Reptile & Amphibian Forums

So I am pretty good with biology and genetics. I can do punnet squares with dominant, co-dom, and recessive genes in my sleep. I Just bought the new "The Complete Boa Constrictor" book written by Vincent Russo, which I HIGHLY recommend to everybody, expert or beginners. But flipping through the genetics section motivated me to further educate myself with punnet squares. I decided to try to teach myself how to do a punnet square with multiple genes in 1 snake. The example I am using in my punnet square actually has 1 dominant gene and 1 recessive gene, which complicated it a little bit more to me. I am doing my favorite pair I have breeding this year which is a sunglow x arabesque 100% het albino. Some ball python guy told me that the sunglow arabesque is a 1 in 16 snake with that pairing. I've done some research online and came up with a website that helped tremendously for anybody trying to figure out punnet squares.... http://reptimania.co.uk/genetics1.html which has everything from normal squares to multiple gene squares.

Anyways, IF I did this right, and I like to think I did, hitting the sunglow arabesque would be a 1 in 8 chance because I'm using a sunglow, not a DH Sunglow. I think if I used a DH it may turn out to be 1 in 16, but I haven't tried that theory on paper yet. But by doing a sunglow x arab het albino, I came up with 16 squares which produced the following results-

Arabesque gene is represented with Rr', het albino is Aa, albino is aa, and hypo is Hh'. Refer to the website I listed above, it explains it really well.

Arabesque het albino genotype- Rr'Aa (RA, Ra, r'A, r'a)
Sunglow Phenotype- Hh'aa (Ha, Ha, h'a, h'a)

each parents alleles posted above are the exact letters I used in my square and below are the way the babies turned out in my square

2 Normal het albinos (RHAa)
2 Hypo het albinos (Rh'Aa)
2 albinos (RHaa)
2 sunglows (Rh'aa)
2 Arabesque het albinos (r'HAa)
2 hypo arabesque het albinos (r'h'Aa)
2 Albino arabesques (r'Haa)
and last but not least
2 sunglow arabesques (r'h'aa)

Thats a total of 16 squares, each one representing a single baby. If I did this right, that makes 2 of 16 sunglow arabesques, which breaks down to 1 in 8. If anybody who knows how to do these kind of squares and wouldn't mind spending a few minutes on it, I would like for someone to also put it down on paper and either confirm what I came up with, or point me in the right direction. I have realized that even with a good backround with punnet squares, teaching myself to do multiple genes, and even worse mixing dominant with recessive all in the same punnet square isn't the easiest thing to do. I think I got it though and if someone confirms that, quite an accomplishment to me!

Thanks in advance to anyone who puts in an effort or input
Tony

Tony, in a perfect world I think you would be right. Genetics never work right.Ive heard of people breeding hypo to normal and getting all normals. its just the luck of the draw. Ricks sunglows were born in a small clutch think 16 or so and he had two from dbl het to het. with only one lone albino.

Tony,
Your numbers are correct. Just keep in mind that the Arabesque and Hypo genes work against perfect numbers. Being CoDom genes, you could get all or nothing with them. Even though the numbers show that 50% of your babies should be Arabesque and that half of those should carry the Hypo gene, you could get just plain Arabesque het Albino.

In working the Punet squares I have found it better to still show all the genes in each animal even if they are not active. For instance, for your Arabesque I would use Rr'AaHH and for the Sunglow I would use RRaaHh'. If you are working with pure BCI species it doesnt really matter if you leave off the unsupporting genes but it makes a difference if you wanted to know what happens to get a Hypo Hogg or what you get when you cross 2 Hypo Hoggs, etc. I have worked thru many of these same senarios. I have a pair of Hypo Hogg X Anery siblings that I'm raising up and I really want to know if I can predict how many of the offspring will exhibit the Hogg Island traits. I should be able to get Anery and Ghost Hoggs from the pairing. And I know darn well the minute I get babies out I'll get a few that look 100% Columbian. I may even put my male Hypo Hogg Het Anery with a female Arabesque het Anery Pos Het Albino just to see what they throw.

Just remember that you still need to do a few breeding trials to prove them out. The first year should tell you what she is capable of producing and a second year with the same pair should solidify the percentages. Good Luck.

Chris

I know these are all rough estimates based on theory, and it never works out the same as on paper. I was just trying to confirm that odds are more along the lines of getting a 1 in 8 sunglow arabesque as opposed to what I was told being a 1 in 16 snake. I don't think breeding trials ever work out perfectly with the percentages, I just wanted to teach myself how to put a multiple gene punnet square down on paper in front of me, not just try to come up with it in my head. But from what I'm being told, my work on paper came out right. It was actually really complicated. The only example I had to work off of in the book was a double recessive x double recessive, which through me off because neither of my snakes are double recessive. If you dont mind, I don't understand why you said you would put the arabesque as RrAaHH? I dont see where the HH comes in to play or what difference it would have made?

>I was just trying to confirm that odds are more along the lines of getting a 1 in 8 sunglow arabesque as opposed to what I was told being a 1 in 16 snake.

You are correct. According to probability theory, the expectation is that 1/8 of the babies from the cross would be sunglow arabesque, not 1/16.

>But from what I'm being told, my work on paper came out right. It was actually really complicated. The only example I had to work off of in the book was a double recessive x double recessive, which through me off because neither of my snakes are double recessive. If you dont mind, I don't understand why you said you would put the arabesque as RrAaHH? I dont see where the HH comes in to play or what difference it would have made?

The answer was right, but the work was not done right. Most of the trouble is that you were trying to apply a two locus model to a three locus problem. In othe words, the albino mutant gene and its normal alternative are located at one place (locus) in the genome (the totality of the genetic information), arabesque and its normal alternative are located at a different place in the genome, and salmon (hypo) and its normal alternative are located at still a different place in the genome. That makes three different locations in the genome. Mushing arabesque and hypo into one locus is not kosher.

For simplicity, both snakes in the mating should have all the loci (plural of locus) listed except the ones where both parents are normal. As the gene pair at the anerythristic locus are normal genes in both snakes, they can be ignored. But one snake has a hypo and a normal gene. Even though the other snake has two normal genes at its hypo locus, those two normal genes have to be listed. And the same goes for the arabesque locus.

I'm not sure of all the abbreviations you used. So here's what I will use:
A = the normal gene at the albino locus
a = albino mutant gene

H = hypo mutant gene
h' = the normal gene at the salmon (hypo) locus

R = arabesque mutant gene
r' = the normal gene at the arabesque locus

Arabesque het albino - h'h' Rr' Aa (sex cells = h'RA, h'Ra, h'r'A, h'r'a)
Sunglow Phenotype - Hh' r'r' aa (sex cells = Hr'a, h'r'a)

I only listed the individual sperm and egg cell genotypes once. It does not matter which h', r', or a gene winds up in a particular sex cell when the source locus is homozygous. And this short cut reduces an 8x8 Punnett square (64 boxes!!!) to a manageable 4x2 Punnett square (8 boxes).

Mixing dominant and recessive genes is not all that difficult. In a sample mating, Aa Bb x Aa Bb, is A the mutant gene or is a the mutant gene? And is B the mutant gene or is b the mutant gene? The genotypes resulting from the mating are
1/16 AA BB
2/16 AA Bb
1/16 AA bb
2/16 Aa BB
4/16 Aa Bb
2/16 Aa bb
1/16 aa BB
2/16 aa Bb
1/16 aa bb

And the phenotypes are
9/16 A_ B_
3/16 A_ bb
3/16 aa B_
1/16 aa bb

(A_ means the phenotype for AA and Aa are the same.)

These results do not change whether A and B are the mutants, A and b are the mutants, a and B are the mutants, or a and b are the mutants.

Paul Hollander

I was hoping to see you respond to this post. I know you know your genetics in and out. So thank you. However i do have a question for you. Towards the bottom of your post you redid the genes of the snakes:
I believe you had the arabesque het albino as h'h'Rr'Aa being that this parent does not carry the hypo gene, why do we need to list it and not just have the genes that we know it carries, being arabesque/normal/albino????

Same thing with the sunglow, you had listed as Hh'r'r'aa, why do we list arabesque as a gene in this snake when he does not carry it?

That part kind of lost me because we are listing genes in the snake that it does not carry, but rather its mate carries. I am just under the frame of mind that we list the genes the snakes carry, and they combine to create those kind of genotypes. If you don't mind Paul, send me an e-mail at Nucci451@ameritech.net I would really like to learn this from someone who seems to have a handle on it, as you do!
Thanks again

I know what Paul was getting at, when running a punnet square you have to account for all four genes between the parents, regardless of what genes each individual has. In Paul's examples the Arabesque albino was shown as h'h'Rr'Aa which is an abbreviation of:

Hypo Locus has two normal genes,
Arabesque locus has one arabesque and one normal gene,
Albino locus has one albino and one normal gene.

Just because a locus does not contain a morph gene does not mean it is not there.

>>That part kind of lost me because we are listing genes in the snake that it does not carry, but rather its mate carries.

The point is both snakes carry genes in that locus, it's just that one only has normal genes while the other has one morph gene paired with a normal.

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Thanks,

Dave Colling

www.rainbows-r-us-reptiles.com

0.1 Wife (WC and still very fiesty)
0.2 kids (CBB, a big part of our selective breeding program)

LOL, to many snakes to list, last count:
24.36 BRB
19.19 BCI
And those are only the breeders

lots.lots.lots feeder mice and rats

I did not pick up on your symbol useage properly. I thought that H = hypo and h' = normal (and the same for R and r'), when it was the other way around. It's a system I hardly ever use. I prefer the mouse system of symbolism (see the link below).

Sorry about that.

Paul Hollander
Rules for Nomenclature of Genes, Genetic Markers, Alleles, and Mutations in Mouse and Rat

I forget who it was who showed me this, but I love it: www.geneticswizard.com
It will make dihybrid crosses a breeze for you.

When using it though there are a few things we need to remember that are often used incorrectly in the herp community. A dominant trait means that there is no visual (phenotypic) difference between the homozygous form and the heterozygous form (though keep in mind, there is no "het" for any of these traits!). A codominant trait means there is a phenotypic difference between the two genotypes, such as in a jungle vs. super jungle. This means that hypomelanism is a DOMINANT trait.

I agree with all you said except "(though keep in mind, there is no "het" for any of these traits!)" refering to a dominant trait.

Of course there is het for any trait whether recessive, dominant, codominant or even multiple possible genes within one allele.

Heterozygous (Het) simply defined is two dis-similar genes within one gene pair. Not possitive but I believe the correct and full way fo saying het albino for example would be heterozygous normal//albino where both of the genes are listed. Just because it's dominant to the normal gene does not mean it's not a het of it has one normal and one mutant gene. Hypo would be heterozygous hypo//normal (totally unsure but I think the dominant gene is listed first but I really could have that backwards)

The one known boa morph that is believed to be related to two different morph genes within one gene pair is paradigm. Technically speaking paradigm by definition has to be a het since there is no paradigm gene and would be listed as heterozygous sharp albino//boawoman caramel (no idea which to list first)
-----
Thanks,

Dave Colling

www.rainbows-r-us-reptiles.com

0.1 Wife (WC and still very fiesty)
0.2 kids (CBB, a big part of our selective breeding program)

LOL, to many snakes to list, last count:
24.36 BRB
19.19 BCI
And those are only the breeders

lots.lots.lots feeder mice and rats

No you are correct. I wasn't clear. I put it in quotes like that just to say you couldn't have a normally appearing boa that was 'het' for salmon for example. In otherwords, the codominant and dominant traits don't visually display heterozygous genotypes the same way as a simple recessive trait does.

Thanks, You re-stated that very well...

you couldn't have a normally appearing boa that was 'het' for salmon for example. In otherwords, the codominant and dominant traits don't visually display heterozygous genotypes the same way as a simple recessive trait does.

I see it time and time again where individuals will mislabel, misrepresent or otherwise just call it wrong saying it can't "be" a het since it is a visual morph. Or one of my biggest pet peeves, "dominant" being use to describe a homozygous ghost, sunglow or hypo. Dominant refers to the method of expression, not the mode of inheritance!
-----
Thanks,

Dave Colling

www.rainbows-r-us-reptiles.com

0.1 Wife (WC and still very fiesty)
0.2 kids (CBB, a big part of our selective breeding program)

LOL, to many snakes to list, last count:
24.36 BRB
19.19 BCI
And those are only the breeders

lots.lots.lots feeder mice and rats

Yeah the punnet square will work but get exponentially more difficult as you increase the number of morphs involved. I calculate the odds this way (and yeah, like everyone stated, single litters tend to not follow punnet rules):

first seperate the individual traits. taking you case a step further to show the ease, how about a DH snow x TH arabesque snow? Or one animal has one albino and one anery gene while the other has one albino, one anery and one arabesque gene.

Three traits, three factors:

Albino (recessive) het x het = 1/4 albino, 3/4 66% het albino
Anery (recessive) het x het = 1/4 anery, 3/4 66% het anery
Arabesque (dominant) het x normal = 1/2 arabesque, 1/2 normal

total phenotypical possible outcomes are the product of each factor. Possible albino phenotypes x anery phenotypes x arabesque phenotypes or 2 x 2 x 2 = 8

From here you can just calculate ythe one(s) you want or all, but have to use the phenotype factor (even if normal) to get the final odds for a given phenotype.

so, the gem would be Arabesque snow or arabesque x albino x anery = 1/2 x 1/4 x 1/4 = 1/32

works the same for any other phenotype
albino 66% het anery (not arabesque). Not arabesque x albino x 66% het anery = 1/2 x 1/4 x 3/4 = 3/32

This branching method can be used for any pairing no matter how complicated. For example, jungle het snow x hypo motley jungle het snow, what are the odds of a hypo, motley, super jungle, snow

5 traits, 5 factors

hypo x normal = 1/2 hypo, 1/2 normal
motley x normal = 1/2 motley, 1/2 normal
jungle x jungle = 1/4 normal, 1/2 jungle, 1/4 super jungle
het albino x het albino = 1/4 albino, 3/4 66% het albino
het anery x het anery = 1/4 anery, 3/4 66% het anery

total number of possible phenotypes is 2x2x3x2x2= 48 WOW

Odds of getting a hypo, motley, super jungle, snow are:

1/2 x 1/2 x 1/4 x 1/4 x 1/4 = 1/128
-----
Thanks,

Dave Colling

www.rainbows-r-us-reptiles.com

0.1 Wife (WC and still very fiesty)
0.2 kids (CBB, a big part of our selective breeding program)

LOL, to many snakes to list, last count:
24.36 BRB
19.19 BCI
And those are only the breeders

lots.lots.lots feeder mice and rats

I think I will get to work on producing that HypoMotleySuperJungleSnow...I can only imagine the name someone will come up with for that combo!!
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Kevin Blumenthal
www.blumenboas.com

Well, if you make the first HypoMotleySuperJungleSnow, you can name it whatever you want!! How about Dave?

and thanks!!
-----
Thanks,

Dave Colling

www.rainbows-r-us-reptiles.com

0.1 Wife (WC and still very fiesty)
0.2 kids (CBB, a big part of our selective breeding program)

LOL, to many snakes to list, last count:
24.36 BRB
19.19 BCI
And those are only the breeders

lots.lots.lots feeder mice and rats

The "Dave" boa it is...checking your math, is it 1/128 or 1/256 odds?
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Kevin Blumenthal
www.blumenboas.com

LOL, you are right, 1/256 is correct, I did it in my head and ya know how that goes, almost as well as my typing skills!!!

-----
Thanks,

Dave Colling

www.rainbows-r-us-reptiles.com

0.1 Wife (WC and still very fiesty)
0.2 kids (CBB, a big part of our selective breeding program)

LOL, to many snakes to list, last count:
24.36 BRB
19.19 BCI
And those are only the breeders

lots.lots.lots feeder mice and rats

One of these days that stuff will sink in. Your explaination seems the easiest to digest. I appreciate you taking the time to go through it in that way.
Thanks.
jsc

np, I love the mind puzzle of genetics.
-----
Thanks,

Dave Colling

www.rainbows-r-us-reptiles.com

0.1 Wife (WC and still very fiesty)
0.2 kids (CBB, a big part of our selective breeding program)

LOL, to many snakes to list, last count:
24.36 BRB
19.19 BCI
And those are only the breeders

lots.lots.lots feeder mice and rats