A plant breeder has determined the following variances for yield of corn in his fields:
Total phenotypic variance - 100
Additive genetic variance - 30
Dominance genetic variance - 50
Environmental variance - 20

Assume that there is not genetic-environmental interaction or other variances.
(a) Calculate the total genetic variance.
(b) Calculate the broad-sense heritability indicated by this data.
(c) Calculate the narrow-sense heritability indicated by this data.
(d) The breeder wishes to improve yield. If the average yield in the starting population is 400 and he selects for breeding plants with an average yield of 500, what will be the expected average yield among the offspring of the selected plants?

A. VG = 80

B. Broad sense heritability = 0.80

C. Narrow sense heritability = 0.30

D. Average yield = 430 Units

Explanation:

A. Given that

VA = 30

VD = 50

VI = assumed not available

Therefore

Total genetic variance (VG) = VA + VD

= 30 + 50

= 80

VG = 80

B. Given that

VP = 100

VG = 80

Broad sense heritability, H2 = VG/VP

= 80/100

= 0.80

C. Given that

VA = 30

VP = 100

Narrow sense heritability, h2 = VA/VP

=30/100

= 0.30

D. The difference in selection = 500 - 400

= 100

Recall,

Selection response is heritability multiplied by selection differential.

That is

R = h2S

Selection differential = 100

Heritability h2 = 0.30

Selection response = 0.30 × 100

= 30units

Therefore, expected average yield = 400 + 30

= 430 Units

is this for the e2020 class