7.03 Genetics: Problem Set 6 Key (2003)

7.03 Problem Sets Fall 2003 7.03 Fall 2003 Problem Set 6 Key 1a) q, the frequency of allele a, is the number of a alleles divided by the total number of alleles. Since there are 1,000,000 individuals in the population, there are 2,000,000 alleles in the population. Each aa homozygote has two a alleles, while each Aa heterozygote has one a allele. We can calculate q as follows: q=f(a)=((2(500)+9000)/2000000=0.005 b) p+q=1, so p=1-q=1-0.005=0.995 c) One generation of random mating is sufficient to bring a population into HardyWeinberg equilibrium, regardless of whether the previous generation was in Hardy-Weinberg equilibrium. Therefore f(aa)=q2=(0.005)2=2.5x10-5, and the total number of aa individuals is 2.5x10-5x(1000000)=25 Likewise, the total number of Aa individuals is 1000000(2pq)=1000000(2x0.995x0.005)=9950 d) no. In the absence of selection and mutation, there will be no change in the frequency of each allele from generation to generation. In generation one, there are 9000+2(500)=10000 a alleles In generation two, there are 9950+2(25)=10000 a alleles e) no. If generation one was in Hardy-Weinberg equilibrium, then the number of homozygous aa albino individuals would be equal to q2. However, q2*1000000=1000000(0.005)2=25, but there were 500 albino individuals in generation one. f) The probability that an unaffected individual is a carrier is p(Aa)/(p(Aa)+p(AA)). For f and g, we can assume that the probability of an unaffected individual being a carrier is just f(Aa), as the number of aa albino individuals is extremely small. The probability of an unaffected individual being a carrier is f(Aa)=2pq=2(0.005)(0.995)=9.95x10-3 The chances that both parents are carriers and both pass on their a allele is (1/4)(9.995x10-3)2=2.48x10-5 g) The frequency of Aa carriers in generation one is 9000/1000000=9x10-3 The chances that both parents are carriers and both pass on their a allele is (1/4)(9.995x10-3)(9x10-3)=2.24x10-5 h) The albino parent is aa, and must therefore pass on an a allele. The probability that the other parent passes on an a allele is (1/2)(9.995x10-3)=5.00x10-3 7.03 Problem Sets Fall 2003 7.03 Fall 2003 Problem Set 6 Key p 2 2a) For a rare dominant disease you can assume that all affected individuals are heterozygous. Therefore the frequency of heterozygotes is f(Aa)=1/1000, and each of those individuals has one q allele, so q=1/2000. For a dominant disorder with selective disadvantage S and mutation rate µ, q=µ/S. Therefore 1/2000=µ/0.2, and µ=0.0001 b) For a recessive mutation, the incidence of disease is equal to q2. Therefore, q=(1/1000)1/2=0.032, plugging that into q2S=µ: (1/1000)x0.2=2 x 10-4 Final answer: µ=2 x 10-4 c) Plugging in the value for q calculated in part b into q*S=h, we find that h=0.032*0.2=6.4x10-3. Heterozygous advantage is phenomena where the heterozygotes of a detrimental allele are better equipped to survive, thus maintaining the detrimental allele in the population. Individuals heterozygous at this particular locus are more fit by a factor of 1+h. Final answer: h=6.4x10-3 3. a) Given µ and S we can calculate q using the equation q = µ / S = 10−5 = 0.005 . 0.4 The incidence of the disease in the population would be q 2 = (0.005) 2 = 2.5×10−5 therefore, one in forty thousand people would be affected by the disease. b) At steady state, ∆q = 0 , and ∆qsel = ∆qmut . Let us call the number of affected individuals nhtotal ; nhtotal is the sum of homozygotes arising from inbreeding and from random mating. The number of homozygotes arising from inbreeding are described by the equation: nhinbreeding = Fq where F is the inbreeding coefficient (in this case 1/16) and q is the allele frequency. The number of homozygotes arising from random mating is described by the familiar equation: nhrandom = q 2 . Adding the respective weights, as random mating corresponds to 90% and cousin marriage corresponds to 10% of the mating, we get a final equation of: nhtotal = 0.9nhrandom + 0.1nhinbreeding = 0.9q 2 + 0.1Fq We plug this modified in the nhtotal , into the ∆qsel equation to get ∆qsel = − S (0.9q 2 + 0.1Fq ) Setting this equal to ∆qmut = µ = 10 −5 and plugging in values for F and S, we get a final equation: ∆q = ∆qmut − ∆qsel = µ − S (0.9q 2 + 0.1Fq ) = 0 10−5 − 0.4[0.9q 2 + 0.1( 1 )q] = 10-5 − 0.36q 2 − 0.0025q = 0 16 q = 0.002839 or q = −0.009784 7.03 Problem Sets Fall 2003 7.03 Fall 2003 Problem Set 6 Key p 3 As we cannot have negative allele frequencies, final answer is q=0.00284. How does this allele frequency compare to part a? To find the incidence of the disease, we plug in our q value into the nhtotal = 0.9q 2 + 0.1Fq equation. This gives us: nhtotal = 0.00002574 . Final answer: incidence=1 in forty thousand. This is the same number obtained due to random mating-why do you think this is the case? The allele frequency is reduced because inbreeding results in a greater proportion of homozygotes, because homozygotes are selected against, those alleles fall out of the population. Why is the incidence the same? Eventhough the allele frequency is low, much more of the inbreeding population is homozygous at this locus relative to the randomly mating population, resulting the same incidence of disease at equilibrium c) The allele frequency can be calculated by qnew = qold + ∆q ≈ q, after one generation ∆q = − Sq 2 + µ = −0.4q 2 +10−5 Assuming that our inbred population was at steady-state, we can use a q = 0.00284 , giving us ∆q = −0.4(0.00284) 2 +10−5 = 6.77 ×10−6 qnew = 0.00284 + 6.77 ×10−6 = 0.00285 2 qnew = (0.00285) 2 = 8.104 ×10−6 Incidence of disease right after cessation of inbreeding is approximately one in 125,000. This is much lower than at Hardy-Weinberg equilibrium. d) The allele frequency would be expected to rise after cessation of inbreeding until q=0.005. This is due to the fact that the mutation rate is creating more alleles than are being selected out in the population. The allele frequency, q, approaches 0.005 as time approaches infinity.