Hardy Weinberg Cystic Fibrosis Problem-please help walk me through:)

Question

1 in 2500 US Caucasian newborns have cystic fibrosis. C is the normal allele, dominant over the recessive c. Individuals must be homozygous for the recessive allele to have the disease.

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

Assuming a Hardy-Weinberg Equilibrium, how many newborns would have cystic fibrosis in a population of 10,000 people?

Answer 1

Let’s walk through this step by step, using the Hardy-Weinberg principle to answer your questions.

Step 1: Identify what is given and what is needed

• 1 in 2500 US Caucasian newborns have cystic fibrosis. This represents q², the frequency of the homozygous recessive genotype (cc).

• The total population is 10,000 people, and we are assuming Hardy-Weinberg equilibrium.

Step 2: Calculate the percentage of individuals with cystic fibrosis (q²)

The frequency of individuals with cystic fibrosis is given by q². This is already stated to be 1 in 2500, so:

To express this as a percentage:

So, 0.04% of the population has cystic fibrosis.

Step 3: Calculate the number of newborns with cystic fibrosis in a population of 10,000

Now, assuming Hardy-Weinberg equilibrium, we can use the percentage from Step 2 to calculate how many newborns would have cystic fibrosis in a population of 10,000 people:

Therefore, in a population of 10,000 people, 4 newborns would have cystic fibrosis.