Download Alexa Resler Homework 3 Epi 516 October 27, 2010 1. Refer to the

Alexa Resler
Epi 516
Homework 3
October 27, 2010
1. Refer to the pedigree in Figure 1. A married couple is unaware they share a common great-great-greatgrandmother, who was herself unaware that she was a carrier for an allele that causes a very rare
autosomal recessive disease that is fully penetrant. What is the probability that the couple’s first child
will be affected? What is the probability that the couple’s first child will be a carrier for the disease? What
is the probability that the couple’s first child will not be a carrier?
Let D be the normal allele and d be the disease allele. Since the great-great-great-grandmother is a
carrier for the allele she would be a Dd heterozygote. We will also assume that everyone else in the
pedigree are DD homozygotes since the d allele would be very rare, as it is stated that the allele causes
a very rare recessive disease.
Also, we know that P(A) 
 P(A | B)P(B)
B
(1)
Probability that the couple’s first child will be affected = P(child is dd)
There isonly one possible mating type that the parents could be in order for the child to be dd, and
that is if both parents are Dd.
P(child is dd) = P(child dd | mom Dd & dad Dd) × P(mom Dd & dad Dd)
= P(child dd | mom Dd & dad Dd) × P(mom Dd) × P(dad Dd)
5
=
5
1  1  1 


4  2  2 
(since each previous generation has a ½ chance of being Dd if all spouses are DD)

(2)
= 0.00024414
Probability that the couple’s first child will be a carrier for the disease = P(child is Dd)
In order for the child to be Dd, there are 3 possible mating types that the parents could have: mom DD
& dad Dd, mom Dd & dad DD, or mom Dd & dad Dd.
P(child is Dd) = P(child Dd | mom DD & dad Dd) × P(mom DD & dad Dd)
+ P(child Dd | mom Dd & dad DD) × P(mom Dd & dad DD)
+ P(child Dd | mom Dd & dad Dd) × P(mom Dd & dad Dd)
= P(child Dd | mom DD & dad Dd) × P(mom DD) × P(dad Dd)
+ P(child Dd | mom Dd & dad DD) × P(mom Dd) × P(dad DD)
+ P(child Dd | mom Dd & dad Dd) × P(mom Dd) × P(dad Dd)
Page 1 of 7
Alexa Resler
Epi 516
Homework 3
October 27, 2010
P(mom DD)  P(dad DD)
(0.5)(0.5)  (0.5)(0)(0.5) 

(0.5)(0.5)  (0.5)(0)(0.5) 



0.5
0.5
(0.5)(0.5)  (0.5)(1)(0) 

(0.5)(0.5)  (0.5)(1)(0) 

 

0.5


(1)


(0.5)(0.5)  (0.5)(0)(0.5) 
(0.5)(0.5)  (0.5)(0)(0.5)  
 
 
0 
1 

(0.5)(0.5)

(0.5)(1)
(1)

(0.5)(0.5)

(0.5)(1)
(1)







 
 
 
 

(0.25)(0.5)  (0.75)(0)0.5
(0.25)(0.5)  (0.75)(0)0.5
 
0.5  
(1)
(0.25)(0.5)  (0.75)(1)0 
(0.25)(0.5)  (0.75)(1)1 
 0.1250.5  0.87500.5  0.1250.5  0.8751(1)
 0.06250.5  0.9375(1)
 0.96925
and
5

1 
P(mom Dd)  P(dad Dd) 
2 
Therefore,

P(child is Dd) = P(child Dd | mom DD & dad Dd) × P(mom DD) × P(dad Dd)
+ P(child Dd | mom Dd & dad DD) × P(mom Dd) × P(dad DD)
+ P(child Dd | mom Dd & dad Dd) × P(mom Dd) × P(dad Dd)
5
1 
1 

 0.96925 
2 
2 
5

1  1 

 0.96925
2  2 

1  1  1 


2  2  2 
5
5
= 0.03077734

(3)
Probability that the couple’s first child will not be a carrier = P(child is DD)
P(child is DD)
=
=
=
1 – [ P(child is dd) + P(child is Dd) ]
1 – [ 0.00024414 + 0.03077734 ]
0.9689785
Page 2 of 7
Alexa Resler
Epi 516
Homework 3
October 27, 2010
2. Laura and Bill meet while attending a lecture about a rare human metabolic disease with an autosomal
recessive mode of transmission. Laura and Bill, as well as their parents, are unaffected by the disease
but each has a sibling with the disease. What is the chance their child is affected?
Let D be the normal allele and d be the disease allele.
Since Laura and Bill are unaffected with siblings that are affected, it means that their parents must be
Dd heterozygotes. Therefore, Laura and Bill could be either DD or Dd since they are unaffected.
However, in order to produce a child that is affected as a homozygous dd Laura and Bill would both
have to be Dd heterozygotes.
P(child is dd) = P(child dd | Laura Dd & Bill Dd) × P(Laura Dd & Bill Dd)
= P(child dd | Laura Dd & Bill Dd) × P(Laura Dd) × P(Bill Dd)
= (1/4) × (2/3) × (2/3)
= 1/9
= 0.111
Page 3 of 7
Alexa Resler
Epi 516
Homework 3
October 27, 2010
3. Refer to the pedigree in Figure 2. This pedigree has 10 individuals, and the father and four offspring are
affected for a rare disease. The genotypes for a locus is given in the figure for all individuals in this large
nuclear family. The genotyped locus has 4 alleles: 1, 2, 3, and 4. Using parametric linkage analysis, it is
of interest to determine if this locus is linked to the disease gene under the assumption that the disease
is an autosomal single gene disease with a fully penetrant dominant allele.
(a) What is the LOD score equation in terms of the recombination fraction θ for this family.
Let D be the disease allele and d be the normal allele for the disease gene.
We can assume that the mother is dd since she is unaffected, and therefore the father must be Dd in
order to produce children that are both affected and unaffected.
This would result in 2 possible phases for the parents, each with a probability of 1/2:
1D 2d 3d 4d and 1d 2D 3d 4d
LOD score
 log 10
P(observed data under  )
P(observed data under 0.5)
 log 10
(0.5) Phase1_ R (1  ) Phase1_ NR  (0.5) Phase2 _ R (1  ) Phase2 _ NR
Phase1_ TOTAL
Phase2 _ TOTAL
(0.5)0.5
 (0.5)0.5
 log 10

1 (1  ) 7   7 (1  )1
(0.5) 8  (0.5) 8
(b) Provide a graph of the LOD score for θ values ranging from 0 to 0.5.
Page 4 of 7
Alexa Resler
Epi 516
Homework 3
October 27, 2010
(c) Approximately what θ value gives the maximum LOD score and what is this corresponding maximum
LOD score?
The LOD score reaches its maximum of approximately 0.798 at approximately θ = 0.125, as shown in
the plot below:
(d) Is there sufficient evidence that this locus is linked to the disease gene under the assumption that the
disease is autosomal dominant and fully penetrant?
The LOD score is much less than 3 and at θ = 0.125 the LOD score approximately reaches its maximum
of 0.798, which tells us that the hypothesis of linkage with 12.5% recombination is only about 100.798 =
6.28 times more likely than the hypothesis of no linkage. Therefore, there is not sufficient evidence that
this locus is linked to the disease gene under the assumption that the disease is autosomal dominant
and fully penetrant.
Page 5 of 7
Alexa Resler
Epi 516
Homework 3
October 27, 2010
4. Redo problem 3 assuming now that you have 4 independent families that are identical to Figure 2.
(a)
The 4 families are independent and identical, which means that their LOD scores could simply
be added. Therefore,
LOD Score
 log 10

1
(1   ) 7   7 (1   )1
(0.5)
8
 (0.5) 8

4

4
 4  log 10
 1 (1   ) 7   7 (1   )1
(0.5)8  (0.5) 8
(b)
(c)
The LOD score reaches its maximum of approximately 3.193 at approximately θ = 0.125, as
shown in the plot below:
Page 6 of 7
Alexa Resler
Epi 516
(d)
Homework 3
October 27, 2010
The LOD score is greater than 3 and at θ = 0.125 the LOD score approximately reaches its
maximum of 3.193, which tells us that the hypothesis of linkage with 12.5% recombination is
about 103.193 = 1559.6 times more likely than the hypothesis of no linkage. Therefore, there is
sufficient evidence that this locus is linked to the disease gene under the assumption that the
disease is autosomal dominant and fully penetrant.
Page 7 of 7