確率の法則

The sum rule calculates the probability of mutually exclusive events by adding their individual probabilities. In genetics, it determines the likelihood of a person inheriting one of several possible genotypes. For example, if a woman's parents are both heterozygous carriers, she has three possible genotypes with equal probabilities. Two result in being a carrier, so her carrier probability is 2/3 (1/3 + 1/3), calculated using the sum rule.

The product rule multiplies the probabilities of multiple independent events to find their combined likelihood, while the sum rule adds probabilities of mutually exclusive events. For inheritance, the product rule applies when both parents must be carriers and both must pass disease alleles to their child. If the mother's carrier probability is 2/3, the father's is 1/120, and the inheritance probability is 1/4, the child's risk is (2/3) × (1/120) × (1/4) ≈ 0.14%.

Theoretical probability is calculated before events occur, predicting the likelihood of outcomes. Empirical probability is based on actual observations after events have happened. A child's calculated 0.14% risk of biotinidase deficiency is theoretical, but if that child actually inherits the disease, the empirical probability becomes 100%. As more pedigrees are studied, theoretical and empirical probabilities converge and align.

Probability laws enable efficient calculations for complex inheritance scenarios where Punnett squares become impractical. A Punnett square for three traits requires 64 possible crosses, making it cumbersome. Probability laws streamline these calculations by using the sum and product rules, allowing geneticists to quickly determine inheritance risks for autosomal recessive diseases like biotinidase deficiency without exhaustive grid construction.

A child's risk of inheriting an autosomal recessive disease depends on whether both parents carry the disease allele. If either parent is not a carrier, the child cannot inherit the disease. For biotinidase deficiency, the mother's carrier probability (2/3) and father's carrier probability (1/120) are multiplied with the inheritance probability (1/4) to calculate the child's overall risk, making parental status critical to accurate risk assessment.

Heterozygous carriers possess one normal allele and one disease allele but do not express the disease phenotype. When both parents are heterozygous (Bb genotype), each has a 50% chance of passing the disease allele to their child. If both parents pass their disease alleles, the child inherits the homozygous recessive genotype (bb) and expresses the disease, such as biotinidase deficiency.

A pedigree shows family history of disease, revealing which relatives are affected or carriers. If an unaffected woman has an affected brother but unaffected parents, both parents must be heterozygous carriers. Using the pedigree and sum rule, the woman's probability of being a carrier can be calculated. Her pedigree eliminates the homozygous recessive genotype, leaving two carrier possibilities out of three equally likely genotypes.