Statistics is a mathematical science. No way to get around it. It is the study of probability. This makes more sense when related to the science of color genetics in rabbits, which is widely misunderstood.

When you have a computer program, or a friend, or a book, tell you that '25% of the offspring (1 in 4) will be black', what this is truly saying is that the probability of getting the double recessive is one chance in four, or 25%. This applies, through independent assortment, to each kit conceived, and only for that particular locus. Thoroughly confused now?

Let's go back to the beginning. In a previous article we discussed color loci versus pattern loci. The loci A and E are pattern loci; B and D are color loci, and C affects both color and pattern. At each of this loci, we have pairs of alleles [uh-leels] that will combine independently from the other loci in large degree (B and D are somewhat linked, but we are not going to discuss that here) when two animals are mated.

This results in a new combination of alleles in each conceptus (kit), and statistics is used to determine the likelihood of anyone kit receiving a particular matchup of alleles, either just at that locus, or on all applicable color loci. This is generally expressed by using letters such as (we'll stick with A locus for now) Aa x Aa = AA, Aa, Aa, aa. Four possible results, three possible genotypes expressed by their portion of the four possibles. One in four, 1/4, AA=25%. 2/4 or 1/2 Aa=50%. 1/4 aa again equals 25%. What does this really mean?

All this means is that, statistically speaking, in a large number of offspring, 25% will be aa, 50% will be Aa, and 25% will be AA. 'Large number' is usually taken to mean hundreds or thousands of offspring. The larger the sample size, the more accurate these proportions become. This is speaking genomically, that is, of the actual allelic combinations. Phenotypically, these numbers change significantly.

Phenotype is the outer appearance of the animal without reference to the genotype. In this same cross of Agouti and Agouti, the only visible result is that 25% of the offspring will be nonagouti. The genotype AA and Aa are identical and may not be differentiated by just looking.

The old saying 'the proof is in the pudding' definitely comes to mind when looking at a 'surprise' litter, where an unexpected color pops up out of lines you would not expect to produce those kits. Recessive genes may 'hide' themselves behind their dominant counterparts for many generations before appearing; bear in mind also that each animal on a pedigree came from a whole litter of kits and may have had siblings exhibiting the recessive allelic combinations.

When looking for the expected number of offspring of a color produced by more than one locus, say, a martenized lilac tortoiseshell (henceforth MLT), one must take into account all the probabilities involved at each locus. So let us do a little exercise. The parents of this MLT are a black otter whose parents were chocolate otter and black and he's thrown blue before. His apparent genotype as far as we know is a(t)a Bb Cc(chd) Dd Ee. We know that bred to a black he threw blue silver marten, and in the same litter a tortoiseshell showed up, so we know that he carries the nonextension ( e) and the chinchilla (c( chd)). The other side of the ML T we want is a tortoiseshell whose parents were chocolate and black. Her history shows blue, REW, tortoiseshell, and chocolate offspring; hence her genome is aa Bb Cc Dd ee.

So. We have the following crosses at the loci:
a(t)a x aa; Bb x Bb; C c(chd) x Cc; Dd x Dd; and Ee x ee.
What we want is art) dd ee.
How often will this show up?

Well, a(t)a x aa = 1/2 a(t)a + 1/2 aa, so 50% chance of what we want at A locus. Bb x Bb will yield 25% BB + 50% Bb + 25% bb; we need bb, so 25% at B locus.

Cc(chd) x Cc = 1/4 CC, 1/4 Cc, 1/4C c(chd), and 1/4 c(chd)c. We need
c(chd)c, so again, 25% is our number. The D locus is the same as the B locus, 25% chance of the double recessive we want. The E locus cross results in Ee x ee = 1/2 Ee + 1/2 ee. 50% chance of the ee we desire. Now how do you bring them all together to make it tell you how many MLT you might get out of 100 kits? Multiply the probabilities from each locus together. A=.5 B=.25 C=.25 D=.25 E=.5

.5 x.25 x.25 x.25 x.5 = .0039 x 100 = .39 percent chance of a MLT. That's 39 kits out of 10,000 that will likely be MLT' s. You might get lucky and get one in the first litter of four or eight...but the odds are really low. One in four thousand means a lot of litters to get one kit!

Remember though that statistics are just another way of saying probably. "Probably", "might", "could"...all of these are affected by interactions we don't yet know of, and it's possible that any combination could turn up at any time, if the parents carry the requisite alleles to make that particular combination.

Some things that affect the expression of genes are environment (feed, water, shelter, husbandry), linkages on the chromosomes, problems with the recombination of the chromosomes at gametogenesis and/or conception, mutations, and unknown factors within the cells.

Heritiability is a term you will hear fairly often, as in 'litter sizes are lowly heritable'. This is another application of statistics in genetics. Highly heritable traits such as weight at 56 days are strongly affected by the traits of the parents; lowly heritable traits such as litter size are not particularly affected by those traits in the parents. That's a very simple way of putting it, but by culling hard for proper values in the highly heritable traits, one may improve one's herd with amazing speed. Selection for lowly heritable traits can come later, or have a lower priority, as it will take longer for the effects of such selection to be seen in the herd as a whole.

Happy Crossing! P A

Note: My grandma wanted to put this article on the website.

~ KSS Rabbitry ~
Katelynn S. Sokolowski
Haskell, Oklahoma
Phone: 915-482-1560
Email: katers20011@yahoo.com