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Assignments Falconer & Mackay, chapters 1 and 2 (1.5, 1.6, 2.5, and 2.6) Sanja Franic VU University Amsterdam 2011 1.4 As an exercise in algebra, work out the gene frequency of a recessive mutant in a randombreeding population that would result in one-third of normal individuals being carriers. aa aA AA p2 2pq q2 1.4 As an exercise in algebra, work out the gene frequency of a recessive mutant in a randombreeding population that would result in one-third of normal individuals being carriers. aa aA AA p2 2pq q2 2pq=1/3 pq=1/6 p+q=1 p=1-q q(1-q)=1/6 -q2+q-1/6=0 -(q2-q+1/6)=0 -(q2-q+…+1/6-…)=0 1.4 As an exercise in algebra, work out the gene frequency of a recessive mutant in a randombreeding population that would result in one-third of normal individuals being carriers. aa aA AA p2 2pq q2 (q-x)2=q2 – 2xq + x2 2x=1 x=.5 (q-.5)2=q2 – q + .52 2pq=1/3 pq=1/6 p+q=1 p=1-q q(1-q)=1/6 -q2+q-1/6=0 -(q2-q+1/6)=0 -(q2-q+…+1/6-…)=0 1.4 As an exercise in algebra, work out the gene frequency of a recessive mutant in a randombreeding population that would result in one-third of normal individuals being carriers. aa aA AA p2 2pq q2 (q-x)2=q2 – 2xq + x2 2x=1 x=.5 (q-.5)2=q2 – q + .52 2pq=1/3 pq=1/6 p+q=1 p=1-q q(1-q)=1/6 -q2+q-1/6=0 -(q2-q+1/6)=0 -(q2-q+.52+1/6-.52)=0 1.4 As an exercise in algebra, work out the gene frequency of a recessive mutant in a randombreeding population that would result in one-third of normal individuals being carriers. aa aA AA p2 2pq q2 (q-x)2=q2 – 2xq + x2 2x=1 x=.5 (q-.5)2=q2 – q + .52 2pq=1/3 -[(q-.5)2-1/12]=0 pq=1/6 -(q-.5)2=-1/12 p+q=1 (q-.5)2=1/12 p=1-q q-.5=√1/12 q(1-q)=1/6 q=√1/12+.5 -q2+q-1/6=0 q=.789 -(q2-q+1/6)=0 p=1-q=.211 -(q2-q+.52+1/6-.52)=0 1.4 As an exercise in algebra, work out the gene frequency of a recessive mutant in a randombreeding population that would result in one-third of normal individuals being carriers. aa aA AA p2 2pq q2 .045 .33 .622 (q-x)2=q2 – 2xq + x2 2x=1 x=.5 (q-.5)2=q2 – q + .52 2pq=1/3 -[(q-.5)2-1/12]=0 pq=1/6 -(q-.5)2=-1/12 p+q=1 (q-.5)2=1/12 p=1-q q-.5=√1/12 q(1-q)=1/6 q=√1/12+.5 -q2+q-1/6=0 q=.789 -(q2-q+1/6)=0 p=1-q=.211 -(q2-q+.52+1/6-.52)=0 1.5 Three allelic variants, A, B, and C, of red cell acid phosphatase enzyme were found in a sample of 178 English people. All genotypes were distinguishable by electrophoresis, and the frequencies in the sample were Genotype AA AB BB AC BC CC Frequency 9.6 48.3 34.3 2.8 5.0 0.0 What are the gene frequencies in the sample? Why were no CC individuals found? 1.5 Genotype AA AB BB AC BC CC Frequency 9.6 48.3 34.3 2.8 5.0 0.0 1.5 Genotype AA AB BB AC BC CC Frequency 9.6 48.3 34.3 2.8 5.0 0.0 Proportion .096 .483 .343 .028 .05 0 1.5 Genotype AA AB BB AC BC CC Frequency 9.6 48.3 34.3 2.8 5.0 0.0 Proportion .096 .483 .343 .028 .05 0 Allele A B C Frequency a b c 1.5 Genotype AA AB BB AC BC CC Frequency 9.6 48.3 34.3 2.8 5.0 0.0 Proportion .096 .483 .343 .028 .05 0 Allele A B C Frequency a b c (a+b+c)2= a2 + b2 + c2 + 2ab + 2ac + 2bc 1.5 Genotype AA AB BB AC BC CC Frequency 9.6 48.3 34.3 2.8 5.0 0.0 Proportion .096 .483 .343 .028 .05 0 Allele A B C Frequency a b c (a+b+c)2= a2 + b2 + c2 + 2ab + 2ac + 2bc AA BB CC AB AC BC 1.5 Genotype AA AB BB AC BC CC Frequency 9.6 48.3 34.3 2.8 5.0 0.0 Proportion .096 .483 .343 .028 .05 0 Allele A B C Frequency a b c (a+b+c)2= a2 + b2 + c2 + 2ab + 2ac + 2bc AA BB CC AB AC BC (but this is assuming Hardy-Weinberg eq.) 1.5 Genotype AA AB BB AC BC CC Frequency 9.6 48.3 34.3 2.8 5.0 0.0 Proportion .096 .483 .343 .028 .05 0 Without assuming anything, we can count: Allele A B C Frequency a b c a=.096+.5(.483+.028)=.3515 b=.343+.5(.483+.05)=.6095 (a+b+c)2= a2 + b2 + c2 + 2ab + 2ac + 2bc c=0+.5(.028+.05)=.039 AA BB CC AB AC BC (but this is assuming Hardy-Weinberg eq.) 1.5 Genotype AA AB BB AC BC CC Frequency 9.6 48.3 34.3 2.8 5.0 0.0 Proportion .096 .483 .343 .028 .05 0 Without assuming anything, we can count: Allele A B C Frequency a b c a=.096+.5(.483+.028)=.3515 b=.343+.5(.483+.05)=.6095 (a+b+c)2= a2 + b2 + c2 + 2ab + 2ac + 2bc c=0+.5(.028+.05)=.039 AA BB CC AB AC BC Now, back to Hardy-Weinberg expectations: (but this is assuming Hardy-Weinberg eq.) c2=.0392=.001521 .001521*178=.27, so less than 1 individual 1.5 Genotype AA AB BB AC BC CC Frequency 9.6 48.3 34.3 2.8 5.0 0.0 Proportion .096 .483 .343 .028 .05 0 Without assuming anything, we can count: Allele A B C Frequency a b c a=.096+.5(.483+.028)=.3515 b=.343+.5(.483+.05)=.6095 (a+b+c)2= a2 + b2 + c2 + 2ab + 2ac + 2bc c=0+.5(.028+.05)=.039 AA BB CC AB AC BC Now, back to Hardy-Weinberg expectations: (but this is assuming Hardy-Weinberg eq.) c2=.0392=.001521 .001521*178=.27, so less than 1 individual Btw: is the system in Hardy-Weinberg eq.? a2=.124 2ab=.428 b2=.371 2ac=.027 c2=.0015 2bc=.048 1.5 Genotype AA AB BB AC BC CC Frequency 9.6 48.3 34.3 2.8 5.0 0.0 Proportion .096 .483 .343 .028 .05 0 Expected pr. .124 .428 .371 .027 .048 .002 Without assuming anything, we can count: Allele A B C Frequency a b c a=.096+.5(.483+.028)=.3515 b=.343+.5(.483+.05)=.6095 (a+b+c)2= a2 + b2 + c2 + 2ab + 2ac + 2bc c=0+.5(.028+.05)=.039 AA BB CC AB AC BC Now, back to Hardy-Weinberg expectations: (but this is assuming Hardy-Weinberg eq.) c2=.0392=.001521 .001521*178=.27, so less than 1 individual Btw: is the system in Hardy-Weinberg eq.? a2=.124 2ab=.428 b2=.371 2ac=.027 c2=.0015 2bc=.048 1.6 About 7% of men are colour-blind in consequence of a sex-linked recessive gene. Assuming Hardy-Weinberg equilibrium, what proportion of women are expected to be (1) carriers, (2) colour-blind? (3) In what proportion of marriages are both husband and wife expected to be colour-blind? 1.6 About 7% of men are colour-blind in consequence of a sex-linked recessive gene. Assuming Hardy-Weinberg equilibrium, what proportion of women are expected to be (1) carriers, (2) colour-blind? (3) In what proportion of marriages are both husband and wife expected to be colour-blind? XX XY 1.6 About 7% of men are colour-blind in consequence of a sex-linked recessive gene. Assuming Hardy-Weinberg equilibrium, what proportion of women are expected to be (1) carriers, (2) colour-blind? (3) In what proportion of marriages are both husband and wife expected to be colour-blind? ♀ Genotype Freq AA Aa ♂ aa A a .07 XX XY 1.6 About 7% of men are colour-blind in consequence of a sex-linked recessive gene. Assuming Hardy-Weinberg equilibrium, what proportion of women are expected to be (1) carriers, (2) colour-blind? (3) In what proportion of marriages are both husband and wife expected to be colour-blind? ♀ Genotype AA Aa ♂ aa Freq (1) female carriers: freq(Aa)=? qm=qf=.07 pm=pf=.93 freq(Aa)=2pq=2*.93*.07=.1302 A a .07 .1302 XX XY 1.6 About 7% of men are colour-blind in consequence of a sex-linked recessive gene. Assuming Hardy-Weinberg equilibrium, what proportion of women are expected to be (1) carriers, (2) colour-blind? (3) In what proportion of marriages are both husband and wife expected to be colour-blind? ♀ Genotype AA Freq Aa ♂ aa .1302 (1) female carriers: freq(Aa)=? qm=qf=.07 pm=pf=.93 freq(Aa)=2pq=2*.93*.07=.1302 A a .07 .1302 XX XY 1.6 About 7% of men are colour-blind in consequence of a sex-linked recessive gene. Assuming Hardy-Weinberg equilibrium, what proportion of women are expected to be (1) carriers, (2) colour-blind? (3) In what proportion of marriages are both husband and wife expected to be colour-blind? ♀ Genotype AA Freq Aa ♂ aa .1302 (1) female carriers: A a .07 freq(Aa)=? (2) colour-blind females: freq(aa)=? qm=qf=.07 freq(aa)=q2=.072=.0049 pm=pf=.93 freq(Aa)=2pq=2*.93*.07=.1302 XX XY 1.6 About 7% of men are colour-blind in consequence of a sex-linked recessive gene. Assuming Hardy-Weinberg equilibrium, what proportion of women are expected to be (1) carriers, (2) colour-blind? (3) In what proportion of marriages are both husband and wife expected to be colour-blind? ♀ Genotype AA Freq ♂ Aa aa .1302 .0049 (1) female carriers: A a .07 freq(Aa)=? (2) colour-blind females: freq(aa)=? qm=qf=.07 freq(aa)=q2=.072=.0049 pm=pf=.93 freq(Aa)=2pq=2*.93*.07=.1302 XX XY 1.6 About 7% of men are colour-blind in consequence of a sex-linked recessive gene. Assuming Hardy-Weinberg equilibrium, what proportion of women are expected to be (1) carriers, (2) colour-blind? (3) In what proportion of marriages are both husband and wife expected to be colour-blind? ♀ Genotype AA Freq ♂ Aa aa .1302 .0049 A a .07 freq(Aa)=? (2) colour-blind females: freq(aa)=? (3) husband and wife colour-blind: prob(♀aa♂a)=prob(♀aa) * prob(♂a) qm=qf=.07 freq(aa)=q2=.072=.0049 prob(♀aa♂a)=.0049*.07=.000343 (1) female carriers: pm=pf=.93 freq(Aa)=2pq=2*.93*.07=.1302 2.5 Medical treatment is, or will be, available for several serious autosomal recessive diseases. What would be the long-term consequences if treatment allowed sufferers from such a disease to have on average half the number of children that normal people have, whereas without treatment they would have no children? Assume that the present frequency is the mutation versus selection equilibrium, that in the long term a new equilibrium will be reached, and that no other circumstances change. 2.5 Medical treatment is, or will be, available for several serious autosomal recessive diseases. What would be the long-term consequences if treatment allowed sufferers from such a disease to have on average half the number of children that normal people have, whereas without treatment they would have no children? Assume that the present frequency is the mutation versus selection equilibrium, that in the long term a new equilibrium will be reached, and that no other circumstances change. Genotype AA Aa aa Freq p2 2pq q2 Sel.coef. 0 0 s Fitness 1 1 1-s Gam.con. p2 2pq q2(1-s) aa Aa AA 1-s 1 fitness 2.5 Medical treatment is, or will be, available for several serious autosomal recessive diseases. What would be the long-term consequences if treatment allowed sufferers from such a disease to have on average half the number of children that normal people have, whereas without treatment they would have no children? Assume that the present frequency is the mutation versus selection equilibrium, that in the long term a new equilibrium will be reached, and that no other circumstances change. Genotype AA Aa aa Freq p2 2pq q2 Sel.coef. 0 0 s Fitness 1 1 1-s Gam.con. p2 2pq q2(1-s) Old s=1 New s=.5 h=0 Aa AA aa 1-s=1-1=0 1 fitness 1-hs=1 2.5 Medical treatment is, or will be, available for several serious autosomal recessive diseases. What would be the long-term consequences if treatment allowed sufferers from such a disease to have on average half the number of children that normal people have, whereas without treatment they would have no children? Assume that the present frequency is the mutation versus selection equilibrium, that in the long term a new equilibrium will be reached, and that no other circumstances change. Genotype AA Aa aa Freq p2 2pq q2 Sel.coef. 0 0 s Fitness 1 1 1-s Gam.con. p2 2pq q2(1-s) Aa AA aa 1-s=1-1=0 1-hs=1 1 fitness Old s=1 Equilibrium: Old equilibrium: New equilibrium: So at the new equilibrium, the New s=.5 u=sq2 q2=u/s q2=u/s frequency of recessive h=0 q2=u/s q2=u q2=u/.5=2u homozygotes (aa) will double. 2.6 Cystic fibrosis is an autosomal recessive human disease with an incidence of about 1 in 2,500 live births among Caucasians. What would be the consequence in the immediately following generation if the mutation rate were doubled? Assume that the present frequency is the mutation versus selection equilibrium, that back-mutation is negligible, and that affected individuals have no children. Express your result as a percentage increase of incidence and as the number of additional cases per million births. 2.6 Cystic fibrosis is an autosomal recessive human disease with an incidence of about 1 in 2,500 live births among Caucasians. What would be the consequence in the immediately following generation if the mutation rate were doubled? Assume that the present frequency is the mutation versus selection equilibrium, that back-mutation is negligible, and that affected individuals have no children. Express your result as a percentage increase of incidence and as the number of additional cases per million births. Genotype AA Aa aa Freq p02 2p0q0 q02 .0004 q0=.02 p0=.98 s=1 v=0 2.6 Cystic fibrosis is an autosomal recessive human disease with an incidence of about 1 in 2,500 live births among Caucasians. What would be the consequence in the immediately following generation if the mutation rate were doubled? Assume that the present frequency is the mutation versus selection equilibrium, that back-mutation is negligible, and that affected individuals have no children. Express your result as a percentage increase of incidence and as the number of additional cases per million births. Genotype AA Aa aa Freq p02 2p0q0 q02 .0004 q0=.02 p0=.98 s=1 v=0 q02=u0/s u0=.0004 2.6 Cystic fibrosis is an autosomal recessive human disease with an incidence of about 1 in 2,500 live births among Caucasians. What would be the consequence in the immediately following generation if the mutation rate were doubled? Assume that the present frequency is the mutation versus selection equilibrium, that back-mutation is negligible, and that affected individuals have no children. Express your result as a percentage increase of incidence and as the number of additional cases per million births. Genotype AA Aa aa Freq p02 2p0q0 q02 .0004 q0=.02 p0=.98 s=1 v=0 q02=u0/s u0=.0004 If u1=2u0, Δq=? 2.6 Cystic fibrosis is an autosomal recessive human disease with an incidence of about 1 in 2,500 live births among Caucasians. What would be the consequence in the immediately following generation if the mutation rate were doubled? Assume that the present frequency is the mutation versus selection equilibrium, that back-mutation is negligible, and that affected individuals have no children. Express your result as a percentage increase of incidence and as the number of additional cases per million births. Genotype AA Aa aa Freq p02 2p0q0 q02 .0004 q0=.02 If u1=2u0, Δq=? p0=.98 s=1 v=0 q02=u0/s u0=.0004 u1=2*.0004=.0008 2.6 Cystic fibrosis is an autosomal recessive human disease with an incidence of about 1 in 2,500 live births among Caucasians. What would be the consequence in the immediately following generation if the mutation rate were doubled? Assume that the present frequency is the mutation versus selection equilibrium, that back-mutation is negligible, and that affected individuals have no children. Express your result as a percentage increase of incidence and as the number of additional cases per million births. Genotype AA Aa aa Freq p02 2p0q0 q02 .0004 q0=.02 If u1=2u0, Δq=? p0=.98 s=1 v=0 u1=2*.0004=.0008 change from mutation: change from selection: Δq=p0u1+q0v Δq=-sq02(1-q0) Δq=p0u1 Δq=-.0004*.98 Δq=.98*.0008 Δq=-.000392 Δq=.000784 q02=u0/s u0=.0004 2.6 Cystic fibrosis is an autosomal recessive human disease with an incidence of about 1 in 2,500 live births among Caucasians. What would be the consequence in the immediately following generation if the mutation rate were doubled? Assume that the present frequency is the mutation versus selection equilibrium, that back-mutation is negligible, and that affected individuals have no children. Express your result as a percentage increase of incidence and as the number of additional cases per million births. Genotype AA Aa aa Freq p02 2p0q0 q02 .0004 q0=.02 If u1=2u0, Δq=? p0=.98 s=1 v=0 u1=2*.0004=.0008 change from mutation: change from selection: total change: Δq=p0u1+q0v Δq=-sq02(1-q0) Δq=.000784-.000392 Δq=p0u1 Δq=-.0004*.98 Δq=.000392 Δq=.98*.0008 Δq=-.000392 Δq=.000784 q1=q0+ Δq q02=u0/s q1=.020392 u0=.0004 q12=.0004158 2.6 Cystic fibrosis is an autosomal recessive human disease with an incidence of about 1 in 2,500 live births among Caucasians. What would be the consequence in the immediately following generation if the mutation rate were doubled? Assume that the present frequency is the mutation versus selection equilibrium, that back-mutation is negligible, and that affected individuals have no children. Express your result as a percentage increase of incidence and as the number of additional cases per million births. So, q02=.0004, q12=.0004158. Genotype AA Aa aa Difference = .0000158 Freq p02 2p0q0 q02 .0000158/.0004=.0395 -> there are 3.95% more aa homozygotes .0000158*1,000,000=15.8 -> around 16 more births per million .0004 q0=.02 If u1=2u0, Δq=? p0=.98 s=1 v=0 u1=2*.0004=.0008 change from mutation: change from selection: total change: Δq=p0u1+q0v Δq=-sq02(1-q0) Δq=.000784-.000392 Δq=p0u1 Δq=-.0004*.98 Δq=.000392 Δq=.98*.0008 Δq=-.000392 Δq=.000784 q1=q0+ Δq q02=u0/s q1=.020392 u0=.0004 q12=.0004158