A Primer of Population Genetics

Transcription

1 CHAPTER ONE 1 A Primer of Population Genetics Answers to Chapter-End Problems Author s Note: Most of the solutions are given to more significant digits than are justified by the data. This has been done for two reasons. First, so that you can reproduce the calculations and be reassured that you are doing them correctly. Second, to minimize the effects of round-off errors when dealing with intermediate results. I have performed all of the calculations using a standard spreadsheet program carrying 9 significant digits in the intermediate results. CHAPTER RFLPs are detected by the use of a labeled probe that hybridizes with homologous DNA sequences present in the sample. A difference in the distance along the DNA between the site of hybridization with the probe and flanking restriction sites determines the size of the labeled restriction fragment, and thus its mobility in the gel. Hence, assuming a unique site of hybridization in the genome, RFLP markers are codominant, which means that heterozygous genotypes produce two bands whereas homozygous genotypes produce only one. The gel data indicate three genotypes say AA, Aa, and aa occurring in the numbers 88, 130, and 32, respectively. [It is completely arbitrary whether the larger (top) or smaller (bottom) band is labeled A. ] The allele frequencies are estimated as p = ( )/(2 250) = and q = ( )/(2 250) = The expected numbers are (0.6120) = , 2(0.6120)(0.3880) 250 = , and (0.3880) = The X 2 is therefore [( ) 2 /93.636] + [( ) 2 / ] + [( ) 2 /37.636] = 2.25, which has 1 degree of freedom. The P value is P = 0.134, which gives no reason to reject the null hypothesis of HWE RAPD fragments are detected by PCR of a genomic region. Generally speaking, RAPD markers are dominant, which means that genotypes that are homozygous or heterozygous for the allele supporting amplification are detected by presence of a band, whereas genotypes that are homozygous for the allele not supporting amplification are detected by absence of a band. This problems posits two RAPDs, call the alleles A +, A and B +, B, so that the lane with two bands is the A + / B + / double heterozygote and the lane with no bands is the A A B B double recessive

2 CHAPTER ONE 2 homozygote. Let A + represent the allele supporting amplification of the smaller fragment (bottom) and B + the allele supporting amplification of the larger fragment (top). a. For A, the genotype frequency of A A is ( )/( ) = For B, the genotype frequency of B B is (14 + 6)/( ) = The estimated allele frequencies are therefore q 1 = = for A and q 2 = = for B. b. The proportion of heterozygous genotypes among all genotypes whose DNA supports amplification is 2pq/(p 2 + 2pq). For the A gene, this quantity is 2(0.8478)(0.1522)/[(0.1522) 2 + 2(0.8478)(0.1522)] = For B, it is c. Solve 2pq/(p 2 + 2pq) = 1/2, from which q = 1/3. d. The expected numbers from left-to-right across the gel are (1 q 2 1 )(q 2 2 ) 250, (q 2 1 )(1 q 2 2 ) 250, (1 q 2 1 )(1 q 2 2 ) 250, and (q 2 2 )(q 2 2 ) 250, which work out to 5.62, , 39.38, and The X 2 = 0.04 with 1 degree of freedom, and P = The fit is very good Let A 1, A 2, and A 3 represent the alleles yielding the three bands, from smallest (bottom) to largest (top), and let the allele frequencies be p 1, p 2, and p 3, respectively. The total number of individuals is the sum of the numbers across the top of the gel, which equals 700, representing 1400 alleles sampled. The estimates of allele frequency are p 1 = [ (2 6)]/1400 = , p 2 = [(2 297) ]/1400 = , and p 3 = [41 + (2 49) + 208]/1400 = With HWE the expected proportions of the phenotypes, from left to right across the gel, are p 2 2, 2p 1 p 3, p 3 2, 2p 1 p 2, p 1 2, and 2p 2 p 3, yielding the expected numbers , 37.67, 43.00, 97.82, 8.25, and The X 2 = 2.98 with 3 degrees of freedom (6 classes of data 1 2 parameters estimated from the data). The P value = 0.395, which gives no reason to reject HWE Assuming HWE, q = ( 1/ 10, 000 ) = The frequency of heterozygous ( carrier ) genotypes is 2pq = 2(1 q)q = , or about 1 in 50 people Assuming HWE, q = ( ) = , and so p = 1 q = The frequency of melanics that are heterozygous equals 2pq/(p 2 + 2pq) = 0.53, or more than half A if and only if B means that A implies B and B implies A, where in this case A stands for HWE and B stands for the statement Q 2 = 4PR. First assume A, which implies that P = p 2, Q = 2pq, and R = q 2 ; the statement Q 2 = 4PR follows immediately. Next assume that Q 2 = 4PR. Note that p = P + Q/2 and q = R + Q/2 by definition; hence P = p Q/2 and R = q Q/2. Substituting for PR we obtain Q 2 = 4(p Q/2)(q Q/2) = 4pq 2(p + q)q + Q 2, or 2(p + q)q = 4pq, which implies that Q = 2pq. It follows from the definitions of the allele frequencies that P = p 2 and R = q 2.

3 CHAPTER ONE Let the allele frequencies be p A, p B, and p O. With HWE the expected frequencies of the blood group phenotypes are p 2 A + 2p A p O for A, p 2 B + 2p B p O for B, p 2 O for O, and 2p A p B for AB, which work out to , , , and , yielding the expected numbers , 94.82, , and The X 2 = 9.58 with 1 degree of freedom, and the P value = Since a deviation as large or larger than that observed would be expected by chance in only samples (1 in 500), there is very good reason to think that this population is not in HWE for this gene. The reason for the discrepancy is not known. One possibility is migration into the population of individuals with allele frequencies significantly different from those among the Basques In this population, the I B allele is either nonexistent or so rare that it can be ignored. The best estimate of the allele frequency of I O is [ 563/( )] = With HWE the genotype frequencies of I A I A, I A I O, and I O I O are therefore expected to be ( ) 2, 2(0.9687)( ), and (0.9687) 2, or , , and Let the alleles be designated A 1, A 2, A 3, and A 4 in order of decreasing frequency. The expected genotype frequencies and, for each genotype, the probability of observing a matching DNA type in an unrelated individual, are as follows: Genotype Expected frequency (HWE) Probability of a match A 1 A 1 (4/10) 2 = 16/100 (16/100) 2 = 256/10,000 A 1 A 2 2(4/10)(3/10) = 24/100 (24/100) 2 = 576/10,000 A 1 A 3 2(4/10)(2/10) = 16/100 (16/100) 2 = 256/10,000 A 1 A 4 2(4/10)(1/10) = 8/100 (8/100) 2 = 64/10,000 A 2 A 2 (3/10) 2 = 9/100 (9/100) 2 = 81/10,000 A 2 A 3 2(3/10)(2/10) = 12/100 (12/100) 2 = 144/10,000 A 2 A 4 2(3/10)(1/10) = 6/100 (6/100) 2 = 36/10,000 A 3 A 3 (2/10) 2 = 4/100 (4/100) 2 = 16/10,000 A 3 A 4 2(2/10)(1/10) = 4/100 (4/100) 2 = 16/10,000 A 4 A 4 (1/10) 2 = 1/100 (1/100) 2 = 1/10,000 The overall probability of a DNA match of this RFLP equals the sum of the righthand column, or 1446/10,000 = 14.46% Let p be the frequency of the allele yielding the smaller DNA fragment and q that of the allele yielding the larger DNA fragment. The sample size is = 250, or 500 alleles sampled. The allele frequencies are estimated as p = ( )/500 = and q = ( )/500 = With HWE the expected numbers of the genotypes, left-to-right across the gel, are (0.6140) 2 250, 2(0.6140)(0.3860) 250,

4 CHAPTER ONE 4 and (0.3860) 2 250, or 94.25, 37.25, and The X 2 = with 1 degree of freedom, for which P = The null hypothesis of HWE must clearly be rejected. The main reason for the poor fit is that there are too few observed heterozygous genotypes, relative to the number expected with HWE. A deficiency of heterozygous genotypes might well be expected from inbreeding, and would be consistent with some degree of self-fertilization. To estimate F, set 2pq(1 F) 250 = 89, or (1 F) = 89, yielding F = With this value of F, the expected numbers agree perfectly with the observed, so there is no opportunity to do a chisquare test. The reason for the perfect fit is that both degrees of freedom were consumed in estimating p and F The RADP marker is X-linked, so (assuming HWE) the female genotype frequencies are p 2, 2pq, and q 2 whereas the male genotype frequencies are p and q. In females q is estimated as [ 102/( )] = , and in males q is estimated as 346/( ) = The average of these is q = The expected numbers in females are RAPD + and RAPD ; in males they are RAPD + and RAPD. The X 2 is 2.53 with 1 degree of freedom, and P = There is no reason to reject the null hypothesis of HWE for this X-linked DNA polymorphism Let A 1 (allele frequency p 1 ) and A 2 (allele frequency p 2 ) be, respectively, the alleles either supporting or not supporting PCR amplification of the smaller band (bottom), and B 1 (allele frequency q 1 ) and B 2 (allele frequency q 2 ) be the alleles either supporting or not supporting PCR amplification of the larger band (top). The total number of haploid gametophytes sampled is = a. The gametic frequencies are estimated as 142/2000 = for A 1 B 2, 611/2000 = for A 2 B 1, 474/2000 = for A 1 B 1, and 773/2000 = for A 2 B 2. Hence p 1 = = and p 2 = , and q 1 = = and q 2 = With LE the expected gametophyte frequencies, left-toright across the gel, are p 1 q 2 = , p 2 q 1 = , p 1 q 1 = , and p 2 q 2 = , yielding the expected numbers , , , and The X 2 = with 1 degree of freedom, and P 0. The null hypothesis of LE is soundly rejected. An alternative method of calculating X 2 makes use of Equations 1.12 and Here D = (0.2370)(0.3865) (0.0710)(0.3055) = , hence the value of ρ = / [( )( )( )( )] = Then X 2 = ρ = , confirming the previous value. b. Since D = is positive, we want to compare it with D max, which is the smaller of p 1 q 2 and p 2 q 1, which is to say the smaller of and , which equals Then D/D max = / = 0.50, or 50% of its maximum possible value.

5 CHAPTER ONE Note that the total sample size tested in the experimental population is 489. a. For each pair of loci, the allele frequencies, linkage disequilibria D, the r and X 2 values from Equations 1.12 and 1.13, and the P value are shown in the following table. Genes (A and B) p 1 p 2 q 1 q 2 D ρ X 2 P E6 versus EC E6 versus Odh EC versus Odh b. The only significant LD is between EC and Odh. The D max is the smaller of p 1 q 2 and p 2 q 1, which is the smaller of and , or Therefore D/D max = 0.227, which is to say that D is about 22.7% of is maximum possible value, given the allele frequencies The total sample size in this case is 34. a. p 1 = (22 + 3)/34 = , p 2 = (4 + 5)/34 = , q 1 = (22 + 4)/34 = , q 2 = (3 + 5)/34 = , and D = [(22)(5) (4)(3)]/(34) 2 = The expected numbers are (versus 22), 6.88 (versus 4), 5.88 (versus 3), and 2.12 (versus 5), for an X 2 = 6.98 and P = (Alternatively, ρ = and X 2 = ρ 2 34 = 6.98.) The LD is statistically significant. b. D max is the smaller of (0.7353)(0.2352) = and (0.2647)(0.7647) = , which equals D/D max = 0.49, which is to say that D is about half of its theoretical maximum For A and B the gametic frequencies are A B 1/4, A b 1/4, a B 1/4, a b 1/4, hence D = 0 and D/D max = 0. For B and C the gametic frequencies are B C 1/4, B c 1/4, b C 1/4, and b c 1/4, hence D = 0 and D/D max = 0. For A and C the gametic frequencies are A C 1/2, A c 0, a C 0, a c 1/2; in this case D = 1/4 and D/D max = 1.00, which means that D is equal to its maximum possible value. This situation indicates that, contrary to what intuition might suggest, LE is not transitive. In other words, if the gene order is A B C, the fact that A is in LE with B and B in LE with C does not imply that A is in LE with C. This is the seeming paradox that the problem demonstrates In calculating inbreeding coefficients, it is usually clearest to redraw the matings pedigree in the gametes format, shown in the accompanying illustration. Then use Equation 1.21.

6 CHAPTER ONE 6 A B C D E I a. F I = (1/2) 5 + (1/2) 5 + (1/2) 5 = 3/32. b. F I = (1/2) 5 (1 + F A ) + (1/2) 5 (1 + F B ) + (1/2) 5 (1 + F D ) With inbreeding the frequency of homozygous recessives is R i = q 2 (1 F) + qf = (0.01) 2 (63/64) + (0.01)(1/64) = , or about 1 in 3,926. With random mating, the frequency of homozygous recessives equals R = q 2 = (0.01) 2 = , or 1 in 10,000. The ratio R i /R = 2.55, so the risk is more than doubled, even though the inbreeding coefficient is only 1/ Equation 1.21 still applies, although in this case it may be more transparent to count the loops around each ancestor rather than the number of ancestors. a. F 1 = 0; F 2 = (1/2) 2 (1 + F M ), where F M is the inbreeding coefficient of the male; and F 3 = (1/2) 3 (1 + F M ). b. F t = (1/2) t (1 + F M ). c. The sex of the common ancestor does not matter in the autosomal case, because both sexes have two copies of each autosome. It does matter in the X-linked case, because males have only one X chromosome and must transmit this chromosome to each daughter. Hence in the X-linked case, the general formula for this pedigree becomes F t = (1/2) t Segregation distortion means that an Aa genotype produces functional A and a gametes in the ratio k : 1 k. a. In an HWE population, each of the 2pq heterozygous genotypes produces k A : (1 k) a gametes, for a total of 2kpq A : 2(1 k)pq a. The allele frequencies among gametes, denoted p for A and q for a, are therefore given by p = p 2 + 2kpq and q = 2(1 k)pq + q 2. b. With random union of gametes, the genotypes in the zygotes of the next generation are formed in the frequencies p 2, 2p q, q 2 for AA, Aa, and aa, respectively. These are in HWE for the allele frequencies p and q A has allele frequencies p = 1 and q = 0, whereas B has allele frequencies p = 0 and q = 1. a. In the metapopulation, the genotype frequencies are 1/2 AA and 1/2 aa.

7 CHAPTER ONE 7 b. After one (or more) generations of random mating, the genotype frequencies are 1/4 AA, 1/2 Aa, and 1/4 aa. c. There is a deficiency of heterozygous genotypes in the admixed metapopulation, relative to the frequency expected with HWE. d. Let A have allele frequencies p 1 and q 1 and B have allele frequencies p 2 and q 2. Then the metapopulation has a frequency of heterozygous genotypes equal to H = (2p 1 q 1 + 2p 2 q 2 )/2. After one (or more) generations of random mating the frequency of heterozygous genotypes equals H 0 = 2[(p 1 + p 2 )/2][(q 1 + q 2 )/2]. Now calculate H H 0 = (p 1 q 1 + p 2 q 2 ) [(p 1 + p 2 )(q 1 + q 2 )/2] = [(1 q 1 )q 1 + (1 q 2 )q 2 ] [(2 q 1 q 2 )(q 1 + q 2 )/2] = q 1 q q 2 q 2 2 q 1 q 2 + q 1 2 /2 2q 1 q 2 /2 + q 2 2 /2 = q 1 2 /2 q 2 2 /2 + q 1 q 2 = (1/2)(q 1 q 2 ) 2. This must be negative unless q 1 = q 2. In the next chapter we shall see that H 0 H equals the variance in allele frequency among A and B.

8 CHAPTER TWO 8 CHAPTER Excision of the target sequence is an irreversible mutation process; hence Equation 2.1 applies, with p 0 = 1 for the initial frequency of the target sequence and µ = The required answer is the smallest value of t for which q t because of HWE. Hence q t or p t , which means that (1 µ) t or t [ln(0.7764)/ln(0.99)] = Hence t = 26 generations is the required answer. A quick check for verification shows that q 25 2 = whereas q 26 2 = The recombinational switch is formally equivalent to reversible mutation. Apply Equation 2.5 with µ /(µ + ν) = and 1 µ ν = For initial p 0 = 0, p 30 = (versus observed 0.16) and p 700 = (versus observed 0.85). For initial p 0 = 1, p 388 = (versus observed 0.88) and p 700 = (versus observed 0.86). The fit to the observed values is very good. The predicted equilibrium frequency of A is given by Equation 2.6 as Haploid selection with constant fitnesses follows p t /q t = (p 0 /q 0 )(1 s) t (Equation 2.27), where 1 s is the fitness of the strain whose frequency is denoted q relative to the fitness of the strain whose frequency is denoted p. In this problem, we are given p 0 and p 35 and required to solve for 1 s. The solution can be obtained from ln(1 s) = [ln(p 0 /q 0 ) ln(p t /q t )]/t. For the gluconate data, ln(1 s) = [ln(0.455/0.545) ln(0.898/0.102)]/35 = , or 1 s = , which indicates very strong selection favoring the strain with the allele gnd(rm43a). For the ribose data, ln(1 s) = [ln(0.594/0.406) ln(0.587/0.413)]/35 = , or 1 s = , which is within experimental error from a relative fitness of Let the relative fitnesses of Cy/Cy, Cy/+, and +/+ be denoted w 11, w 12, and w 22, respectively, where w 22 = 1 and w 11 = 0 is implied by the lethality of Cy/Cy. The initial frequency of the Cy allele, p = 0.168, is given. The frequency of the Cy allele in the next generation is given by Equation a. Here w 12 = 1 and p = [(0.168) 2 (0)+(0.168)(0.832)(1)]/[(0.168) 2 (0) + 2(0.168)(0.832)(1) + (0.832) 2 (1)] = b. Here w 12 = 0.5 and p = [(0.168) 2 (0)+(0.168)(0.832)(0.5)]/[(0.168) 2 (0) + 2(0.168)(0.832)(0.5) + (0.832) 2 (1)] = Melanic moths result from a dominant allele. Therefore, with HWE, the frequency of the recessive allele can be estimated as the square root of the proportion of

9 CHAPTER TWO 9 nonmelanics. The data given are that t = 50 generations ( ), q 0 = ( ) = , and q 50 = ( ) = a. The change in frequency for a favored dominant allele is given by Equation 2.33, hence st = ln(p t /q t ) + (1/q t ) ln(p 0 /q 0 ) (1/q 0 ) = ln(0.7764/0.2236) + (1/0.2236) ln(0.9950/0.0050) ln(1/0.0050) = , yielding s = b. If the melanic allele were recessive, we would have p 0 = 001. = and p t = 095. = ; s = is given as the same selection coefficient as above. The change in frequency of a favored recessive allele is given by Equation 2.35, so we can write st = ln(p t /q t ) (1/p t ) ln(p 0 /q 0 ) + (1/p 0 ) = ln(0.9747/0.0253) (1/0.9747) ln(0.1000/0.9000) + (1/0.1000) = Therefore, t = generations A favored additive allele changes according to Equation 2.34, which with slight rearrangement reads ln(p 0 /q 0 ) = ln(p t /q t ) st/2. In this application we set the mutant allele frequency as q, and we are given that q t = in To solve the problem we must infer the initial frequency q 0 from the frequency in For an initial year 1950, which corresponds to t = 0, 1965 corresponds to t = 15 because we are told that that there is one generation per year. We also know that s = Consequently, ln(p 0 /q 0 ) = ln(0.9920/0.0080) (0.20)(15)/2 = , from which it follows that q 0 = (The observed value in 1950 was ) The initial year 1940 makes t = 25, hence ln(p 0 /q 0 ) = , from which q 0 = (The observed value in 1940 was 0.111) Let the ST inversion be represented as A, and its allele, the AR inversion, be represented as a. The relative fitnesses of AA, Aa, and aa are given as 0.47, 1.00, and 0.62 (or w AA = 0.53 and w aa = 0.38, using the symbols in Equation 2.36). The expected equilibrium frequency of A is = 0.38/( ) = , and the average fitness in the population at equilibrium is given by Equation 2.30 as (0.4176) 2 (0.47) + 2(0.4176)(0.5824)(1) + (0.5824) 2 (0.62) = a. In the presence of warfarin the relative fitnesses of SS, SR, and RR are 0.68, 1.00, and When written as 1 w SS, 1.00, 1 w RR, then w SS = 0.32 and w RR = The stable equilibrium with overdominance is given by Equation 2.36 as p = w RR /(w SS + w RR ) = , which is the frequency of the S allele. The frequency of the R allele is therefore b. In the absence of warfarin the fitnesses of SS, SR, and RR are given by 1.00, 0.77, and 0.46, which with additive selection are written as 1.00, 1 s/2, and 1 s. The fitness of the heterozygous genotype yields s/2 = 0.23 or s = 0.46, whereas the fitness of the homozygous R genotype yields s = This is a reasonable approximation of additive selection when s is taken as the mean, or s = ( )/2 = 0.50, which yields relative fitnesses of 1.00, 0.75, and The change

10 CHAPTER TWO 10 in frequency of an additive allele is given by Equation 2.34; hence t =[ln(p t /q t ) ln(p 0 /q 0 )]/(s/2) = [ln(0.99/0.01) ln(0.6632/0.3368)]/(0.25) = generations. 2.9 a. At equilibrium for a complete recessive, q = µ /s = ( 5 10 )/ = b. At equilibrium for a partial dominant, q = µ/hs = ( )/( ) = Even though the relative fitness of the heterozygous genotype is 0.975, the partial dominance reduces the equilibrium allele frequency by a factor of more than Use Equation 2.9 with F 0 = 0. For 1 F ST = 0.88, t is given by 0.88 = [1 (1/2N)] t, or t = ln(0.88)/ln[1 (1/2N)]. When N = 20, t = 5.0 generations, and when N = 100, t = 25.5 generations Equation 2.9 with F 0 = 0 applies again. The solution for N can readily be obtained by rearranging the equation to read ln[1 (1/2N)] = (1/t)ln( 1 F ST ). We are given t = 45 and three values of F ST. For Hawaii F ST = and N = ; for Australia F ST = and N = ; and for the combined data F ST = and N = These estimates have relatively large sampling errors. Easteal (1985) has estimated the 95% confidence intervals as (119, 812), (104, 719), and (112, 770), respectively Use the harmonic mean 1/N e = (1/t) (1/N i ) as in Equation The estimated N e = 1/ = Apply Equation 2.49 with N m = 2 and N f = 200. In this case N e = 7.92, or only 3.9% of the actual population size Let q W, q C, and q E be the average frequency of the blue-color allele in the West, Central, and East regions, respectively, and q T be the overall average allele frequency, weighting each population equally. Then q W = 3.092/6 = , q C = 0.254/11 = , q E = 0.755/4 = , and q T = 4.101/21 = a. The average subpopulation heterozygosity is the weighted average of the regional heterozygosities, where the weight for each region equals the number of subpopulations in the region. In effect, we treat the number of individuals in each region as proportional to the number of subpopulations sampled. Therefore, H S = [6 2(0.5153)( ) (0.0231)( ) + 4 2(0.1888)( )]/21 = The total heterozygosity is the heterozygosity that would be expected were the population one large unit in HWE, or H T = 2(0.1953)(0.8047) = b. F ST = (H T H S )/H T = , which means that the population substructure decreases the average frequency of heterozygous genotypes by almost 30% as

11 CHAPTER TWO 11 compared with a single population in HWE For a deleterious recessive allele the average fitness at equilibrium equals 1 q 2 s where q = µ /s ; hence the reduction in average fitness from its value in the absence of the mutation equals q 2 s = µ. For a partially dominant deleterious allele the average fitness at equilibrium equals 1 2pqhs q 2 s where q = µ/hs. To a very good approximation p 1 and q 2 0 at equilibrium, hence the reduction in average fitness from its value in the absence of the mutation equals approximately 2qhs = 2µ Equation 2.27 suggests how to proceed. a. Since p t /q t = (p t 1 /q t 1 )(1/w t 1 ), the method of successive substitutions yields p t /q t = (p t 2 /q t 2 )(1/w t 1 )(1/w t 2 ) = = (p 0 /q 0 )(1/w t 1 )(1/w t 2 ) (1/w 0 ) = p 0 /(q 0 w 0 w 1 w t 1 ). b. Set w t = w 0 w 1 w t 1, from which it follows that w = (w 0 w 1 w t 1 ) 1/t. This kind of average is called the geometric mean For a recessive lethal allele a, w AA = 1, w Aa = 1, and w aa = 0. Then p is given by p = (p 2 + pq)/(p 2 + 2pq) = 1/(1 + q). Replacing p by 1 q we obtain q = 1 1/(1 + q) = q/(1 + q). Writing q t for q and q t 1 for q yields q t = q t 1 /(1 + q t 1 ) or 1/q t = 1 + 1/q t 1. Since the relation between q t 1 and q t 2 is the same as that between q t and q t 1, it follows that 1/q t = 2 + 1/q t 2. Continuing in this manner we obtain 1/q t = t + 1/q 0 or, reinverting, q t = q 0 /(1 + tq 0 ) From Equation 2.29, p = [p 2 + pq(1 s)]/[p 2 + 2pq(1 s) + q 2 (1 s) 2 ] = p[p + q(1 s)]/[p + q(1 s)] 2 = p/[p + q( 1 s)], and therefore q = q(1 s)/[p + q(1 s)]. It follows that p /q = p/q(1 s), which is identical to Equation 2.27 for a haploid with relative fitnesses 1 : 1 s Set = 2(0.75 1)w 12 /(1 2w 12 ) and solve for w 12, which evaluates to At equilibrium φ(x, t)/ t = 0 because the function φ(x, t) no longer changes with time. Therefore we need to show that φ (x) solves dm(x)φ(x)/dx + (1/2)d 2 V(x)φ(x)/dx 2 = 0. Here is the spirit of how Sewall Wright did it in 1938 (Wright 1938). For compactness we will write φ, M, and V instead of φ(x), M(x), and V(x), and use prime ( ) and double prime ( ) to denote first and second derivatives with respect to x. In what follows C 1, C 2, C 3, and C 4 are constants of integration. The argument is that, if φ solves (Mφ) + (1/2)(Vφ) = 0, then (1/2)(Vφ) = (Mφ) or, integrating both sides over x, (1/2)(Vφ) + C 1 = Mφ + C 2. But C 1 = 0 and C 2 = 0 because φ(x) = 0 for x = 0 and x = 1. Now divide both sides by Vφ, yielding (1/2)(Vφ) /Vφ = M/V, then multiply both sides by 2 to obtain (Vφ) /Vφ = 2M/V. But (Vφ) /Vφ is the derivative of ln(vφ); hence, integrating over both sides leads to

12 CHAPTER TWO 12 ln(vφ ) + C 3 = C (M/V)dx, or ln(vφ ) = (C 4 C 3 ) + 2 (M/V)dx, from which we have Vφ = C exp[2 (M/V)dx], where exp means e raised to the power of and C, which equals exp(c 4 C 3 ), is a constant of integration chosen so that φ(x)dx = 1. Therefore φ = (C/V) exp[2 (M/V)dx]. In this specific application, M/V = [(1 x)ν xµ]/[x(1 x)/2n] and hence integrating 2 (M/V)dx = {(4Nν/x) [4Nµ/(1 x)]}dx = 4Nν ln(x) + 4Nµ ln(1 x). Therefore exp[2 (M/V)dx] = x 4Nν( 1 x) 4Nµ, and finally, φ(x) = (C/V) exp[2 (M/V)dx] = C x 4Nν 1 (1 x) 4Nµ 1. Alternatively, one can start with the solution and do the differentiations by brute force to show that (M φ) + (1/2)(Vφ) = 0. This approach is perfectly legitimate, but not as elegant as Wright s. (Incidentally, the value of the constant is C = Γ(4Nµ + 4Nν )/[Γ(4Nµ ) Γ(4Nν )] where Γ( ) is the gamma function.)

13 CHAPTER THREE 13 CHAPTER Consult the following table. Site a. Pairwise mismatches b. Sample configuration c. Phylogenetically informative 1 6 (3, 2) Yes 2 6 (3, 2) Yes 3 4 (4, 1) No 4 6 (3, 2) Yes 5 6 (3, 2) Yes 6 4 (4, 1) No 7 4 (4, 1) No 8 4 (4, 1) No 9 4 (4, 1) No 10 6 (3, 2) Yes 11 4 (4, 1) No 12 6 (3, 2) Yes 13 6 (3, 2) Yes 14 6 (3, 2) Yes 15 6 (3, 2) Yes 3.2. Note that the total number of nucleotides in the aligned sequences is = 448. a. Apply Equations with a 1 = [1 + (1/2) + (1/3) + (1/4)] = and a 2 = [1 + (1/2) 2 + (1/3) 2 + (1/4) 2 ] = Then the estimate of θ per nucleotide site based on the number of polymorphic sites is S/a 1 = (15/448)/ = and the variance of the estimate is ( )/[(448)(2.0833)] + (1.4236)( ) 2 /(2.0833) 2 = , yielding a standard deviation of the estimate of b. Apply Equations with b 1 = 6/12 = and b 2 = 66/180 = Then the estimate of θ per nucleotide site based on the number of pairwise differences is [(6 4) + (9 6)]/(448 10) = Note that the numerator in this expression is the sum of the pairwise mismatches in the answer to Problem 3.1. The variance of the estimate is (0.5000)( )/448 + (0.3667)( ) 2 =

14 CHAPTER THREE , yielding a standard deviation of the estimate of Use Equations with c 1 = / /(2.0833) 2 = and c 2 = 1/[(2.0833) 2 + (1.4236)] [(0.3667) 7/[(2.0833)(5)] + (1.4236)/(2.0833) 2 ] = The numerator in Equation 3.11 equals π S/a 1 = = (see Problem 3.2) and the denominator is the long radical shown in the next line {( )( 15/ 448) + ( )( 15/ 448)[( 15/ 448) ( 1/ 448)]}, which equals Therefore Tajima s D = , which is very small and indicates no significant difference between the estimates of θ based on S/a 1 or based on π. (The data set is too small to make much of, in any case.) The fact that D is positive indicates a slight excess of pairwise differences, suggesting that the polymorphic nucleotides are a little more equally frequent than expected. Of course this is merely an illustrative example, and the excess is well within the range of experimental error For 12 sequences there are (12 11)/2 = 66 pairwise comparisons per aligned nucleotide site. a. The minimum number of pairwise mismatches equals 11/66 when the configuration is (11, 1); the maximum number of pairwise mismatches equals 36/66 when the configuration is (6, 6). b. Minimum 21/66 with configuration (10, 1, 1); maximum 48/66 with configuration (4, 4, 4). c. Minimum 30/66 with configuration (9, 1, 1, 1); maximum 54/66 with configuration (3, 3, 3, 3) The sequences of 18 aligned amino acids differ at 10 amino acid sites, hence D in Equation 3.27 is given by D = 10/18 = , and therefore the corrected number of replacements per amino acid site is estimated as K a = ln(1 D ) = The estimated rate of amino acid replacement per amino acid site per year equals K a /( ) = (The divergence time is doubled because an evolutionary branch of years extends to each species.) The standard deviation of K a is given in Equation 3.27 as [ /( )] = The separation of kangaroos and dogs took place 135 MYA. We are given the value λ = amino acid replacements per amino acid site per year. The proportion of amino acid sites in α-globin that are expected to differ is given by Equation 3.26 as D t = 1 e 2λt = (The observed proportion is 0.234, so the fit to prediction is excellent.) 3.7 a. The corrected amount of nucleotide divergence is given by Equation 3.28 as K n = (3/4)ln[1 (4/3)D] where D = 9/60 = is the proportion of differences,

15 CHAPTER THREE 15 hence K n = The variance of K n is given by (0.1500)(0.8500)/[(60)(1 4D/3) 2 ] = , which yields a standard deviation of For amino acids, the corrected divergence is given by Equation 3.27 as K a = ln(1 D), where D = 2/20 = ; hence K a = The standard deviation is given by [ D/( 20 ( 1 D)] = b. Estimate λ from λ = K/2t where t is the divergence time, in this case 80 MY. For nucleotide substitutions the estimated rate is λ = K n /2t = /( ) = substitutions per nucleotide site per year. For amino acid replacements the estimated rate is λ = K a /2t = /( ) = replacements per amino acid site per year. 3.8 a. The degeneracies are tabulated below. The amino acid sequence is given only for E. coli ; the asterisks denote where it differs from that of S. typhimurium. Differences in the nucleotide sequences are underlined. K12: V* A P I F I C P P N A D D D L L R Q I* A K12: GTCGCACCTATCTTCATCTGCCCGCCAAATGCCGATGACGACCTGCTGCGCCAGATAGCC LT2: ATCGCGCCGATCTTCATCTGCCCGCCAAATGCGGATGACGATCTTCTGCGCCAGGTCGCA N----S--S S S--S N-N--S b. Altogether there are 38 nonsynonymous sites, 12 twofold degenerate sites, and 10 fourfold degenerate sites. The number of synonymous sites is calculated as the number of fourfold degenerate sites plus 1/3 the number of twofold degenerate sites, or 10 + (1/3)(12) = 14. The number of nonsynonymous sites is defined as the number of nondegenerate sites plus 2/3 the number of twofold degenerate sites, or 38 + (2/3)(12) = 46. The nucleotide substitutions are classified as synonymous (S) or nonsynonymous (N) according to the criteria given in the problem in the line below the degeneracy assignments. There are 3 nonsynonymous substitutions (even though there are only 2 amino acid replacements) and 6 synonymous substitutions. c. The 3 nonsynonymous substitutions yield D n = 3/46 = , and the 6 synonymous substitutions yield D s = 6/14 = Using Equation 3.28, which is appropriate for nucleotide data, the estimates are K n = nonsynonymous substitutions per nonsynonymous site and K s = synonymous substitutions per synonymous site. (It should be noted that there is some loss of reliability when the correction for multiple hits is as large as it is in this example for synonymous sites.) 3.9. The problem specifies that the rate of divergence 2λ = nucleotide substitutions per site per year, and that the average number of differences between mtdna molecules is D = Equation 3.28 yields K = (the correction is

16 CHAPTER THREE 16 actually unnecessary because D is so small), and because 2λt = K, the estimates of time are t = /( ) 180,000 years ago to /( ) 360,000 years ago. That is, the present mtdna genetic diversity in this sample of individuals could have originated from a single mitochondrion present in a female as recently as 9,000 18,000 generations ago, assuming 20 years per generation. One possible explanation of this finding is that the human population might have undergone a bottleneck in population size at about this time. In the popular press, this conclusion has been interpreted as evidence for the existence of Eve (and, by implication, Adam ). However, Equation 2.45 indicates that, among neutral alleles destined to be fixed, the average time to fixation equals 4N e generations. Assuming that a single mitochondrial genome became fixed in 9,000 18,000 generations, and assuming that these values bracket the average, the corresponding range of estimates of N e for mitochondrial DNA (and therefore the N e of females) is , which is not unreasonable for small societies of the sort that have characterized most of human evolution. Moreover, the average effective size of human populations in regard to mtdna may not be representative of the average effective size in regard to nuclear DNA a. With Y = , w = b. (0.004/Z) = (1/ ) (0.130/1.00) = , hence Z = hence 1 Z = This means that the β-galactosidase activity has to be reduced by 89% to produce the same fitness as that produced by a 20% decrease in permease acivity. c. The double mutant has a relative fitness of w = If the fitness reductions of the single mutants ( ) were exactly additive, the fitness of the double mutant would be expected to be , which is quite close to the calculated value a. There are 4146 nucleotides in the aligned region and 34 indels. The estimate required is K = ln[1 (34/4146)] = indels per nucleotide site. b. For t = 34 MY, the rate of incorporation of indels equals λ = K/2t = indel substitutions per nucleotide site per year. c. A rate of nucleotide sequence divergence of 1% per 2.2 MY corresponds to λ = 0.01/( ) = substitutions per nucleotide site per year, which is about 38-fold higher than the rate of indel substitution D is given as which, from Equation 3.28 corresponds to K = , and 2λt = K, where λ is the rate of sequence evolution given as λ = per site per year. Hence the range of t values implied is t = /( ) 16.7 MY to t = /( ) = 23.3 MY.

17 CHAPTER THREE a. For IS1, the distribution in Equation 3.29 is given by p 0 = 1/5 and p i = (4/5)(1/6)(5/6) i 1 for i 1. Hence the expected proportions are p 0 = , p 1 = , p 2 = , p 3 = , p 4 = , and p 5 = 1 p 0 p 1 p 2 p 3 p 4 = , yielding the expected number for 71 strains of 14.20, 9.47, 7.89, 6.57, 5.48, and 27.39, respectively. The X 2 = 3.58 with 3 degrees of freedom and P = 0.311, so the fit is acceptable. b. For IS2, p 0 = 2/5 and p i = (3/5)(1/3)(2/3) i 1 for i 1. The expected numbers from 0 copies to 5 copies are 28.40, 14.20, 9.47, 6.31, 4.21, and 8.41, respectively. The X 2 = 6.31 and P = 0.097, which is also acceptable. c. For IS4, the distribution is given by p 0 = 2/3 and p i = (1/3)(1/4)(3/4) i 1 for i 1. The expected numbers from 0 copies to 5 copies are 47.33, 5.92, 4.44, 3.33, 2.50, and 7.49, respectively. The X 2 = 4.00 and P = 0.261, which is, once again, acceptable The X 2 statistic for the chi-square test for independence ( homogeneity ) in a 2 2 table with top-row entries a and b and bottom-row entries c and d can be shown from Equation 1.13 in Chapter 1 to equal (ad bc) 2 N/[(a + b)(c + d)(a + c)(b + d)], where N = a + b + c + d. Alternatively, the expected values can be worked out from the marginal totals and the X 2 calculated as (observed expected) 2 /expected in the usual way. a. Here a = 71, b = 106, c = 8, and d = 13, and X 2 = with 1 degree of freedom for P = In these data the ratio of amino acid fixations to amino acid polymorphisms is not significantly different from the ratio of synonymous fixations to synonymous polymorphisms. b. In this case, a = 60, b = 75, c = 20, and d = 44, yielding X 2 = 3.14 and P = Based on these data there is no reason to suggest that the ratio of synonymoussite fixations to polymorphisms is any different between boss and Adh a. The relevant quantity from the PRF model with s = 0 is the ratio of (4, 1) to (3, 2) configurations. Recall that the (4, 1) configuration includes both 4 ancestral : 1 mutant and 4 mutant : 1 ancestral, because without additional information one cannot identify which nucleotide is mutant and which ancestral. Similarly, the (3, 2) configuration includes both 3 ancestral : 2 mutant and 2 mutant : 3 ancestral. Hence the overall expected ratio of (4, 1) to (3, 2) is given by ( ) : ( ) or 1.25 : This implies that, among all polymorphic nucleotide sites in 5 aligned sequences, the expected proportion with the (4, 1) configuration is and that with the (3, 2) configuration , assuming s = 0. For 27 polymorphisms, the expected numbers are of (4, 1) and of (3, 2), compared with the observed 17 and 10, respectively. The X 2 = with 1 degree of freedom, and P = The synonymous polymorphisms in this case do fit a model with no selection (s = 0), but it must be emphasized that the

18 CHAPTER THREE 18 sample size is small. b. This can be solved by trial and error with the expected numbers : Significance at the 5% level requires X 2 > 3.841, which means that a fit with observed numbers 11 : 16 or worse, as well as 22 : 5 or worse, will yield a significant X a. For Θ = 1, Pr{K 2 = i} is given by (1/2) i+1 for i = 0, 1. Hence, Pr{K 2 = 0} = 1/2, Pr{K 2 = 1} = (1/2) 2 = 1/4, Pr{K 2 = 2} = (1/2) 3 = 1/8, and so forth. b. The probability that two randomly chosen sequences are identical is Pr{K 2 = 0} = 1/(1 + Θ). c. The probability that two randomly chosen sequences differ at one of more sites is Pr{K 2 1} = 1 Pr{K 2 = 0} = Θ/(1 + Θ). Equation 2.12 states that the equilibrium heterozygosity in the infinite-alleles model is given by the expression H = θ/(1 + θ). The results are completely consistent because the infinite-alleles model assumes that each mutation in a gene is unique, whereas the infinite-sites model assumes that each mutation in a sequence of nonrecombining nucleotides occurs at a different site. d. Recombination increases the probability of one or more mismatches, since it increases the numbers of ways in which two randomly chosen sequences can differ. This can be appreciated by considering the case of free recombination. Suppose the sequence consists of n nucleotides and that the mutation rate per site is µ in a population of effective size N. Then the probability of identity betweeen two random sequences at any one nucleotide is 1/(1 + 4Nµ), and therefore with independence the probability of identity betweeen two random sequences at all nucleotides is [1/(1 + 4Nµ)] n. Now (1 + 4Nµ) n = 1 + 4Nnµ + [n(n 1)/2](4Nµ) NU, where U = nµ is the aggregate mutation rate across all nucleotides. Since (1 + 4Nµ) n 1 + 4NU, it follows that [1/(1 + 4Nµ)] n 1/(1 + 4NU). This means that recombination decreases the probability of identity between randomly chosen pairs sequences, which implies that it increases the probability of one or more mismatches Equation 3.3 states that T i = 4N/[(i(i 1)] (the overbar on T i has been dropped for typographical convenience). a. T i for i = 2 to 2N equals 4N{1/(1 2) + 1/(2 3) + 1/(3 4) + + 1/[(2N 1) 2N]} = 4N{1/2 + 1/6 + 1/ /[(2N 1) 2N]}. b. If there are 2N allele lineages in the population, then there are 2N coalescences tracing all alleles back to a single common ancestral allele. c. The sum 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + goes to 1 as 2N becomes large. For N = 50 it already equals 99/100, and for N = 500 it is 999/1000. [The general formula for the sum is (2N 1)/2N].

19 CHAPTER FOUR 19 CHAPTER The mean µ = 100 and σ = 15. Note that µ + 2σ = 130. Since 95% of a normal distribution is within the range µ ± 2σ, and since the normal distribution is symmetrical, 2.5% have phenotypic values µ 2σ and 2.5% have phenotypic values µ + 2σ. Hence 2.5% have phenotypic values 130. Note that 85 is µ ± 1σ. Since 68% of the phenotypic values lie in the range µ ± 1σ, because of symmetry 16% are µ 1σ = 85. Hence 100% 16% = 84% of the phenotypic values are The variance in the genetically heterogeneous population equals σ g 2 + σ e 2 = 40, whereas that of the inbred lines is theoretically σ e 2 = 10. a. The genotypic variance σ g 2 is estimated as (σ g 2 + σ e 2 ) σ e 2 = = 30, hence the environmental variance is estimated as σ e 2 = 10. The broad-sense heritability H 2 is therefore estimated as H 2 = 30/40 = 75%. b. The F 1 from a cross of inbred lines is also genetically homogeneous, hence its variance is theoretically σ e 2 = 10. In fact, the variance of the F 1 tends to yield a better estimate of σ e 2 than the variance of inbred lines, since inbred lines tend to have a somewhat inflated variance due to the absence of heterozygosity. 4.3 a. The F 1 generation yields an estimate of σ 2 e. b. The F 2 generation yields and estimate of σ 2 g + σ 2 e The estimated σ e 2 = 1.46 and the estimated σ g 2 = = The estimated H 2 = 4.51/5.97 = 75.5% 4.5. From Equations 4.9 and 4.6, the additive genetic variance σ 2 a = 2pqα 2 = 2pq[a + (q p)d] 2. To apply this formula directly we must express the relative fitnesses of ST/ST, ST/AR, and AR/AR as +a, +d, and a, respectively. It will be convenient to do this more generally, so let the relative fitnesses be written as w AA, w Aa, and w aa, in the same order as above. The quantity a equals the deviation of the phenotypic value of AA from the mean of that of the homozygous genotypes, hence a = w AA (w AA + w aa )/2 = (w AA w aa )/2. Similarly, a is the deviation of the phenotypic value of aa from the mean of that of the homozygous genotypes, or a = w aa (w AA + w aa )/2 = (w AA w aa )/2. Likewise, d is the deviation of the phenotypic value of the heterozygous genotype from the mean of that of the homozygous genotypes, or d = w Aa (w AA + w aa )/2 = (2w Aa w AA w aa )/2. In this specific case, w AA = 0.47, w Aa = 1.0, and w aa = 0.62; hence a = (it is all right for a to be negative and a

20 CHAPTER FOUR 20 positive) and d = a. or p = 0 and p = 1, σ a 2 = 2pqα 2 = 0. b. For p = 0.2, σ a 2 = 2(0.2)(0.8)[ ( )(0.455)] 2 = , and for p = 0.8, σ a 2 = c. For p = 0.38/( ) = , σ a 2 = 0. d. What is special about p = 0.38/( ) is that it is the stable equilibrium with overdominance given by Equation 2.36 in Chapter 2, namely, p = (w Aa w aa )/[(w Aa w AA ) + (w Aa w aa )] = 0.38/( ). What the result means is that, at a genetic equilibrium, the genotypic variance can be 0 while the additive variance can = 0. In fact, if the additive variance were 0 at an equilibrium, the point would not be an equilibrium, because the allele frequencies would change due to natural selection The values of a and d are found as explained in the answer to Problem 4.5. a. For the proportions, a = 0.87 ( )/2 = and d = 0.44 ( )/2 = 0.005, and the degree of dominance equals d/a = ; hence the alleles are nearly additive. b. For the arcsin x transformation, a = , d = 7.115, and d/a = On this scale the c r pigmentation allele is partially dominant. Which scale of measurement is the one to use? Whichever one yields a normal distribution of the phenotypic values With this distribution, the phenotypic values 0 8 occur with probabilities , , , , , , , , and , respectively. a. The mean equals ( )(0) + (0.0313)(1) + = 4. The direct calculation is unnecessary because the mean of a binomial distribution of n elements with respective probabilities p and 1 p equals np = (8)(1/2) = 4. The variance equals ( ) 2 (0) + ( ) 2 (1) + = 2. No direct calculation is necessary here, either, because the variance of a binomial distribution of n elements with respective probabilities p and 1 p equals np(1 p) = (8)(1/2)(1/2) = 2. b. Among individuals with phenotypic values of 6, 7, or 8, the relative proportions are : : These must be normalized to equal 1 by dividing by their total ( ), yielding the probabilities (phenotypic value 6), (phenotypic value 7), and (phenotypic value 8), which now sum to 1. The mean of this group equals (0.7568)(6) + ( )(7) + ( )(8) = ; hence the selection differential S = µ S µ = = c. With a narrow-sense heritability of h 2 = 50%, the expected response to selection R is given by R = µ µ = h 2 S = (0.50)( ) = ; hence the expected value of µ =

21 CHAPTER FOUR a. Compare R = h 2 S with R = iσ p h 2, from which it is clear that iσ p = S or i = S/σ p. Hence i equals the selection differential S expressed in multiples of the phenotypic standard deviation σ p. b. Any normal distribution can be transformed into a standard normal distribution with mean 0 and variance 1 by means of the transformation y = (x µ)/σ. A truncation point of T in the original distribution is transformed into the truncation point t = (T µ)/σ in the standard normal distribution. However, the proportion of the population saved for breeding, B, remains the same under the transformation. Hence S = (µ S µ)/σ p depends only on B The relevant means are µ = of the G 1 generation, µ S = of the selected parents, and µ = of the G 2 generation. Therefore, S = = is the selection differential and R = = is the response, yielding a realized heritability of h 2 = R/S = a. The mean of the variances of the parental inbred lines equals ( )/2 = , which is an estimate of σ 2 e, hence an estimate of σ 2 g = = The difference between the means of the inbred lines is D = 0.609, and so the Wright-Castle index is n = D 2 /(8σ 2 g ) = b. Estimating σ 2 e as the F 1 variance yields σ 2 e = , from which σ 2 g is estimated as In this case, n = This principle can be demonstrated by the method of successive substitutions using µ µ = Sh 2. Because we are dealing with a population across multiple generations, it is best to use generational subscripts rather than µ and µ. The initial generation has mean µ 0 and the selection differential in this generation is S 0 ; hence µ 1 µ 0 = S 0 h 2. By the same logic µ 2 µ 1 = S 1 h 2. Adding these two equations yields µ 2 µ 0 = (S 0 + S 1 )h 2. Note that the µ 1 terms have cancelled. Now µ 3 µ 2 = S 2 h 2, and adding this to the previous equation yields µ 3 µ 0 = (S 0 + S 1 + S 2 )h 2, where now the µ 2 terms have cancelled. Continuing in this manner yields the desired formula µ n µ 0 = (S 0 + S S n 1 )h 2. This equation does not hold if h 2 changes during the course of the selection, because then we would have µ n µ 0 = S 0 h S 1 h S n 1 h n Use the formula given in Problem We are given µ 76 = 18.8% and µ 0 = 4.8%. The statement that the cumulative selection differential increased by an approximately constant amount of 1.1 per generation implies that S 0 1.1, S 1 1.1, S 2 1.1, and so forth. After 76 generations of selection the cumulative selection differential equals = 83.6 and the total response equals = The realized heritability across 76 generations is therefore h 2 = 14.0/83.6 = 16.75%. It should be noted that the selection was for entire ears, not individual kernels, based on the

22 CHAPTER FOUR 22 average oil content of the kernels in each ear. Since kernels on the same ear are related as half-siblings (because they have the same mother plant but different pollen donors), the realized heritability is actually the realized heritability of halfsib family means This kind of problem is well-suited to calculations using any standard spreadsheet program, as the calculations are rather tedious when carried out by hand. The values needed are the variance of midparent values, which equals mm 2, and the covariance between midparent values and offspring means, which equals mm 2. The narrow-sense heritability h 2 is estimated as the regression coefficient b = / = 0.634, or 63.4%. There is some loss of accuracy from grouping the data into categories. The regression coefficient for the ungrouped data is b = It should also be noted that there is substantial assortative mating for shell breadth, which inflates the estimate of the narrow-sense heritability a. w = p 2 w AA + 2pqw Aa + q 2 w aa and so dw /dp = 2pw AA 2pw Aa + qw Aa 2qw aa = 2[p(w AA w Aa ) + q(w Aa w aa )]. The numerator in Equation 2.31 is pq[p(w AA w Aa ) + q(w Aa w aa )]. Using the chain rule given in the problem, dw /dt = dw /dp dp/dt = 2pq[p(w AA w Aa ) + q(w Aa w aa )] 2. b. See the answer to Problem 4.5, where it is shown that a = w AA (w AA + w aa )/2 = (w AA w aa )/2 and d = w Aa (w AA + w aa )/2 = (2w Aa w AA w aa )/2. c. α = a + (q p)d = [(p + q)(w AA + w Aa )/2] [(q p)(2w Aa w AA w aa )/2] = p(w AA w Aa ) + q(w Aa w aa ). d. dw /dt = 2pqα 2, which says that the increase in average fitness in the population at any time equals the additive genetic variance in fitness at that time. This is Fisher s principle of the fundamental theorem of natural selection For a recessive allele, d = a, so α = a + (q p)a = 2qa. Therefore σ a 2 = 2pqα 2 = 8pq 3 a 2 and σ d 2 = (2pqd) 2 = 4p 2 q 2 a 2. Then set h 2 = σ a 2 /(σ a 2 + σ d 2 ), which reduces to the expression 2q/(1 + q) In this case, d = a. Then σ a 2 = 8p 3 qa 2 and σ d 2 = (2pqd) 2 = 4p 2 q 2 a 2. Then set h 2 = σ a 2 /(σ a 2 + σ d 2 ), which reduces to 2(1 q)/(2 q) The narrow-sense heritability depends on the parent-offspring correlation and not on the sibling correlation (see Table 4.4). For a trait determined by a rare recessive allele, the parent-offspring correlation goes to 0 as q goes to 0, because parent and offspring will not both be affected. This is because affected offspring will come only from heterozygous heterozygous matings. For a trait determined by a rare dominant allele, on the other hand, the parent-offspring correlation increases to a