# Numerical Problems (Heritability  & Genetic Gain)

Question 1: The inbred lines of wheat are crossed. In the F1 the variance in wheat yield is 1.5, the F1 is selfed, in the F2 the variance in wheat yield is 6.5. Estimate the H in broad sense of wheat yield in the experiment.

Solution:

Given,

Variance in F1 is due to environment (Ve) = 1.5

Phenotypic variance in F2 (Vp) = 6.5

Heritability in broad sense (H) =Vg / Vp

Vp = Vg + Ve

6.5 = Vg + 1.5

Vg = 5

Heritability in broad sense (H) = Vg / Vp

= 5 / 6.5

= 0.76923

Question 2: With the help of following data, calculate H and interpret the result.

Solution:

Heritability (H) = Vg /Vp

Vg = (MSS due to genotype – MSS due to error) / R

Vg = (10.8 – 0.13)/ 0.13

= 3.5566

Vp = Vg + Ve

Vp = 3.5566 + 0.13

Vp = 3.6866

Heritability (H) = Vg /Vp

= 3.5566 / 3.6866

= 0.96474

Question 3: Calculate the H from the following information.

Solution:

Heritability (H) = Vg / Vp

Vp = Vg + Ve

Vp = Variance in F2 = 90.35

Ve = (P1 + P2 + F1) / 3

= (11+10.32+5.23) / 3

= 8.85

Vp = Vg + Ve

Vg = 90.35 – 8.85

= 81.5

Heritability (H) = Vg / Vp

= 81.5 / 90.35

= 0.90204

Question 4: Estimate H through parent offspring regression method from the following available data:

Solution:

H = b (if mid-parents given)

H = 2b (if single parent value given)

Question 5 : For a quantity trait in a random population, mean is 100 and the variance is 240. The narrow sense heritability is 0.5. What is the expected mean of the next generation, if the top 10 % plants are used as parents.

Solution:

Mean of base population (Xo) = 100

Variance (Va) = 240

Narrow sense heritability (h2) = 0.5

Mean of the progeny (Xp) = ?

Selection intensity = 10 %

We know that,

R = Xp – Xo

And,

σP = √Va

= 15.5

R= k. h2 σp

= 1.76 * 0.5 * 15.5

= 13.64

R = Xp – Xo

Xp = 13.64 + 100

= 113.64

Question 6 : In the same population above, if the breeder wants the next generation to have a mean is at least 120, what population of the top plants should be used for breeding?

Solution:

Mean of the progeny (Xp) = 120

K = ?

We know,

R = Xp – Xo

= 120 – 100

= 20

σP = √Va

= 15.5

R= K. h2 σp

K = R / h2 σp

= 2.58

So, selection intensity = near 1 %

Question 7 : For a quantitative trait in a RPM, the mean is 100 and the variance is 240. The regression of the offspring on mid parent value is 0.25. Truncation selection is practiced with a selection differential of 32. What is the expected mean in the next generation?

Solution:

Mean of base population (Xo) = 100

b = h2 = 0.25

Selection differential (S) = 32

Expected mean in next gen. (Xp) =?

R = Sh2

= 32 * 0.25

= 8

R = Xp – Xo

Xp = 8 + 100

= 108

Question 8 : The mean days to maturity and variance are 120 and 144 respectively. A plant breeder selected the top 5% plants from base population and found mean days to maturity 110 in the next generation. Find the genetic gain and the heritability of the trait.

Solution:

Mean days to maturity (Xo) = 120

Variance (Va) = 144

Selection intensity = 5%

Value of K at 5% selection intensity = 2.06

Mean days to maturity in next gen. (Xp) = 110

Genetic gain (R) =?

Heritability (h2) =?

R = Xp – Xo

= 110 – 120

= -10

(-ve sign indicated it is the condition of genetic gain. That means the variety matures 10 days   earlier)

Now,

σP = √Va

= 12

h2 = R / K.σp

= 10 / 2.06 * 12

= 0.40