Genetics Prelab If the human population includes 23 variations in the length of

Question

Genetics Prelab

If the human population includes 23 variations in the length of the VNTR that you analyzed (i.e., there are 23 alleles), how many combinations are possible, considering that everyone has two identical chromosomes and thus two of this particular VNTR?

Then, what is the probability that a given individual will have a specific one of those combinations?

2) Explain this statement: “While an analysis of a single VNTR is not sufficient to identify DNA as belonging to an individual, it may exclude someone as a possibility.”

Explanation / Answer

1. 23 different alleles for a given VNTR

Humans are diploid. So, they can contain only two alleles.

= AB

So, there are 23 different alleles for A locus + there are 23 different alleles for B locus

Product rule: Total number of different allele combination = 23 X 23 = 529

For a person to have any one of the given 529 allele combinations, the probability = 1/529 = 0.00189 = 1.89%

2. While an analysis of a single VNTR is not sufficient to identify DNA as belonging to an individual, it may exclude someone as a possibility.

The given statement is true because a person can possess a given VNTR by chance due to his parental genotypic combinations. But, if he does not contain the given VNTR, he can be excluded from the suspect list. A single VNTR may be present in all the suspects. So, we need to use the data from multiple VNTRs to find out the culprit.

If a suspect does not contain a VNTR that is found at the crime scene, he can be excluded from the suspect list. This is because the presence of a VNTR but not the absence can be attributed to chance/probability.

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