Question

1. A researcher is interested in increasing the yield of new variety of lentil plants. His starting population had a mean of 80 seeds/plant, with a total variance (Vc) of 30. From this initial population, he chose parents with an average yield of 112 seeds/plant. When the progeny are grown in the same test fields, they had a mean yield of 100 seeds/plant. What is additive genetic variance (Va) of the initial population? (a) 31.0 (b) 18.8 (c) 16.0 (d) 15.5 (e) 5.8

Answer 1

Answer: So this is the question related to heritablity; especially narrowed type heritability as narrow-sense heritability is used to measure how variation among individuals is influenced by genetic differences that are, on average, passed from parents to offspring.

Narrow type heritability is defined as the ratio of additive genetic variance and phenotypic genetic variance. for this case as we have the mean given so for calculating heritability(h2) we use the formula:

h2= selection response/selection differential.

Selection Differential= 112-80; Selection response= 100-80 (because we have selected the parents with 112 seeds/plant from the initial population having mean 80 seeds/plant).

so h2= 20/32; which gives 0.625 or 62.5%.

Now h2= Va/Vp.

Vt=Vp.

so putting the value of all the variables we get

0.625=Va/30.

So Va=0.625*30= 18.75.

So additive genetic variance of the initial poulation is 18.75.