Conditional Probability - No Penetrance - Answers


 

1.

 A certain genetics professor has brown hair and brown eyes. His father and mother have brown hair and brown eyes. His sister and e of his two brothers have brown hair and brown eyes. However, his other brother has blond hair and blue eyes (how's that for a Mendelian family!).

His wife has brown hair and blue eyes. Her three sisters and two brothers also have brown hair and blue eyes.  Her four children by a previous marriage all had blue eyes but half of them had blond hair.

You may assume that brown eyes (E) is dominant to blue eyes (e)  and that brown hair (B) is dominant to blond hair (b)
   

a.

 What is the probability any child that they decide to have would have brown eyes?
   

b.

 What is the probability any child that they decide to have would have both blond hair and brown eyes?

What do you know?

For Part a,  we can ignore hair color. since the question only asks about eye color.

First, the wife's genotype can be specified.  It is given that blue eyes (e) are recessive to brown eyes (E) and the wife has blue eyes.  Therefore she must be a homozygote (ee).

The husband has brown eyes so we know 1/2 of his genotype. That is, he has at least one E allele.  His genotype is E_.   Furthermore, he has a brother with blue eyes so both his parents were carriers. That is, both his parent had the genotype Ee.

This means that he has a 1/3 chance of having the genotype BB and a 2/3 chance of having the genotype Bb. (review probability of being a carrier here) .


Calculations and Solutions

There are two possibile marriages each with a different probability of having a child with brown eyes.
 
Possibility 1 Wife is ee and husband is EE. This occurs with probability of 1/3.  The probability of a child with brown eyes = 1.
 
Possibility 2 Wife is ee and husband is Ee. This occurs with probability of 2/3.  The probability of a child with brown eyes = 1/2.

Therefore, the overall probability of having a child with brown eyes is:

 [1/3*1 ]+ [2/3*1/2] = 1/3 + 2/6 = 2/3

What do you know?

b.

 You  have now established that the probability of brown eyes is 2/3.
   
  Let B be the allele for brown hair and b be the allele for blond hair
   
  From the information given, the wife has brown hair but she has two blond children.  Therefore she must be heterozygous for hair color (Bb).
   
  The husband has brown hair and he has a brother with blond hair eyes so both his parents were carriers. That is, both his parent had the genotype Bb. 
.
  This means that he has a 1/3 chance of having the Genotype BB and 2/3 probability of having the genotype Bb (see above).
 


Calculations and Solutions
 

There are two possible marriages  each with a different probability of having a child with blond hair.
 
Possibility 1 Wife is Bb and husband is BB. This occurs with probability of 1/3.  The probability of a child with blond hair  = 0.
 
Possibility 2 Wife is Bb and husband is Bb. This occurs with probability of 2/3.  The probability of a child with blond hair = 1/4.
 


Therefore, the overall  probability of having a child with blond hair is:

 [1/3*0 ]+ [2/3*1/4] = 0 + 2/12 = 1/6

 
   
Finally, the probability of the child having both blond hair and brown eyes (where and means multiply) is:
 
  Pr (blond hair) * Prob (brown eyes)
 
  1/6 * 2/3 = 2/18 = 1/9


 

2. Emmett was blue-green color deficient (a sex-linked, recessive trait).  His daughter, Mildred, had normal color vision.  Mildred married a man with normal color vision.  Mildred's son, Fergus, had one sister and two brothers.  Fergus and his brother Rick had normal color vision.  His older brother Curtis had blue-green color deficiency.   His sister Ann married a normal man with normal color vision. 
   
  What is the probability that Ann's son would have blue-green color-deficiency? 
   

What do you know?

Blue-green color deficiency is sex-linked and recessive.  Let A be the normal allele and a be the allele for color-deficiency.

 Since Fergus's brother Curtis is affected, his mother Mildred must have be a carrier (heterozygote - Aa).  Since her father has normal color vision, this means that his sister Ann has a 50% chance of being a carrier (genotype Aa).  She also has a 50% chance of not carrying the trait (genotype AA).

If Ann is Aa she will pass this trait to 50% of her sons.  If Ann is AA, she will pass this trait to none of her sons.
 

Calculations and Solutions

From the available information,  the probability of Ann's son being affected is:

    1/2(Aa) * 1/2(affected son) + 1/2(AA) * 0 (affected son) =  1/4 + 0 = 1/4

 

3. A young couple, Susan and Bill,  comes to you - a family physician - and tells you that they want to start a family but that they are very worried about having a child with cystic fibrosis.
   
  Cystic fibrosis is an autosomal recessive genetic disease affecting approximately 30,000 children and adults in the United States. One in 30 Caucasians (of mixed Northern European descent) is an unknowing, symptom-less carrier (heterozygote) of the defective gene. 
   
  The only information that you have is that that both Susan and Bill are healthy and that they are Caucasians of mixed Northern European descent. 
   

a.

 Based on this limited information, what is the probability that they will have a child with cystic fibrosis?
 

For convenience let the allele for cystic fibrosis be a; and the normal allele be A. For the specified situtation let the probability of being AA be 29/30 and of being Aa be 1/30
   

b.

 After you tell the couple the probability from Part a,  Susan tells you that while her parents are normal her brother Ted has cystic fibrosis.  She wants to know how this would change the probability.   What can you tell her?
   
 

What do you know?

Neither the woman nor the man has CF.  Both the man and the woman are American Caucasians of mixed ethnicity. To their knowledge, they are not related to each other.

At this point you have minimal information.  You know that both partners are phenotypically normal so their individual genotypes are ieither AA or Aa (carrier). 

Furthermore, the only way they could have a child with CF (cystic fibrosis) is if both the man and woman were carriers.

 

Calculations and Solutions

Assuming that marriage occurs at random according to the CF gene, the chance of an AA x AA marriage  is (29/30 * 29/30).  The chance of an Aa * AA marriage is (1/30 * 29/30) and so forth. 

If there are three genotypes in the general population, then there are nine possible marriages possible in the population (shown below).  Five of these marriages do not apply to this couple since neither one of them has CF.  The four marriages that apply are given in blue.

Also given is the probability that each marriage could produce a child with CF.


 

Woman's Genotype
Man's
Genotype
 Prob. of Condition for this couple Prob. of CF Child
AA AA (29/30)*(29/30) = 841/900 0
AA Aa (29/30)*(1/30)= 29/900 0
AA aa 0 0
Aa AA (1/30)*(29/30)= 29/900 0
Aa Aa (1/30)*(1/30)=1/900 1/4
Aa aa 0 1/2
aa AA 0 0
aa Aa 0 1/2
aa aa 0 1

The probability of a CF child is the weighted probabilities of the three marriages with no more than one carrier (841/900 + 29/900 + 29/900 = 899/900) times the probability that that marriage would produce an affected child (in this case zero) and the probability of a carrier x carrier (Aa x Aa) marriage (1/900) times the probability that that marriage would produce an affected child (1/4).

 
Conditional Probability of a CF child = (899/900* 0)+ (1/900 *1/4) = 1/3600

Part B. If Susan's brother has CF then everything changes.  Susan's parents must be carriers (Aa) and the probability that Susan is a carrier (Aa) changes from  the populational probability of 1/30 to the family probability of  2/3 (just like the Brown eye example in Problem 1 above).  The probability that Susan is AA is 1/3, the probability that she is Aa is 2/3.

The marriage chart with these new probabilities is below:

Woman's Genotype
Man's
Genotype
 Prob. of Condition for this couple Prob. of CF Child
AA AA (1/3)*(29/30) = 29/90 0
AA Aa (1/3)*(1/30)= 1/90 0
AA aa 0 0
Aa AA (2/3)*(29/30)= 58/90 0
Aa Aa (2/3)*(1/30)=2/90 1/4
Aa aa 0 1/2
aa AA 0 0
aa Aa 0 1/2
aa aa 0 1
 

As stated before the probability of a CF child is the weighted probabilities of the three marriages with no more than one carrier (29/90 + 1/90 + 58/90 = 88/90) times the probability that that marriage would produce an affected child (in this case zero) times the probability of a carrier x carrier (Aa x Aa) marriage (2/90) and the probability that that marriage would produce an affected child (1/4).

Conditional Probability of a CF child due to new information is = (88/90* 0)+ (2/90 *1/4) = 2/360 = 1/180

The risk of child with CF has increased from 1/3600 to 1/180 (a 20 fold increase) as a result of new information.