1. Hardy-Weinberg came up with five basic reasons why a population would stay at genetic equilibrium: 3. no mutations occur in the DNA of any organisms within this population. Under these conditions it is obvious that evolution would not occur. There are no mechanisms for change acting on the population, so the process cannot happen--the gene pool frequencies will remain unchanged. 5. natural selection is not occurring 1. no catastrophes/disasters occur. 4. all mating is totally random (No sexual selection) 2. there is no migration in or out of the population

2. However, since it is highly unlikely that any one of these five conditions, let alone all of them, will happen in the real world, evolution is inevitable. Mutations will occur at random….creating new alleles… Organisms will move in and out of a population, bringing with them or taking away alleles from the gene pool. Reproduction will not be random, organisms will choose mates based on certain traits….. Catastrophes and predation will occur that can change allelic frequencies. There will be traits that are more beneficial to have than others….

3. The definition of evolution was developed in the early 20th century by Godfrey Hardy, an English mathematician, and Wilhelm Weinberg, a German physician. Hardy Weinberg “Evolution is simply a change in frequencies of alleles in the gene pool over time.”

4. Hardy and Weinberg developed an equation we can use to calculate how many individuals in a population are: BB (Homozygous Dom.) Bb (Heterozygous) bb (Homozygous Rec.) Or in other words, determine the allelic frequencies and genotypic frequencies in a particular gene pool. This is used to track allelic frequencies from generation to generation in a population to monitor evolution OR changes in the gene pool.

5. This is the Hardy-Weinberg equilibrium equation. p² + 2pq + q² = 1 p is defined as the frequency of the dominant allele p = B q is defined as the frequency of the recessive allele q = b

6. Because there are only two alleles in this case, “B” and “b” the frequency of one plus the other must equal 100%, so… p + q = 1 (or 100%)

7. p² + 2pq + q² = 1 In this equation: p² = homozygous dominant (BB) organisms in a population. 2pq = heterozygous (Bb) organisms q² = homozygous recessive (bb) ones p = Bq = b

8. Albinism is a rare genetically inherited trait that is only expressed homozygous recessive individuals (aa). The most characteristic symptom is a marked deficiency in the skin and hair pigment melanin. This condition can occur among any human group as well as among other animal species. The average human frequency of albinism in North America is only about 1 in 20,000.

9. The Hardy-Weinberg equation (p² + 2pq + q² = 1), and the frequency of homozygous recessive individuals (aa) in a population is q². Therefore, in North America the following must be true for albinism: q² = 1/20,000 =.00005 By taking the square root of both sides of this equation, we get: q =.007 (rounded) Knowing one of the two variables (q) in the Hardy-Weinberg equation, it is easy to solve for the other (p). p = 1 – q p = 1 -.007 p =.993

10. The frequency of the dominant, normal allele (A) is, therefore,.99293 or about 99 in 100. The next step is to plug the frequencies of p and q into the Hardy-Weinberg equation: p² + 2pq + q² = 1 (.993)² + 2 (.993)(.007) + (.007)² = 1.986 +.014 +.00005 = 1 This gives us the frequencies for each of the three genotypes for this trait in the population: p² = AA =.986 = 98.6% 2pq = Aa =.014 = 1.4% q² = aa =.00005 =.005%

11. You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following: 1. The frequency of the "aa" genotype. The frequency of the “aa” genotype is given in the problem as 36%!

12. 2. The frequency of the "a" allele. The frequency of aa is 36%, which means that q 2 = 0.36, by definition. If q 2 = 0.36, then q = 0.6, again by definition. Since q equals the frequency of the a allele, then the frequency is 60%. 3. The frequency of the "A" allele. Since q = 0.6, and p + q = 1, then p = 0.4; the frequency of A is by definition equal to p, so the answer is 40%.

13. 4. The frequencies of the genotypes "AA" and "Aa." The frequency of AA is equal to p 2, and the frequency of Aa is equal to 2pq. So, using the information above, the frequency of AA is 16% (p 2 = 0.4 x 0.4 = 0.16) and Aa is 48% (2pq = 2 x 0.4 x 0.6 = 0.48)

14. 5. The frequencies of the two possible phenotypes if "A" is completely dominant over "a." Because "A" is totally dominate over "a", the dominant phenotype will show if either the homozygous "AA" or heterozygous "Aa" genotypes occur. The recessive phenotype is controlled by the homozygous aa genotype. Therefore, the frequency of the dominant phenotype equals the sum of the frequencies of AA and Aa, and the recessive phenotype is simply the frequency of aa. Therefore, the dominant frequency is 64% and, in the first part of this question above, you have already shown that the recessive frequency is 36%.

15. PRACTICE HARDY-WEINBERG!