Transcript Problem 5

Problem 1
• Consider the populations that
have the genotypes shown in
the following table:
• A. Which of the populations
are in Hardy-Weinberg
equilibrium
• B. What are the p and q in
each population?
• In population 5, it is discovered
that the A to a mutation rate is
5x10-6 and that the reverse
mutation is negligible. What
must be the fitness of the a/a
pehnotype? (q2=u/s where u is
the mutation rate and s is the
coefficient of selection)
population
A/A
A/a
a/a
1
1.0
0.0
0.0
2
0.0
1.0
0.0
3
0.0
0.0
1.0
4
0.5
0.25
0.25
5
0.986049
0.013902
0.000049
Answer Problem 1
A. Which of the populations are in HardyWeinberg equilibrium
•
B. What are the p and q in each population?
•
For each the p and q must first be calculated
–
–
–
–
–
•
•
•
•
•
•
#1 p=1+1/2(0)=1 q=1-1=0
#2 p=0+1/2(1)=.5 q=1-.5=.5
#3 q=1+1/2(0)=1 p=1-1=0
#4 p=0.5+1/2(0.25)=0.625 q=1-0.625=0.375
#5 p=0.986049+1/2(0.013902)=0.993 q=10.993=0.007
From the p and q simply calculate A/A= p2,
A/a=2pq, a/a=q2
#1 A/A= p2=1x1=1 so yes
#2 A/a=2pq=2x.5x.5=0.5 so no
#3 a/a=q2=1x1=1 so yes
#4 A/A=p2=0.625x0.625=0.391 so no
#5 A/A=p2=0.993x0.993=0.986
A/a=2pq=2x0.993x0.007=0.014
a/a=q2=0.007x0.007=0.000049 so yes
populatio
n
A/A
A/a
a/a
Equilibriu
m
1
1.0
0.0
0.0
yes
2
0.0
1.0
0.0
no
3
0.0
0.0
1.0
yes
4
0.5
0.25
0.25
no
5
0.986049
0.013902
0.000049
yes
Problem 1
• In population 5, it is discovered
that the A to a mutation rate is
5x10-6 and that the reverse
mutation is negligible. What
must be the fitness of the a/a
phenotype? (q2=u/s where u is
the mutation rate and s is the
coefficient of selection)
• we know that the fitness is
related to the coefficient of
selection by the equation f=1s. So s=1-f. Therefore
0.000049=5x10-6/s s=0.102
and f=0.898
• So the fitness of the a/a
phenotype is 0.898
population
A/A
A/a
a/a
1
1.0
0.0
0.0
2
0.0
1.0
0.0
3
0.0
0.0
1.0
4
0.5
0.25
0.25
5
0.986049
0.013902
0.000049
Problem 2
•
•
•
•
•
A woman (II2 in the pedigree) wishes to know the probability that she is a carrier of Duchenne muscular dystrophy.
a. What is the probability if she has another affected male child
b. What is the probability if she has another unaffected male child
c. What is the probability if she has another unaffected female child
d. If she has another unaffected male child what is the probability she will have another affected child
I
II
III
Answer Problem 2
•
A woman (II2 in the pedigree) wishes to know the
probability that she is a carrier of Duchenne
muscular dystrophy.
This problem is best solved using Bayes’ Theorem
It is easiest to make a table
I
Ancestral information
Hypothesis 1
II-2 Is a carrier
Hypothesis 2
II-4 Is Not a
carrier
Prior probability
1/2
1/2
Conditional
probability
(½)3=1/8
1
Joint probability
1/8X1/2=1/16
1/2x1
Posterior
probability
(1/16)/(1/16+1/2)
=1/9
1/2/(1/16+1/2)
=8/9
II
Considers children
III
So the probability she is a carrier is 1/9
Answer Problem 2
a. What is the probability if she has another affected
male child
If she has an affected child then she is a carrier =1
I
II
III
Answer Problem 2
•
b. What is the probability if she has another
unaffected male child
This problem is best solved using Bayes’ Theorem
It is easiest to make a table
I
II
Hypothesis 1
II-2 Is a carrier
Hypothesis 2
II-4 Is Not a
carrier
Prior probability
1/2
1/2
Conditional
probability
(½)4=1/16
1
Joint probability
1/8X1/2=1/32
1/2x1
Posterior
probability
(1/32)/(1/32+1/2)
=1/17
1/2/(1/32+1/2)
=16/17
If she has another
unaffected male child
then the conditional
probability will change
III
So the probability she is a carrier is 1/17
Answer Problem 2
•
c. What is the probability if she has another
unaffected female child
This problem is best solved using Bayes’ Theorem
It is easiest to make a table
I
II
If she has a female
child no additional
information is gained
because the disease is
X-linked recessive
Hypothesis 1
II-2 Is a carrier
Hypothesis 2
II-4 Is Not a
carrier
Prior probability
1/2
1/2
Conditional
probability
(½)3=1/8
1
Joint probability
1/8X1/2=1/16
1/2x1
Posterior
probability
(1/16)/(1/16+1/2)
=1/9
1/2/(1/16+1/2)
=8/9
III
So the probability she is a carrier is 1/9
Answer Problem 2
•
d. If she has another unaffected male child what is
the probability she will have another affected child
This problem is best solved using Bayes’ Theorem
It is easiest to make a table
I
II
Hypothesis 1
II-2 Is a carrier
Hypothesis 2
II-4 Is Not a
carrier
Prior probability
1/2
1/2
Conditional
probability
(½)4=1/16
1
Joint probability
1/8X1/2=1/32
1/2x1
Posterior
probability
(1/32)/(1/32+1/2)
=1/17
1/2/(1/32+1/2)
=16/17
If she has another
unaffected male child
then the conditional
probability will change
III
So the probability she is a
carrier is 1/17 if she had four
unaffected boys. To pass on
the recessive allele is ½ and
the chance the child is male
is 1/2. So the chance she
has an affected child is
1/17X1/2X1/2=1/68
Problem 3
•
What is the probability that the a healthy member of the general population is a carrier of cystic
fibrosis if she tests negative on the common mutation screening analysis (DF508 mutation). It is
known that the incidence of cystic fibrosis in the general population is 1 in 1600 and the common
mutation test detects 75 per cent of all cycstic fibrosis alleles.
Answer Problem 3
•
•
•
•
•
First you should make a table
The prior probability.
– Is a carrier
• We know that the incidence of cystic fibrosis in this case is
1/1600. Because cystic fibrosis is an autosomal recessive
disease a/a=q2=1/1600 so q=.025 and p=0.975 (p+q=1).
The possibility of being heterozygous is
2pq=2x.025x0.975=0.049
– Is not a carrier
• 1-0.049=0.951
Conditional probability
– Is a carrier
• fequals chance she does not have the mutation detected by
the test which 25%
– Is not a carrier
• equal 1
Joint probability
– just priorxconditional
– Is a carrier
• 0.25x0.049=0.012
– is not carrier
• 1x0.951
Posterior
–
Is not carrier
•
–
0.951/(0.012+0.951)=0.987
Is a carrier
•
0.012/(0.012+0.951)=0.012
Hypothesi
s1
Is a
carrier
Hypothesi
s2
Not a
carrier
Prior
probability
~1/20
~19/20
Condition
al
probability
1/4
1
Joint
probability
1/20x1/4=
1/80
1x19/20=
19/20
Posterior
probability
1/80(19/2
0+1/80)=1
/77
19/20/(19/
20+1/80)=
76/77=0.9
87