Question

A bluebonnet (type of wild flower) population along a roadside is genetically variable. A botanist counts the seeds produced by 100 plants and calculates a variance of 20. Selecting one of these plants, the botanist cultivates a genetically identical population of bluebonnets in a greenhouse. The botanist transplants these genetically identical bluebonnets back to the roadside where the original plants were found. The next year, when the seeds are counted for these genetically identical plants, the variance is 3. What is the broad-sense heritably of seed number for the roadside population of bluebonnets? Enter the answer with up to 4 decimal places (0.xxxx). Show your work in the next question box.

b) Show your work for the last question here: A bluebonnet (type of wild flower) population along a roadside is genetically variable. A botanist counts the seeds produced by 100 plants and calculates a variance of [x]. Selecting one of these plants, the botanist cultivates a genetically identical population of bluebonnets in a greenhouse. The botanist transplants these genetically identical bluebonnets back to the roadside where the original plants were found. The next year, when the seeds are counted for these genetically identical plants, the variance is [y]. What is the broad-sense heritably of seed number for the roadside population of bluebonnets?

Answer 1

a) Broad-sense heritability = 0.8500

b)Broad-sense heritability is the proportion of variation in a phenotype explained/contributed by inherited genetic variation.

H² = V_G/V_P where H² is broad-sense heritability ,V_G is genetic variance , V_P is phenotypic variance.

Phenotypic variance includes genetic variance and variance due to environmental factors.

so, V_P = V_G + V_E

Here it is given that initial phenotypic variance is [x], genetically identical cloned plants have a variance of [y] which would be variance due to environmental factor

V_G = V_P - V_E = x-y

Hence H² = [x-y]/x

Given x=20 and y=3

H² = (20-3)/20 = 0.8500