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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 greatgrandparents (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
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(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.
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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.
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Probabilities in pedigrees are calculated using the same basic, methods as are used in other fields. The first formula is the product, rule: the joint probability of two independent events is the product of, their individual probabilities; this is the probability of one event AND, another event occurring. For example, the probability of a rolling a “five”, with a single throw of a single six-sided die is 1/6, and the probability, of rolling “five” in each of three successive rolls is 1/6 x 1/6 x 1/6 =, 1/216. The second useful formula is the sum rule, which states that, the combined probability of two independent events is the sum of their, individual probabilities. This is the probability of one event OR another, event occurring. For example, the probability of rolling a five or six in a, single throw of a dice is 1/6 + 1/6 = 1/3.
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Assuming the disease has an AR pattern of inheritance, what is the, probability that individual 14 will be affected? We can assume that, individuals #1, #2, #3 and #4 are heterozygotes (Aa), because they, each had at least one affected (aa) child, but they are not affected, themselves. This means that there is a 2/3 chance that individual #6 is, also Aa. This is because according to Mendelian inheritance, when two, heterozygotes mate, there is a 1:2:1 distribution of genotypes AA:Aa:aa., However, because #6 is unaffected, he can’t be aa, so he is either Aa or, AA, but the probability of him being Aa is twice as likely as AA. By the, same reasoning, there is likewise a 2/3 chance that #9 is a, heterozygous carrier of the disease allele.
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If individual 6 is a heterozygous for the disease allele, then there is a, ½ chance that #12 will also be a heterozygote (i.e. if the mating of, #6 and #7 is Aa × AA, half of the progeny will be Aa; we are also, assuming that #7, who is unrelated, does not carry any disease, alleles). Therefore, the combined probability that #12 is also a, heterozygote is 2/3 x 1/2 = 1/3. This reasoning also applies to, individual #13, i.e. there is a 1/3 probability that he is a heterozygote, for the disease. Thus, the overall probability that both individual #12, and #13 are heterozygous, and that a particular offspring of theirs, will be homozygous for the disease alleles is 1/3 x 1/3 x 1/4 = 1/36.
