Question

"Six single nucleotide polymorphism (SNP) loci are known to contribute to a tendency to contract coronary heart disease. The risk of heart disease is high if 7 or more of these alleles are of the ""contributing"" variety. Upper case indicates a contributing allele and lower case a non-contributing allele. A couple wants to know the probability of producing a child who is at high risk of heart disease.Their genotypes were determined by micro-array analysis and are: AA Bb cc DD EE Ff and Aa Bb CC Dd Ee FF. Notice that each of the parents is susceptible to heart disease. What is the probability their first child will have 7 or more total contributing alleles and be susceptible to coronary heart disease?"

		0.7125
		0.6563
		0.8906
		0.2344
		0.3438
		0.5

Answer 1

Answer - The different gametes genotype produced by

Parent 1 - AA Bb cc DD EE Ff are                               

ABcDEF, ABcDEf, AbcDEF,AbcDEf

Parent 2- Aa Bb CC Dd Ee FF are

ABCDEF,ABCDeF,ABCdEF,ABCdeF,AbCDEF,AbCDeF,AbCdEF,AbCdeF,aBCDEF,aBCDeF,aBCdEF,aBCdeF,abCDEF,abCDeF,abCdEF and abCdeF

The offspring - punnett square

As there are 57 offspring from 64 that will have 7 or more total contributing alleles and be susceptible to coronary heart disease.

Thus, the probability their first child will have 7 or more total contributing alleles and be susceptible to coronary heart disease = 57/64

= 0.8960

Hence, the correct option is 0.8906