c fibrosis is the most common fatal genetically inberited disease in Cucasian in

Question

c fibrosis is the most common fatal genetically inberited disease in Cucasian individuals being carriers for a mutant version of the CFTR gene that causes common cystic fibrosis mutant allele (APS08) is a loss-of-function mutation that results in a non- functional transmembrane transporter normally responsible for Individuals who have cystic fibrosis have consequently have respirafory and pancreatic issues and are typically sterile. 1. Based on the genotype coll cystic fibrosis. The most thickened mucous because of an imbalance in chloride ions and information collected from a random sampling of 3500 people the U.S Caucasian population, calculate the observed genotypic and allelic frequencies for the CFTR gene as well as those expected if the population were in Hardy-Weinberg equilibrium. (2 pts total) F' wild-type, F508 = mutant allele with single codon deletion at position 508 Number of individuals with cach genotype: F 3359 F/AF508 140 a What is the observed frequency ofF Ft individuals in the population, F)? 0.1 pt) b. What is the observed frequency of FYAF508 individuals in the population, fF /AFsd8)? (0.1 pt What is the observed frequency of AF508/AF508 individuals in the population, frAF508) AF508)? (0.1 pt) c. d. What is the observed frequency of F alleles in the population, p? (0.1 pt) What is the observed frequency of AF508 alleles in the population, f(AF508) q? (0.1 pt) e.

Explanation / Answer

Answer

As given in question F+ - wild type allele, sorry i am using phone so can not write delta so using ¤ this symbol plz dont mind

So ok ¤F508- for mutant type allele

Part d) Frequency of allele (F+)- (3359×2+140)/3500×2=0.9797

Part e)

Frequency of allele ¤F508- (1×2+140)/3500×2= 0.0202

Part a)

Total individuals - 3500

So f(F+/F+)- total wild type individuals/total individuals of population=3359/3500=0.9598

Or can calculate this way also

f(F+/F+)= square of allele frequency F+= 0.9797×9797= 9598

Part b) f( F+/¤F508)-140/3500=0.04,

Or 2×0.9797×0.0202=0.04

Part c) f(¤F508/¤F508)- 1/3500= 0.0003

Or

f(¤F508/¤F508)= square of allelic frequency of ¤F508= 0.0202= 0.0004

The expected frequency of wild type individuals(F+/F+)=0.9598×3500=3359

Expected frequency of carrier type= 0.04×3500=140,

Expected frequency of diseased individuals= 0.0003×3500=1

This population is in Hardy-weinberg law.

Thanks

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