Genetic Load

The reduction in fitness in a population is proportional to the deleterious mutation rate (Haldane 1937). The genetic load for a single gene is

[math]\bar{w}=1-\mu[/math]

So, if the reduction in fitness is independent among genes and multaplicative for n genes it is

[math]\bar{w}=\left(1-\mu\right)^n \approx e^{-n\mu}[/math].

Define the genomic deleterious mutation rate, over all n basepairs in the genome, as

[math]U = n\mu[/math]

then

[math]\bar{w} \approx e^{-U}[/math].

If we have three to eight deleterious mutations per genome per generation (I need to build the argument with Kong et al. 2012 and other references) then

[math]\bar{w} \approx e^{-U} = e^{-3} \approx 0.04979[/math]

and

[math]\bar{w} \approx e^{-U} = e^{-8} \approx 0.00033546 [/math].

At equilibrium the rate of input of new mutations must be equal to the rate of removal by selection. This implies that on average to have two offspring with no additional mutations a couple must on average produce

[math]\frac{2}{e^{-U}} \approx 40\mbox{ to }5962[/math]

offspring per generation.

Notes

Being able to generate a wide range of complexity from a small number of underlying genes may be a way to evolution to deal with some of the problems of genetic load. There are less genetic targets for mutations but the phenotype is very flexible and evolvable (versus many genes to fine tune various details of a phenotype).

http://blog.wolfram.com/2009/03/25/minimum-inventory-maximum-diversity/

https://www.nature.com/news/2003/030402/full/news030331-3.html

http://journals.plos.org/plosone/article?id=10.1371/journal.pone.0029324

https://phys.org/news/2018-09-chaos-inducing-genetic-approach-stymies-antibiotic-resistant.html