Transcript Assignments Ch 1 & 2

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