12.8
합의 법칙(sum rule)과 곱의 법칙(product rule)을 사용하여 형질을 물려받을 확률을 계산할 수 있습니다. 합의 법칙은 상호 배타적(mutually exclusive) 사건의 확률을 계산하는 데 사용됩니다. 곱의 법칙은 여러 독립된 사건의 확률을 예…
- [강사] 확률에 대한 합과 곱의 법칙은누군가의 특성의 표출에 대한 가능성을 결정하는 데 사용됩니다.예를 들어, 추정 여성의 가계도에서바이오티니다제 결핍과 같은 질병이 나타나며그녀와 배우자가 해당될 지 모르는열성 바이오티니다제 결핍 대립 유전자 보균자의 확률은해당 장애에 대한 자녀의 위험 정도를 결정합니다.이 여성의 가계도에서 한 형제가 질병에 걸린 것을 볼 수 있는데영향이 없는 부모는 이종접합이 분명합니다퍼니트스퀘어는 이 여성이 동형접합으로정상 대립 유전자를 보유하거나 유전으로둘 중 한 부모가 질병 대립 유전자 전했다는 것을 보여줍니다동형접합의 열성 선택권은 무시되는데그녀가 보균자가 될 수 있는 방법이 두 가지이기 때문입니다양쪽 모두 3분의 1 확률입니다그 확률의 합은이 여성의 이질접합 가능성이며, 합의 법칙입니다반대로, 아버지의 확률은 120분의 1입니다임의의 개인에 대한 경우로서 가능성은이형접합입니다추정 부모 둘 다 보균자로서확률이 3분의 2와 120분의 1이며병든 대립 유전자의 전달은4 분의 1의 확률입니다따라서, 자녀가 바이오티니다제 결핍이 생길 가능성은이 확률의 결과입니다대략 0.14 %로서곱의 법칙입니다이는 계산된 이론적 확률이지만일부 커플에게는 자녀만 해당되어바이오티니다제 결핍이 생길 수 있고실증적 확률은 100%로 관측됩니다그러나, 다수의 가계도를 연구한다면이러한 확률은 적용됩니다
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Q1: What is the sum rule of probability and when is it used in genetics?
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.
Q2: How does the product rule differ from the sum rule in probability calculations?
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%.
Q3: What is the difference between theoretical and empirical probability?
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.
Q4: How do probability laws improve genetic analysis compared to Punnett squares?
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.
Q5: Why is determining parental carrier status essential for calculating child disease risk?
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.
Q6: What does it mean when both parents are heterozygous carriers of a recessive allele?
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.
Q7: How does a pedigree help determine whether someone is likely to be a carrier?
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.