## Campbell Biology (11th Edition)

Published by Pearson

# Chapter 23 - 23.2 - Concept Check - Page 491: 1

#### Answer

The frequency of allele A is 0.351 The frequency of allele a is 0.649

#### Work Step by Step

For this question, we will need to know two equations for the Hardy Weinberg Equilibrium: $p^{2} + 2pq + q^{2}=1$ $p + q=1$ $p$= the frequency of the A allele, $p^{2}$= the frequency of homozygous AA individuals, $2pq$= the frequency of heterozygous Aa individuals, $q$= the frequency of the a allele, $q^{2}$= the frequency of homozygous aa individuals. Using this information, we will first find the value of $q^{2}$ by finding the proportion of the population that is aa: $q^{2}=\frac{295}{700}=0.4214$ We then use this value to find the frequency of the a allele, $q$ $q=\sqrt 0.4214=0.649$ Finally, we use the equation $p + q=1$ to find the frequency of the A allele, $p$ $p + 0.649=1$ $p =1-0.649=0.351$

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