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TL;DR

What's the relationship between the following?

  • Mutation rates (expressed as "mutations per year" for a virus like e.g. COVID-19 (SARS-CoV-2). E.g. this article mentions "The rate looks to be about 24 mutations per year"
  • Actual virion replication error rates in a single host cell, measured as e.g. error rates per released virion from the infected cell, i.e. (# of mutated released virions) per infected cell.

Assumptions

I would assume that mutations happen at a single virion-cell level because there are differences in e.g. nucleotides in RNA between:

  • the virion infecting a cell (e.g. ACE2 for SARS-CoV-2)
  • the released virions from that same single cell

If so, wouldn't one argue that:

  • replication errors (at the nucleotide and thus released virion level) can easily happen, potentially billions of times during the infection of a single host organism
  • there is no "single successful" mutation or RNA variant within a host, rather, we have a distribution of RNAs, many of which could be simultaneously transmitted to another host organism?

Further context

I quote from e.g. this paper:

On a per-site level, DNA viruses typically have mutation rates on the order of 10E-8 to 10E-6 substitutions per nucleotide site per cell infection (s/n/c). RNA viruses, however, have higher mutation rates that range between 10E−6 and 10E−4 s/n/c

Again, wouldn't that mean that a single host could 1) easily generate many mutations of the same virus and 2) pass most of those uniquely generated mutations to another host? If so, how come we can simply say "24 mutations (over time) per year"?

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    $\begingroup$ Only those mutations which are themselves viable can be passed on. Most mutations are deleterious — evolutionary dead-ends, which do not reproduce. Thus you get a substitution rate calculated from the population of mutants that can reproduce. $\endgroup$ Commented Mar 18, 2020 at 5:09
  • $\begingroup$ Thanks @AlexReynolds but isn't it fair to say that on any single organism, you would still have a population (distribution) of viral RNAs, and not a single (even if dominant) mutation? $\endgroup$ Commented Mar 18, 2020 at 18:40
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    $\begingroup$ @AlexReynolds That sounds more like an answer than a comment. $\endgroup$
    – Bryan Krause
    Commented Mar 18, 2020 at 19:47
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    $\begingroup$ Hi Amelio, I'd suggest taking a look at the difference between mutation rates and substitution rates, which is a subtle difference in terminology but which may help get a better comment to address your question. Hope this helps! $\endgroup$ Commented Mar 19, 2020 at 4:29
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    $\begingroup$ +1 I did here Thanks @AlexReynolds $\endgroup$ Commented Mar 20, 2020 at 15:17

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Since no one else has done so, just gonna expand on what occurred here in the comments (mostly from @Alex Reynolds):

tl;dr: natural selection exists.

Longer version:

  1. Mutations/replication errors/whatever will happen in the creation of new virus particles within a host cell. There is some distribution of number of mutational events that will occur as a function of the number of new RNA molecules created per cell and the replication error rate. Some subset of these RNA molecules will be packaged into new particles, yielding a new generation of viruses that get a chance to infect. Yes, this is potentially a large number of new mutations per infection event.
  2. Some large proportion of these new particles will contain non-functional RNAs, due to inactivating mutations of one kind or another. Most of the new mutations will be worse at being a virus than the originally infecting virus (their immediate parent). So therefore, they will be outcompeted by non-mutated forms.
  3. Among all those viruses which continue successfully infecting throughout the course of a year, on average we observe 24 substitution mutations in the genome over the course of that year. Every transmission chain included in the group of viruses at the end of the year must have remained viable for infection of hosts. In other words, it has survived natural selection.

In population genetics, this phenomenon is known as "mutation-selection balance"- e.g. new mutations arise in the population at a rate that is limited by the action of natural selection (at equilibrium).

Genotype-phenotype maps

What of course I am leaving unstated is exactly why "...new mutations will be worse at being a virus than the originally infecting virus..."

This is very hard to predict in many cases. In some cases it is very obvious why a mutant is bad at being a virus, if e.g. an essential gene is deleted from the RNA genome. But for many single mutations we don't really understand what their effect will be on virus function, we just observe that they drop out of the population due to negative selection.

If you are interested in this question, I recommend looking into genotype-phenotype maps. I know Jesse Bloom has done some work on influenza in this field, here is one example of such a paper. It is generally extremely laborious to do this kind of work, so relatively little is known about the functional effect of any specific mutation in any specific genome.

Hope it is helpful to have all that written out.

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  • $\begingroup$ Thanks! This is perfect, and confirms exactly what I thought (the terminology here was confusing me so this is great!). Does the literature really use the term "mutation rate" to refer to both: 1) "population genome" mutation rates (for lack of a better term) and 2) "per replicating virion" mutation rates? $\endgroup$ Commented Mar 22, 2020 at 21:45
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    $\begingroup$ There are many different mutation rates in the literature, and each of them has its purposes. For example, there is the "population mutation rate", en.wikipedia.org/wiki/Watterson_estimator, the rate at which new mutations arise at a site in a population. There is also the "mutation rate" which is closer to the "replication error rate", which is the rate at which mistakes/new nucleotide variants occur. There is also what is sometimes called the "substitution rate", handled elsewhere. And more. All have their proper domains, but figuring out which one to use can be confusing. $\endgroup$ Commented Mar 22, 2020 at 22:19

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