H-Weezy
Problem:
Population: 1000
q2= .47 (Given)
q=?
p2=?
p=?
2pq=?
Steps:
1. The given was .47 which goes into the q2 value.
2. You then square root .47 to get q by itself, which results is .69
3. After you find q, you can find p. The way to do that is by finding out what number can be added to .69 to find 1. (p+q=1) Therefore it will be .31+.69=1
4. Once you figure out what p (.31) is, you need to find p2. To do that, you need to square the p variable which will result to .10.
5. To figure out 2pq, you need to substitute the p(.31) and the q(.69) and multiply 2(.31)(.69)=.43
5. To figure out 2pq, you need to substitute the p(.31) and the q(.69) and multiply 2(.31)(.69)=.43
Problem Solved:
Population: 1000
q2: .47 (homozygous recessive individual)
q: .69 (recessive allele frequency)
p2: .10 (homozygous dominant individual)
p: .31 (dominant allele frequency)
2pq: .43 (heterozygous individual)
Details:
To figure out the number of people for each variable, you need to multiply the variable x 1000. The first one is going to be .47(q2) x 1000=470. This represents that there are 470 individuals in the population that are homozygous recessive individuals. The next one is going to be .10(p2) x 1000= 100. That means that there are 100 homozygous dominant individuals in the population. The final one is to find the heterozygous individuals. To do that, .43(2pq) x 1000=430, which means there are 430 heterozygous individuals in the population.
