© 1997
David H.A. Fitch
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 Lecture notesNull hypothesis:  Genetic basis of variation at single loci I.  Evolution in a Mendelian populationA.  Mendelian population:  a community of (diploid) sexually interbreeding organisms in which each individual has (equal) access to every other, and Mendelian laws apply:1.  Law of Segregation:  a gene pair segregates into the (haploid) gametes2.  Law of Independent Assortment:  genes independently (randomly) come together in zygotesB.  Important characteristics of a Mendelian population are:1.  Mating structure (randomness of mating)2.  Population size (number of individuals that contribute gametes)3.  Degree of intactness (degree of gene flow in or out of the breeding population)C.  "Genetic composition" of a population = "gene" frequencies (actually, proportions of different alleles of a gene locus in a population)D.  Evolution = change in "gene" (actually allele) frequencies (i.e., change in genetic composition)II.  Concepts of "gene" (actually allele) frequency and genotype frequency in diploid populationsA.  Allele frequency = the proportion of a particular kind of allele relative to copies of all alleles of that locus in the populatione.g., p = f(A) [frequency of allele A]B. For a population (of size N) with 2 alleles at a locus (A, a), then the genotype frequencies are:
 Genotype AA Aa aa Number of Individuals n(AA) n(Aa) n(aa) Genotype Frequency D H R 1 = D    + H   + R
 C.  The allele frequencies can be calculated given these genotype frequencies:
 D.  But genotype frequencies cannot be directly calculated just from allele frequencies (i.e., you cannot calculate 3 unknowns [D, H, R] from only 2 equations [p=D+H/2; q=R+H/2])...unless we assume random mating.III.  The "gene pool" approachA.  This approach results from the Mendel's "Law of Independent Assortment"B.  The "gene pool" is the sum total of the gene copies in a randomly breeding population:
 C.  Results in a Hardy-Weinberg ratio of genotypes:1.  Freq. of AA in next generation = p22.  Freq. of Aa in next generation = 2pq3.  Freq. of aa in next generation = q24.  The Hardy-Weinberg Theorem:  p2 + 2pq + q2 = (p+q)(p+q) = 15.  Also, p+q = 1 and D+ H + R = 1D.  The Hardy-Weinberg ratio (i.e., the HW "equilibrium") is achieved after only one generationIV.  The mating table approachA.  Again, assume random matingB.  Remember:  p = D + (H/2);  q = (H/2) + R

Frequency of genotypes in progeny

 Mating FrequencyofMating AA Aa aa AA x AA D2 D2 0 0 AA x Aa andAa x AA 2DH 1/2(2DH)= DH 1/2(2DH)= DH 0 AA x aa andaa x AA 2DR 0 2DR 0 Aa x Aa(Dihybrid cross) H2 1/4(H2)= H2/4 1/2(H2)= H2/2 1/4(H2)= H2/4 Aa x aa andaa x Aa 2HR 0 1/2(2HR)= HR 1/2(2HR)= HR aa x aa R2 0 0 R2 Sum Totals: (D+H+R)2= 1 D2 +DH +H2/4= [D+(H/2)]2 DH + (H2/2) + 2DR + HR= 2[D+(H/2)] [(H/2)+R] H2/4 +HR +R2= [(H/2)+R]2 Substitute: 1  = p2     + 2pq    + q2
 Note:  This approach is nice if one has to consider the effects of non-random mating!V.  The Hardy Weinberg Equilibrium as a null hypothesisIf a population sample is not in HWE (i.e., if the null hypothesis is falsified), then:A.  One or more of the assumptions of the model is incorrect:1.  Mating is not random:  could result...a.  From the alleles themselves (e.g., if they affect flowering time), orb.  If inbreeding occurs often2.  Gene flow has occurred between the sampled population and another population3.  Selection has operated between sampling and zygote formationB.  There has been a sampling error because:1.  Sample is not from a single population, or2.  Genotypes have different likelihoods of being included in the sample
 Exercises 500 mice are trapped on a farm and classified for the fast (F) and slow (S) electrophoretically detectable alleles at a locus, with the following results:
 Genotype FF FS SS Total Number 91 208 201 500
 Assuming a Mendelian population (i.e., what specifically are these assumptions?), predict the frequencies of genotypes that you would expect to see in the next generation.  If you trapped another 500 mice in the next generation, how many would you expect to get of each genotype? Are the results of the trapping and analysis experiment consistent with the population existing at a Hardy-Weinberg equilibrium (HWE) at this locus?  If not, what explanations are there for the deviation?  What additional observational tests can you suggest?  (This exercise and the next two are modified slightly from ones in J. Maynard Smith's Evolutionary Genetics, on reserve in Bobst Library.)   In London, 5% of all male cats are ginger-colored.  What is the expected frequency of ginger females and of tortoiseshell-colored females?  (Ginger is an X-linked recessive trait and tortoiseshell is the heterozygous phenotype.)  What assumptions have you made? At the ABO locus, 3 alleles determine 4 phenotypes according to the following scheme:  genotypes AA and AO specify the A blood group phenotype, BB and BO specify the B phenotype, AB specifies the AB phenotype (A and B are codominant alleles), and OO specifies the O group.  In a randomly mating population, the allele frequencies are 0.3 A, 0.2 B, and 0.5 O.  What blood group frequencies are expected?  A child with A blood has an A father and a B mother.  What is the probability that a full sibling has A blood? (This problem comes from Suzuki, Griffiths and Lewontin's text An Introduction to Genetic Analysis (2nd ed., 1981).  A study mad in 1958 in the mining town of Ashibetsu in Hokkaido, Japan, revealed the frequencies of MN blood-type genotypes for individuals and for married couples, as shown below:
 Genotype No. of individuals or couples MM 406 MN 744 NN 332 total individuals: 1482 MM x MM 58 MM x MN 202 MN x MN 190 MM x NN 88 MN x NN 162 NN x NN 41 total couples: 741
1. Determine if the population is in HWE with respect to the MN blood types.  (Use a chi-square test to provide statistical support to your conclusion.)

2. Determine if mating is random with repsect to MN blood types.  (Again, apply a chi-square analysis.)

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