Evolution and Genetics Is there a mathematical model that can show that evolution occurs in a population? Why does Bob have to show off so much? I’m different from the others! My alleles are so special. I am special!

Evolution is a change in the frequency of alleles of a population. If evolution happens, then there should be a way to show that populations are changing. Can we find an equation that will detect changes in the gene pool of a population?

The Null Hypothesis If populations do not change, then the allele frequencies will remain constant; there will always be an equal amount of each allele in the population. 50 % A and 50 % a Individuals will be: AA or Aa or aa

Hardy and Weinberg also suggested that some populations might not change at all if certain criteria were met. These include: 1. large population 2. random mating 3. no mutation 4. no migration 5. no selection That makes no sense! What population would meet all those criteria? Oh right...this is the null hypothesis. These populations would not evolve at all.

So let’s get back to the equation So let’s get back to the equation. In order to solve it, you need 1 single bit of information about your population. Something observable and countable…. You need to know how many individuals in your population display the recessive trait. I just gotta be me!

The Hardy Weinberg Equation p2 + 2pq + q2 = 1.0 and p + q = 1

Another way to look at the equation.

Steps for solving the equation: 1. Determine the number of individuals that are homozygous recessive. Express this number as a decimal. 1 / 10 = .10 This number is q2 You will need this number for the next step.

2. Take the square root of q2 to determine q 3. To find p, use this part of the equation p + q = 1 If q2 = .1 then q = .316 p + .316 = 1 the p = .684

5. You can also find the number of heterozygotes (Aa): 4. Now that you know both p and q, you can determine the number of individuals a that are homozygous dominant (AA) p2 5. You can also find the number of heterozygotes (Aa): 2pq p = .684 p2 = .468 .468 f the 10 original population means that ~ 5 penguins are AA 2pq = 2 * .684 * .316 = .432 .432 of the 10 original population means that ~ 4 penguins are Aa

Let’s check our math. .468 + .432 + .1 = 1

q2 = ___ q = ____ p = ____ p2 = ____ 2pq = ____ Sample Problem 1: In lions the allele for the albino trait is recessive over the normal tawny-striped coloration. A sample of 100 wild lions was examined, and it was determined that 9 of these lions were white (aa). How many lions in this population would you expect to be heterozygous for the albino trait? How many homozygous and tawny colored? q2 = ___ q = ____ p = ____ p2 = ____ 2pq = ____ This formula is so easy! Let’s do another one! What’s the formula again?

The frequency of the recessive allele (q) ________ Sample Problem #2 There are 100 students in a class. Ninety-six did well in the course whereas four blew it totally and received a grade of F. Sorry. In the highly unlikely event that these traits are genetic rather than environmental, if these traits involve dominant and recessive alleles, and if the four (4%) represent the frequency of the homozygous recessive condition, calculate the following: The frequency of the recessive allele (q) ________ The frequency of the dominant allele (p) ________ The frequency of heterozygous individuals (2pq). ______