Problem S-12


(To do BEFORE class on Friday, Nov. 11, and bring to class.)

This is a simple algebra and graphing problem to begin our study of population genetics.

Back in Part I of the course, we used "p" and "q" as the probabilities of two mutually exclusive events that add up to 1 (100%). We used p=q=1/2 in the context of gender (girl/boy) calculations, and we used p=3/4, q=1/4 for calculating dominant/recessive phenotype probabilities in specific-size families from two parents who are both heterozygous for a simple Mendelian gene alpha.

Now we are going to look at LARGE POPULATIONS rather than individuals, and we are going to find relevance to the situations of p and q being any values between zero and one, as long as p+q=1.

Starting from p+q=1,
squaring both sides gives ...... (p+q) squared = 1,
which is          p squared + 2 p q + q squared = 1

For all of the possible values of p and q, in tenths, calculate each individual term in the binomial equation. That is, start with {p=1, q=0}, and you get p squared = 1, 2pq=0, and q squared =0. Now do the same for {p=0.9, q=0.1}, then for {p=0.8, q=0.2}, etc., down to {p=0, q=1}.

For graphing, make the horizontal axis be p (starting at zero and going up to 1 in tenths) {these same horizontal axis points should also be labelled as q, starting at 1 and going DOWN to zero in tenths}. Make the vertical axis be the calculated numbers for p squared, 2pq, and q squared separately. Connect the points to give smooth curves for each of the three terms. By the nature of the basic algebra, the graph should have left-right symmetry and, at any point along the horizonal axis, the three vertical values should add up to one.

Solve algebraically for the exact values of p and q at points where the 2pq curve crosses another curve.

Bring your graph to class with you on Friday, November 11.