In a population of pumpkins, the mean seed weight is 100 mg. Since a major supplier of pumpkin seeds will pay double for seeds that weigh 125 mg, you begin a selective breeding program. You select seed that average 140 mg as parents for a selective breeding program. s2e 95, s2g 605, s2d 205What is the additive genetic variance of this trait?a.400b.700c.1000d.500e.Not enough dataWhat is the Broad Sense heritability (H2) for this trait?a.0.7833b.0.29c.0.86d.0.57e.0.01What is the mean weight of the next generation of pumpkin seeds?a.100b.111.7c.122.9d.125e.140
yeah its cool to use that stuff,i used it,it did not work but it did not damage anything,
I’ve omitted the s2 for simplicity and just used the subscripts (A, D, E, P and G) instead. 1. Genetic variance is the sum of additive variance and dominance variance. So G A + D, and we know the values of G and D so just sub in the numbers and solve for A. 2. The formula for broad sense heritability is H2G/P. Phenotypic variance is the sum of genetic and environmental variance, so PG +E. 3. The first step in solving this problem is to calculate narrow sense heritability, h2. This is simply the additive variance divided by the phenotypic variance, so h2 A/(A + D + E). Next, you have to use the formula h2 selection response/ selection differential. Selection response is the mean weight of the F1 (this is what you’re solving for) minus the original population mean (100mg). Selection differential is the mean weight of the selected parents (140mg) minus the original population mean. Plus everything in, and then use simple algebra to solve.
