A geneticist studying the number of bristles on the second leg of Drosophila determined that a wild-type strain has a mean number of 486.3 bristles per leg. A sample of males and females from this population with 420.0 were bred and the offspring had a mean bristle number of 432.0.
a. What is the h2 for this population?
b. If we were to select a group of flies with 500 bristles and breed them, what would be the mean bristle number in the progeny?
c. If we wished to generate a population of flies that had a mean leg bristle number of 450, what should we do?
a. To calculate the heritability (h2) for this population, we will first find the selection differential (S) and the response to selection (R).
Selection differential (S) = X_selected - X_population = 420.0 - 486.3 = -66.3
Response to selection (R) = X_offspring - X_population = 432.0 - 486.3 = -54.3
Now, we can find heritability (h2) using the formula:
h2 = R / S
h2 = (-54.3) / (-66.3) = 0.819
Thus, the heritability (h2) for this population is 0.819, or 81.9%.
b. To determine the mean bristle number in the progeny of selected flies with 500 bristles, we will use the formula:
X_progeny = X_population + (h2 * (X_selected - X_population))
X_progeny = 486.3 + (0.819 * (500 - 486.3))
X_progeny = 486.3 + (0.819 * 13.7)
X_progeny = 486.3 + 11.2
X_progeny = 497.5
If we were to select a group of flies with 500 bristles and breed them, the mean bristle number in the progeny would be approximately 497.5.
c. To generate a population of flies with a mean leg bristle number of 450, we will use the same formula as before but now solve for X_selected:
450 = 486.3 + (0.819 * (X_selected - 486.3))
Rearranging the equation, we have:
X_selected = ((450 - 486.3) / 0.819) + 486.3
X_selected = (-36.3 / 0.819) + 486.3
X_selected = -44.3 + 486.3
X_selected = 442.0
To generate a population of flies that has a mean leg bristle number of 450, we should select and breed flies with a mean leg bristle number of approximately 442.