Heh guys,
quick question about genetics:
1Cystic fibrosis in humans is by a recessive allele. The first child of a phenotypically normal couple has the disease. Among all their children, what proportion is expected to have the disease?

The answer is 1/4 ? Why! How do you approach this?

2 When the base compostion of double-stranded DNA from new species of bacteria was determined, 20 % of the base were found to be adenine, What is the percentage of the thymine and the cytosine in the DNA of the organism?

The answer is thymine 20%, cytosine 30% Why?

1. Mother's gene is XXc ( as she is phenotypically normal ) .
Father's gene is XY ( as he is phenotypically normal).

So possible offsprings are XY,XcY,XX,XcX . This give only 1/4 of genotypic abnormality.

2. A binds with T , G with C . So the amount of A must be equal to T and that of G to C. Given A is 20, obviously T is 20. That gives remaining 60 , which must be divided equally so 30, 30.

Hope some one says this is right!!

1. Mother's gene is XXc ( as she is phenotypically normal ) .
Father's gene is XY ( as he is phenotypically normal).

So possible offsprings are XY,XcY,XX,XcX . This give only 1/4 of genotypic abnormality.

2. A binds with T , G with C . So the amount of A must be equal to T and that of G to C. Given A is 20, obviously T is 20. That gives remaining 60 , which must be divided equally so 30, 30.

Hope some one says this is right!!


From what I know, the answers you provided are correct.

1/2 chance of inheriting a mutant allele from a parent.
1/2 X 1/2 chance of inheriting mutant alleles from both parents.
0.5 x 0.5 = 0.25