fh Topic 18 Quiz x\\e 2)A Person ls Genotype C Secure ! https://moodle.oakland.e
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fh Topic 18 Quiz xe 2)A Person ls Genotype C Secure ! https://moodle.oakland.edu/mod/quiz/review.php?attempt:2796287 My Courses Students Faculty Faculty- Get Help a -Felicia Robinson d. Neither possible father can be excluded. Question 5 Incorrect 0.0 points out of 1.0 P Flag question A person is genotyped for the STR marker D3S1358 and the result is tht their genotype for this marker is 1216, i.e. they are heterozygous for alleles 12 and 16. Using the allele frequency data in the table below, calculate the probability of a randomly selected person having this genotype. Round your answer to the nearest ten-thousandth. four decimal places, in other words if you calculated 0.001192, you would enter 0.0012 in the answer box.) Locus Allele Frequency D3S1358 12 D3S1358 13 0.01 D3S1358 14 D3S1358 15 D3S135816 D3S1358 17 016 D3S1358 18 10 1616 D3S1358 19 015 015 1341 0.2896 0.2287 1616 0152 Answer 0.2387 Question 6 Incomect 0.0 points out of 1.0 P Flag question The image below represents a gel showing the PCR products for a the results of analysis of a DNA marker in a patemity dispute. The bars represent the bands that show up in the gel. To the right are the population allele frequencies, with A, indicating the allele represented by the lowest band. For this marker, what is the Paternity Index (PI)? Round your answer to the nearest tenth Alleged Father Mom Child De 7:41 PM 7/25/2017 3 Type here to search ^4 725 2017Explanation / Answer
5) Probability of selecting allele 12 is 0.015
Probability of selecting allele 16 is 0.2287
Therefore, the probability of selecting both the alleles 12 and 16 are,
= 0.015 x 0.2287
= 0.0034
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