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# anthropology equilibrium equation?

orange eyes are dominant. black are recessive.

there are 1000 individuals. 910 have orange.

determine the frequency of the recessive allele in this group.

how many buzzers are heterozygous for eye color.

my numbers arent adding up 1. can someone do this?

### 2 Answers

- Anonymous9 years agoFavorite Answer
My answers:

The frequency of the recessive allele is 0.3.

420 individuals are heterozygous for eye color.

Work:

Hm...Hardy-Weinberg...

If 910 have orange, 90 must have black eyes.

p = fraction of population w/ orange eye allele

q = fraction of population w/ black eye allele

p^2 + 2pq + q^2 = 1

q^2 represents the portion of the population with black eyes. So:

q^2 = 90/1000 = 0.09

Since q^2 = 0.09, q = 0.3.

Therefore, the frequency of the recessive allele in this group is 0.3.

p^2 + 2pq + q^2 = 1

(p + q)^2 = 1

(p + 0.3)^2 = 1

p + 0.3 = 1

p = 0.7

To find how many individuals are heterozygous for eye color, solve for the 2pq term:

2pq = 2(0.7)(0.3) = 0.42

Since there are 1000 individuals, multiply 0.42 by 1000:

(0.42)(1000 individuals) = 420 individuals

So, 420 individuals are heterozygous for eye color.

Just to check:

p^2 + 2pq + q^2

= (0.7)^2 + (0.42) + (0.09)

= (0.49) + 0.42 + 0.09

= 1

That means that all our calculations were correct.

~~EDIT~~Clarification: When taking the square root, only the positive square root is considered because probabilities are always positive.

- 9 years ago
I'll not do your work for you, but I'll help

to solve this you want to use the Weinberg formulas

p^2 + 2pq +q^2 = 1

p + q = 1

these equations assume no evolutionary force is active and that the traits are simple (even though eye color actually isn't)

your p=frequency of dominant allele

your q=frequency of the recessive allele

common mistakes people make when solving one of these problems are:

-not taking the square root correctly

-round off error

an example (different numbers from yours)

Cerumen (ear wax) has two phenotypes: Sticky, yellow/brown (dominant) and Grey flaky/dry (recessive). Assume a population of 47. 45 have the dominant trait and 2 have the recessive trait

first calculate your phenotype frequencies:

45/(45+2) = .9574 or ~.96

2(45+2) = .0426 or ~ .04

that .04 is your q^2 value

to find p you take the square root

when you do you get q = .2 (frequency of your recessive allele)

now plug in to your equations to find p

p + .2 = 1

p= 1 - .2

p = .8 (frequency of your dominant allele)

now your are trying to find the frequencies of your genotypes

p^2 (homozygous dominant) = .8^2 = .64 or 64%

q^2 (homozygous recessive) = .04 or 4%

2pq (heterozygous) = 2 (.8)(.2) =.32 or 32%

.64+.32+.04 = 1