Additional STR allele frequency information can be added to impro...

Question

Additional STR allele frequency information can be added to improve the analysis in Problem 8. The frequency of D8S1179₁₂ = 0.12. The frequency of D16S539₁₈ = 0.08 and of D16S539₂₀ = 0.21. Lastly, D18S51₁₉ = 0.13 and D18S51₂₀ = 0.10. Combine the allele frequency information for these three STR genes with the information used in Problem 8 to calculate the frequency of the genotype for six of the STR genes.

Accepted Answer

Welcome back, everybody. Here's the next problem. Consider the following genotype frequencies for three S T R markers. And we have three different markers listed with their genotype frequencies. And then our question says, how many people are expected to have these genotypes choice? A two people per million B nine people per 1000 C nine people to per 10 million or D five people per 100,000. Well, if we're looking at how many people are expected to have these genotypes, we're talking about the combined probability of inheriting all three of those genotypes, we know they independently assort. So that means that the probability of inheriting all three equals the product of the genotype frequencies of all three. So you're going to multiply the three genotype frequencies together because of course, the genotype frequency is essentially a probability being the number of times that genotype occurs as a percent of the entire population. So if we multiply these three numbers together, 0.12, multiplied by 0.4, multiplied by 0.18. The product of all three of those is equal to 8. times 10 to the minus seventh and what expressed as how many people expected to have these genotypes? That would then be choice C nine people per million. We can see that 10 to the minus seventh leads us to the million as the pool and we have nine people out of those million. So choice C as our answer to how many people are expected to have those three particular genotypes. See you in the next video.

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