Re: Reproductive Selection

On Dec 17, 6:55 pm, Treus <treusd...@xxxxxxxxx> wrote:
John Harshman wrote:
Treus wrote:

leland.mcin...@xxxxxxxxx wrote:

On Dec 16, 7:33 pm, Treus <treusd...@xxxxxxxxx> wrote:

<snip>

The two allele model does not sufficiently account for the complete
set of traits uniquely necessary to the RC of a population arbitrarily
different from whatever ancestor is taken as the starting point.

Presuming I understand your jargon right, yes, it does. You can
simply apply the two allele model repeatedly at different loci to
generate as large a difference as you like. Equally, once the
reproductive incompatability has been introduced (via, for example,
the two allele model) there's nothing keeping the populations from
diverging drastically by drift, and/or environmental and sexual
selection -- which is to say diverging arbitrarily.

Okay so far, but how are you retaining RC within the respective
populations over time without reproductive selection? Without the
(selection driven) co-evolution of R(Sn,x), where R(Sn,Sn) is at least
equal to R(Sn-1,Sn-1), for every Sn, you have a slow degradation of RC
in each successive generation.

Wrong question. The right question is why we would expect RC not to be
retained,

Absent reproductive selection, RC is not being selected for, on the
other hand with reproductive selection, backward compatibility is
maintained. If you want to simultaneously retain RC and decrease
backward compatibility, then some other "force" is necessary.

But backward compatability only needs to be with respect to the last k
generations (for some value of k) since compatability with ancestors
further back than that is simply compatability with dead critters who
are not potential mates; there is no reason to keep such
compatability. Seeing as you seem to like couching things in
mathematical notation, and mathematics is a more natural domain of
discourse for me, let's put this in math terms.

Let r(x) denote the reproductive "type" of an individual x: something
which for all intents and purposes can be assumed to take values in
some continuous range such that individuals x and y are reproductively
compatible if and only if their difference in type is below some
threshold; that is, if and only if |r(x) - r(y)| < epsilon. Now, lets
consider an ancestral line varying through time such that critter x_t
was born at time t. Reproductive compatability, as a selection
criteria, is local: you need only be able to mate with those who are
close in time to you (you can't mate with critters that are dead, nor
those that aren't born yet, s there is no pressure to be compatible
with them. Putting that into math terms, we are saying that if |t1 -
t2| < delta, then we require that |r(x_t1) - r(x_t2)| < epsilon. A
short glance at that and you see that as long as r is a continuous
function (with respect to time) we can meet that requirement (for any
value of epsilon you choose no less; and we can simply rescale things
as required to deal with any apparent discrete granularity in
generations or "reproductive type"). You seem to be arguing that
continuous functions are constant, or, at least, bounded: you are
claiming that there exists some M such that |r(x_0) - r(x_t)| < M for
all t. But that's patently nonsense; continuous functions need not be
bounded functions.

Does that help? It certainly makes the whole thing clear to me, but
then I am a mathematician. Locally (in time) reproductive type always
falls within a narrow range, thus the existent population at any given
time is all quite compatible with one another. That is only a
continuity requirement, not a bounding requirement; globally the
reproductive type can vary over an arbitrarily large range (as long as
continuity, which is a local property, is maintained).

and why we would respect a slow degradation of RC in each
successive generation without some active factor preventing it. If you
remember the 2-locus model, there is in fact no degradation of RC at all
between successive generations, yet we end up with total incompatibility
after a number of generations, and all without any mention of
reproductive selection.

You were right about that, however, without reproductive selection, RC
within successive generations will decrease to zero over
macroevolution. This is because the physical prerequisites (whatever
they may be, positive or negative) uniquely necessary to the RC of
each generation are not being selected for.

But reproductive compatability is a local force, not a global one. It
keeps compatability with regard to the current cohort, not in general,
and global divergence/change is quite possible under that schema.
Let's go back to the model from the previous thread, and couch it in
your new longo and symbology.

We have source population A_0, with successive populations A_i for i =
1...n; We have, as before, mutations X1 and X2 such that X1 confers no
reproductive incompatability, but X2 is incompatible with individuals
not carrying X1. The two allele model neatly bears out such an
arrangement; we can also make far more complex (and less "fragile")
arrangements that will amount to the same thing.

Now, one member of A_1 is born with mutation X1. This confers no
incompatability so R(A_0,A_1) = R(A_0,A_0) = R(A_1,A_1). Over
successive generations individuals that happen to have X1 are more
successful for whatever reason, and each successive generation has a
greater percentage that carries X1. This goes on for m generations to
the point where the entire A_m all carry X1. Now since X1 confers no
incompatability, we still have R(A_0,A_m) = R(A_0,A_0) = R(A_m,A_m).
One member of A_{m+1} is born with mutation X2. Now since this doesn't
present a problem if the partner has X1, we have R(A_0,A_0) =
R(A_m,A_{m+1}) = R(A_{m+1},A_{m+1}). On the other hand, the individual
with X2 is incompatible with earlier generations -- the fewer the
number of carriers of X1 in the population, the less compatible A_{m
+1} is with that population. That is,
R(A_m,A_{m+1}) > R(A_{m-1},A_{m+1}) > R(A_{m-2},A_{m+1}) > ... >
R(A_0,A_{m+1}) >> 0
with the last "strictly greater than zero" coming from the fact that
most of A_{m+1} doesn't carry X2 and so is still compatile with A_0.
The important point, however, is that R(A_{m+1},A_{m+1}) = R(A_0,A_0):
A_{m+1} is just as reproductively compatilble amongst itself as A_0
was, so there's nothing to be selected against here. Now, over another
n successive generations individuals with X2 are, for whatever reason,
more successful to the point where A_{m+n} is entirely composed of
individuals that carry X2. This can certainly happen, as X2 confers no
incompatability with any population back to A_m, and since these
represent the only populations that will ever be bred with, there is
no selection pressure against X2 -- it is completely neutral: R(A_{m
+i},A_{m+j}) = R(A_0,A_0) for all i,j > 0. Of course with each
successive generation, each with a higher percentage of X2 carriers,
the compatability with A_0 will decrease; that is
R(A_0,A_{m+1}) > R(A_0,A_{m+2}) > ... > R(A_0,A_{m+n}) = 0
but that isn't a problem because those populations will never need to
interbreed; R(A_{m+n},A_{m+n}) = R(A_0,A_0) as required, and all is
well or A_{m+n} continuing to have a long lineage into the future,
despite the fact that we have introduced an incompatability. There was
no need to "generate compatability"; it was simply there with each
successive generation. Rather, we simply introduced incompatability
with ancient long dead generations that can no longer exert any
selective pressure.

I have to admit to be a little frustrated with the fact that the old
example from the old thread perfectly adequately counters your "new"
claims, but that you can't apparently recognise that unless I dress it
up in drag (your new terminology and symbology). To me this says that
you still don't have a clear idea in your own head of what you think
your claims really are, nor what the objections actually are; you are
mired in attempting to formalise things, rather than actually
understand the root issues.

Consider, again, the other model, where "incompatability" at each
generation is small (within the average variation of the population)
but, by progressive increments over many generations, large
incompatbilities can be introduced. This is the "great dane can't
breed with chihuahua" model. Such divergence, with small incremental
steps, doesn't have any clear bounding on it, thus, given enough time
for enough small increments, incompatabilities can be arbitrarily
large. What you really need to demonstrate is some reason why there is
a finite upper bound on how many small increments can accrue. You
haven't done that.

Sure I have, and just did so again in this very post.

No, you have merely claimed that there is such a bound. Notice that with
the great dane/chihuahua example there is absolutely no incompatibility
between successive generations, yet we end up with major incompatibility
after many generations.

Setting aside "absolutely no incompatibility" for now, how do you know
a great dane and chihuahua are reproductively incompatible?

Explain to me how they will successfully mate without active human
intervention to allow it? The actual act of mating seems to be beyond
what they can physically manage; in other words, their reproductive
systems are physically incompatible.

.

Relevant Pages

  • Re: Reproductive Selection
    ... But backward compatability only needs to be with respect to the last k ... successive generation without some active factor preventing it. ... reproductive incompatability, but X2 is incompatible with individuals ... alive today that require RC in order to reproduce can be traced as far ...
    (talk.origins)
  • Re: Memory
    ... NO he doesn't need more RAM; ... It -might- be an incompatability or just a bad ... install. ... Look up Compatability Mode; might make it work ...
    (microsoft.public.windowsxp.basics)