A genetically heterogeneous population of rice has a mean in the number of days to maturation of 30. Selection for decreased period of maturation is carried out for one generation. The average period to maturation among the plants selected as parents for the next generation is 25 days. F1 plants mature on average in 27 days. Estimate the narrow sense heritability.

h² = 0.6

Explanation:

Before answering the question, we need to know a few concepts.

Artificial selection is the selecting practice of a specific group of organisms in a population -that carry the traits of interest- to be the parents of the following generations.

Parental individuals carrying phenotypic values of interest are selected from the whole population. These parents interbreed, and a new generation is produced.

The selection differential, SD, is the difference between the mean value of the trait in the population (X₀) and the mean value of the parents, (Xs). So,  

SD = X₀ - Xs

Heritability in the strict sense, h², is the genetic component measure to which additive genetic variance contributes. The heritability might be used to determine how the population will respond to the selection done, R.  

h² = R/SD

The response to selection (R) refers to the metric value gained from the cross between the selected parents. R can be calculated by multiplying the heritability h², with the selection differential, SD.  

R = h²SD  

R also equals the difference between the new generation phenotypic value (X₁) and the original population phenotypic value (X₀),  

R = X₀ - X₁

Now that we know these concepts and how to calculate them, we can solve the proposed problem.

Available data:  

According to what we sow previously, we need to find out the value of h².

We know that h² = R/SD, so we need to get R and SD first.  

R = X₁ - X₀

R = 27 - 30

R = -3

SD = Xs -  X₀  

SD = 25 - 30

SD = -5

Knowing this, we can calculate h²

h² = R/SD

h² = -3/-5  

h² = 0.6

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explanation: