
Lecture notes Deviations from the null hypotheses: Inbreeding I. Inbreeding coefficient A. Autozygous: not only homozygous, but both gene copies are identical by descent (i.e., can be traced to the same ancestral gene copy) B. Inbreeding coefficient: F is the probability that an individual is autozygous (or the fraction of the population that is autozygous); i.e., F is the probability that a gene copy is Fixed. C. For example, what is F for an individual, I, that results...
 from a sib mating?
F = Prob. (I is autozygous for one allele A* from a grandparent) = Prob.(both parents have an allele A* that is identical by descent) And Prob.( I has inherited A* homozygously) = (1/2 * 1/2) * 1/4 = 1/16 for A* There are 4 such gene copies in the grandparents, so F = 4 * 1/16 = 1/4  from selfing?
F = Prob. (I is autozygous for one allele A* from the single selfing parent) = the Mendelian 1/4 for A* There are 2 such gene copies in the parent, so F = 2 * 1/4 = 1/2  from cloning?
 from random mating in a Mendelian population?
D. Allozygous fraction of the population is the nonautozygous fraction of the population (1  F) (Return to top of page.) II. Correction of the Hardy Weinberg theorem for Inbreeding A. Assume the population is divided into an Autozygous fraction and an Allozygous fraction
 Of the allozygous fraction (1  F), the frequency of genotype AA (i.e., D) in the next generation is predicted by HW: (1  F)p^{2}Of the autozygous fraction (F), D is the same as the probability that an individual carries an A allele (because they are autozygous, such individuals will only contribute A alleles to the next generation)
Thus, D = (1F)p^{2} + Fp  Similarly, R = (1F)q^{2} + Fq
 However, heterozygotes can only result from matings in the allozygous fraction of the population, so
H = (1F)2pq  Thus, D + H + R = [(1F)p^{2} + Fp] + (1F)2pq + [(1F)q^{2} + Fq] = 1
= [p^{2} + Fpq] + (1F)2pq + [q^{2} + Fpq] = 1
B. Predictions:
 If F = 0, then D = p^{2} ; H = 2pq ; R = q^{2} (i.e., the HW equilibrium!)
 If F = 1, then D = p ; H = 0 ; R = q
 If F = 1/2, then D = (1/2)p^{2} + (1/2)p ; H = (1/2)2pq ; R = (1/2)q^{2} + (1/2)q
 If there is inbreeding (F > 0), then the genotype frequencies change in successive generations and homozygosity increases
 But the allele frequencies do NOT change (inbreeding by itself does not cause evolution)
(Return to top of page.) III. Heterozygosity A. Heterozygosity is the proportion of heterozygotes in a population inbred to an extent F: H_{F} = (1F)2pq B. Predictions:
 For exclusively selffertilizing populations, H decreases by 1/2 each generation
 If there is some fraction of matings among unrelated individuals, F and H will reach equilibrium values
(Return to top of page.) IV. Application to 2 loci A. For 2 loci, any existing linkage disequilibrium will decay at a slower rate in an inbred population B. This is because the frequency of heterozygotes is lower (and recombinants can only result from recombination in heterozygotes) (Return to top of page.) V. Application to quantitative loci A. For additive loci, inbreeding does not change the phenotypic mean, but increases the phenotypic variance around the mean (there are more alleles distributed in the extreme homozygotes) B. For dominant (or overdominant loci), inbreeding not only increases the phenotypic variance, but also changes the phenotypic mean = Inbreeding Depression 1. If recessive alleles cause lower fitness, they are eliminated more rapidly from an inbreeding population 2. This selection will result in lower variation in inbred populations (Return to top of page.) 
