A natural interpretation of the inversion operator

The inversion operator may have a different meaning in natural evolution, which has earlier been described by Kjellström, 1996, see references. On this page, two different cases will be discussed. Both cases are based on the assumption that the ontogenetic program is a modified replay of evolution, seen as a random walk inside some region of acceptability; the starting point is the fertilized egg and the end point the adult individual. While the evolution over about four billion years has to make a number of time-consuming tests for each feasible step, the ontogenetic program may carry out the feasible steps already tested in a couple of years.

A problem is that it is not known if a gene may represent a step, so, in order to avoid confusion we assume that steps are in some way stored along a chromosome. In the first case it will be assumed that steps are expressed in random order and in the second case in consecutive order.

The row below is an example of a random sequence of ten steps along a chromosome, i. e. the first step (denoted by 1) is number 2 in the sequence, the second step (denoted by 2) is number 8 and so on.

7 1 6 5 9 4 3 8 2 10

Assuming, for example, that that the crossing-over operator will first read 5 steps from maternal string followed by 5 steps from paternal string. This is a strong linkage, but if we look at the time sequence, the crossing-over is purely random:

Steps from maternal string 1 5 6 7 9

Steps from paternal string 2 3 4 8 10

For a rough estimate of the effects of such a process, we may assume that the phenotypic region of acceptability ( A ) is a rectangle such that the range of parameter x₁ is three times larger than the range of the x₂. Assuming further, that steps can contribute to the parameter values by +1 (+) or –1 (-) only and that parents are, after some time, uniformly distributed over the rectangle. We therefore expect that there should be three times as many differences between maternal and paternal strings in x₁ (6 differences, for instance) as in x₂(2 differences), on average. Because crossing-over at random will give no variability from homozygous loci, we may focus attention on the heterozygous loci only. The order of + and – steps is irrelevant here, because we are only interested in the summation of steps. A possible distribution of + and – steps might be:

Parameter x₁ x₂

Maternal string + + + + + + + + - - + + + + + + - - - -

Paternal string + + - - - - - - - - + + + + - - - - - -

The expected variability in offspring from such parents, assuming that phenotypes are expressed by single strings, may therefore be seen from the Pascal triangle lines 6 and 2:

Line two 1 2 1

Line six 1 6 15 20 15 6 1

If scaling gives the variance equal to 6 for x₁, it becomes equal to 2 for x₂. But the ratio between standard deviations is equal to the square root of 3 and not equal to 3 if the adaptation had been made using the theorem of Gaussian adaptation. Thus, we have got a certain adaptation – better than nothing, but not as good as it could be.

For the second case suppose, that the inversion operator may (like in the traveling salesman example) alter the order in which steps are expressed, such that the first 5 steps in ascending order comes from maternal string and the rest from paternal string. Then our sequence may look as follows:

1 2 3 4 5 6 7 8 9 10

Then, if crossing-over still occurs after the fifth step, the sum of those steps from the maternal string contributes to a vector, and likewise for the 5 paternal steps. The sum of those vectors constitutes a Gaussian distributed vector, which more probably falls inside A, avoiding less suitable directions.

In a high-dimensional region and with each vector as a sum of more steps, the efficiency of the process may probably be increased by many orders of magnitude.