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### Hardy-Weinberg

In a species of fish, a single gene controls color. In a population of 100 fish, there are 60 red fish, 30 purple fish, and 10 blue fish.

What is the allele frequency of the allele for red color?

What is the allele frequency of the blue allele?

PR = f(RR) + ½ f(Rr) = .60 + ½ (.30) = .75

- Qr = f(rr) + ½ f(Rr) = .10 + ½ f(.30) = .25

The Hardy-Weinberg principle is often stated as:

- In the absence of other influences, phenotype and allele frequencies in a population remain constant over time
- Thus, the principle describes the behavior of alleles in a large, stable population

There were 2 types of candy you will chose from the bag at random – no peeking or trading!

- The two different colors represent 2 different alleles, and the bag represents a gene pool.
- The silver chocolates which we will call “A” has a frequency of .47
- The gold chocolates, “a” has a frequency of .53
- Here’s the bet: If I know the frequencies of alleles in the bag, I can predict how many of you got two silvers, how many got 1 of each, and how many got 2 golds.

About 22% of you selected 2 silver chocolates.

- About 50% of you selected 1 of each color.
- About 28% of you got 2 gold chocolates.

Consider the following Punnet square, showing a cross between 2 heterozygous individuals for allele A:

Remember, P + Q = 1In the population as a whole, the chance of a new individual receiving an “A” allele is equal to that allele’s frequency in the population, represented as p.

- Thus, the chance of receiving 2 “A” alleles is p x p, or p2, and the chance of receiving two “a” alleles is q2.

The probability for producing a heterozygote = p x q, or pq.

- Since there are 2 possible ways to produce a heterozygote, the total probability is 2pq

In any population in equilibrium,

- p + q = 1
- Therefore, (p + q)2 = 1
- Expanded… p2 + 2pq + q2 = 1

In a population in equilibrium,

- f(AA) = p2
- f(Aa) = 2pq
- f(aa) = q2
- Remember: p = f(A) and q = f(a)

Like any formula with 2 variables, if one is known, the other can be determined.

- For this type of problem, use the simple form P + Q = 1
- Example: The allele for tongue rolling has a frequency of .95. What is the frequency of the allele for non-tongue rolling?

A more useful way to use the Hardy-Weinberg equation is to convert between phenotype frequencies in a population and allele frequencies in a gene pool.

- Use p2 + 2pq + q2 = 1
- Remember that:
- p2 = f(AA)
- 2pq = f(Aa)
- q2 = f(aa)

- In chickens, having brown feathers is dominant to having white feathers. In a population of 200 birds, 180 have brown feathers. What are the allele frequencies of B and b in the population?
- Step 1 – figure out the value of p and q

Q2 = f(bb) these are the white birds

- Q2 = 20/200
- Q2 = .10
- Q = .32
- Since P + Q = 1 P = 1 – Q
- P = 1 - .32 = .68

In the same population of birds, how many of the brown birds are heterozygous for the trait?

- What term in the Harvey-Weinberg equation gives us the frequency of heterozygotes?
- 2pq

- f(Bb) = 2pq
- p = .68
- q = .32
- f(Bb) = 2(.68)(.32)
- f(Bb) = .44
- 44% of the birds are heterozygotes!

So… in our population of 200 birds, 44% are Bb, 10% are bb, and the remainder, 46% should be BB!

- Let’s check the equation. Since f(BB) = p2, do we get the same answer from the equation and subtraction?
- P2 should = .44 (the frequency of BB)
- .682 = .46
- It works!

If 9% of an African population is born with a severe form of sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria because they are heterozygous(Ss) for the sickle-cell gene?

Another example…- Q2 = .09
- Q = .3
- Therefore, P = 1 – Q = 1 - .3 = .7
- f(Ss) = 2pq = 2(.7)(.3) = .42
- 42% have increased resistance to malaria (Ss)

- Identify what you know, remembering
- P2 = f(AA)
- 2pq = f(Aa)
- Q2 = f(aa)
- Then, solve for what you don’t know – there’s only 5 different quantities listed above: 3 genotype frequencies, and 2 allele frequencies
- p2 + 2pq + q2 = 1

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