In Variation under Domestication, Darwin makes several references to the concept of "correlation of variation":

I will here only allude to what may be called correlated variation. Important changes in the embryo or larva will probably entail changes in the mature animal.


Some instances of correlation are quite whimsical; thus cats with blue eyes are invariably deaf.

In the chapter's ending, he also calls this "correlation of growth":

Variability is governed by many unknown laws, of which correlated growth is probably the most important.

I gather that this is Darwin's humble way of admitting that there are many unknown unknowns, at least as far as correlation is concerned.

But, since 1859 until present day, what did we find out about this quirky aspect of evolution? Are all cats with blue eyes deaf?

  • $\begingroup$ It seems obvious: one characteristic or change is associated with another, as with blue eyed cats being deaf. $\endgroup$ – jamesqf Nov 29 '19 at 18:00
  • $\begingroup$ Yes, I was more curious about the modern understanding of the topic, that is, if the theory is as sound as it was perceived back then and, if not, in what regards. $\endgroup$ – Paul Razvan Berg Nov 29 '19 at 18:12

One must not confound a phenotypic variance-covariance matrix (usually called $P$) and a genetic variance-covariance matrix (usually called $G$) for a set of phenotypes. The mutational variance-covariance matrix is also a common concept (usually called $M$). There are a lot of papers on the evolution of the G-matrix (and related matrices).

In the abstract of Roff (2000), you can find the equuation

$$\Delta\bar z = GP^{-1}-S$$

, where $\Delta\bar z$ is the vector of mean responses, $G$ (often just called the G matrix) is the matrix of additive genetic variances and covariances, $P$ is the matrix of phenotypic variances and covariances and $S$ is the vector of selection differentials.

You will find much more by just googling "evolution of G-matrix"

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  • 2
    $\begingroup$ It might be helpful to add a lay summary to this answer, since I think it's on a bit of a different level than the original question. $\endgroup$ – Bryan Krause Nov 29 '19 at 19:59

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