Figure E.1 illustrates the results of an electrophoretic analysis of 13 CODIS STR markers on a DNA sample and identifies the alleles for each gene. Table E.2 lists the frequencies for alleles of three of the STRs shown in the figure. Use this information to calculate the frequency of the genotype for STR genes FGA, vWA, and D3S1358 given in Figure E.1.

Question

Accepted Answer

Hi, everybody. Let's look at our next question. Suppose for a short tandem repeat str marker tho one ails 15, 16, 17 and 18 have frequencies of 0.0150 0.1340 0.16 and 0.21 respectively. What is the frequency of 15 slash 18 heterozygosity choice A 0.30%. Choice B 0.75%. Choice C 0.99% or choice D 0.63%. So when we're talking about these str markers, the alleles are numbers of repeats. So that's why our alls have numbers 15, 16, 17 80 18. And of course, every individual inherits one ally from each parent. So our question is asking us, what is the frequency of a 15, 18 heterozygous genotype? Well, we can recall from our Hardy Weinberg equations that if the frequency of one ally is P and the frequency of a second all is Q. So these are the all frequencies in a population, then the frequency of the heterozygous genotype of both of these alls is equal to two PQ. So in this case, that would equal two, multiplied by the frequency of our first all. So the frequency of allele 15, which is going to be 0.015 multiplied by the frequency of our second all. So the frequency of all 18 which is 0.21. So we have two multiplied by 0.015 multiplied by 0.21. And that is going to come out to 0.0063. And if we convert that into a percent, we end up with 0.63% as the frequency of that heterozygous genotype. So when we look at our answer choices, that is choice D, so given those all frequencies, the frequency of the 15 18 heterozygote is choice D 0.63%. See you in the next video.

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Key Concepts

Short Tandem Repeats (STRs)

Allele Frequency and Genotype Frequency

Hardy-Weinberg Equilibrium