Question

For the cross shown in the Punnett square, assume genotype AA has a phenotypic value of 1.0 and aa has a phenotypic value of 0.0. Assume there is no environmental effect on heritability.

The cross is of Aa x Aa

Situation 1: If (A) is completely dominant over (a), what will be the phenotypic value frequencies of the offspring? (adding up to 1.0)

1.0 =

0.5 =

0.0 =

Situation 2: If neither allele is dominant and (A) and (a) contribute equally to phenotype, what will be the phenotypic value frequencies? (adding up to 1.0)

1.0 =

0.5 =

0.0 =

How does the broad sense heritability of situation 1 compare with situation 2?

A) Situation 1 is higher

B) Situation 2 is higher

C) The broad sense heritability is the same for both situations

D) Neither situation exhibits broad sense heritability

Please show work and explain. Also please define the difference between broad and narrow sense heritability. I was really confused on this. Thank you!

Answer 1

The broad-sense heritability is the ratio of total genetic variance to total phenotypic variance. The narrow-sense heritability is the ratio of additive genetic variance to the total phenotypic variance.

This is different between broad sense heritability and narrow sense heritability .

The 1st part mcq answer is ia C ( the board sence heridity is same for both situation ) because

For situation i this is heterogenous accourding to hardy weinberg equation q2+2pq+p2

Here 2pq is heterogenous and calculate from p+q=1

So in case 1.0 condition we calculate from this equation p=1-q

= 1-1.0

=0

And the value on the 0.5

p=1- q

=1- 0.5

= 0.5

And the value of 0.0

p= 1- q

=1 - 0.0

= 1

In the case situation 2

The value on the 1.0

p=1 - q

= 1 -1.0

= 0

and the value on the 0.5

p= 1 - q

= 1 - 0.5

= 0.5

On the value 0.0

p= 1 - q

= 1 - 0.0

= 1

So two situation is equal .