Biology....Hardy Weinberg equation

Question

1 out of 32 births will have a serious autosomal recessive genetic disease

 

The way I solve the problem

1/32...√q²= √.03125= .18

 

• What is the frequency of children with this disease?  .82
• What is the frequency of the normal allele?  .18
• What percent of the population will be carriers for this recessive disease?

                                2pq=2(.82)(.18)=9.4%

 

 

Answer 1

Pamela,

 

I think you mixed up your answers somewhat.  On the one hand you answer that the frequency of children with this disease is 0.82 (or 4 of 5) whereas the problem states that only 1 of 32 are observed with this disease. Can't be.

 

You calculated q correctly as 0.18  (the abnormal allele).

Since (p+q) = 1 always, then p = 0.82 (the Normal allele)

And then, you have an arithmetic error in 2pq.  It's really = 0.295 or 30%, not 9.4%.

 

BTW, Hardy Weinberg get credit for a simple formula in a population at equilibrium.

Since (p + q) = 1 by definition for all p & q, then (p + q)² = (1)² = 1 

But (p + q)² = p² + 2pg + q²  , so out comes there formula that p² + 2pq + q² = 1.

 

See how easy it is to gain fame, so work at it.