assignments falconer & mackay, chapters 1 and 2
DESCRIPTION
Assignments Falconer & Mackay, chapters 1 and 2. (1.5 , 1.6, 2.5, and 2.6). Sanja Franic VU University Amsterdam 2011. 1.4 - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/1.jpg)
AssignmentsFalconer & Mackay, chapters 1 and 2
(1.5, 1.6, 2.5, and 2.6)
Sanja FranicVU University Amsterdam 2011
![Page 2: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/2.jpg)
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.
aa aA AAp2 2pq q2
![Page 3: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/3.jpg)
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.
aa aA AAp2 2pq q2
2pq=1/3pq=1/6p+q=1p=1-qq(1-q)=1/6-q2+q-1/6=0-(q2-q+1/6)=0-(q2-q+…+1/6-…)=0
![Page 4: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/4.jpg)
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.
aa aA AAp2 2pq q2
2pq=1/3pq=1/6p+q=1p=1-qq(1-q)=1/6-q2+q-1/6=0-(q2-q+1/6)=0-(q2-q+…+1/6-…)=0
(q-x)2=q2 – 2xq + x2
2x=1x=.5(q-.5)2=q2 – q + .52
![Page 5: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/5.jpg)
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.
aa aA AAp2 2pq q2
2pq=1/3pq=1/6p+q=1p=1-qq(1-q)=1/6-q2+q-1/6=0-(q2-q+1/6)=0-(q2-q+.52+1/6-.52)=0
(q-x)2=q2 – 2xq + x2
2x=1x=.5(q-.5)2=q2 – q + .52
![Page 6: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/6.jpg)
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.
aa aA AAp2 2pq q2
2pq=1/3pq=1/6p+q=1p=1-qq(1-q)=1/6-q2+q-1/6=0-(q2-q+1/6)=0-(q2-q+.52+1/6-.52)=0
-[(q-.5)2-1/12]=0-(q-.5)2=-1/12(q-.5)2=1/12q-.5=√1/12q=√1/12+.5q=.789p=1-q=.211
(q-x)2=q2 – 2xq + x2
2x=1x=.5(q-.5)2=q2 – q + .52
![Page 7: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/7.jpg)
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.
aa aA AAp2 2pq q2
.045 .33 .622
2pq=1/3pq=1/6p+q=1p=1-qq(1-q)=1/6-q2+q-1/6=0-(q2-q+1/6)=0-(q2-q+.52+1/6-.52)=0
-[(q-.5)2-1/12]=0-(q-.5)2=-1/12(q-.5)2=1/12q-.5=√1/12q=√1/12+.5q=.789p=1-q=.211
(q-x)2=q2 – 2xq + x2
2x=1x=.5(q-.5)2=q2 – q + .52
![Page 8: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/8.jpg)
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?
Genotype AA AB BB AC BC CCFrequency 9.6 48.3 34.3 2.8 5.0 0.0
![Page 9: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/9.jpg)
1.5 Genotype AA AB BB AC BC CCFrequency 9.6 48.3 34.3 2.8 5.0 0.0
![Page 10: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/10.jpg)
1.5 Genotype AA AB BB AC BC CCFrequency 9.6 48.3 34.3 2.8 5.0 0.0Proportion .096 .483 .343 .028 .05 0
![Page 11: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/11.jpg)
1.5 Genotype AA AB BB AC BC CCFrequency 9.6 48.3 34.3 2.8 5.0 0.0Proportion .096 .483 .343 .028 .05 0
Allele A B CFrequency
a b c
![Page 12: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/12.jpg)
1.5 Genotype AA AB BB AC BC CCFrequency 9.6 48.3 34.3 2.8 5.0 0.0Proportion .096 .483 .343 .028 .05 0
Allele A B CFrequency
a b c
(a+b+c)2= a2 + b2 + c2 + 2ab + 2ac + 2bc
![Page 13: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/13.jpg)
1.5 Genotype AA AB BB AC BC CCFrequency 9.6 48.3 34.3 2.8 5.0 0.0Proportion .096 .483 .343 .028 .05 0
Allele A B CFrequency
a b c
(a+b+c)2= a2 + b2 + c2 + 2ab + 2ac + 2bc
AA BB CC AB AC BC
![Page 14: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/14.jpg)
1.5 Genotype AA AB BB AC BC CCFrequency 9.6 48.3 34.3 2.8 5.0 0.0Proportion .096 .483 .343 .028 .05 0
Allele A B CFrequency
a b c
(a+b+c)2= a2 + b2 + c2 + 2ab + 2ac + 2bc
(but this is assuming Hardy-Weinberg eq.)
AA BB CC AB AC BC
![Page 15: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/15.jpg)
1.5 Genotype AA AB BB AC BC CCFrequency 9.6 48.3 34.3 2.8 5.0 0.0Proportion .096 .483 .343 .028 .05 0
Allele A B CFrequency
a b c
(a+b+c)2= a2 + b2 + c2 + 2ab + 2ac + 2bc
(but this is assuming Hardy-Weinberg eq.)
AA BB CC AB AC BC
Without assuming anything, we can count:
a=.096+.5(.483+.028)=.3515b=.343+.5(.483+.05)=.6095c=0+.5(.028+.05)=.039
![Page 16: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/16.jpg)
1.5 Genotype AA AB BB AC BC CCFrequency 9.6 48.3 34.3 2.8 5.0 0.0Proportion .096 .483 .343 .028 .05 0
Allele A B CFrequency
a b c
(a+b+c)2= a2 + b2 + c2 + 2ab + 2ac + 2bc
(but this is assuming Hardy-Weinberg eq.)
AA BB CC AB AC BC
Without assuming anything, we can count:
a=.096+.5(.483+.028)=.3515b=.343+.5(.483+.05)=.6095c=0+.5(.028+.05)=.039
Now, back to Hardy-Weinberg expectations:
c2=.0392=.001521.001521*178=.27, so less than 1 individual
![Page 17: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/17.jpg)
1.5 Genotype AA AB BB AC BC CCFrequency 9.6 48.3 34.3 2.8 5.0 0.0Proportion .096 .483 .343 .028 .05 0
Allele A B CFrequency
a b c
(a+b+c)2= a2 + b2 + c2 + 2ab + 2ac + 2bc
(but this is assuming Hardy-Weinberg eq.)
AA BB CC AB AC BC
Without assuming anything, we can count:
a=.096+.5(.483+.028)=.3515b=.343+.5(.483+.05)=.6095c=0+.5(.028+.05)=.039
Now, back to Hardy-Weinberg expectations:
c2=.0392=.001521.001521*178=.27, so less than 1 individual
Btw: is the system in Hardy-Weinberg eq.?
a2=.124 2ab=.428b2=.371 2ac=.027c2=.0015 2bc=.048
![Page 18: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/18.jpg)
1.5 Genotype AA AB BB AC BC CCFrequency 9.6 48.3 34.3 2.8 5.0 0.0Proportion .096 .483 .343 .028 .05 0Expected pr. .124 .428 .371 .027 .048 .002
Allele A B CFrequency
a b c
(a+b+c)2= a2 + b2 + c2 + 2ab + 2ac + 2bc
(but this is assuming Hardy-Weinberg eq.)
AA BB CC AB AC BC
Without assuming anything, we can count:
a=.096+.5(.483+.028)=.3515b=.343+.5(.483+.05)=.6095c=0+.5(.028+.05)=.039
Now, back to Hardy-Weinberg expectations:
c2=.0392=.001521.001521*178=.27, so less than 1 individual
Btw: is the system in Hardy-Weinberg eq.?
a2=.124 2ab=.428b2=.371 2ac=.027c2=.0015 2bc=.048
![Page 19: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/19.jpg)
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?
![Page 20: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/20.jpg)
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
![Page 21: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/21.jpg)
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 A a
Freq .07
XX XY
![Page 22: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/22.jpg)
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 A a
Freq .07
(1) female carriers:freq(Aa)=?qm=qf=.07pm=pf=.93
freq(Aa)=2pq=2*.93*.07=.1302
.1302 XX XY
![Page 23: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/23.jpg)
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 A a
Freq .1302 .07
(1) female carriers:freq(Aa)=?qm=qf=.07pm=pf=.93
freq(Aa)=2pq=2*.93*.07=.1302
.1302 XX XY
![Page 24: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/24.jpg)
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 A a
Freq .1302 .07
(1) female carriers:freq(Aa)=?qm=qf=.07pm=pf=.93
freq(Aa)=2pq=2*.93*.07=.1302
(2) colour-blind females:freq(aa)=?
freq(aa)=q2=.072=.0049
XX XY
![Page 25: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/25.jpg)
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 A a
Freq .1302 .0049 .07
(1) female carriers:freq(Aa)=?qm=qf=.07pm=pf=.93
freq(Aa)=2pq=2*.93*.07=.1302
(2) colour-blind females:freq(aa)=?
freq(aa)=q2=.072=.0049
XX XY
![Page 26: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/26.jpg)
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 A a
Freq .1302 .0049 .07
(1) female carriers:freq(Aa)=?qm=qf=.07pm=pf=.93
freq(Aa)=2pq=2*.93*.07=.1302
(2) colour-blind females:freq(aa)=?
freq(aa)=q2=.072=.0049
(3) husband and wife colour-blind:
prob(♀aa♂a)=prob(♀aa) * prob(♂a)
prob(♀aa♂a)=.0049*.07=.000343
![Page 27: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/27.jpg)
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.
![Page 28: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/28.jpg)
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 AAAa
1-s 1fitness
![Page 29: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/29.jpg)
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 AAAa
1-s=1-1=0
1fitness
Old s=1New s=.5h=0
1-hs=1
![Page 30: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/30.jpg)
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 AAAa
1-s=1-1=0
1fitness
Old s=1New s=.5h=0
1-hs=1
Equilibrium:u=sq2
q2=u/s
Old equilibrium:q2=u/sq2=u
New equilibrium:q2=u/sq2=u/.5=2u
So at the new equilibrium, the frequency of recessive homozygotes (aa) will double.
![Page 31: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/31.jpg)
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.
![Page 32: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/32.jpg)
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 q0
2
.0004
q0=.02p0=.98s=1v=0
![Page 33: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/33.jpg)
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 q0
2
.0004
q0=.02p0=.98s=1v=0
q02=u0/s
u0=.0004
![Page 34: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/34.jpg)
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 q0
2
.0004
q0=.02p0=.98s=1v=0
q02=u0/s
u0=.0004
If u1=2u0, Δq=?
![Page 35: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/35.jpg)
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 q0
2
.0004
q0=.02p0=.98s=1v=0
q02=u0/s
u0=.0004
If u1=2u0, Δq=?
u1=2*.0004=.0008
![Page 36: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/36.jpg)
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 q0
2
.0004
change from mutation:Δq=p0u1+q0vΔq=p0u1
Δq=.98*.0008Δq=.000784
q0=.02p0=.98s=1v=0
q02=u0/s
u0=.0004
If u1=2u0, Δq=?
u1=2*.0004=.0008
change from selection:Δq=-sq0
2(1-q0)Δq=-.0004*.98Δq=-.000392
![Page 37: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/37.jpg)
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 q0
2
.0004
change from mutation:Δq=p0u1+q0vΔq=p0u1
Δq=.98*.0008Δq=.000784
q0=.02p0=.98s=1v=0
q02=u0/s
u0=.0004
If u1=2u0, Δq=?
u1=2*.0004=.0008
change from selection:Δq=-sq0
2(1-q0)Δq=-.0004*.98Δq=-.000392
total change:Δq=.000784-.000392Δq=.000392
q1=q0+ Δqq1=.020392q1
2=.0004158
![Page 38: Assignments Falconer & Mackay, chapters 1 and 2](https://reader036.vdocuments.site/reader036/viewer/2022062501/56816571550346895dd80812/html5/thumbnails/38.jpg)
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 q0
2
.0004
change from mutation:Δq=p0u1+q0vΔq=p0u1
Δq=.98*.0008Δq=.000784
q0=.02p0=.98s=1v=0
q02=u0/s
u0=.0004
If u1=2u0, Δq=?
u1=2*.0004=.0008
change from selection:Δq=-sq0
2(1-q0)Δq=-.0004*.98Δq=-.000392
total change:Δq=.000784-.000392Δq=.000392
q1=q0+ Δqq1=.020392q1
2=.0004158
So, q02=.0004, q1
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