Hardy-Weinberg Problem 1 (step by step)
p2 + 2pq + q2 = 1
p + q = 1

Problem: 1 in 1700 US Caucasian newborns have cystic fibrosis. C for normal is dominant over c for cystic fibrosis. Instructions

1. When counting the phenotypes in a population why is cc the most significant? Answer

2. What percent of the above population have cystic fibrosis (cc or q2)? Answer

3. From the above numbers you should be able to calculate the expectant frequencies of all the following (assuming a Hardy-Weinberg equilibirum):


Why calculate "q" first? Answer
()c = q = ? Answer

Why is it now easy to find "p"? Answer
()C = p = ? Answer

5. Now that you know that p =.976 and q = .024. The following genotypes can be found.


()CC- Normal homozygous dominant = p2 = ? Answer
()Cc -carriers of cystic fibrosis = 2pq = ? Answer

6. How many of the 1700 of the population are homozygous Normal? Answer
7. How many of the 1700 in the population are heterozygous (carrier)? Answer

8. It has been found that a carrier is better able to survive diseases with severe diarrhea. What would happen to the frequency of the "c" if there was a epidemic of cholera or other type of diarrhea producing disease?
Would "c" Increase? Answer or would it Decrease? Answer

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