Please show work :) Question 1 Cystic fibrosis (CF) is a homozygous recessive di
ID: 133998 • Letter: P
Question
Please show work :)
Question 1 Cystic fibrosis (CF) is a homozygous recessive disorder that causes the body to produce abnormally thick mucus due to defective or absent chloride ion (CI-) transport channels in cellular membranes. This mucus builds up in the pancreas, lungs, and digestive tract, causing difficulty breathing, poor absorption of nutrients in the intestines, and life-threatening bacterial infections. 1 in every 1,700 children in the U.S. are born with cystic fibrosis (Hint: this is the frequency of the recessive phenotype). Answer the following questions based on this information. a) What are the frequencies of the dominant (C) and recessive (c) cystic fibrosis alleles? b) What are the frequencies of the homozygous dominant, heterozygous, and homozygous recessive genotypes? c) Although 1 in every 1,700 children born in the U.S. have cystic fibrosis, how many of these 1,700 children are carriers of the disease (i.e., heterozygous)? Give your answer as a number of people, not a percentage.Explanation / Answer
According to hardy-Weinberg law
P2+q2+2pq = 1, also p+q = 1, where p is dominant allele frequency and q is recessive allele frequency. p2= frequency of homozygous dominant, q2= frequency of recessive homozygote and 2pq= is frequency of heterozygotes.
a) According to the question patient suffering from cystic fibrosis is 1 in 1700, therefore frequency of recessive allele (q) is (1/1700) *100 = 0.058 and from hardy- Weinberg's law frequency of dominant allele (p)=1-0.058= 0.942.
b) Frequency of homozygous dominant = p2= (0.942)2 = 0.88
frequency of recessive homozygote = q2= (0.058)2 = 0.003
frequency of heterozygotes = 2pq = 2*0.942*0.058 = 0.054
c) frequency of heterozygotes can also be calculated as f(2pq) = (number of heterozygotes/carrier)/ total number of individuals
hence, number of heterozygotes = f(2pq) *total number of individuals
=0.054*1700= 91.8