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 =
1/32.
(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) = 0.52 = 1/4. The probability that she
will pass it to none of three offspring is 0.53 = 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..