# A bit of Punnett square confusion!



## KELK CORNS (Aug 25, 2012)

Hi guys! I'm just starting to learn how to work out Punnett squares and could just do with a little help.

I have a male khal sunklow and a female motley 100% het khal albino, now I know that the motley and the hypo in the sunglow are co-dominant and it's this that is throwing me off a little, basically how would these be carried around the Punnett square? Could someone tell me how to and what letters to put in the Punnett square? All help is appreciated I'm trying my best to learn lol! Thanks Sam.


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## paulh (Sep 19, 2007)

All genetics problems are worked out the same way:
1. Identify the parents' genes. Symbolize them to save bandwidth.
2. Determine the gene(s) in the parents' possible sperm and eggs.
3. Determine the genes in all possible fertilized eggs. Use your preferred method -- the Punnett square, branching system, etc.
4. Combine identical genotypes, as necessary.
5. Assign a phenotype to each genotype.
6. Combine identical phenotypes and delete every unnecessary "normal".

male Kahl sunglow mated to female motley 100% het Kahl albino

This problem has two gene pairs:
male gene pair #1: Salmon mutant gene (AKA hypo) (Sa) and normal gene (+) = Sa//+
male gene pair #2: two albino mutant genes (a) = a//a 

female gene pair #1: motley mutant gene (Sa^m) and normal mutant gene = Sa^m//+
female gene pair #2: normal gene (+) and albino mutant gene (a) = +//a

// = a pair of chromosomes. The +//a gene pair is made up of a normal gene in one chromosome and an albino mutant gene in the same location in the other chromosome.

Sa^m stands for Sa with m as a superscript. As far as we know, the salmon (AKA hypo) mutant gene, the motley mutant gene, and the corresponding normal gene reside at the same location in the chromosomes. This allows the formation of 6 gene pairs:
1) 2 normal genes
2) two salmon mutant genes
3) two motley mutant genes
4) a salmon mutant gene and a normal gene
5) a motley mutant gene and a normal gene
6) a salmon mutant gene and a motley mutant gene

Each sperm or egg gets one gene from each gene pair in the parents.

Half the sperm get an Sa mutant gene and the other half get the + gene from the first gene pair. All the sperm get an a gene from the second gene pair. So there are two types of sperm -- Sa a and + a. These go on the top of the Punnett square.

Half the eggs get an Sa^m mutant gene and the other half get a + gene from the first gene pair. Half of the eggs get a + gene and the other half get an a gene from the second gene pair. This produces 4 types of eggs --
Sa^m +
Sa^m a
+ +
+ a

The four types of eggs go down the left side of the Punnett square.

That's the first two steps in working out the problem. I'll let you finish the problem's last 4 steps.


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## paulh (Sep 19, 2007)

http://www.reptileforums.co.uk/foru...cs-outcomes-albino-anery-18.html#post10081993

Posts 173-178 have an assortment of boa matings. The parent and offspring genotypes are given, but the work is not shown. Try seeing if you can work some of the problems and get the same answers.

Good luck.


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## KELK CORNS (Aug 25, 2012)

Cheers for the advice Paul, ill try and get mi head around it lol!


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## KELK CORNS (Aug 25, 2012)

Right buddy lets see If I'm along the right lines, after doing the Punnett square here is what I've come up with,

Overall there is a 1 in 8 chance of getting any of these morphs.

Sa^m + // Sa a = HYPO MOTLEY 100% het albino.
Sa^m + // + a = MOTLEY 100% het albino.
Sa^m a // Sa a = SUNGLOW MOTLEY.
Sa^m a // + a = ALBINO MOTLEY.
++ // Sa a = HYPO 100% het albino.
++ // + a = NORMAL 100% het albino.
+a // Sa a = SUNGLOW.
+ a // + a = ABINO.

I hope at least a part of this is correct as it did take a little while to work out, thanks again Paul h and correct me if I'm totally wrong!


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## paulh (Sep 19, 2007)

Mostly correct. Pretty good for a beginner!

Correct, overall there is a 1 in 8 chance of getting any of the morphs below.

100% het albino is a long way of writing het albino.

The Punnett square results are correct, except each gene pair is written separately. Because as far as we know, the albino gene and the salmon gene are in different chromosome pairs. Hypo motley het albino should be written Sa^m//Sa +//a instead of Sa^m + // Sa a. Corrections in red.

Sa^m + // Sa a = HYPO MOTLEY 100% het albino = Sa^m//Sa +//a
Sa^m + // + a = MOTLEY 100% het albino = Sa^m//+ +//a
Sa^m a // Sa a = SUNGLOW MOTLEY = Sa^m//Sa a//a
Sa^m a // + a = ALBINO MOTLEY = Sa^m//+ a//a
++ // Sa a = HYPO 100% het albino = Sa//+ +//a
++ // + a = NORMAL 100% het albino = +//+ +//a
+a // Sa a = SUNGLOW = Sa//+ a//a
+ a // + a = ABINO = +//+ a//a


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## KELK CORNS (Aug 25, 2012)

Aaaaah right I see what you mean buddy I've had a go at a few others and compared them to your predictions and I'm not 100% there but nearly I think, hopefully one day il finally understand genetics! Do you know of an easier way than Punnett squares?


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## paulh (Sep 19, 2007)

KELK CORNS said:


> ... Do you know of an easier way than Punnett squares?


Yes. The branching system. Eliminates the additions required for the Punnett square.

http://www.reptileforums.co.uk/forums/genetics/2-learning-genetics-7.html

See post 67.

Must log out. Back tomorrow.


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## KELK CORNS (Aug 25, 2012)

Cheers buddy ill let you know how I get on regards Sam.


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