Re: The Living Dead

In article <1137099883.858771.24030@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> on
January 12th in talk.origins "Seanpit"
<seanpitnospam@xxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:

> David Wilson wrote:

... [snip] ...

> > For your information, for these models (under various assumptions, some of
> > which I listed in my previous article), the mean population fitnesses
> > _relative to a genotype with no deleterious mutations_ are (approximately):
> >
> > 1) for sexually reproducing diploids in the case when the deleterious
> > mutations are all dominant, e^(-Ud); (this approximation holds even when
> > the dominance is partial, as long as the degrees of dominance of the
> > mutations are not too close to 0)
>
> So, for this scenario, a breeding couple (Ud = 3) would have to produce
> about 40 offspring, 2e^Ud, to keep up - right? ...

Not necessarily. As I tried to emphasize in my preceding article, it does
_not_ immediately follow from the above fact that _every_ breeding pair has
to produce 2*e^Ud offspring to maintain a constant population size. This
conclusion only follows if one makes some extra _extremely unrealistic_
assumptions over and above those which are needed to derive the estimate
of e^(-Ud) for the mean relative fitness.

You really do have to keep in mind that this estimate comes from a naive,
very simplistic genetic model. It merely assumes that fitnesses _relative
to some arbitrarily chosen base_ interact multiplicitavely; it makes no
assumptions whatever about _absolute_ viabilities or fertilities, except
to the extent that they have to combine together in such a way that they
reproduce the correct values for the assumed _relative_ fitnesses.

You appear to be under the misapprehension that unless the number of
offspring per breeding pair is enough to maintain a constant population
size then deleterious mutations will accumulate indefinitely until the
population becomes extinct. _This is simply not true for this model_,
at least not when it's applied to sexually reproducing diploids, the case
I'm discussing at the moment. It _may_ be true _in reality_ for all I
know---I'm a mathematician, not a geneticist---but if the behaviour of
the _model_ departs from _reality_ as much as that, then you can't expect
the estimate, e^(-Ud), of the mean relative fitness you get from it to tell
you anything much at all about what's _really_ happening.

What happens in _the model_ (though maybe not in reality) is that the
distribution of the number of deleterious alleles carried by an individual
will approach an equilibrium in which the losses of deleterious alleles
through selection exactly balance the gains through mutation. This
remains true, _regardless_ of what assumptions one eventually makes about
_absolute_ viabilities and fertilities. These latter assumptions will
determine whether the population increases, decreases, remains constant,
or fluctuates up and down; but _regardless_ of which of these actually
does occur, the distribution of deleterious alleles will still reach an
equilibrium and _remain_ there either indefinitely, or at least until the
population becomes extinct.

Until one actually does make some assumptions about absolute viabilities
and fertilities, one can only say that _if_ these are chosen in such a way
as to maintain or increase the population size, _then_ any mating pair _who
carry no deleterious mutations between the two of them_ _must_ produce at
least 2*e^(Ud) offspring on average who themselves then survive to
reproduce. However, one can't say anything specific about how many
offspring _other_ members of the population will produce _until_ one makes
some more specific assumptions aout their absolute viabilities and
fertilities.

Nachman and Crowell, in the paper of theirs you have previously cited
(Genetics, 136 (Sept 2000), 297-304), say "For U=3, the average fitness
is reduced to 0.05, or put differently, each female would have to produce
40 offspring for 2 to survive and maintain the population at constant
size." The last part of this statement assumes implicitly (and, as far
as I can see, with no justification whatever) that _every_ female has to
produce the _same_ number of offspring on average---i.e. that there is _no_
fertility differential among females. But even _with_ that assumption,
the statement is _still_ not true unless one makes some further extremely
unrealistic assumptions (such as that absolute viabilites are constant,
independent of population size; or that the population size is many many
orders of magnitude larger than the current human population of the earth).

Suppose, for instance, one instead assumes that the absolute _viability_
(rather than the absolute _fertility_) of every individual is the same, V,
say---i.e. selection occurs entirely through differences in fertility.
While this assumption is also unrealistic, I can't see why it is any
more so than the one Nachman and Crowell have implicitly made. At
any rate, if this assumption _is_ made, then it is _only_ matings between
individuals carrying very small numbers of deleterious mutations between
them that will need to produce at least 2*e^U offspring on average for
the population size to remain constant . In fact, under this assumption,
the _mean_ number of offspring per mating pair, taken over the whole
population need only be 2/V to maintain a constant population size.
With V = 2/3, for instance, this gives a mean of only 3 offspring per
mating pair (though matings between individuals with no deleterious
mutations will have to produce 2*e^U/V offspring, on average, or 60
per pair, when U=3).

> ... [snip] ...
> > 2) for sexually reproducing diploids in the case when the deleterious
> > mutations are all completely recessive, e^(-Ud/2);
>
> Now, I'm not exactly understanding why recessiveness would reduce the
> required offspring in this manner. For Ud = 3 the required offspring
> needed to keep up with such a recessive mutation rate would only be
> about 9? I'm not sure I conceptually understand why this would be? ...
>

But do you in fact at all "conceptually understand" why the mean fitness is
e^(-Ud) when the mutations are all dominant? You seemed willing to raise the
value of what you "understood" this to be from e^(-Ud) (in your articles
<http://groups.google.com/group/talk.origins/msg/292ff89cf5511e87> and
<http://groups.google.com/group/talk.origins/msg/40677f70ae8740f2>) to
2*e^(-Ud) (in your article
<http://groups.google.com/group/talk.origins/msg/b9ceb1ffd90d4350>---where
you inappropriately cited it as if it had some relevance to a case of
truncation selection) and then drop it back down to e^(-Ud) in response to
my previous article. While I can sympathise with your difficulty in
understanding why these expressions might be so different, I'm afraid I
can't fathom why you automatically assume you "understand" why e^(-Ud)
should be about right for dominant mutations rather than why e^(-Ud/2) might
be about right for recessive ones.

> .. [snip] ...
> > 5) in the second case, each mating pair has to produce 2*e^(Ud/2)
> > offspring on average;
>
> Again, a more conceptual explanation as to why recessiveness reduces
> the needed reproductive rate would be helpful.

There are two main factors which account for the difference between the
two cases:

1) If the deleterious mutations are recessive, _both_ alleles at a given
locus have to be deleterious for them to affect the fitness of their
carrier, whereas if they're dominant only _one_ of them has to be
deleterious to have an affect. This factor is however partially offset
by the fact that:

2) If the mutations are recessive then their frequency at mutation-
selection equilibrium is much larger than it would be if they were
dominant with the same selection coefficients (the former is in fact the
square-root of the latter).

The upshot of these two factors combined is that when the mutations
are recessive, the frequency of loci at which _both_ alleles are
deleterious is very nearly half that of the loci at which _at least
one_ allele is deleterious when the mutations are dominant with the
same selection coefficients. That is, the frequency of loci contributing
to a reduction in fitness when the mutations are recessive is only half
what it is when they are dominant with the same selection coefficients.
It follows from this that the total fitness in the former case is only
the square root of what it is in the latter, because these fitnesses
are determined by multiplying together the fitness factors for each
locus.

--------------------------------------------------------------------------
David Wilson

SPAMMERS_fingers@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
(Remove underlines and upper case letters to obtain my email address.

.

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