assignments falconer & mackay, chapters 1 and 2

AssignmentsFalconer & Mackay, chapters 1 and 2

(1.5, 1.6, 2.5, and 2.6)

Sanja FranicVU 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.

aa aA AAp2 2pq q2

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

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

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

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

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

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

1.5 Genotype AA AB BB AC BC CCFrequency 9.6 48.3 34.3 2.8 5.0 0.0

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

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

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

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

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

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

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

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

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

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

AA Aa aa A a

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?

♀ ♂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

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

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

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

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

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 AAAa

1-s 1fitness

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

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.

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 q0

2

.0004

q0=.02p0=.98s=1v=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 q0

2

.0004

q0=.02p0=.98s=1v=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 q0

2

.0004

q0=.02p0=.98s=1v=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 q0

2

.0004

q0=.02p0=.98s=1v=0

q02=u0/s

u0=.0004

If u1=2u0, Δq=?

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 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

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

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