430 INTRODUCTION TO EVOLUTION 



parents we note that |^ of the effective sperm cells are produced by MM 

 individuals; these sperm cells will all contain gene M. Two-fourths of the 

 effective sperm cells are produced by Mm individuals; half of these (or ^4 

 of the total number of sperms) will contain gene M. As a result, half the 

 total number of sperm cells will contain M (\^ from MM males plus ^ 

 from Mm males). The other half of the sperm cells will contain gene m (Y^ 

 from Mm males plus ^ from mm males). In the female parents the situa- 

 tion is exactly comparable. Half the ova will contain gene M (y^ derived 

 from MM females plus % from Mm females), and half of the ova will 

 contain gene m (^ from Mm females plus |4 from mm females). 



As a result we can considerably simplify our checkerboard diagram as 

 follows : 



Sperms 



Ova 



The fraction in each square is obtained by multiplying the fraction of ova 

 having the gene in question by the fraction of sperms having it. 



Assembling the results from the chart we find that the offspring occur in 

 the proportions Y^MM, -/\Mm, Y^ff'^m — the same result we obtained with 

 the more complicated diagram. 



We may even go one step further in our simplification. It is not really 

 necessary to separate the sperms from the ova in our thinking. Grouping 

 the two together we have a "gene pool" in which half the genes (regardless 

 of whether they are in sperms or ova) are recessive, m, half are dominant, 

 M. In such a gene pool the equilibrium of Y-^^^> YiMm, y^mm will be 

 maintained as long as random mating occurs (i.e., as long as choice of 

 mates is entirely a matter of chance and hence obeys the mathematical 

 laws of probability). 



An instructive model of such a gene pool is afforded by a box containing 

 red and blue beads (corresponding to gene M and gene m respectively) 

 in equal numbers. If without looking you reach into the box and pick up 

 two beads at a time, you may, obviously, pick up two red ones (MM), a 

 red and a blue one (Mm), or two blue ones (mm). If you do this enough 

 times you will obtain a good approximation of the ratio: Y4 both beads of 

 the pair red, % one bead red and one blue, Yt both beads blue. 



