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.
| SV | df | SS | MSS |
| Genotype | 48 | 419.4 | 10.8 |
| Replication | 2 | 0.08 | 0.04 |
| Error | 96 | 12.3 | 0.13 |
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
or, Vp = 3.5566 + 0.13
thus, Vp = 3.6866
Heritability (H) = Vg /Vp
= 3.5566 / 3.6866
= 0.96474
Question 3: Calculate the H from the following information.
| Population | Variance |
| P1 | 11 |
| P2 | 10.32 |
| F1 | 5.23 |
| F2 | 90.35 |
| BC1 | 47.35 |
| BC2 | 54.29 |
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:
| Mid parent value(X) | Individual offspring (y) |
| 20 | 25 |
| 18 | 21 |
| 25 | 20 |
| 17 | 20 |
| 21 | 26 |
| 22 | 25 |
Solution:
| Mid parent value(X) | Individual Offspring(Y) | XY | ∑X2 |
| 20 | 25 | 500 | 400 |
| 18 | 21 | 378 | 324 |
| 25 | 20 | 500 | 625 |
| 17 | 20 | 340 | 289 |
| 21 | 26 | 546 | 441 |
| 22 | 25 | 550 | 484 |
| ∑X = 123 | ∑Y = 137 | ∑XY = 2814 | ∑X2 = 2563 |
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
| Selection intensity % | Value of K |
| 1 | 2.64 |
| 2 | 2.42 |
| 5 | 2.06 |
| 10 | 1.76 |
| 20 | 1.4 |
= 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 % | Value of K |
| 1 | 2.64 |
| 2 | 2.42 |
| 5 | 2.06 |
| 10 | 1.76 |
| 20 | 1.4 |
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