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Hardy-Weinberg

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?.

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Hardy-Weinberg

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  1. 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?

  2. PR = f(RR) + ½ f(Rr) = .60 + ½ (.30) = .75 • Qr = f(rr) + ½ f(Rr) = .10 + ½ f(.30) = .25 P = f(RR) + ½ (Rr)Q = f(rr) + ½ (Rr)

  3. 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 Hardy-Weinberg Principle

  4. 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. Here’s What It’s Good For

  5. 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. My hypothesis

  6. How it works…

  7. Consider the following Punnet square, showing a cross between 2 heterozygous individuals for allele A: Remember, P + Q = 1

  8. In 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. Defining P2 and Q2

  9. 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 Heterozygotes

  10. In any population in equilibrium, • p + q = 1 • Therefore, (p + q)2 = 1 • Expanded… p2 + 2pq + q2 = 1 The Hardy-Weinberg Equation

  11. In a population in equilibrium, • f(AA) = p2 • f(Aa) = 2pq • f(aa) = q2 • Remember: p = f(A) and q = f(a) Relationships between allele frequency and phenotype frequency

  12. 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? Simple Application

  13. 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) Predictive ability

  14. Example problem: • 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 Determining heterozygote frequencies

  15. 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 Finding Allele Frequencies

  16. 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 Hidden Heterozygotes

  17. Since • f(Bb) = 2pq • p = .68 • q = .32 • f(Bb) = 2(.68)(.32) • f(Bb) = .44 • 44% of the birds are heterozygotes! Genotype frequency

  18. 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! Completing the Bird

  19. Love the Symmetric Beauty…

  20. 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…

  21. f(ss) = .09 = q2 • 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) Why Sickle Cell?

  22. When solving problems: • 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 Your Problems

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