# Boa Genetics for Idiots



## dazt2801 (Dec 3, 2009)

Can someone please help me here. Despite me searching on here and around the net, I cant for the life of me work out Boa Genetics.
I am after what genes are Recessive, Double Recessive, Co-Dominant and Dominant. I am having a blonde moment and cant seem to find anything.

I am also after help with the punnet table in working the genetics out. Is it like this. If one snake has 3 genes and the other 2 genes, you put 3 across the top and 2 down the side, and work it out from there. How would you then get the percentage for each type of snake.

Any help would be greatly appreciated as I am starting to tear my hair out, and there aint much left. :whistling2:


----------



## james boa (Jan 24, 2009)

hi mate try genetics at the top is the boa calculater.
http://www.reptileforums.co.uk/forums/genetics/258989-boa-genetics-outcomes-albino-anery.html


----------



## BecciBoo (Aug 31, 2007)

JnB Boas - Genetics

JnB Boas - Morph List

: victory:


----------



## paulh (Sep 19, 2007)

dazt2801 said:


> I am after what genes are Recessive, Double Recessive, Co-Dominant and Dominant. I am having a blonde moment and cant seem to find anything.
> 
> I am also after help with the punnet table in working the genetics out. Is it like this. If one snake has 3 genes and the other 2 genes, you put 3 across the top and 2 down the side, and work it out from there. How would you then get the percentage for each type of snake. :whistling2:


Use BecciBoo's link to the JnB Morph List

Have you gone through the stickies in the RFUK genetics forum. Here's another web site that is helpful:
Main Menu

Both parents have the same gene pairs. If the first parent has three gene pairs containing mutant genes and the second parent has two of the same gene pairs containing mutants, then the second parent's third gene pair contains a pair of wild type (AKA normal) genes.

by the way, it is best to start with a problem with one gene pair before trying to do problems with two or more gene pairs.

With the Punnett square, the adult snakes' genes are in pairs, but only one gene from each pair goes into a sperm or egg. Each gene pair either has two identical genes or two nonidentical genes. If the two genes in a gene pair are identical, then all the sperm (or eggs) have one copy of that gene. If the two genes in a gene pair are differentl, then half the sperm (or eggs) gene have one copy of one gene and half get a copy of the other gene. The list of possible sperm goes across the Punnett square's top, and the list of possible eggs goes down the Punnett square's left side. Or vice versa; it doesn't really matter which is across the top. So if there are three gene pairs in the problem and the genes in each gene pair are different, then there are 2 x 2 x 2 = 8 possible gene combinations in the sperm (or eggs). When a sperm fertilizes an egg, the gene pairs are reestablished.

To get the percentage of each kind of baby, you make a list of all the different kinds of babies in the Punnett square. And by each kind of baby, you put the number of boxes that kind of baby appears in. Then you devide the number of boxes that baby appears in by the total number of boxes in the square.


----------



## dazt2801 (Dec 3, 2009)

*cheers*

Thanks for all the replies and help. I was having one of those moments last night where I just couldn't seem to do anything right. My brain is kind of in gear this morning so we will soon see. Thank you.


----------



## dazt2801 (Dec 3, 2009)

*is this a moonglow maker*

Tried to use my head here and did the punnet square for the following pair of snakes. 

Hypo het albino het anery boa male with a hypo het albino het anery boa female. 

I got some results but I don't think I did it right as I have come out with a lot of combinations. Can anyone see what they would get and put it on this thread along with the % chance. 

Cheers all.


----------



## Caz (May 24, 2007)

There was an excell program for boa genetics that i lost in a computer crash - anyone know where it's listed at all?


----------



## paulh (Sep 19, 2007)

dazt2801 said:


> Hypo het albino het anery boa male with a hypo het albino het anery boa female.
> 
> I got some results but I don't think I did it right as I have come out with a lot of combinations.


I did this a while back.

Both snakes have a salmon (AKA hypo) mutant gene (symbol = Sa) paired with a normal gene (symbol = +). So each gamete (collective term for sperm and eggs) has either an Sa gene or a + gene.

Both snakes have an albino mutant gene (symbol = a) paired with a normal gene (symbol = +). So each gamete has either an a gene or a + gene.

Both snakes have an anerythristic mutant gene (symbol = an) paired with a normal gene (symbol = +). So each gamete has either an an gene or a + gene.

This produces 8 possible gene combinations in the sperm or eggs:
Sa a an
Sa a +
Sa + an
Sa + +
+ a an
+ a +
+ + an
+ + +

8 sperm across the top by 8 eggs down the side produces a Punnett square with 64 boxes.

By the way, "//" stands for a pair of chromosomes. So Sa//+ means a pair of chromosomes with an Sa mutant gene in one chromosome and a + (normal) gene in the same location in the other chromosome.

Final result:
Sa//+ +//a +//an (salmon het albino het anerythristic) X
Sa//+ +//a +//an (salmon het albino het anerythristic) -->
1/64 (1.5625%) Sa//Sa a//a an//an (super moonglow)
2/64 (3.125%) Sa//Sa a//a +//an (super sunglow het anerythristic)
1/64 (1.5625%) Sa//Sa a//a +//+ (super sunglow)
2/64 (3.125%) Sa//Sa +//a an//an (super ghost het albino)
4/64 (6.25%) Sa//Sa +//a +//an (super salmon het albino het anerythristic)
2/64 (3.125%) Sa//Sa +//a +//+ (super salmon het albino)
1/64 (1.5625%) Sa//Sa +//+ an//an (super ghost)
2/64 (3.125%) Sa//Sa +//+ +//an (super salmon het anerythristic)
1/64 (1.5625%) Sa//Sa +//+ +//+ (super salmon)
2/64 (3.125%) Sa//+ a//a an//an (moonglow)
4/64 (6.25%) Sa//+ a//a +//an (sunglow het anerythristic)
2/64 (3.125%) Sa//+ a//a +//+ (sunglow)
4/64 (6.25%) Sa//+ +//a an//an (ghost het albino)
8/64 (12.5%) Sa//+ +//a +//an (salmon het albino het anerythristic)
4/64 (6.25%) Sa//+ +//a +//+ (salmon het albino)
2/64 (3.125%) Sa//+ +//+ an//an (ghost)
4/64 (6.25%) Sa//+ +//+ +//an (salmon het anerythristic)
2/64 (3.125%) Sa//+ +//+ +//+ (salmon)
1/64 (1.5625%) +//+ a//a an//an (snow)
2/64 (3.125%) +//+ a//a +//an (albino het anerythristic)
1/64 (1.5625%) +//+ a//a +//+ (albino)
2/64 (3.125%) +//+ +//a an//an (anerythristic het albino)
4/64 (6.25%) +//+ +//a +//an (normal het albino het anerythristic)
2/64 (3.125%) +//+ +//a +//+ (normal het albino)
1/64 (1.5625%) +//+ +//+ an//an (anerythristic)
2/64 (3.125%) +//+ +//+ +//an (normal het anerythristic)
1/64 (1.5625%) +//+ +//+ +//+ (normal)

See mating 129 in three loci


----------



## dazt2801 (Dec 3, 2009)

*genetics*

Many thanks for that. That was what I got. I didn't think it was right due to the amount of combinations. Cheers.


----------



## dazt2801 (Dec 3, 2009)

*super genes*

Cheers paulh for the amazing explanation in this thread. I am upto speed now on the punnet squares. The only thing that is throwing me now, is how would you express a super gene in the punnet table. For instance a super sunglow. This has 2 hypo genes but how would you express this in the table. 
Once this is cleared up, that's it. I can work any morph combination out then.


----------



## paulh (Sep 19, 2007)

dazt2801 said:


> The only thing that is throwing me now, is how would you express a super gene in the punnet table. For instance a super sunglow.


Super is herper slang for a homozygous pair of dominant or codominant mutant genes. In other words, super salmon = homozygous salmon. Treat the gene pair the same as any other homozygous pair of genes.

Super sunglow is not produced by a single gene pair. A super sunglow boa has two salmon genes and two albino genes (Sa//Sa a//a). All of its sperm (or eggs) would contain an Sa mutant gene and an a mutant gene. 

If two super sunglow boas are mated, then the Punnett square would have a single column with "Sa a" at the top and a single row with "Sa a" at the left side. This gives a single cell containing Sa//Sa a//a. In other words, all the babies would be super sunglows.

By the way, R.C. Punnett (note spelling) was an early British geneticist at the UK University of Cambridge. Among other things, he wrote a good early genetics text named "Mendelism". There have been minor changes in terminology since then and the addition of wild type as the standard of comparison. But the book is still good, as long as one can live without the molecular genetics in modern texts. It can be downloaded from the Project Gutenberg web site or one of its mirrors.


----------



## paulh (Sep 19, 2007)

Wilmer Miller has a downloadable genetics text named A Survey Of Genetics at 
Wilmer Jay Miller's web site

Starting on page 271, there are some genetics problems that might stretch your abilities.


----------

