Question

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9. You are conducting an observational study of the effect of directional selection on the size of the wattle in the New Zealand wattlebirds. In an initial study, you measure the mid-parent/mid-offspring regression of this trait (below) and found the slope of the regression line to be 0.35. You have a population of wattlebirds with an average wattle size of 28 cm2. You want to breed for decreased wattle size the wattlebirds so that the mean wattle size for the next generation is 21 cm2. To do so, you will need to take advantage of the breeder's equation (R = h2 x S). (Tip: you may or may not be able to complete these questions in order). Find/calculate the value for “R” for the above example: Find/calculate the value for “h2" for the above example: Find/calculate the value for “S” for the above example In order to breed for this decreased trait size, you would need to artificially select for breeders with what average wattle size?

Answer 1

9) We know that, R = h² x S where, R = Response to selection, h² = narrow sense heritability, S = Selection differential

According to the question, h² for the above example is 0.35.

Given, average wattle size of the population = 28 cm², average wattle size for the next generation = 21 cm²

Now, average wattle size of progeny = Average wattle size of the population + Response to selection

So, 21 = 28 + Response to selection

So, response to selection (R) = (21 - 28) cm² = -7 cm²

So, -7 = h² x S = 0.35 x S

So, selection differential (S) = -7/0.35 = -20 cm²

Now, S = Average wattle size of selected parents - Average wattle size of the population

So, -20 cm² = Average wattle size of selected parents - 28 cm²

So, average wattle size of selected parents = (28 - 20) cm² = 8 cm²