# Is a wild-type strain more “fit” than its mutant strains?

## Question's context

In the context of a presentation on mathematical models for antibiotic-resistance bacteria research, someone mentioned that wild type strain (WT) are expected to be more fit than mutant strains.

Assume that there is one WT introduced to an environment E (no drugs present), and consider the WT mutates, yielding a certain number of fit-enough mutant strains.

## Question

Is WT expected to be the ONLY strain to persist in the long run?

How common is this scenario?

Examples and references for beginners will be very appreciated.

My context: Please note that this may be a very naive question, and I may even be using the wrong terminology, given my background on mathematical modeling and a very, very basic knowledge on evolution.

Thinking about dynamical systems, I would have guessed that coexistence would be possible, so I didn't understand how strong was such an assumption.

• There is no such thing as a "wild-type" strain. – user37894 Apr 18 '18 at 9:04

I will assume the mutation is not under some kind of balancing selection such as negative frequency dependent selection. As fitness is conditional upon the environment, I will assume a constant environment (heterogeneous environment can also lead to balancing selection).

Is WT expected to be the ONLY strain to persist in the long run?

Yes. In fact, even if the mutation was beneficial, as it starts out at low frequency it still has a low probability to actually fix (fixation = reaching a frequency of one).

How common is this scenario?

Mutations happen constantly. For example, every new human baby carry something like 10-100 new mutations that were not present in any parent. Most mutations are neutral or deleterious, only very few are beneficial. New mutations have a low probability to reach fixation even if the mutation is beneficial. The exact probability depends upon the selection coefficient $s$, the dominance coefficient $h$ and the effective population size $N_e$.