)
The roadside population is genetically variable and as a consequence, the variance must be a combination of genetic and environmental components. The VP here is 20.
VP = VG + VE
20 = VG + VE -------- (2)
In the identical population, the variance must be environmental as all the individuals are of same genotype and thus VG must be 0. Here, the VP is 5.
VP = VG + VE
5 = 0 + VE
Thus, VE = 5
Now substitute, the value of VE = 5 in equation (2)
Thus, 20 = VG + 5
VG = 20-5
VG = 15
Now put the values of VG and VE of the roadside population in equation 1:
H2 = VG/VP
H2 = 15/20 = 0.75
Thus, the broad-sense heritability of seed number of the roadside population of bluebonnets is 0.75.