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Minchanka [31]
3 years ago
6

Cystic fibrosis is a genetic disorder in homozygous recessives that causes death during the teenage years. If 9 in 10,000 newbor

n babies have the disease,
what are the expected frequencies of the dominant (A1) and recessive (A2) alleles according to the Hardy-Weinberg equation?
(a) f(A1) = 0.9997, f(A2) = 0.0003(b) f(A1) = 0.9800, f(A2) = 0.0200(c) f(A1) = 0.9700, f(A2) = 0.0300(d) f(A1) = 0.9604, f(A2) = 0.0392
Biology
1 answer:
padilas [110]3 years ago
5 0

Answer:

f(A1) = 0.9700, f(A2) = 0.0300

Explanation:

The frequency of homozygous recessive genotype of children is  

\frac{9}{10000} \\= 0.0009

As per Hardy weinberg's equation, frequency of homozygous recessive genotype is represented by  

q^2

Here q^2 = 0.0009

Thus, frequency of recessive allele would be  

\sqrt{q^2}\\ = \sqrt{0.0009}\\ = 0.03

as per first law of Hardy weinberg

p+q = 1\\p = 1- 0.03\\p = 0.97

Hence, option C is correct

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