H-Weezy Problem

 The q² I was given was .17 which means that 17% of the population carry's the recessive trait. The population size was 1000.The steps to get q, p, p ², and 2pq are:

1. Since we are given q² then we can find q and in order to do that you take the square root of .17 which gives you .41 so q=.41
2. Now that we have q we can get p since p+q=1. So what you do next is you do q-1=p, so its .41-1=.59, p=.59
3. Next your able to find p² by squaring p which is .59, so .59²=.35
4. Next you can fin 2pq you're able to do this step before finding p² but I prefer to do after I find p². To find 2pq you do 2×.59×.41=.48;  2pq=.48
5. You can check your work by using the Hardy-Weinberg equation which is                    p²+2pq+q²=1, you would plug in the numbers corresponding the variable and it has to equal 1 and if it doesn't then you did something wrong; .35+.48+.17=1

Now to find out what number of the population that are homozygous recessive individuals, homozygous dominant individuals, and heterozygous individuals you have to: 

1. We know that q² are the homozygous recessive individuals so we do 1000×17%(which is the percent of the population that are homozygous recessive individual) equals 170; so 170 people from the population of 1000 are homozygous recessive individuals.
2. p² are the homozygous dominant individuals so you do the same as in step 1. You do 1000×35%=350; so 350 people in the population are homozygous dominant individuals.
3. 2pq are the heterozygous individuals so then you do the same as in step 1 and 2. You do 1000×48%=480; so 480 people from the population of 1000 are heterozygous individuals
4. You can check your work buy adding all the numbers of the people that have these three trait and it has to equal 1000 since that's the population we started with; 170+350+480=1000