Explanation Given info: Frequency of heterozygous carriers 2 p (1 − p ) Concept used: Finding maximum or minimum value by making square a 2 + 2 a b + b 2 = ( a + b ) 2 a 2 − 2 a b + b 2 = ( a − b ) 2 Calculation: Let y = 2 p ( 1 − p ) = 2 p − 2 p 2 = − 2 p 2 + 2 p = − 2(p 2 − p) ( taking − 2 as common ) = − 2 ( ( p ) 2 − 2 . ( p ) . 1 2 + ( 1 2 ) 2 − ( 1 2 ) 2 ) (Adjusting to make square) = − 2 ( ( p − 1 2 ) 2 − 1 4 ) ( a 2 -2ab+b 2 = ( a-b ) 2 ) = − 2 ( p − 1 2 ) 2 + 1 2 Square of real expression ≥ 0

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 1.2, Problem 48E

To determine

If the alleles A and B of the cystic fibrosis gene occur in a population with frequencies

(p) and (1−p)(where p is a fraction between 0 and 1) Then the frequency of heterozygous carriers (Carriers with both alleles) is 2p(1−p), which value of p Gives the largest frequency of heterozygous carriers.

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Chapter 1.2, Problem 47E

Chapter 1.2, Problem 49E

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Solution Summary: The author calculates the frequency of heterozygous carriers based on the alleles A and B of the cystic fibrosis gene.

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