Looking for some advise on morph calculations/predictions

[shaberry]

Hey guys, so I was planning on trying to breed some corns within the next year, but I don't know anything about the subject. And before I did any breeding, I wanted to at least be a little bit knowledgeable on the subject. (Don't know how to do the punnet square, and don't quite understand how morphs are made. I'm trying to look through some threads, but I just get so confused. >.<) Naturally, I figured the best place to try and understand the trade is here. So, how are the offspring of a corn pairing calculated? Where can I go to look up information in determining offspring outcome between paired corns?

For example, I want to try and make pewter corns. Now, according to what I've read, pewters are made from a pairing of a bloodred and charcoal (anery B) corn. However, when I tried out the morph generator/calculating program, it calculated that the offspring would be normal het. charcoal and bloodred. This is what I don't understand. So how exactly are pewters made then? Does that mean that I have to breed the offspring to get the pewter to show?

Also, I want to breed my normal corn, Arshess, to see what kind of hets she has. And, from what I've gathered from looking around the forum so far, in order to check her hets, I have to breed the offspring to see what results. Is that correct?

Gah...>.<;;; This is all just so confusing. Please help an extreme newbie out and point me in the right direction. unch:

 

[airenlow]

Bloodred X Charcoal will give you Normals het Pewter (Bloodred Charcoal)...so then you would pair these offspring together to have a chance at Pewters.

The most common hets are Amel and Anery A...so you would pair your normal with an animal who is proven to carry these genes. If your Normal is het for these, you should get some visual morphs from the resulting offspring.

Hope that helps a little...

 

[shaberry]

Yeah, you helped me out a bunch! Thanks airenlow! I really appreciate your help. =)

So, does that mean that I have to breed Arshess to a numerous amount of male morphs to figure out what hets she has? Wow, that sounds really expensive. >.<;;;

Also, are there other sources where I can look to find out how to calculate/predict the outcome offspring of pairings? Is there a "morph calculation/prediction for dummies" manual floating around somewhere? If there is, I'd really like to know.

 

[airenlow]

Or you can find a couple males with a bunch of hets. I have one that is het Amel, Anery A, Charcoal and Motley...that would make it much easier. If Arshess was a male you could breed it to multiple females every year but females can only go once or twice.

Good site...
http://iansvivarium.com/cornmorphs.html
Great book...
http://www.cornguide.com/
Good predictor...
http://www.cornsnakes.nl/GenericGenWiz/MPGW.htm

 

[bitsy]

So, does that mean that I have to breed Arshess to a numerous amount of male morphs to figure out what hets she has? Wow, that sounds really expensive. >.<;;;

Afraid so. This is one of the benefits of buying from an established breeder - they'll be able to provide you with all of the info about a Corn's parents which will give you the likely hets.

However, as Josh says, your lass is most likely to be carrying Amel and Anery A, unless you have reason to believe that she came from a breeder producing more exotic morphs. You could try mating her with a Snow. If she's het for those two morphs, you'd statistically get one-quarter Normals, one-quarter Amels, one-quarter Anerys and one-quarter Snows. However, this is all just theoretical.

It probably isn't worth investing in males of rarer morphs, just to look for hets in a Normal that may not exist. Bear in mind that you can only pair her up once a year. You (and she) could work for years, only to reach the conclusion that she has no hets at all!

Now that you have the breeding bug, you're better off deciding what morphs you're most interested in, then buying new males and females that will give those in the first generation of offspring. Don't worry, I'd guess that many of us bought our first Corns as pets and then realised we didn't know the first thing about their genetic history when we decided to breed. How do you think we justify buying all the other ones?! LOL!

If you use the Marcel Poots Genetics Wizard program, it actually shows you the punnet squares for any morph combo. That might give you a head start on learning the process.

If you want a good book, then try Charles Pritzel's Cornsnake Morph Guide. But do beware that this also doubles as the most expensive wishlist in history!

 

[Vinman]

Bloodred X Charcoal will give you Normals het Pewter (Bloodred Charcoal)...so then you would pair these offspring together to have a chance at Pewters.
Hope that helps a little...

very well said

 

[paulh]

Also, are there other sources where I can look to find out how to calculate/predict the outcome offspring of pairings? Is there a "morph calculation/prediction for dummies" manual floating around somewhere? If there is, I'd really like to know.

There is a good corn snake genetics tutorial at http://www.serpwidgets.com/Genetics/genetics.html

Also, check out Mick's Progeny Predictor. He's just gotten version 3 out. The announcement is near the top of the list of threads in this forum as I write this.

 

[shaberry]

How do you think we justify buying all the other ones?! LOL!

Lol! Now, that's a great explanation of why you guys have so many! *sigh* But, I just wish that I had a bigger income to accommodate for my "wanting to breed" bug. It kinda limits my possibilities at the moment.

There is a good corn snake genetics tutorial at http://www.serpwidgets.com/Genetics/genetics.html

Also, check out Mick's Progeny Predictor. He's just gotten version 3 out. The announcement is near the top of the list of threads in this forum as I write this.

Oh yeah, I downloaded the program, but I'm having some issues with it.

But, thanks for the link for the tutorial paulh! And thanks everyone for helping an extreme newbie find her way around. I really appreciate it! =) If you guys have any more great advice on this topic, please don't hesitate to endow me with your knowledge. =)

 

[paulh]

Oh yeah, I downloaded the program, but I'm having some issues with it.

... If you guys have any more great advice on this topic, please don't hesitate to endow me with your knowledge. =)

I downloaded the program today but have not installed it yet. I will probably have some issues with it, too.

Serpwidgets has a booklet named "Genetics for Herpers". I have a copy. Of the introductory genetics stuff aimed at herpers, it's the best thing I've seen so far. See
http://cornguide.com/genetics.php

There is a genetics text online for free download named A Survey of Genetics (in 4 parts):
http://www.ringneckdove.com then go to the Contents page and look on the right side of the page under the Genetics Notes heading.

Many genetics problems boil down to identifying the relevent genes in the parents and then recombining them in the offspring. Mick's program and the others available just do the combining part really fast. Once you learn how to work these problems, you just need to practice until it is second nature. And practice is something you have to do yourself.

 

[ZoologyGirl]

However, when I tried out the morph generator/calculating program, it calculated that the offspring would be normal het. charcoal and bloodred. This is what I don't understand. So how exactly are pewters made then?

Some basics that will hopefully help you in your understanding (that is, if I can say it right without tripping over words, and I hope it's not too long winded).

Most morphs in the cornsnake world are controlled by recessive genes.

For every trait, a snake has a pair of alleles...one inherited from the mother, one inherited from the father. On paper, punnet squares, etc... these alleles are usually donated by letters (AA, Bb, CC, dd, etc...). Each letter represents an allele (1 from father, 1 from mother)

For the following examples I will use a generic AA combo, not meant to represent any particular trait.

With these you can have:
Homozygous Dominant: AA
Heterozygous: Aa
Homozygous Recessive: aa

As I first stated, most cornsnake morphs are governed by simple recessive genetics. This means that for a trait to be present, the snake must be homozygous recessive for that trait. If it is heterozygous, then it is possible for its offspring to show that trait, but it will not show on the snake itself.

Lets say for this example that the A allele is for the amel trait. It's phenotype (how the snake will look) will be the following

AA=Normal.....Aa=Normal.....aa=Amel.

Now, when you want to breed for a specific trait, it is important to remember that BOTH parents MUST have at least 1 recessive allele for that trait. That is, both parents must either be homozygous recessive, or heterozygous, for every trait you want in the offspring. This is because like I stated, an offspring gets 1 allele from each parent. If a parent is homozygous dominant for a trait (AA) then the only allele that can be given from that parent is an (A)...And you cannot create a homo. recessive combo (aa) with that.

In your example of wanting a pewter, you tried to breed charcoal to bloodred in the calculator and did not get a pewter. That is because both parents did not have at least 1 recessive allele for both charcoal and bloodred. The charcoal was homozygous recessive (cc) for for charcoal and homozygous dominant for bloodred (BB). And just the opposite with the bloodred parent. The offspring therefore come out as normals het for pewter (CcBb). And yes, breed the offspring together and you will possibly get some pewters, because each parent has the possibility of giving a recessive allele for each trait to its offspring.

Rather, to created pewters you would need both parents to either be pewter, bloodred het charcoal, charcoal het bloodred, or anything het bloodred charcoal....Both parents don't have to be the same, but they must be one of these so that each parent has the possibility of passing along a recessive (c and b) allele.

And I'll close with a quick note on breeding hets and the like...

AA x AA will always give AA offspring
aa x aa will always give aa offspring

Aa x Aa will give AA, Aa, Aa, and aa. A simple method to figuring this out is the FOIL method (which possibly you may have also learned in math class). FOIL stands for First, Outside, Inside, Last. For simplicity I will use 4 different letters to represent this,

ab x cd

First stands for the first digit in each combo (ac)
Outsdie is for the outside digits (ad)
Inside is for the inside digits (bc)
Last is for the last digit in each combo (bd)

So using foil, we can figure AA x Aa to result in AA, Aa, AA, and Aa. Then apply all the other info and hets and homo dom/recessive to figure out the traits. (Note the FOIL method is used for crossing a single trait)

Multiple traits work a little different, take the common example of the snow which is a combo of amel and anery. A snow, therefore, must be homo recessive for both amel (aa) and anery (nn)....snow =aann (I arbitrarily used these letters)

snow x snow (aann x aann) will always = snow (aann)
however, amel x anery (aaNN x AAnn) will always equal normals het for amel and anery (AaNn)

amel het anery x snow (aaNn x aann) =
Because both parents are recessive for amel (aa) the only allele the offspring can get from each parent is (a) and therefore every offspring will be recessive for amel (aa).
The anery part is a little different. Nn x nn = Nn, Nn, nn, nn....or in other words 50% Nn (het for anery), and 50% nn (homo recessive for anery).
Therefore, 50% of the offspring will be amel het anery (aaNn), and 50% will be snow (aann).

Alright, I think I typed too much, but hopefully you can find some of this helpful.

 

[shaberry]

WOWWWWWWWWWWWW!!! Now THAT was an AWESOME explanation. (Please type more if it'll help me!) :bowdown:

Still a little confused, but I think I'm starting to get it. Punnet squares get me all kinds of confused, for some reason. But when you said foil method, and gave the example, it clicked! =) Well, kinda, I'm still confused. I got it when I was doing your examples, but when I tried to do it with the genotypes presented for morphs on websites, I didn't know how to apply it.

I tried to apply it to the bloodred x charcoal pairing for pewter, and the furthest I got was: (Got the genotypes from Ians Vivarium)
bloodred: DDDD[SIZE=-2]
[/SIZE]charcoal: chcchc
pewter: chcchcDDDD

(DDDD x chcchc)
Since, there are no matching alleles, the expected outcome should be normals. For hets, I wasn't quite sure how to do it. Since, you said the foil method was only for crossing a single trait. And from what I gathered, one allele is given from each parent. So, the het would be (DDchc)? But, the genotype for pewter has two DD and chc components. So, how was that figured out? I tried the punnet square method and got DDchc (in different orders) in all four squares. So, this is where I got really confused.

But, thanks ZoologyGirl for taking the time and effort for trying to explain to me the methodology for predicting the outcome of pairings. Even though I'm still confused, I feel like you clarified the topic a little bit for me.

 

[Ssthisto]

Actually, if you express Diffused as "D/D" and Charcoal as "ch/ch" (I use the split to indicate one copy comes from one parent and the other comes from the other parent) then your charcoal is:

d/d ch/ch (Does not carry Diffused, is homozygous for Charcoal)

Your Bloodred would be:

D/D CH/CH (Homozygous Diffused, does not carry Charcoal).

All offspring of that pair would be:

D/d CH/ch (het Diffused, het Charcoal).

That's because each baby is getting one copy of EACH gene from its parents - they're inheriting the "not-diffused" from the charcoal parent as well as "diffused" from the bloodred. If you bred them together, you'd have a one in sixteen chance per egg of:

D/D ch/ch - homozygous diffused, homozygous charcoal AKA Pewter.

Think of the genes like seats on a train - there's pairs of seats up and down the train, but if you've got one reserved, you don't pay attention to who's sitting in the other seats, right? So, because we're only interested in what's sitting in the "Diffused" and "Charcoal" family seats, we're not paying attention to who's sitting in the Amel family seat (that D/D ch/ch could be D/D ch/ch A/a) or who's sitting in the Motley family seat (D/D ch/ch A/a M/ms*) ... do you see where I'm going with this?

My personal suggestion to really test out your normal female's hets would be a Snow Motley or Snow Stripe - because there's a fair few pattern-morph hets floating around out there!

* You may have a virtual cookie if you tell me what you think that snake looks like - and what genes it is carrying.

 

[shaberry]

Actually, if you express Diffused as "D/D" and Charcoal as "ch/ch" (I use the split to indicate one copy comes from one parent and the other comes from the other parent) then your charcoal is:

d/d ch/ch (Does not carry Diffused, is homozygous for Charcoal)
But where did you get the d/d? Sorry, if it's a no-brainer answer that I'm looking for. It's just that when I look up the genotype for the morph, I don't see that allele expressed. >.<;;; So, how do I know what morph carries what allele (if it's not expressed in the text)? Like, I use vmsherp and Ians Viv. for genotype reference and I don't see the alleles that you use. Perhaps maybe I should just go by a morph guide book then? Will it be expressed there?
Your Bloodred would be:

D/D CH/CH (Homozygous Diffused, does not carry Charcoal).
Also, the same for here. How do you know that the bloodred has an allele for CH/CH? Do all bloodreds have this allele?
All offspring of that pair would be:

D/d CH/ch (het Diffused, het Charcoal).
When you put the genotypes like above, I really understand how you get the results. =)
That's because each baby is getting one copy of EACH gene from its parents - they're inheriting the "not-diffused" from the charcoal parent as well as "diffused" from the bloodred. If you bred them together, you'd have a one in sixteen chance per egg of:

D/D ch/ch - homozygous diffused, homozygous charcoal AKA Pewter.

Think of the genes like seats on a train - there's pairs of seats up and down the train, but if you've got one reserved, you don't pay attention to who's sitting in the other seats, right? So, because we're only interested in what's sitting in the "Diffused" and "Charcoal" family seats, we're not paying attention to who's sitting in the Amel family seat (that D/D ch/ch could be D/D ch/ch A/a) or who's sitting in the Motley family seat (D/D ch/ch A/a M/ms*) ... do you see where I'm going with this?
Yes, for the most part, I think so. ^_^;;;

My personal suggestion to really test out your normal female's hets would be a Snow Motley or Snow Stripe - because there's a fair few pattern-morph hets floating around out there!

* You may have a virtual cookie if you tell me what you think that snake looks like - and what genes it is carrying.

*pout* I want a cookie. But, I'm confused as to which snake you're referring to. >.<;;; But, I'll take a crack at it anyway. If you're talking about the snows, aren't they homo. for anery A and amel? And if they're motley or stripe, they should be carrying the motley or stripe alleles too, right?

Oh, and thanks for the further explanation Ssthisto! =) I really appreciate your help in trying to guide me through this procedure.

 

[Ssthisto]

An animal who is homozygous "not diffused" is d/d (not diffused-not diffused) - just like, if it's not carrying Amel, it's A/A (homozygous not-amel). Yes, the Bloodred has a "charcoal" gene pair - it just happens to have two copies of "Not Charcoal" sitting in that gene pair instead of having "charcoal" sitting there.

ALL animals have "train seats" for ALL of the genes that are possible... the only time they share seats is if you have members of the same "family" that have different visual appearances (like Motley and Stripe, which are family members and share the same set of seats). So when you show Charcoal as ch/ch you're shortening it from the real chain of what's there:

A/* E/* Ca/* ch/ch d/d H/* Ld/* Lv/* ... and so on.

You'd be going on for ages trying to write down everything your snake doesn't carry - so you only pay attention to the "seats" on the train that you're particularly interested in (the "charcoal family" seats) - but this is where the confusion pops up. Unless you know that EVERY snake has the WHOLE train (but that you're only looking at the parts you're interested in) you'll wind up asking "where did the D/D and Ch/Ch come from?"

The chain of letters I gave you translate to:

"Homozygous Diffuse, Homozygous Charcoal, heterozygous amel, heterozygous stripe".

The animal looks like a Pewter, but carries the genes to produce striped and/or amelanistic offspring if bred to another stripe or amel carrier (and motleys if you breed to a motley carrier!)

 

[Susan]

Here's a link to the genetics tutorial I wrote up a few years ago to help another newbie. Sometimes, having several people explaining the same thing in their own way will help you get a better grasp of the whole idea.

http://www.cornsnakes.com/forums/showthread.php?t=41890

 

[shaberry]

ALL animals have "train seats" for ALL of the genes that are possible... the only time they share seats is if you have members of the same "family" that have different visual appearances (like Motley and Stripe, which are family members and share the same set of seats). So when you show Charcoal as ch/ch you're shortening it from the real chain of what's there:

A/* E/* Ca/* ch/ch d/d H/* Ld/* Lv/* ... and so on.

That's right! Heh, your explanation just jogged my memory on what I learned from the corn snake morph guide website. I totally forgot that all corns have all the genes possible, it's just that they're not all "activated" (as I like to put it). Alright! So, I finally understand one thing! Now, the rest should be a little bit easier to understand, hopefully. ^_^;;;

The chain of letters I gave you translate to:

"Homozygous Diffuse, Homozygous Charcoal, heterozygous amel, heterozygous stripe".

Wow, I was wayyy off base when I tired to answer your question. Guess, I was too confused with trying to process the other stuff. >.<;;;

But yeah, thanks soooo much Ssthisto! I totally understand the genotypes thing now. =) THANK YOU!

And, thanks susan for the link. =) I'll be reading your tutorial after I submit this post. Hopefully, I'll be able to make it click. =)

Once again, thanks to everyone who's taking the time and effort to help me understand this material. I deeply appreciate all of your efforts.