Reptile & Amphibian Forums

Can someone explain to me the difference between an anaconda, a super anaconda and a normal hognose? Thinking about starting a hognose project but don't want to go into battle unarmed....thanks
-----
Dustin Smith

Here is the big differences:

normal hognose: anywhere from $20 - $40

anaconda: more

super anaconda: even more.
-----
Chris - TX

A normal hognose is your typical wild type normal looking hognose. The Anaconda is a co-dom pattern morph created by Brent Bumgardner. The Superconda is a result of breeding two Anacondas together to create a "patternless" morph.

Normal x Normal will be 100% normals.
Anaconda x Normal will be roughly 50% Anacondas 50% Normals
Anaconda x Anaconda will be roughly 50% Anacondas, 25% Supercondas and 25% normals
Superconda x Normal will be 100% Anacondas

-----
Jessica @ Gecko Babies
Gecko Babies Website
Facebook Page

Thanks I was looking for this too.

.
-----
Jessica @ Gecko Babies
Gecko Babies Website
Facebook Page

The Anaconda is a co-dom pattern morph created by Brent Bumgardner.

Let's be a little more clear about this. Brent didn't create the Anaconda morph, he discovered and proved the Anaconda gene. To create something, you have to have intent. Being that the Anaconda gene came into being unexpectedly, there was no intent. Now, Brent did create morphs like the Amelanistic Superconda, he had intent and did selective breeding trials to create that specific morph.

In regards to the Anaconda Codominant trait, it is the heterozygous form of the Superconda (Patternless). It's not much different than having Recessive hets except there is a visual phenotype for the Codominant het which is the Anaconda phenotype.

In Recessive traits, there can only be two phenotypes, that of the Normal (which is the typical dominant trait) and the Recessive trait (We'll use Amelanistic as example) :
Normal = AA
Amelanistic = aa
Normal heterozygous for Amelanism = Aa

When we breed a Normal to a Recessive trait animal, we get 100% hets (Aa) but they all look Normal. This is because the Normal trait is dominant and will override the Recessive gene in the heterozygous form.

However, in Codominant traits (We'll use the Anaconda/Superconda trait as an example), there are three phenotypes:
Normal = AA
Superconda = aa
Anaconda (Codominant heterozygote)= Aa

When we breed a Normal to a Superconda, we also get 100% heterozygous animals (Aa), the big difference is that both traits are dominant and therefore are competing against each other to show their individual traits creating the Anaconda phenotype.

You can use Punnett squares or the FOIL (First, Outside, Inside, Last) method to determine heritable ratios. Let's use the FOIL method to determine Superconda/Anaconda ratios:

AA x aa = Aa, Aa, Aa, Aa = 4 Anaconda.
AA x Aa = AA, Aa, AA, Aa = 2 Normal, 2 Anaconda.
Aa x Aa = AA, Aa, aA (Same as Aa), aa = 1 Normal, 2 Anaconda, 1 Superconda.
Aa x aa = Aa, Aa, aa, aa = 2 Anaconda, 2 Superconda
-----
Troy Rexroth
Rextiles

Great post as usual Troy!
I enjoy to read your explanations every time...please continue this work
-----
Greetings from Berlin
Stefan & Raimo

Reptiles-Breeding-Enterprise.com

As always Stefan, thanks for your support!
-----
Troy Rexroth
Rextiles

Troy,
I like the way you break it down so most people can understand it easily... Fine post my friend...

Thank you kindly Gregg, I appreciate it!
-----
Troy Rexroth
Rextiles

Troy can you do a breakdown of a recessive gene?

Troy can you do a breakdown of a recessive gene?

Sure, it's the exact same method for what I did with the Superconda/Anaconda Codominant traits except that we're defining the genes differently.

Here's the FOIL method for Recessive traits:
(We're using Amelanism as an example but any Recessive trait will work)
Normal = AA
Amelanistic = aa
Normal 100% het for Amelanism = Aa

#1. AA x aa = Aa, Aa, Aa, Aa = 4 Normal 100% hets.
#2. AA x Aa = AA, Aa, AA, Aa = 2 Normal, 2 Normal 100% hets.
#3. Aa x Aa = AA, Aa, aA (Same as Aa), aa = 1 Normal, 2 Normal 100% hets, 1 Amelanistic.
#4. Aa x aa = Aa, Aa, aa, aa = 2 Normal 100% hets, 2 Amelanistic

This time I numbered the combinations because this is where it can get a little confusing as this is where we also learn about possible hets and understanding where 50% and 66% hets come from.

In examples #1 and #4, you are producing all Normals that are 100% het for Amelanism. There's no surprises here. Examples #2 and #3 are where the fun begins.

In example #2 you are producing all Normals but half are het for Amelanism and half are not. The problem is, you can't usually tell which ones might be carrying the gene or not. So we use ratios to determine the possibility of each one to carry the gene. If the method determines that we produce 2 out of 4 hets and 2 out of 4 non-hets, that gives us 50% odds whether each individual is het or not. The truth is, these are just statistical ratios though. In real life, anything is possible. You could actually end up with all hets or none at all.

In example #3 we actually produce an Amel out of the 4, but the problem is still determining the Normal non-het from the Normal hets. Being that there are 3 Normals, 1 non-het and 2 hets, we simply use ratios again. Since 1 out of 3 is 33%, we simply add up how many percentage points for each het which gives us 66%. So in this clutch, we'll produce one actual visual Amelanistic and 3 Normals which are 66% het for Amelanism.
-----
Troy Rexroth
Rextiles

You can find all the Anaconda and Superconda info at Superconda.com.

Brent Bumgardner

-----
Brent Bumgardner
bwbumgardner@aol.com
703.431.1776
Superconda Website

OH YEAH!! The really red ones and such are the B.E.S.T. and are prety great TO!!!! I love HOG NOSED SNAKES to!