Question

# Using the Hardy Weinberg Equilibrium equation, if you are given 25% are recessive, would the total...

Using the Hardy Weinberg Equilibrium equation, if you are given 25% are recessive, would the total amount of recessive alleles be 50% or something else, mostly wanting to know how to do this equation if anything but with real numbers. -Thanks

If 25% is given it is the frequency of the recessive genotype(q2 = 25/100 = 0.25) The frequency of recessive allele can be calculated as

q = √.25 = 0.5 p= 1- 0.5 = 0.5

Let us assume a population of 200 people in which p= 0.5 and q = 0.5

According to Hardy Weinberg equation, genotype frequency can be calculated from allele frequency by using the equation

p2 +q2 +2pq = 1

Where p2 = frequency of homozygous dominant AA = 0.5*0.5 = 0.25

Number of homozygous dominant individuals = 0.25 *200 = 50

q2 = frequency of homozygous recessive aa = 0.5 * 0.5 = 0.25

Number of homozygous recessive individuals = 0.25 * 200 = 50

2pq = frequency of heterozygous Aa= 2*0.5*0.5 = 0.5

Number of heterozygous individual = 0.5 *200 = 100

The allele frequency can be calculated by genotype frequency

p = (2*Number of homozygous dominant individuals + Number of heterozygous individuals)/2* total population

(2*50+100)/2*200 = 200/400 = 0.5

q = (2*Number of homozygous recessive individuals + Number of heterozygous individuals)/2* total population

(2*50+100)/2*200 = 200/400 = 0.5

p+q = 1

(Yes you can say 50% alleles are recessive in the population (0.5 * 100 = 50%), both dominant and recessive alleles have the same frequency in this case 0.5)

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