Lets start with some definitations then get into the whole subject of Genetics:
Allele = Of the two paired genes affecting an inherited trait. Example: One from male and one from female.
Codominant = An allele that causes the homozygous form to look different than wild type and the heterozygous form to have traits of both. Example: All three look different from each other.
Dominant = an allele that causes the homozygous form and the heterozygous form to look the same as each other, but different than wild type.
Recessive = an allele that affects a animal's appearance if it's present in the homozygous state only. An animal that's heterozygous for a mutant, recessive gene looks wild type, but that gene can be passed on to offspring.
Wild Type = A normal appearing animal.
Het=Hetrozygous Heterozygous = the two members of a gene pair are not identical.
A Salmon boa is a codominant with three phenotypes -- Sa//Sa = homozygous salmon (supersalmon), Sa//+ = heterozygous salmon (salmon), +//+ = normal. IOW, in this case the heterozygote shows the effect of both members of the gene pair.
Homo=the two members of a gene pair are identical. All recessive traits must be Homo to show in the animal.
"All recessive traits must be Homo to show in the animal." A
possibly better alternative could be -- A recessive mutant gene must be homozygous to visibly affect the animal's appearance.
A trait is what you see (phenotype). A gene is invisible to
the naked eye. A "difference from wild type" may be better than "trait."
Male Homo Hypo X Normal Female= 100% het Hypo
Male Homo Hypo X Het Hypo Female= 50% Homo Hypo and 50% het Hypo
Male Homo Hypo X Female Homo Hypo= 100% Homo Hypo
Male Homo Hypo X Female Stripe=
100% het Hypo & Het Stripe
Now for the fun one
Male Homo Hypo Het Albino X Female Het Hypo
Het Albino= 12.5% Het Hypo
25% Het Hypo & Het Albino
12.5% Het Hypo & Homo Albino
12.5% Homo Hypo
25% Homo Hypo & Het Albino
12.5% Homo Hypo & Homo Albino or
25% Sunglow 25% Hypo
50% Hypo Het for Albino(sunglow)
Must realize in all of these examples that the Hypo trait will show through because it is an incomplete dominant trait.
Thanks to Bryan for the above!
Genetic Principles:
This was originally written by rtdunham@earthlink.net
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
Linear Mode
