2.  Read Chapter 6 of text

3.  Brachydachtyly displays the classic 3:1 pattern of inheritance (for a cross between heterozygotes) that mendel described.

5.  In the early 1900’s there was confusion about how such simple patterns of inheritance affected populations.  Why, aren’t 3 of every 4 people brachdactylyous?  Why don’t dominant alleles replace recessive alleles?

6.  Problem stems from confusing what happens at the individual level with what occurs at the population level.  Individual-level thinking enables us to figure out the result of particular crosses.

8.  Population level thinking is used to figure out how the genetic characteristics of populations change quantitatively over time.

10.  Population genetics is the study of the frequency distribution of alleles and genotypes in populations and the causes of allele frequency changes

11.  Diploid individuals carry two alleles at every locus › Homozygous: alleles are the same › Heterozygous: alleles are different

12.  Hardy-Weinberg equilibrium model serves as the fundamental null model in population genetics

13.  The Hardy-Weinberg model examines a situation in which there is one gene with two alleles A 1 and A 2.  There are three possible genotypes A 1 A 1, A 2 A 2,and A 1 A 2

14.  Hardy and Weinberg used their model to predict what would happen to allele frequencies and genotype frequencies in a population in the absence of any evolutionary forces acting on the population.  Their model produced three important conclusions

15.  In the absence of evolutionary processes acting on them:  1. The frequencies of the alleles A 1 and A 2 do not change over time.  2. If we know the allele frequencies in a population we can predict the equilibrium genotype frequencies (frequencies of A 1 A 1, A 2 A 2,and A 1 A 2 ).

16.  3. A gene not initially at H-W equilibrium will reach H-W equilibrium in one generation.

17.  1. No selection. › If individuals with certain genotypes survived better than others, allele frequencies would change from one generation to the next.

18.  2. No mutation › If new alleles were produced by mutation or alleles mutated at different rates, allele frequencies would change from one generation to the next.

19.  3. No migration › Movement of individuals in or out of a population would alter allele and genotype frequencies.

20.  4. Large population size. › Population is large enough that chance does not affect allele frequencies. › If assumption is violated and by chance some individuals contributed more alleles than others to next generation allele frequencies might change. This mechanism of allele frequency change is called Genetic Drift.

21.  5. Individuals select mates at random. › Individuals do not prefer to mate with individuals of a certain genotype. › If this assumption is violated allele frequencies will not change, but genotype frequencies might.

25.  Assume two alleles A 1 and A 2 with known frequencies (e.g. A 1 = 0.6, A 2 = 0.4.)  Only two alleles in population so their allele frequencies add up to 1.

26.  Can predict frequencies of genotypes in next generation using allele frequencies.  Possible genotypes: A 1 A 1, A 1 A 2 and A 2 A 2

27.  Assume alleles A 1 and A 2 enter eggs and sperm in proportion to their frequency in population (i.e. 0.6 and 0.4)  Assume sperm and eggs meet at random (one big gene pool).

28.  Then we can calculate genotype frequencies.  A 1 A 1 : To produce an A 1 A 1 individual, egg and sperm must each contain an A 1 allele.  This probability is 0.6 x 0.6 or 0.36 (probability sperm contains A 1 times probability egg contains A 1 ).

29.  Similarly, we can calculate frequency of A 2 A 2.  0.4 x 04 = 0.16.

30.  Probability of A 1 A 2 is given by probability sperm contains A 1 (0.6) times probability egg contains A 2 (0.4). 0.6 x 04 = 0.24.

31.  But, there’s a second way to produce an A 1 A 2 individual (egg contains A 1 and sperm contains A 2 ). Same probability as before: 0.6 x 0.4= 0.24.  Overall probability of A 1 A 2 = 0.24 + 0.24 = 0.48.

32.  Genotypes in next generation:  A 1 A 1 = 0.36  A 1 A 2 = 0.48  A 2 A 2 = 0.16  Adds up to one.

33.  General formula for Hardy-Weinberg.  Let p= frequency of allele A 1 and q = frequency of allele A 2.  p 2 + 2pq + q 2 = 1.

34.  If three alleles with frequencies P 1, P 2 and P 3 such that P 1 + P 2 + P 3 = 1  Then genotype frequencies given by:  P 1 2 + P 2 2 + P 3 2 + 2P 1 P 2 + 2P 1 P 3 + 2P 2 P 3

35.  1. Allele frequencies in a population will not change from one generation to the next just as a result of assortment of alleles and zygote formation.

36.  2. If the allele frequencies in a gene pool with two alleles are given by p and q, the genotype frequencies will be given by p 2, 2pq, and q 2.

37.  3. The frequencies of the different genotypes are a function of the frequencies of the underlying alleles.  The closer the allele frequencies are to 0.5, the greater the frequency of heterozygotes.

39.  You need to be able to work with the Hardy-Weinberg equation.  For example, if 9 of 100 individuals in a population suffer from a homozygous recessive disorder can you calculate the frequency of the disease causing allele? Can you calculate how many heterozygotes are in the population?

40.  p 2 + 2pq + q 2 = 1. The terms in the equation represent the frequencies of individual genotypes.  P and q are allele frequencies. It is vital that you understand this difference.

41.  9 of 100 (frequency = 0.09) of individuals are homozygotes. What term in the H-W equation is that equal to?

42.  It’s q 2.  If q 2 = 0.09, what’s q? Get square root of q 2, which is 0.3.  If q=0.3 then p=0.7. Now plug p and q into equation to calculate frequencies of other genotypes.

43.  p 2 = (0.7)(0.7) = 0.49  2pq = 2 (0.3)(0.7) = 0.42  Number of heterozygotes = 0.42 times population size = (0.42)(100) = 42.

44.  There are three alleles in a population A 1, A 2 and A 3 whose frequencies respectively are 0.2, 0.2 and 0.6 and there are 100 individuals in the population.  How many A 1 A 2 heterozygotes will there be in the population?

45.  Just use the formulae P 1 + P 2 + P 3 = 1 and P 1 2 + P 2 2 + P 3 2 + 2P 1 P 2 + 2P 1 P 3 + 2P 2 P 3 = 1 Then substitute in the appropriate values for the appropriate term 2P 1 P 2 = 2(0.2)(0.2) = 0.08 or 8 people out of 100.

46.  Hardy Weinberg equilibrium principle identifies the forces that can cause evolution.  If a population is not in H-W equilibrium then one or more of the five assumptions is being violated.