What is the narrow-sense heritability of seed weight?

[jena]

Hi,
My Question:

Assume that all phenotypic variance in seed weight in beans is genetically determined and is additive. From a population in which the mean seed weight was 0.88g, a farmer selected two seeds, each weighing 1.02g. He planted these and crossed the resulting plants to each other, then collected and weighed their seeds. The mean weight of their seeds was 0.96g. What is the narrow-sense heritability of seed weight?

Work

I know that I'm supposed to use this equation:

h^2= Va/Vt, where

Va is the variance due to additive alleles

Vt is the total variance, so I did this

h^2= .88/.96 =.92

Is this correct??

Thank You

 

[Moonbear]

jena, I'm afraid I have no recollection of this area of genetics (you've finally stumped me). I hope someone else will come along who can better help you, but in the meantime, perhaps we can figure this out together. In your equation, why did you use the mean weight for the variance? How is the term "variance" being used? Also, what does the term h^2 mean? Is that heritability? And do you need to take the squareroot of your final answer to get h (if that's the right term to use there)?

Maybe if you can answer my questions, it will help you find your own answers to how to solve this problem to double-check your work.

 

[iansmith]

I think you are using the wrong the formula. You do not have enough information to calculate the Va/Vt. I think you should calculate the R/S.

Look at this page
http://www.evotutor.org/Selection/Sl4A.html

 

[jena]

Could this work

Selection Differential(difference between the mean phenotype of the selected parents and the mean phenotype of the population before selection)
1.02g-.88g= .14g

Selection Response(Amount phenotype change in one generation)
.96g-.88g=.08 g

then divide the selection reponse response over the selection differential

HN2 =.08/14

Finally,

HN2 =.57