Question

You are researching a population of brittle stars that varies in the number of legs that they produce. You survey the local population, and come up with the numbers below. Assume that the phenotypes you encounter represent the full range of possible phenotypes in this population.

1. Calculate the phenotypic variance in this trait.
2. Do the traits in this population appear to be normally distributed? Give an explanation of why this pattern might be observed.

3. Through a series of breeding experiments, you calculate that h2 for this population is 0.3. If no dominant alleles contribute to this trait, and there are no epistatic interactions involving the gene loci that control this trait, what is VE for this population?

#of legs: # of individuals:

5 ; 420

6 ; 189

7 ; 72

8 ; 5

9 ; 3

Answer 1

Answer 1:)

	X	(X – X’)2
5	420	79636.84
6	189	2621.44
7	72	4329.64
8	5	17635.84
9	3	18171.04
Total	689;	122394.8

Calculate mean of X:

Mean, X’ = 689/5

Mean X’ = 137.8

Variance S2= Σ(X – X’)2 / (n-1)

n is the number of observations, which is 689.

Σ(X – X’)2 = 122394.8

Variance S2= Σ(X – X’)2 / (n-1)

Variance S2= 122394.8 / (689 – 1)

Variance S2= 177.9

Therefore, the phenotype variance (VP) is 177.9.

Answer 2:)

Standard deviation = √S2

Standard deviation = √177.9

Standard deviation = 13.3

There are 3 species (leg#7, 8, 9) come under the variance value while 2 does not.

Fraction of three species = 3/5 x 100

Fraction of three species = 60%

60% population should come under 1 standard deviation. There are 13.3 standard deviations. Therefore, the population is not evenly distributed.

Answer 3:)

VE attributes in VP that is inversely proportional to heritability.

Therefore,

VE = VP / h2

VE = 593

Therefore, the environmental variance (VE) should be 593.