Question

Suppose you test a population of poison ivy (my favorite plant!) and find that out of 800 plants 32 of them are not poisonous, 768 are poisonous. What are the gene frequencies for the poisonous and non-poisonous alleles?

Answer 1

Total number of Poison Ivy population =800

Total number of poison Ivy which are not poisonous= 32

Total number of Poison ivy which are poisonous=768

Let p be the dominant allele ie. poisonous and q be the recessive allele ie. non-poisonous allele.

by Hardy Weinberg equation,

p² + 2pq + q² =1, p+q=1 where p² = percentage of homozygous dominant individual

q² = percentage of homozygous recessive individual

2pq= percentage of heterozygous individual

p= frequency of dominant allele in the population and   

q=frequency of recessive allele in the population.

The percentage of prevelence of non-poisonous allele aa = (32/800)X 100

=4%

Which means that q²=0.04 ( by definition)

If q² is 0.04 then q= 0.2 ( by taking square root)

So by using the above formula,

p+ q=1

p= 1-q

p= 1-0.2

p= 0.8

In conclusion, the gene frequency for poisonous allele is 0.8 and the gene frequency of non-poisonous allele is 0.2.