Explaining how and explaining why: developmental and evolutionary explanations of dominance

Abstract

There have been two different schools of thought on the evolution of dominance. On the one hand, followers of Wright [Wright S. 1929. Am. Nat. 63: 274–279, Evolution: Selected Papers by Sewall Wright, University of Chicago Press, Chicago; 1934. Am. Nat. 68: 25–53, Evolution: Selected Papers by Sewall Wright, University of Chicago Press, Chicago; Haldane J.B.S. 1930. Am. Nat. 64: 87–90; 1939. J. Genet. 37: 365–374; Kacser H. and Burns J.A. 1981. Genetics 97: 639–666] have defended the view that dominance is a product of non-linearities in gene expression. On the other hand, followers of Fisher [Fisher R.A. 1928a. Am. Nat. 62: 15–126; 1928b. Am. Nat. 62: 571–574; Bürger R. 1983a. Math. Biosci. 67: 125–143; 1983b. J. Math. Biol. 16: 269–280; Wagner G. and Burger R. 1985. J. Theor. Biol. 113: 475–500; Mayo O. and Reinhard B. 1997. Biol. Rev. 72: 97–110] have argued that dominance evolved via selection on modifier genes. Some have called these “physiological” versus “selectionist,” or more recently [Falk R. 2001. Biol. Philos. 16: 285–323], “functional,” versus “structural” explanations of dominance. This paper argues, however, that one need not treat these explanations as exclusive. While one can disagree about the most likely evolutionary explanation of dominance, as Wright and Fisher did, offering a “physiological” or developmental explanation of dominance does not render dominance “epiphenomenal,” nor show that evolutionary considerations are irrelevant to the maintenance of dominance, as some [Kacser H. and Burns J.A. 1981. Genetics 97: 639–666] have argued. Recent work [Gilchrist M.A. and Nijhout H.F. 2001. Genetics 159: 423–432] illustrates how biological explanation is a multi-level task, requiring both a “top-down” approach to understanding how a pattern of inheritance or trait might be maintained in populations, as well as “bottom-up” modeling of the dynamics of gene expression.

Appendix

Mutation Selection Balance (From Maynard Smith 1989)

Consider a large population of size N, where a is the wildtype allele at frequency q, and A is the mutant allele, at frequency p. So:

Genoätypes	aa	Aa	AA
Fitäness	1	1-hs	1-s
Numäber of zygotes	Nq 2	2Npq	Np 2

There are 2Npq Aa zygotes, of which a proportion hs die, eliminating one A gene. Further, there are Np ² AA homozygotes, of which s die in each generation, eliminating two A genes with each death. So, the number of A genes lost by selection in each generation will be

$$ 2N_{pqhs} + 2N_{p{\hbox{2}}s} = 2Nps(qh + p) $$

Further, in each new generation, new A genes arise by mutation. Let the mutation rate be u. Thus, since there are 2Nq a genes in each population, there will be 2Nqu new A genes each generation. At equilibrium, (mutation–selection balance), the number of new mutations will equal the number eliminated. Or:

$$ 2N_{qu} = 2N_{ps} (qh + p) $$

Or,

$$ qu = \,{\hbox{ }}ps(qh + p) $$

If A is fully recessive to a, or h = 0, then qu = p ² s, so, since p approximately = 1

$$ p = \surd (u/s) $$