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HYB: Chromosome Counts

From: Sharon McAllister <73372.1745@compuserve.com>

Linda Mann wrote:

>  Very wild.  What's the deal with all these dudes having an extra
>  chromosome or two or being short one?  Several famous ones seem to be
>  that way.  Are there implications for inheritance of traits or does it
>  mostly just make for sterility problems?

and  Donald Eaves wrote:

>  Questions here.  If 48 is the normal chromosone count and PURISSIMA has
>  47 but shows up in several pedigrees, LADY MOHR has 45 and is known to
>  ave very limited fertility, COPPER LUSTRE has 49 - any fertility here? 
>  there a guideline for using a count such as this to increase the chances
>  for
>  setting seeds?  Can you match equal counts and expect an increased
>  of takes?  If the count is only off plus or minus one chromosone, is
>  still
>  a good likelihood of setting seeds?  Do your chances decrease to very
>  unlikely
>  as you move the parent's count away from each other?

I think all of this will make more sense if we start with some background. 

Today, we tend to think of TBs as an artificial race of autotetraploids
[n=12].  It's a convenient model that works for most purposes, but we must
remember that it is just an approximation.  Reality is much more
complicated, especially when we go back to the days of the
diploid-to-tetraploid conversion.  

The vast majority of iris species are autoploids.  [For the moment, I'm
going to ignore the exceptions.]  This means that whether the species is
diploid or tetraploid, its various sets of chromosomes are homologous.  NOT
that they carry identical DNA, merely that they have the same structure,
can be identified and numbered, and that chromosomes bearing the same
number can pair on meiosis.  

Different species often have some homologous pairs, and the more closely
related ones have more pairs in common.  This is not only a indication of
evolutionary relationships, but of breeding compatibility.  It is the
behavior of homologous and non-homologous pairs at the time of meiosis that
brings both breeding potential and sterility barriers.

Looking at the two extremes:

1.      Some species are so closely allied that most of their pairs are
homologous and an inter-species cross is as easy to make as an
intra-species cross.  [In the case of the oncos, it's is actually much
easier because of the inavailibility of  adequate breeding stock.] 

2.      In true amphidiploids, the ancestral species have no homologous
chromosomes so that in each generation a complete set of each type is
transmitted from each parent to each offspring.

Both of these extremes result in completely fertile offspring.  Most
interspecies crosses, however, fall somewhere between the two extremes and
-- ironically -- have reduced fertility.  The two ancestral types have some
homologous chromosomes, some non-homologous.  The homologous ones pair on
meiosis.  The non-homologous ones sort themselves out at more-or-less at

The first generation therefore has a relatively low incidence of viable
seedlings.  If the two ancestral species share ploidy, both alloploids and
aneuploids prove useful.  An aneuploid tends to have less fertility than a
balanced sib but, fortunately, fertility is not an all-or-nothing
proposition and even limited fertility can be enough to produce an
important breakthrough.  

If one of the ancestral species is diploid and the other tetraploid, the F1
seedlings tend to be triploid -- but sometimes the diploid parent produces
an unreduced gamete that produces a tetraploid offspring.  Triploids often
have just enough fertility to be useful, but a tetraploid is considered a
breakthrough because it tends to be compatible with other available

The race of modern tetraploids is the result of the effort of many
hybridizers to capitalize on such breakthroughs.  Homologous pairs have
been retained. The least-compatible of the non-homologus pairs have been
eliminated.  The result is an artificial race.  

I believe that this answers all of the questions Linda & Jeff posed -- plus
some they didn't -- but I must admit I've been studying this problem so
long that I find it difficult to explain its complexities in simple terms. 
If some of the answers aren't obvious, just pose those particular questions
again and I'll expand the explanation.

Sharon McAllister

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