In Northern European populations, the incidence of cystic fibrosis is about 1/2500 or 0.0004. What is the frequency of carriers (heterozygous individuals that have a disease-associated allele, but don’t have the disease) in the population?
Added by Pamela H.
Step 1
0004, we can calculate the recessive allele frequency (q) using the formula q^2 = 0.0004. q = √0.0004 q = 0.02 Show more…
Show all steps
Your feedback will help us improve your experience
Shaiju T and 95 other Biology educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Cystic fibrosis is caused by an autosomal recessive mutation. Assuming H-W equilibrium, if the frequency of individuals with cystic fibrosis is 0.005, what is the expected frequency of carriers in the population?
Anand J.
In a given population, approximately 0.0025 of the individuals are phenotypically suffering cystic fibrosis (a recessive autosomal disorder). What is the frequency of the heterozygous genotype? 95% 9.5 0.9 0.095
Joanna Q.
If the recessive allele for an X-linked recessive disease in humans has a frequency of 0.02 in the population, what proportion of individuals in the population will have the disease? Assume that the population is 50: 50 male:female.
Recommended Textbooks
Biology for AP Courses
Objective Biology for NEET
Introduction to General, Organic and Biochemistry
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD