HUMANGENETICS

for 1st YEAR STUDENTS


MENDELIAN INHERITANCE

X-LINKED RECESSIVE INHERITANCE

Everyone has heard of some X-linked recessive disease even though they are, in general, rare. Hemophilia, Duchenne muscular dystrophy, Becker muscular dystrophy, and Lesch-Nyhan syndrome are relatively rare in most populations, but because of advances in molecular genetics they receive attention in the media. More common traits, such as glucose-6-phosphate dehydrogenase deficience or color blindness, may occur frequently enough in some populations to produce a few affected females. However, their effect on individuals is rarely life threatening and medical intervention is not needed. Pedigree 7 shows one typical inheritance pattern for a rare X-linked recessive disease.

Pedigree 7

In Pedigree 7 above, which of the following females is least likely to be a heterozygote for the rare X-linked recessive gene, III-1, III-3, or III-5? The answer of course is III-3. III-1 and III-5 each have a 1/2 chance of being a carrier but III-3 has almost a 0 chance of being a carrier. Why? Let's look at the Punnett Squares for X-linked recessive inheritance.

    Affected Father's Genotype       Normal Father's Genotype
    XA Y       X Y
Normal
Mother's
Genotype
X XXA XY   Carrier
Mother's
Genotype
XA XAY XAY
X XXA XY   X XX XY
All daughters carriers, all sons normal.   Half of sons affected, half of daughters carriers.

In Pedigree 7, II-2 and II-5 are both carriers, their father was affected and passed on his only X chromosome to his daughters. II-3 cannot be a carrier for two reasons. First, males are either affected or normal, never carriers. Second, he didn't inherit his father's X chromosome. He inherited his father's Y chromosome. III-3 couldn't have been a carrier since neither her father nor her mother had the mutant gene.

What are the hallmarks of X-linked recessive inheritance?

Counseling in X-linked recessive diseases is a bit more complex than it is in autosomal recessive diseases. In X-linked recessive diseases, Bayes theorem, or Bayesian probability must be used to accurately calculate carrier probabilities. In some pedigrees these probabilities change as new information appears. Sometimes we use Bayesian probability without recognizing it. Consider the following:

A patient of yours is getting married and comes to you for counseling. She has a brother with a rare X-linked recessive disease. Her mother's father also had the disease. She wants to know the probability of her being a carrier of the disease and the probability that she will pass the disease to her children. What is your advice?

Being a reasonably good human geneticist, you tell her that her mother was a carrier and that she has a one chance in two of being a carrier, depending upon which of her mother's X chromosomes she inherited. You also explain that if she is a carrier she will pass the affected X to her son one half of the time, but that her daughters will not be affected because they will always get a normal X from their father. Excellent advice!

But now suppose the conditions change. Her first child is a boy and he is affected. Now what is the probability that she is a carrier? Of course this probability changes from 1/2 to 1. That is the basis for Bayesian probability. The individual undergoing counseling has an a priori calculated probability of being a carrier (not certainty) and later developments alter this probability.

Now suppose this patient who came to you for counseling had a normal son, not an affected son. Shouldn't her probability of being a carrier be changed? Yes it should, but it is more complex than moving from 1/2 to 1 as it did with the affected son. The following is an example of how probabilities change with normal sons.

  Probability of
  Carrier Non-Carrier
At Birth 1/2 1/2

At birth she had an equal chance of being a carrier or a non-carrier, her mother was a known carrier.

What is the probability of 1 normal son, given that she is a carrier? Answer: 1/2

What is the probability of 1 normal son, given that she is a non-carrier? Answer: 1

  Probability of
  Carrier Non-Carrier
At Birth 1/2 1/2
One Normal Son 1/2 1

The joint probability of both events happening is the product of each separate event. The probability of being a carrier and also having one normal son is 1/2 x 1/2. The probability of being a non-carrier and having a normal son is 1/2 x 1.

  Probability of
  Carrier Non-Carrier
At Birth 1/2 1/2
One Normal Son 1/2 1
Joint Probability 1/4 1/2

Since your patient is either a carrier or a non-carrier, we need to calculate the relative probability of each possibility. To do this we must make the joint probabilities add to 1. This is done by dividing each joint probability by the sum of the joint probabilities.

  Probability of
  Carrier Non-Carrier
At Birth 1/2 1/2
One Normal Son 1/2 1
Joint Probability 1/4 1/2
Relative Probability (1/4)/(1/4+1/2)=1/3 (1/2)/(1/4+1/2)=2/3

The relative probability of your patient being a carrier after the birth of one normal son is 1/3 and the relative probability of her not being a carrier is now 2/3. Just as the probability changed after the birth of an affected son, it also changes after the birth of a normal son.

Now, if she has another son and he is normal, her probability of being a carrier and producing two normal sons would be 1/2 x 1/2 x 1/2. The joint probability is now 1/8 of being a carrier. The joint probability of not being a carrier would be 1/2 x 1 x 1. It would remain at 1/2. The relative probability of being a carrier would drop to 1/5. The relative probability of not being a carrier would increase to 4/5.

However, if she were to have a third son and he were affected, all of this Bayesian probability calculation would not be necessary. She would be a known carrier. If she were a non-carrier, her affected son would have to be a new mutation. This is not an impossibility for X linked recessive diseases because it only requires one mutational event to be expressed in males, but when it is factored in the Bayesian probability tables above as a probability of 10-5, it shifts the probability of being a carrier to almost certainty.

Pedigree 8 shows a family with a rare X-linked recessive trait. For which of the following females should Bayesian probability be used to calculate the carrier probability, II-2, II-5, III-2, III-5, or III-7? II-2 and II-5 are both certain carriers, their father was affected and they produced affected sons. III-2 and III-5 each had a 1/2 chance of being carriers at birth. One of them has one normal son, the other has had two normal sons. These are Bayesian probabilities. III-7 is related to the affected individuals of the family through her father. She has an almost 0 chance of being a carrier.

Pedigree 9

Pedigree 9 demonstrates the one other common use of Bayesian probability in counseling. Two requirements must be met. The disease must be a genetic lethal, that is, individuals with the disease cannot reproduce. An example of such a disease is Duchenne muscular dystrophy. The proband must be the only affected individual in the family. When these two conditions are met, as they are in Pedigree 9, there is a reasonable possibility that the affected individual, III-1, is a new mutation.

What is the probability that any female in the population where there is no history of the disease is a carrier of the Duchenne muscular dystrophy gene? She could be a new mutation. If the mutation rate is u, her probability of being a new mutation would be 2u because she has two X chromosomes and each could be mutated. Her mother could have been a new mutation (2u) with a 1/2 chance of passing it to her daughter. Her grandmother could have been the new mutation (2u), with a 1/4 chance of passing the mutant allele to her granddaughter. (Note that the father has no part in the equation since he is normal - affected males cannot reproduce.) Her great-grandmother could have been the new mutation (2u) with a 1/8 chance of passing the mutation to her great-granddaughter, etc. Since these are all independent ways to produce a carrier female, the probabilities sum:

2u + 1/2(2u) + 1/4(2u) + 1/8(2u) + 1/16(2u) + ... = 4u

A female with no history of the disease in her family has a 4u chance of being a carrier before her first son is born. She has a 1-4u chance of not being a carrier. This is close enough to 1 for the purposes of our Bayesian calculations. The Bayesian table for II-2 in Pedigree 9 looks like this:

  Probability of
  Carrier Non-Carrier
At Birth 4u 1-4u=1
Probability of one affected child 1/2 u
(III-1 was a new mutation)
(III-1)    
Joint Probability 2u u
Relative Probability 2/3 1/3

Even though II-3 has produced an affected son, under the above two conditions she still has only a 2/3 chance of being a carrier. 1/3 of all Duchenne muscular dystrophy children born into families with no history of the disease must be new mutations. This is true only of X-linked recessive diseases where the affected males cannot reproduce.


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