please explain the answer is B 5 loci are known to affect the human utilization

Question

please explain the answer is B

5 loci are known to affect the human utilization of complex carbohydrates. These include the amylases in the saliva, the pancreatic secretions, and secretions of the stomach. Individuals can be tested for these genetic loci and the alleles can be classified as contributing or non-contributing to the digestion of carbohydrates. A person with fewer than 4 contributing alleles at these 5 loci has a real problem with carbohydrate digestion and requires a special diet. A micro-array was used to find the genotype of two parents at these 5 loci. What is the probability g parents will have a child with 3 or fewer contributing alleles? Aa bb Cc dd EExaa BB cc Dd Ee 0.156 0.3125 0.0781 0.25 0.0625

Explanation / Answer

According to the information, individuals having genotype carrying less than 4 contributing sites will have problem with carbodydrate digestion and require special diet. Thus, the genotype of the parents will be contributing towards the genotypes of the offsprings. Accordingly, the genotype of the parents are:

Parent 1: AabbCcddEE

Parent 2: aaBBccDdEe

Thus, it can be clearly seen from the genotypes of the parents that both parents carry recessive allele at 2 loci and thus, the recombinant offsprings too shall have atleast 2 offsprings carrying recessive alleles at 2 loci.

Also, it must be noted that the parents are heterozygous for 2 positions each. Thus, a recombination at these loci will results in atleast 8 offsprings which will be homozygous recessive in nature. Thus, the total number of individuals with more than 1 recessive allelic loci will be given by 2+8=10.

Now it can be seen that there are total 5 alleles associated with this trait so number of phenotypic classes that could be formed will be given by 25 = 32.

Thus, the total frequency of indivudals carring more than 1 (i.e. 3 or more) contributing alleles will be given by:

10/32 = 0.3125.

Thus,choice B is correct.

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