Medical Genetics by G. Bradley Schaefer

Author:G. Bradley Schaefer
Language: eng
Format: epub
Publisher: McGraw-Hill Education
Published: 2014-09-04T16:00:00+00:00

Figure 9-7. A pedigree to evaluate as a sample problem (see text).

Now knowing the manner of transmission, we can begin to assign genotypes to some individuals. For example, the first generation parents in the right-hand side of the pedigree (the female I-3 and male I-4) must both be heterozygous since they produce a homozygous recessive daughter. For convenience, let us assign the symbol A for the dominant and a for the recessive alleles (Figure 9-7b). On the left-hand side, a phenotypically normal AA female I-1 and affected aa male I-2 have a phenotypically normal, thus heterozygous, daughter (II-3). In the absence of any conflicting evidence, we always assume that individuals, like male II-4, marrying into the pedigree are genetically normal for the trait. The mating that gives rise to male III-1 is, therefore, Aa × AA, and there is a ½ chance that male III-1 is heterozygous Aa.

Returning again to the right-hand side of the pedigree, let’s consider the genotype of male II-5. In order for the child of interest (IV-1) to be homozygous for the a allele, the allele must be passed on by male II-5. What is his probability of his being heterozygous? The answer is 2/3. This number might initially surprise you (a common error is to say the probability is ½), but the logic is simple. Of the four possible outcomes of a mating between two heterozygous parents, one is eliminated by the pedigree; the male II-5 is not aa since he does not express the recessive trait. Thus, among the remaining three possible outcomes involving phenotypically normal offspring, two are Aa heterozygotes and the third is homozygous normal (AA). Then, if II-5 is heterozygous, there is a 0.5 chance of his passing the recessive allele to his daughter, III-2. For the yet-to-be-born child, IV-1, to show the recessive trait (a ¼ chance if both parents are heterozygotes), then all of these transmission events must have occurred. The overall probability requires applying the product rule.

The product rule applied to probabilities is simply that the likelihood of two or more events occurring together is the product of their individual probabilities. For example, the probability of flipping two nickels and getting a head both times is ½ (the probability of a head the first time) times ½ (the probability of getting a head the second time) = ¼. The other three outcomes are: head + tail; tail + tail; and tail + head. An assumption is, of course, that the events in question are independent.

Each member in a pedigree is the product of an independent fertilization event. The probability of inheriting the mutant a allele from a Aa heterozygote is, therefore, ½. The overall probability of a given outcome can be calculated by multiplying the probabilities of each required step leading to that hypothetical outcome. We can multiply the required steps in any order we wish, as long as all are included in the calculation. Ignoring the certainties (a probability of 1.0) and moving from generation II through

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