In a population of cattle, the average body fat percentage is 10.5%. You select parents with an average body weight percentage of 20.8%. The offspring they produce have an average body weight percentage of 17.6%. Based on these data, what is the narrow-sense heritability of this trait

Answers

Answer 1

Heritability in the narrow sense -h²- refers to the proportion of the phenotypic variability that is influenced by genetic factors. In the exposed example, the narrow-sense heritability of this trait is h² = 0.689 ≅ 0.7

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Before answering the question, let us first review a few concepts.

The selection differential, SD, is calculated by getting the difference between the mean value of the desired trait in the population (X₀) and the mean value of the same trait expressed by parents, (Xs). So,

                                                     SD =  Xs - X₀

Heritability in the narrow sense -h²- is the measure of the genetic component to which additive genetic variance contributes.  

It referes to the proportion of the phenotypic variability that is influenced by genetic factors.

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 = X₁ - X₀

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

Available data:

Population average body fat ⇒ 10.5% ⇒ X₀  Selected parents average body weight ⇒ 20.8% ⇒ XsOffspring average body weight ⇒ 17.6% ⇒ X₁

We ned to get the narrow-sense heritability of this trait, h².

h² = R/SD

R  = X₁ - X₀ = 17.6% - 10.5% = 7.1%SD = Xs - X₀ = 20.8% - 10.5% = 10.3%

     

h² = R/SD =  7.1%  /   10.3% = 0.689 ≅ 0.7

The narrow-sense heritability of this trait is h² = 0.689 ≅ 0.7

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Question 15
2 pts
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Answers

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Question 11
1 pts
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