# Assignments Falconer & Mackay, chapters 1 and 2 (1.5, 1.6, 2.5, and 2.6) Sanja Franic VU University Amsterdam 2011.

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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 random- breeding population that would result in one-third of normal individuals being carriers. aaaAAA p2p2 2pqq2q2

1.4 As an exercise in algebra, work out the gene frequency of a recessive mutant in a random- breeding population that would result in one-third of normal individuals being carriers. aaaAAA p2p2 2pqq2q2 2pq=1/3 pq=1/6 p+q=1 p=1-q q(1-q)=1/6 -q 2 +q-1/6=0 -(q 2 -q+1/6)=0 -(q 2 -q+…+1/6-…)=0

1.4 As an exercise in algebra, work out the gene frequency of a recessive mutant in a random- breeding population that would result in one-third of normal individuals being carriers. aaaAAA p2p2 2pqq2q2 2pq=1/3 pq=1/6 p+q=1 p=1-q q(1-q)=1/6 -q 2 +q-1/6=0 -(q 2 -q+1/6)=0 -(q 2 -q+…+1/6-…)=0 (q-x) 2 =q 2 – 2xq + x 2 2x=1 x=.5 (q-.5) 2 =q 2 – q +.5 2

1.4 As an exercise in algebra, work out the gene frequency of a recessive mutant in a random- breeding population that would result in one-third of normal individuals being carriers. aaaAAA p2p2 2pqq2q2 2pq=1/3 pq=1/6 p+q=1 p=1-q q(1-q)=1/6 -q 2 +q-1/6=0 -(q 2 -q+1/6)=0 -(q 2 -q+.5 2 +1/6-.5 2 )=0 (q-x) 2 =q 2 – 2xq + x 2 2x=1 x=.5 (q-.5) 2 =q 2 – q +.5 2

1.4 As an exercise in algebra, work out the gene frequency of a recessive mutant in a random- breeding population that would result in one-third of normal individuals being carriers. aaaAAA p2p2 2pqq2q2 2pq=1/3 pq=1/6 p+q=1 p=1-q q(1-q)=1/6 -q 2 +q-1/6=0 -(q 2 -q+1/6)=0 -(q 2 -q+.5 2 +1/6-.5 2 )=0 -[(q-.5) 2 -1/12]=0 -(q-.5) 2 =-1/12 (q-.5) 2 =1/12 q-.5=√1/12 q=√1/12+.5 q=.789 p=1-q=.211 (q-x) 2 =q 2 – 2xq + x 2 2x=1 x=.5 (q-.5) 2 =q 2 – q +.5 2

1.4 As an exercise in algebra, work out the gene frequency of a recessive mutant in a random- breeding population that would result in one-third of normal individuals being carriers. aaaAAA p2p2 2pqq2q2.045.33.622 2pq=1/3 pq=1/6 p+q=1 p=1-q q(1-q)=1/6 -q 2 +q-1/6=0 -(q 2 -q+1/6)=0 -(q 2 -q+.5 2 +1/6-.5 2 )=0 -[(q-.5) 2 -1/12]=0 -(q-.5) 2 =-1/12 (q-.5) 2 =1/12 q-.5=√1/12 q=√1/12+.5 q=.789 p=1-q=.211 (q-x) 2 =q 2 – 2xq + x 2 2x=1 x=.5 (q-.5) 2 =q 2 – q +.5 2

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 What are the gene frequencies in the sample? Why were no CC individuals found? GenotypeAAABBBACBCCC Frequency9.648.334.32.85.00.0

1.5 GenotypeAAABBBACBCCC Frequency9.648.334.32.85.00.0

1.5 GenotypeAAABBBACBCCC Frequency9.648.334.32.85.00.0 Proportion.096.483.343.028.050

1.5 GenotypeAAABBBACBCCC Frequency9.648.334.32.85.00.0 Proportion.096.483.343.028.050 AlleleABC Frequencyabc

1.5 GenotypeAAABBBACBCCC Frequency9.648.334.32.85.00.0 Proportion.096.483.343.028.050 AlleleABC Frequencyabc (a+b+c) 2 = a 2 + b 2 + c 2 + 2ab + 2ac + 2bc

1.5 GenotypeAAABBBACBCCC Frequency9.648.334.32.85.00.0 Proportion.096.483.343.028.050 AlleleABC Frequencyabc (a+b+c) 2 = a 2 + b 2 + c 2 + 2ab + 2ac + 2bc AABBCCABACBC

1.5 GenotypeAAABBBACBCCC Frequency9.648.334.32.85.00.0 Proportion.096.483.343.028.050 AlleleABC Frequencyabc (a+b+c) 2 = a 2 + b 2 + c 2 + 2ab + 2ac + 2bc (but this is assuming Hardy-Weinberg eq.) AABBCCABACBC

1.5 GenotypeAAABBBACBCCC Frequency9.648.334.32.85.00.0 Proportion.096.483.343.028.050 AlleleABC Frequencyabc (a+b+c) 2 = a 2 + b 2 + c 2 + 2ab + 2ac + 2bc (but this is assuming Hardy-Weinberg eq.) AABBCCABACBC Without assuming anything, we can count: a=.096+.5(.483+.028)=.3515 b=.343+.5(.483+.05)=.6095 c=0+.5(.028+.05)=.039

1.5 GenotypeAAABBBACBCCC Frequency9.648.334.32.85.00.0 Proportion.096.483.343.028.050 AlleleABC Frequencyabc (a+b+c) 2 = a 2 + b 2 + c 2 + 2ab + 2ac + 2bc (but this is assuming Hardy-Weinberg eq.) AABBCCABACBC Without assuming anything, we can count: a=.096+.5(.483+.028)=.3515 b=.343+.5(.483+.05)=.6095 c=0+.5(.028+.05)=.039 Now, back to Hardy-Weinberg expectations: c 2 =.039 2 =.001521.001521*178=.27, so less than 1 individual

1.5 GenotypeAAABBBACBCCC Frequency9.648.334.32.85.00.0 Proportion.096.483.343.028.050 AlleleABC Frequencyabc (a+b+c) 2 = a 2 + b 2 + c 2 + 2ab + 2ac + 2bc (but this is assuming Hardy-Weinberg eq.) AABBCCABACBC Without assuming anything, we can count: a=.096+.5(.483+.028)=.3515 b=.343+.5(.483+.05)=.6095 c=0+.5(.028+.05)=.039 Now, back to Hardy-Weinberg expectations: c 2 =.039 2 =.001521.001521*178=.27, so less than 1 individual Btw: is the system in Hardy-Weinberg eq.? a 2 =.1242ab=.428 b 2 =.3712ac=.027 c 2 =.00152bc=.048

1.5 GenotypeAAABBBACBCCC Frequency9.648.334.32.85.00.0 Proportion.096.483.343.028.050 Expected pr..124.428.371.027.048.002 AlleleABC Frequencyabc (a+b+c) 2 = a 2 + b 2 + c 2 + 2ab + 2ac + 2bc (but this is assuming Hardy-Weinberg eq.) AABBCCABACBC Without assuming anything, we can count: a=.096+.5(.483+.028)=.3515 b=.343+.5(.483+.05)=.6095 c=0+.5(.028+.05)=.039 Now, back to Hardy-Weinberg expectations: c 2 =.039 2 =.001521.001521*178=.27, so less than 1 individual Btw: is the system in Hardy-Weinberg eq.? a 2 =.1242ab=.428 b 2 =.3712ac=.027 c 2 =.00152bc=.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? ♀♂ GenotypeAAAaaaAa Freq.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? ♀♂ GenotypeAAAaaaAa Freq.07 (1)female carriers: freq(Aa)=? q m =q f =.07 p m =p f =.93 freq(Aa)=2pq=2*.93*.07=.1302.1302 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? ♀♂ GenotypeAAAaaaAa Freq.1302.07 (1)female carriers: freq(Aa)=? q m =q f =.07 p m =p f =.93 freq(Aa)=2pq=2*.93*.07=.1302.1302 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? ♀♂ GenotypeAAAaaaAa Freq.1302.07 (1)female carriers: freq(Aa)=? q m =q f =.07 p m =p f =.93 freq(Aa)=2pq=2*.93*.07=.1302 (2) colour-blind females: freq(aa)=? freq(aa)=q 2 =.07 2 =.0049 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? ♀♂ GenotypeAAAaaaAa Freq.1302.0049.07 (1)female carriers: freq(Aa)=? q m =q f =.07 p m =p f =.93 freq(Aa)=2pq=2*.93*.07=.1302 (2) colour-blind females: freq(aa)=? freq(aa)=q 2 =.07 2 =.0049 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? ♀♂ GenotypeAAAaaaAa Freq.1302.0049.07 (1) female carriers: freq(Aa)=? q m =q f =.07 p m =p f =.93 freq(Aa)=2pq=2*.93*.07=.1302 (2) colour-blind females: freq(aa)=? freq(aa)=q 2 =.07 2 =.0049 (3) husband and wife colour-blind: prob( ♀ aa ♂ a )=prob( ♀ aa ) * prob( ♂ a ) prob( ♀ aa ♂ a )=.0049*.07=.000343

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. GenotypeAAAaaa Freqp2p2 2pqq2q2 Sel.coef.00s Fitness111-s Gam.con.p2p2 2pqq 2 (1-s) aaAA Aa 1-s1 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. GenotypeAAAaaa Freqp2p2 2pqq2q2 Sel.coef.00s Fitness111-s Gam.con.p2p2 2pqq 2 (1-s) aaAA Aa 1-s=1-1=01 fitness Old s=1 New s=.5 h=0 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. GenotypeAAAaaa Freqp2p2 2pqq2q2 Sel.coef.00s Fitness111-s Gam.con.p2p2 2pqq 2 (1-s) aaAA Aa 1-s=1-1=01 fitness Old s=1 New s=.5 h=0 1-hs=1 Equilibrium: u=sq 2 q 2 =u/s Old equilibrium: q 2 =u/s q 2 =u New equilibrium: q 2 =u/s q 2 =u/.5=2u So at the new equilibrium, the frequency of recessive 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. GenotypeAAAaaa Freqp02p02 2p 0 q 0 q02q02.0004 q 0 =.02 p 0 =.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. GenotypeAAAaaa Freqp02p02 2p 0 q 0 q02q02.0004 q 0 =.02 p 0 =.98 s=1 v=0 q 0 2 =u 0 /s u 0 =.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. GenotypeAAAaaa Freqp02p02 2p 0 q 0 q02q02.0004 q 0 =.02 p 0 =.98 s=1 v=0 q 0 2 =u 0 /s u 0 =.0004 If u 1 =2u 0, 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. GenotypeAAAaaa Freqp02p02 2p 0 q 0 q02q02.0004 q 0 =.02 p 0 =.98 s=1 v=0 q 0 2 =u 0 /s u 0 =.0004 If u 1 =2u 0, Δq=? u 1 =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. GenotypeAAAaaa Freqp02p02 2p 0 q 0 q02q02.0004 change from mutation: Δq=p 0 u 1 +q 0 v Δq=p 0 u 1 Δq=.98*.0008 Δq=.000784 q 0 =.02 p 0 =.98 s=1 v=0 q 0 2 =u 0 /s u 0 =.0004 If u 1 =2u 0, Δq=? u 1 =2*.0004=.0008 change from selection: Δq=-sq 0 2 (1-q 0 ) Δq=-.0004*.98 Δq=-.000392

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. GenotypeAAAaaa Freqp02p02 2p 0 q 0 q02q02.0004 change from mutation: Δq=p 0 u 1 +q 0 v Δq=p 0 u 1 Δq=.98*.0008 Δq=.000784 q 0 =.02 p 0 =.98 s=1 v=0 q 0 2 =u 0 /s u 0 =.0004 If u 1 =2u 0, Δq=? u 1 =2*.0004=.0008 change from selection: Δq=-sq 0 2 (1-q 0 ) Δq=-.0004*.98 Δq=-.000392 total change: Δq=.000784-.000392 Δq=.000392 q 1 =q 0 + Δq q 1 =.020392 q 1 2 =.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. GenotypeAAAaaa Freqp02p02 2p 0 q 0 q02q02.0004 change from mutation: Δq=p 0 u 1 +q 0 v Δq=p 0 u 1 Δq=.98*.0008 Δq=.000784 q 0 =.02 p 0 =.98 s=1 v=0 q 0 2 =u 0 /s u 0 =.0004 If u 1 =2u 0, Δq=? u 1 =2*.0004=.0008 change from selection: Δq=-sq 0 2 (1-q 0 ) Δq=-.0004*.98 Δq=-.000392 total change: Δq=.000784-.000392 Δq=.000392 q 1 =q 0 + Δq q 1 =.020392 q 1 2 =.0004158 So, q 0 2 =.0004, q 1 2 =.0004158. Difference =.0000158.0000158/.0004=.0395 -> there are 3.95% more aa homozygotes.0000158*1,000,000=15.8 -> around 16 more births per million

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