Question

Fill in the blank with a whole number (no decimals). The chance that somebody would be heterozygous for the two most common alleles at both the D13S317 and D16S539 loci is approximately 1 in ___.

a. 16
b. 32
c. 64
d. 128

Answer 1

The chance that someone is heterozygous for both the D13S317 and D16S539 loci is 1 in 4, which is answer choice (a). This is found by using the Mendelian inheritance probabilities, assuming no specific allele frequencies are given.

Step-by-step explanation:

The question asks for the chance that somebody would be heterozygous for the two most common alleles at both the D13S317 and D16S539 loci. To determine this, we can use the product rule of probability. For one locus, the chance of being heterozygous (having one dominant and one recessive allele) is 2pq, where 'p' is the frequency of the dominant allele and 'q' is the frequency of the recessive allele.

Since the problem does not provide specific frequencies, we can use the standard Mendelian inheritance for a single trait, which gives us a 3/4 chance of a dominant phenotype (either homozygous dominant or heterozygous) and a 1/4 chance of a homozygous recessive phenotype. The chance of being heterozygous for one locus is therefore 2 * (1/2 * 1/2) = 1/2.

For two loci, we multiply the probabilities for each locus. So, (1/2) for the first locus times (1/2) for the second locus gives us (1/2 * 1/2) = 1/4, which simplifies to a 1 in 4 chance for being heterozygous at both loci. Therefore, the correct answer to fill in the blank is a. 16.