The Theory of Population Genetics [ 111 



( Note that the bb adaptive vahie is .40 but the selection coefficient 

 is .60, since 1 — 5 ecnials the adaptive value. ) The maintenance of 

 the equilibrium at B = .545, b = .455 by this selective system is 

 shown in Table 6.9. 



Suppose that a climatic change suddenly increases the rainfall 

 in the area occupied by our hypothetical locusts, encouraging the 

 growth of a mold which is fatal to BB and Bb eggs but to which the 

 bb eggs are relatively immune. The adaptive values W are now 

 BB = .00; Bb = .00; bb = 1.00. The survival of the population now 

 depends entirely on the presence of bb eggs. If a prerain (poly- 

 morphic) adult population of 100 pairs and an average egg produc- 

 tion of 100 eggs per female are assumed, there would be 10,000 

 eggs exposed to the mold. Of these, 2,070 (.207X10,000) would 

 be of the resistant kind, presumably giving the population a reason- 

 able chance of survival. 



On the other hand, if the prerain population had been mono- 

 morphic (all individuals BB, perhaps because of strong selection 

 against Bb and bb individuals), the outcome would almost cer- 

 tainly be different. If a mutation rate B — > Z? of 10~'' is assumed, 

 only one egg in 10 billion would be of the surviving genotype. ( The 

 chance of both members of a pair of alleles being mutant in a single 

 individual is the product of the chances of either one being mutant: 

 10-^ X 10-5 ^ 10-1" ^ 1/10,000,000,000.) The advantage of bal- 

 anced polymorphism to the population is obvious. 



Table 6.9 | Balanced Polymorphism 



Succeeding generations continue this pattern as long as assumptions hold. 

 * Random mating is assumed in the calculation of this row, giving the follow- 

 ing genotype frequencies: B" = .545" = .297; ZBb = 2 ( .545) ( .4,55) = .496; 

 b^ = .4552 = .207. 



f The selection pressure is included by multiplying each genotype frequency 

 by its adaptive value: .297 (.50) = .148; .496 ( 1.00) = .496; .207 (.40) = .083. 

 tB gene frequency is given by (D + Vs H)/N. Half of .496 = .248, thus 

 (.148 + .248)7.727 = .545. Gene frequency oi b is 1 - .545 = .455. 



