4 years ago

PLEASE HELP. THANKS

Biology

1 answer:

vitfil [10]4 years ago

4 0

A is the answer. Easy

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Hardy Weinberg Equilibrium formula for allelic frequency is: $p+q=1$

Hardy Weinberg Equilibrium formula for genotypic frequency is: $p^2+q^2+2pq=1$

Here, $p$ =frequency of dominant allele

$q$ =frequency of recessive allele

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here,affected individuals means the one's having Cystic Fibrosis and since it is recessive disease so we can write frequency of recessive genotype as: $q^2$ =0.0004

If $q^2$ =0.0004 $q$ will be 0.02;frequency of recessive allele will be 0.02

Using the formula $p+q=1$

$p+0.02=1\\p=1-0.02=0.98$

hence,frequency of dominant allele will be 0.98

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