1. 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 2 3 4
0.5
0.25 5

2. Answer Problem 1
A. Which of the populations are in Hardy-Weinberg 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= =0.375
#5 p= /2( )=0.993 q= =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= so yes
population
A/A
A/a
a/a
Equilibrium 1
1.0
0.0
yes 2 no 3 4
0.5
0.25 5

3. 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=1-s. So s=1-f. Therefore =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 2 3 4
0.5
0.25 5

4. 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

5. 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 II
III
Ancestral information
Hypothesis 1
II-2 Is a carrier
Hypothesis 2
II-4 Is Not a carrier
Prior probability
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
Considers children
So the probability she is a carrier is 1/9

6. Answer Problem 2 If she has an affected child then she is a carrier =1
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

7. 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
III
Hypothesis 1
II-2 Is a carrier
Hypothesis 2
II-4 Is Not a carrier
Prior probability
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
So the probability she is a carrier is 1/17

8. 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
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
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
III
So the probability she is a carrier is 1/9

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
Hypothesis 1
II-2 Is a carrier
Hypothesis 2
II-4 Is Not a carrier
Prior probability
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 II
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

10. 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.

11. 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/ 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
=0.951
Conditional probability
fequals chance she does not have the mutation detected by the test which 25%
equal 1
Joint probability
just priorxconditional
0.25x0.049=0.012
is not carrier
1x0.951
Posterior
Is not carrier
0.951/( )=0.987
0.012/( )=0.012
Hypothesis 1
Is a carrier
Hypothesis 2
Not a carrier
Prior probability
~1/20
~19/20
Conditional probability
1/4 1
Joint probability
1/20x1/4=1/80
1x19/20=19/20
Posterior probability
1/80(19/20+1/80)=1/77
19/20/(19/20+1/80)=76/77=0.987