Question

. The mean grain yield of an open pollinated rye breeding population is determined to be 72 bushels/acre. Your goal is to improve grain yield in this population. The mean grain yield of selected individuals from this population is 93 bushels/acre. Selected individuals are intermated to form a new population. The new population mean is determined to be 84 bushels/acre. With this information, calculate the selection intensity, response to selection and heritability of grain yield. (3) Report your answers for A and B as whole numbers and your answer for C to 2 decimal places.

Answer 1

Answer:

a) the selection intensity is 21bushels/acre

b) the response to selection is 12bushels/acre

c) The heritability is 0.57 (two decimal places)

Explanation:

Given:

mean grain yield population=72bushels/acre

mean grain yield selected individuals=93bushels/acre

mean grain yield new population=84busgels/acre

a) The selection intensity is given by:

SI=mean grain yield selected individuals-mean grain yield population=93-72=21bushels/acre

b) The response to selection is:

RS=mean grain yield new population-mean grain yield population=84-72=12bushles/acre

c) The heritability of grain yield is:

H=RS/SI=12/21=0.57