1. Hardy-Weinberg Equilibrium When mating is completely random, the zygotic frequencies expected in the next generation may be predicted from the knowledge of the allelic frequencies in the gene pool of the parental population If there are no perturbations in the system, allele and genotype frequencies will remain constant through time If p is the frequency of R alleles in the gene pool, and if q is the frequency of W alleles in the gene pool, then p + q = 1.

2. Assumptions –Infinite population size –No mutation –No selection –Closed population (no gene flow) –Random mating

3. Genotypic (zygotic) Frequencies We can use the Punnett square to produce all possible combinations of these gametes Segregation of alleles and random mating means that probability of different genotypes (x, y, z) is solely determined by allele frequencies (p, q) At equilibrium we expect genotypic (zygotic) frequencies of: (p + q)[from females] × (p + q)[from males] = 1 i.e. p 2 + 2pq + q 2 = 1 (Hardy-Weinberg Rule) RR RW WW q × q = q 2 q × pq p × qp × p = p 2 females p q males p Gametes From: R R W W RW RR WW

4. Verifying Hardy-Weinberg Equilibrium From our previous problem: p(R) = 0.6 q(W) = 0.4 Expected genotypes in next generation: x(RR) = p 2 = 0.6 × 0.6 = 0.36 y(RW) = 2pq = 2 × 0.6 × 0.4 = 0.48 z(WW) = q 2 = 0.4 × 0.4 = 0.16 These are the values we got through the step-by-step analysis in the previous problem

5. Does HWE hold for more than two alleles? YES Example: Three alleles A, B, & C with frequencies p, q, & r (respectively) p + q + r = 1 (p + q + r) × (p + q + r) = 1 p 2 + pq + pr + qp + q 2 + qr + rp + rq + r 2 = 1 p 2 + 2pq + 2pr + q 2 + 2qr + r 2 = 1 AA AB AC BB BC CC

6. Prediction and Evaluation of Crosses ● How can we tell if a population is in equilibrium? Since experiments involve a finite number of observations, we anticipate deviations from the expected numbers and the observed numbers (sampling error). If the population is in equilibrium, observed and expected values should be the same Null Hypothesis: The assumption that there is no difference between the observed results and the expected results of an experiment is referred to as the Null Hypothesis.

7. Question: In a given experiment, how much of a deviation from the expected results should be allowed before the Null Hypothesis is rejected Statistics can never render absolute proof of a hypothesis. It merely sets limits to the uncertainity. Conventionally, the null hypothesis in most biological experiments is rejected when the deviation is so large that it could be accounted for by chance of less than 5% of the time. Such results are termed significant. It is clear that large sample sizes reduce the deviations proportionally

8. Degree of Freedom Example: An experiment involves 4 phenotypes (classes). We can assign numbers to three of the classes at random, but the number in the fourth class must constitute the remainder of the total number of individuals observed. Therefore we have (4 – 1) = 3 degrees of freedom (written as df = 3). For genetic experiments, the df will be 1 less than the number of phenotypic forms. If the number of phenotypic forms is n, then df = (n – 1)

9. Chi – Square Test ● Converts deviations of observed values from expected values into the probability of such inequalities occuring by chance ● Takes into account sample size and the number of variables (df = n – 1) 1. Let n = number of classes 2. e(i) be the number expected in the i-th class according to the hypothesis under test 3. o(i) represent the number observed in the i-th class then

10. The value of the chi-square is then converted into the probability that the deviation is due to chance by using the Table of Chi-Square distribution at the appropriate number of degrees of freedom. Limitations: 1.The chi-square test must be used only on numerical data and never on percentages, ratios, etc 2. It cannot properly be used when the expected frequency within any class is less than 5.

11. EXAMPLE: A coin is tossed 20 times and lands heads up 12 times. Are these results consistent with the expected 50:50 ratio ? (df =1) 0.44-2108Tail 0.4421012Hea d (D ^ 2) / CD ^ 2 D O(i) – e(i) Ce(i)Ce(i) Bo(i)Bo(i) A Class

12. ACCEPT OR REJECT NULL HYPOTHESIS Using Chi-Square table, for df = 1, we have 0.46 < 0.8 < 1.07. The corresponding probabilities are 0.5 and 0.3. Thus the magnitude of the deviation could be anticipated by chance alone in more than 30% but less than 50% of an infinite number of experiments. This is above the critical probability value of 5%. Hence, we accept the Null Hypothesis, and conclude, the coin is conforming to the expected probabilities. EXAMPLE: A coin is tossed 100 times and lands heads up 60 times. Are these results consistent with the expected 50:50 ratio?