Calculating probability and risks in pedigree analysis:
The pedigree shows the
occurence 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,
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.
the probability that III-1
is Aa is the probabilityy that
II-2 is heterozyogous and passed
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
calculations in this example involved a distinction between a priori and a posteriori
probability, and are often gotten wrong in elementary genetics
textbooks (Griffiths et al.
2002 on p. 137 is wrong).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,
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 have at least one head will be HT (or TH) 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,
other is H or T with
equal probability. It remains true that, of all two penny combinations
involving at least one head, 2/3 are HT
a priori; in analyzing any particular experiment, one is
asking about that event a posteriori.
Thus, in the above example, we reasoned that since I-1 and I-2 are heterozygotes and II-2 shows the dominant phenotype,
we know a posteriori that the
chance he is heterozygous is 1/2. It is incorrect to reason that, since
2/3 of all unaffected (that is, non-aa)
are heterozygotes a priori,
his risk is also 2/3. Stated another way, by knowing the nature of one
allele, we have lost one statistical degree
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)
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 homozygous (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. With a family of ten
children, the odds are < 0.1% that none would have received the a allele if II-5 were a heterozygote**. In other
words, this is strong a posteriori
evidence that II-5 is a homozygote, which if so 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
calculation to 1/16, as in (4) above.
does not mean that the outcome of previous births change the
probability of the next birth. The occurence of a run of heads does not
alter the probability of the next H
as 1/2. Given that the occurence of
a family with ten AA followed
by one Aa child is extremely
unlikely, the probability that a twelfth child will be Aa remains 1/2.