BREEDING EXPECTATIONS

Writer Dave Barry says scientists can alter the genetic structures of living things using genetic knowledge "and a pair of very tiny pliers."

If only it were that easy for herpetoculturists.

Instead, our tools are the animal husbandry practices that keep our snakes healthy and strong while we apply the genetic principles esssential to our success.

Here are the principles herpetoculturists need to understand, and the breeding results that follow:

(Once you've read this, I'd welcome E-mail from you telling me where I lost you, what I could explain better (and how), where I've made mistakes, or what other information you'd like to see here.)

Genetic Principles:

  • Traits occur on gene pairs

  • "Alternate forms of the same gene are called alleles" (Medical Genetics: Principles & Practice, by Nora & Fraser)

  • All the popular Honduran morphs are recessive gene mutations

    (such a mutation is called recessive because if paired with another, "dominant" gene, the recessive gene recedes into the backround--it is overpowered by the dominant gene and the dominant gene determines the color of the animal)

  • An animal with two different genes or alleles, one recessive and one "normal" or dominant, looks "normal" and is called heterozygous or het ("hetero" = "different") for that recessive mutation

  • An animal with two genes for the same trait shows the trait and is called homozygous ("homo" = "same") for that mutation

  • An animal with two genes for the same recessive trait shows the recessive trait and is called a homozygous recessive, for that mutation

  • When parents breed, the father's sperm joins the mother's ovum, with each parent contributing one gene from each pair to each baby

  • A homozygous parent has two of the recessive genes, so it always passes on a recessive gene to every baby it produces

  • A heterozygous parent (with one recessive gene and one dominant or "normal" gene) will contribute a recessive gene to half its babies, and a "normal" gene to the other half

Breeding Expectations

Remember these generalizations apply to large samples. In small numbers, such as the total number of eggs a female might produce in a clutch--or even in a lifetime--the results can vary greatly, just as the odds of tossing heads or tails with a coin are 50-50 but you'll not always throw two heads in four throws, or ten in twenty.

  • Homozygous X homozygous (albino x albino, for example)

    =All homozygous (albinos, in this example)

    Every baby is going to get a recessive (albino) gene from each parent, because that's all each parent has to contribute, so all the babies will end up with TWO albino genes, making all of them albinos

  • Homozygous X heterozygous (hypo x het/hypo, for example)

    =1/2 homozygous (hypos, in this example)
    =1/2 heterozygous (het/hypos)

    Every baby will get a recessive (hypo) gene from the homozygous (hypo) parent, because that's all that parent has to contribute; the het parent will contribute hypo genes half the time, and normal genes half the time, So half the babies will get two hypo genes and will be hypos, and the other half will get one hypo gene and one normal gene and will be hets)

  • Heterozygous x heterozygous (het/anerythristic x het/anerythristic, for example)

    =1/4 homozygous (anerythristics, in this example)
    =1/2 heterozygous (het/anerythristics)
    =1/4 normal

    The father, who has one anerythristic gene and one "normal" gene, will give anerythristic genes to half the babies and normal genes to the other half; the mother will do likewise. Half the babies that got anerythristic genes from the father will get an anerythristic gene from the mother, too, so will have two anerythristic genes and will BE anerythristics. They'll make up 1/4 of the offspring (half of half). Half of the babies that got normal genes from the father will get normal genes from the mother, too, so will have two normal genes, and will be normal. They'll make up 1/4 of the offspring. The other 2/4--or one-half--of the babies get an anerythristic gene from one parent and a normal gene from the other, so are hets. Remember the hets and the normals look alike and, combined, make up 3/4 of the production. But two out of three of those normal-looking babies are actually hets, so we call those normal-looking babies possible hets with 2/3 chance of being het, or "2/3 chance possible hets". They're either hets or they're not, but we don't know which are which, so we use terminology that correctly characterizes the likelihood of their being hets.

Double-mutation breeding expectations
    Ohmigod! Double-hets! (And a question for visitors to this site)

  • Double-het x double-het

    • (double-het for snow, for example)

    You're right. Things get a little trickier here. Maybe a lot.

    Think of this pairing as TWO het x het pairings combined (one, of het/albino x het/albino, and the other, of het/anery x het/anery): Remember the het x het explanation above? The het/alb x het/alb portion of this breeding will produce 1/4 albinos, 2/4 het/albinos, and 1/4 normals. Remember, too, that the het/anery x het/anery pairing will produce babies, 1/4 of which are anerythristic (and 2/4 of which are het/anery, and 1/4 of which are normal). Consider a sample of 16 babies, which is necessary to illuminate the results here: FOUR of the babies will be albinos, as I've just explained; 1/4 of all the babies will be anerythristic, so ONE (one-fourth) of those four albino babies will be anerythristic as well. Anerythristic-albinos are snows, so 1/16 of the babies (1/4 of 1/4) will be snows. Continuing that logic and those calculations results in the full breakdown below:

    =1/16 Homozygous (snows)

    =2/16 albino het/anerythristic

    =1/16 albino

    =2/16 anerythristic het/albino

    =4/16 double-hets for albino and anerythristic

    =2/16 het/albino

    =1/16 anerythristic

    =2/16 het/anerythristic

    =1/16 normal

    Notice that there will be 3/16 that are visibly albino: one is an albino with no anerythristic "blood" and two are albinos that are also het/anery. Since 2/3 of these albinos are het/anerythristic and 1/3 is not, I refer to these as "albino 2/3 chance het/anerythristics" elsewhere on this website (2/3 will be het/anery but you can't tell which). Similarly, 2/3 of the animals that are visibly anerythristic are also het/albino, so the visible (homozygous) anerythristics would be called "anerythristics 2/3 chance het/albino". Nine of 16 would be normal-looking: of those nine, four would be double-hets; two would be het/albino; two would be het/anerythristic and one would be perfectly normal. I call those "possible double hets".

  • Double-homozygous x double-het
    (snow x double-het for snow, for example)

    This pairing shows how understanding genetics can make a big difference in your results:

    =1/4 double-hets

    =1/4 albino het/anerythristic

    =1/4 anerythristic het/albino

    =1/4 snows

    A reader of this page prompted me to include the example above, when he wrote to discuss breeding alternatives:

    "This would be the most efficient pairing for any large or even small scale breeder," e-mailed Dwight Good, an Elaphe obsolete ssp breeder in Guthrie, KY. "It would allow the production of all four color anomalies from one pair of snakes, and would eliminate (the production of) possible hets."

    Dwight's right: You can see just how efficient it is by comparing it to the results of the double-het x double-het pairing described earlier. Out of 16 babies, that double-het pairing produces NINE different kinds of snakes genetically, with FOUR different appearances, and the genetic makeup of only ONE of those 16--the snow--is known with certainty.

    When a double-homozygous animal is substituted for one of the double-hets in the pairing, however, only FOUR different kinds of snakes genetically are produced, and ALL FOUR are distinguishable visually. Furthermore, the production of snows is quadrupled and the production of albinos and anerythristics is increased, and those animals have the added advantage of always being definite hets for the other mutation.

    Instead of 15 snakes out of 16 whose genetic composition is uncertain, there are none. Still, the double-het x double-het pairings are necessary to GET the double-homozygous animals that produce such improved results, or the even more dramatic but possibly less interesting results, below.

  • Double-homozygous x double-homozygous
    (snow x snow, for example)

    =ALL snow babies

Triple het breeding expectations

This has to be one of the most fantastic genotypes imaginable. A single pair of triple hets will produce albinos, anerythristics, and hypos, ghosts, snows, and hybinos!

I won't go into the precise percentages. But think of these snakes as being three different kinds of double hets. So you'll get roughly one-sixteenth ghosts, one-sixteenth snows, one-sixteenth hybinos, three-sixteenth albinos, three-sixteenth anerythristics, three-sixteenth hypos--and the small remaining number of normal-looking babies, which will be possible triple hets.

Imagine watching those clutches as they begin to hatch!

I'd like to thank my friend Bob Montoya for suggesting the triple het breeding. I'd like to thank Dwight for sharing his thoughts. And thanks to Dr. Howard A. Bessen, Director of the Emergency Medicine Residency Program at Harbor-UCLA Medical Center, Torrance, CA, for the time and ideas he's shared with me as I worked to improve this section. Lastly, I'd like to again urge all of you to E-mail me to let me know where I lost you, what I could explain better (and how?!), where I've made mistakes, or what other information would be useful here.

Back To Main Page

Power