Genetics. Please show work. Thanks! The question I\'m posting is for Part D Part

Question

Genetics. Please show work. Thanks! The question I'm posting is for Part D

Part C

In a given plant population, a study was undertaken to determine the average number of seeds per plant. Different variances were calculated and are shown in this table.

Calculate the broad sense heritability. Enter your answer to two decimal places. (example 0.85)

0.610

All attempts used; correct answer displayed

The equation for broad sense heritability is H2 = VG / VP. However, VG is not provided in this problem. You can rearrange the formula VP = VG + VE to solve for VG (VG = VP - VE).

Combining the two formulas gives you:

H2 = (VP-VE) / VP = (20.2-7.8) /20.2 = 0.61.

Part D

Now calculate the narrow sense heritability for the plants in Part C.

Enter your answer to two decimal places. (example 0.85)

Explanation / Answer

Answer:

Part C

In outbreeding species evolutionary rates are affected by narrow sense heritability. It is the ratio of additive or fixable genetic variance to the total or phenotypic variance. Narrow Sense Heritability is also known as degree of genetic resumblance. It is calculated by the folowing formula:

Heritability (h2): VA/VP

Where, VA or D is additive genetic variance

              VP is phenotypic variance

h2 = VA/VP

h2 = 3.22/20.2

h2 = 0.159 or (0.16 approx)

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