Question

Given that : the most common ATP7B mutation in the USA is the ATP7BH1069Qmutation, which has...

Given that : the most common ATP7B mutation in the USA is the ATP7BH1069Qmutation, which has an allele frequency of 0.0026

Using the overall Wilson disease allele frequency calculated (when we assumed that only one mutant allele caused all Wilson disease cases), what proportion of all ATP7B mutations are represented by the ATP7BH1069Qallele in the USA?

Homework Answers

Answer #1

Ans- Wilson disease is an autosomal recessive. Therefore affected individual will have q2 .   ( ATP7B / ATP7B ) Genotype .

* Allele frequency (q) = 0.0026 = 26/10000

Genotypic frequency (q2) = (26/10000)2   = 676/100000000

* Out of every 100000000 individuals , proportion of affected individuals will be = 676 * So out of every 1000 individuals, proportion of affected individuals will be = 676 * 1000. ---------------    100000000

= 0.00676 ~ 0.007

* 0.007 affected individuals per 1000 or 7 individuals per 100,0000 (ten lakes) .        

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