Calculating probability and risks in pedigree analysis:

Elementary principles

The pedigree
shows the occurrence of an autosomal
recessive
trait, where the black squares have genotype aa. We wish to
calculate the
probability that IV-1
(shown
as ?) will be
either affected (aa),
or a carrier heterozygote (Aa).

(1) For IV-1 to be an affected recessive homozygote, s/he must inherit an a allele from the father**and**** **the
mother. Given that II-1
must be aa, both
great-grandparents (I-1
and I-2) must be Aa. II-2 shows the dominant
phenotype,
and therefore has at least one A
allele: the probability that the other is a is 1/2.
II-3 is from outside
the
affected pedigree and can be assumed to be AA. Like his
father, III-1 shows
the dominant phenotype,
and therefore has at least one A.
Then,
the probability that III-1
is Aa is the
probability that
II-2 is heterozygous **and**** **passed
the
a allele to III-1 : (1/2) x (1/2) =
1/4. The
same reasoning leads to the conclusion that III-2 is heterozygous
with a
probability of 1/4. Thus, for IV-1
to be aa, both
parents must be
Aa, **and**** **they
must both pass the a
allele to their offspring: 1/4 x
1/4 x 1/4 = 1/64

(2) Alternatively, for IV-1 to be a heterozygous, carrier, either s/he most inherit an a allele from the father, or from the mother. The chance of either parent being a heterozygote is 1/4, as calculated above. Then, the probability that both parents are heterozygotes, and the probability that two heterozygotes will have a heterozygous child, is 1/4 x 1/4 x 1/2 = 1/32.

(3) Finally, the probability that IV-1 is a dominant homozygote is 1 - 1/64 - 1/32 = (64 - 1 - 2)/64 = 61/64. This can also be calculated more tediously by summing the alternative probabilities at each of the steps above.

The calculations in this example involved a distinction between a priori and a posteriori probability, which are often presented incorrectly in elementary genetics textbooks. To take a simple case: the a priori probability of getting heads on a single toss of a penny is 1/2, since there are two equal possibilities, H or T. Then, given two pennies tossed at random, HH, HT, TH, and TT are all equally likely. The a priori probability of getting at least one head is**3/4**. The
**a priori**
probability that any
combination with at least one head will have two tails (**HT
**or **TH **vs HH)
is **2/3**.

However, consider an experiment in which I have tossed two pennies. I show you that one is H, and ask, What is the probability that the other is also H? The a posteriori probability is 1/2 : given the knowledge that one coin is H, the other is H or T with equal probability. In anticipation of the experiment, the*a
priori* probability of **HT **given **H-** is
2/3. In analyzing the
results of any particular
experiment, the added information changes
probabilities a
posteriori.

In the above example, we know that I-1 and I-2 are heterozygotes and II-2 shows the dominant phenotype. We therefore know a posteriori that he has inherited a dominant alleles from one parent, and the probability that he will inherit a recessive alleles from the other parent and be heterozygous is 1/2. It is incorrect to reason that, since 2/3 of all unaffected children (that is, all non-aa) are heterozygotes a priori, his individual risk is also 2/3. Stated another way, by knowing the nature of one allele, we have lost one statistical degree of freedom.

[In the simplest case: the probability that the next child of** I-1** and** I-2** will be a
boy is *a priori* 1/2: once the child is born, the
probability *a posterior* is either 0 or 1].

Two further extensions of this idea. For this scenario, assume that a genetic test is available to distinguish AA from Aa, but II-5 is deceased and III-2 will not take the test.

(4) Suppose III-2 has a heterozygous sibling. How does this change the calculation IV-I's risk? This would mean that II-5 must be a heterozygote with a probability of 1, not 1/2 as before. Then, the probability that III-2 is a heterozygote is 1/2, the probability that the father (III-1) is a heterozygote remains 1/4, and the probability that IV-1 is aa is 1/2 x 1/4 x 1/4 = 1/16.

(5) Suppose III-2 has one or more siblings who test as unaffected homozygotes (AA). How does this change the calculation of IV-1's risk? Note that, whereas the birth of a heterozygous sibling proves that the mother (II-5) is a heterozygote, the birth of unaffected homozygous offspring cannot prove that she is a homozygote. However, multiple births of unaffected siblings do decrease the probability that she is a heterozygote, as follows. The probability that a heterozygote will not pass the a allele to an offspring is 1/2. Then, the probability that she will not pass it to either of two offspring is (1/2)(1/2) = 0.5^{2}
= 1/4. The probability that
she will pass it to none of three offspring is 0.5^{3}
= 1/8, to none of four is 1/16, and so on. Less than 0.1% of
all families with ten children would have known with an **a**
alleles, if II-5
were a heterozygote. In other
words, this is strong a
posteriori
evidence that II-5
is a *homozygote*,
which if true means
that IV-1 cannot be affected. Of
course, the
birth of an eleventh child who is Aa
immediately proves that II-5
is heterozygous, and returns IV-1's
risk
calculation to 1/16, as in (4) above..

(1) For IV-1 to be an affected recessive homozygote, s/he must inherit an a allele from the father

(2) Alternatively, for IV-1 to be a heterozygous, carrier, either s/he most inherit an a allele from the father, or from the mother. The chance of either parent being a heterozygote is 1/4, as calculated above. Then, the probability that both parents are heterozygotes, and the probability that two heterozygotes will have a heterozygous child, is 1/4 x 1/4 x 1/2 = 1/32.

(3) Finally, the probability that IV-1 is a dominant homozygote is 1 - 1/64 - 1/32 = (64 - 1 - 2)/64 = 61/64. This can also be calculated more tediously by summing the alternative probabilities at each of the steps above.

The calculations in this example involved a distinction between a priori and a posteriori probability, which are often presented incorrectly in elementary genetics textbooks. To take a simple case: the a priori probability of getting heads on a single toss of a penny is 1/2, since there are two equal possibilities, H or T. Then, given two pennies tossed at random, HH, HT, TH, and TT are all equally likely. The a priori probability of getting at least one head is

However, consider an experiment in which I have tossed two pennies. I show you that one is H, and ask, What is the probability that the other is also H? The a posteriori probability is 1/2 : given the knowledge that one coin is H, the other is H or T with equal probability. In anticipation of the experiment, the

In the above example, we know that I-1 and I-2 are heterozygotes and II-2 shows the dominant phenotype. We therefore know a posteriori that he has inherited a dominant alleles from one parent, and the probability that he will inherit a recessive alleles from the other parent and be heterozygous is 1/2. It is incorrect to reason that, since 2/3 of all unaffected children (that is, all non-aa) are heterozygotes a priori, his individual risk is also 2/3. Stated another way, by knowing the nature of one allele, we have lost one statistical degree of freedom.

[In the simplest case: the probability that the next child of

Two further extensions of this idea. For this scenario, assume that a genetic test is available to distinguish AA from Aa, but II-5 is deceased and III-2 will not take the test.

(4) Suppose III-2 has a heterozygous sibling. How does this change the calculation IV-I's risk? This would mean that II-5 must be a heterozygote with a probability of 1, not 1/2 as before. Then, the probability that III-2 is a heterozygote is 1/2, the probability that the father (III-1) is a heterozygote remains 1/4, and the probability that IV-1 is aa is 1/2 x 1/4 x 1/4 = 1/16.

(5) Suppose III-2 has one or more siblings who test as unaffected homozygotes (AA). How does this change the calculation of IV-1's risk? Note that, whereas the birth of a heterozygous sibling proves that the mother (II-5) is a heterozygote, the birth of unaffected homozygous offspring cannot prove that she is a homozygote. However, multiple births of unaffected siblings do decrease the probability that she is a heterozygote, as follows. The probability that a heterozygote will not pass the a allele to an offspring is 1/2. Then, the probability that she will not pass it to either of two offspring is (1/2)(1/2) = 0.5

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©2014 by Steven M. Carr