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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?.

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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?


P f rr rr q f rr rr

  • 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)


Hardy weinberg principle

  • 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


Here s what it s good for

  • 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


My hypothesis

My hypothesis



Remember p q 1

Remember, P + Q = 1


Defining p 2 and q 2

Defining P2 and Q2


Heterozygotes

Heterozygotes


The hardy weinberg equation

  • In any population in equilibrium, receiving an “A” allele is equal to that allele’s frequency in the population, represented as p.

  • p + q = 1

  • Therefore, (p + q)2 = 1

  • Expanded… p2 + 2pq + q2 = 1

The Hardy-Weinberg Equation


Relationships between allele frequency and phenotype frequency

  • In a population in equilibrium, receiving an “A” allele is equal to that allele’s frequency in the population, represented as p.

  • f(AA) = p2

  • f(Aa) = 2pq

  • f(aa) = q2

  • Remember: p = f(A) and q = f(a)

Relationships between allele frequency and phenotype frequency


Simple application

Simple Application


Predictive ability

Predictive ability


Determining heterozygote frequencies

  • Example problem: convert between phenotype frequencies in a population and allele frequencies in a gene pool.

  • 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


Finding allele frequencies

  • Q convert between phenotype frequencies in a population and allele frequencies in a gene pool.2 = 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


Hidden heterozygotes

Hidden Heterozygotes


Genotype frequency

  • Since are heterozygous for the trait?

    • f(Bb) = 2pq

    • p = .68

    • q = .32

  • f(Bb) = 2(.68)(.32)

  • f(Bb) = .44

  • 44% of the birds are heterozygotes!

Genotype frequency


Completing the bird

Completing the Bird


Love the symmetric beauty
Love the Symmetric Beauty… bb, and the remainder, 46% should be BB!


Another example

Another example…


Why sickle cell

  • f( sickle-cell anemia (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?


Your problems

  • When solving problems: sickle-cell anemia (

    • 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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