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.
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?
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