# dominant & co - dominant genes



## atum (Jun 1, 2009)

to all the leo enthusiasts 

okay i kinda have my head round the whole genetics thing, very basically, i get the whole recessive gene with albino, patternless etc and line bred with carrot tails, hypos etc. if only leo's laid 4 eggs at a time. would make the whole punnet square a lot easier

double recessive genes are causing a bit more of a headache but i kinda grasp it (or pretend i do)

and dominant / co dominant genes. a morph that, when bred to a normal for instance, the morph gene is dominant over the normal gene. 

anyways, at the week end i got an enigma leo. i've not been keeping up with the reptile world for a while so this is a new one for me. my girlfriend picked him / her out and i must admit it is rather stunning. i've tried to understand what makes an enigma an enigma but i don't seem to be grasping it. what are the traits?

and if i was to to breed a co - dominant enigma (Ne) to a normal (NN) on the punnet square we have a chance of getting (Ne). but because the enigma is a dominant gene that would be the phenotype of the leo correct?

so if we breed co - dominant enigma (Ne) to another co - dominant enigma (Ne) there's a chance that you may be looking at two eggs both holding enigma phenotypes (Ne). but there's also a chance one of these eggs is (ee) then making it dominant? 

and if you we're to breed this dominant enigma (ee) to a normal (NN) you would be looking at a certainty of all co - dominant enigmas (Ne). but if you we're to breed the dominant (ee) with a co - dominant (Ne) there's a chance that both egg's are dominant enigmas

is there any way of telling you have a dominant enigma ie. appearance? 

and i've seen others such as tangerine enigmas. this is a stab in the dark is this starting to look into mixing a morph with a line bred tangerine? as the enigma is dominant that gene it will be the phenotype so how does the tangerine come into it?


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## forgottenEntity (Sep 7, 2008)

As far as I know, there's no visual difference between a Leo carrying a single Engima gene and a Leo carrying two.


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## HadesDragons (Jun 30, 2007)

Tangerine is - like you said - just mixing a line-bred trait in with a simple trait. Think of it that a leo is either Enigma (e) or not (N) like you said, then once that's determined the amount of tangerine can be "added" to alter the effects of the Enigma gene.

As far as has been shown so far, the Engima trait shows full dominance - ie a het Enigma (Ne) looks the same as a **** Enigma (ee) - ie both look like visual Enigmas. The only way to tell them apart is by breeding trials. There has been suggestions made that Enigmas produced from an Enigma x Enigma pairing are more prone to circling and other abnormal behaviour, but from what I can tell this is just anecdotal, and no proper survey has been done.

The rest of what you said is pretty much correct:

Ne x NN:
50% Ne
50% NN

Ne x Ne:
25% ee
50% Ne (bearing in mind these will look the same as the ee ones, so technically it's 75% "Enigma", of which 1/3 are het, 2/3 ****)
25% NN

ee x NN:
100% Ne

ee x Ne:
50% Ne
50% ee (so all visual Enigmas, but on average half should come out as being **** - you'd have to prove them by breeding)


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## atum (Jun 1, 2009)

okay so it's best to presume that if he / she was to be bred that it is (Ne) i.e het and not ****?

all to do now is try and get my head fully round double recessives. 

i've also noticed that the enigma gene has been used with albinos and other recessive genes. any quick overview as to how this works?

thanks


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## HadesDragons (Jun 30, 2007)

atum said:


> okay so it's best to presume that if he / she was to be bred that it is (Ne) i.e het and not ****?
> 
> all to do now is try and get my head fully round double recessives.
> 
> ...


The majority of Enigmas around will have been produced from Ne x NN pairings, so unless the breeding trials prove otherwise I'd assume it's a het Enigma.

Double recessives can be worked out exactly the same as single recessive genes, except the punnet square would have 4 options at the top and on the side, giving 16 squares rather than the four from a single gene.


Alternatively, work out one gene first, then work out the second gene for *each* of the possible outcomes for the first gene:

e.g. het Tremper Albino (Tt) het Blizzard (Bb) x Blizzard (bb) het Talbino (Tt)

If we work out the effects of the Talbino gene first:
*Tt x Tt:*
25% TT
50% Tt
25% tt

Then add in the effects of Blizzard; *Bb x bb* will give 50% Bb and 50% bb, so for each of the possible outcomes above, you add those:

So, of the *25% TT* babies, 50% will be Bb (het Blizzard), and 50% will be bb (visual Blizzard). So Overall it's 12.5% Normal het Blizzard, 12.5% visual Blizzard. Do the same for the others:

*50% Tt* = 25% Tt Bb and 25% Tt bb

*25% tt* = 12.5% tt Bb and 12.5% bb

Therefore overall you'd get:

12.5% TT Bb (normal het Blizzard)
12.5% TT bb (Blizzard)
25% Tt Bb (double het Talbino and Blizzard)
25% Tt bb (Blizzard het Talbino)
12.5% tt Bb (Talbino het Blizzard)
12.5% tt bb (Talbino Blizzard)


The hets would look the same as the non-hets, so unless you proved them out, all non Albino babies would be "66% poss. het Talbino", so you add all of the het Talbino and non-het Talbino babies together (e.g. 12.5% TT bb + 25% Tt bb would give you 37.5% of the clutch being visual Blizzards, with each having a 66% chance of being het for Talbino).

So, overall:

37.5% het Blizzard (66% poss. het Talbino)
37.5% Blizzard (66% poss. het Talbino)
12.5% Talbino het Blizzard
12.5% Talbino Blizzard


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## atum (Jun 1, 2009)

okay it's slowly sinking in. think i'll draw up some of my own punnet squares later as well. it'll make more sense when it's right in front of me


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## HadesDragons (Jun 30, 2007)

atum said:


> okay it's slowly sinking in. think i'll draw up some of my own punnet squares later as well. it'll make more sense when it's right in front of me


It just takes practice to get it to sink in properly - I'd recommend having a look on Google to find out which genes do what, then coming up with some completely random pairings and working out the odds for all the babies.


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## paulh (Sep 19, 2007)

As enigma is dominant to normal, it would be better to to use E for enigma and either n or + for normal. The + character, either alone or as a superscript, is the internationally recognized symbol for the normal allele at each location in the genome. The convention is that if a mutant gene's symbol is written all in lower case, then the gene is recessive to the normal gene. And all the letters of the symbol for a mutant gene that is dominant or codominant to the normal gene are lower case except the first letter, which is upper case.

Once you get your head around double recessives, you can do any two gene pair problem. Two gene pairs containing dominant mutant genes or a gene pair with a recessive mutant and a gene pair with a dominant mutant, etc. All of them are done the same way.

1. Identify all the genes and assign a unique symbol that is easy to remember. For example, a is for Tremper albino, and + is for Tremper albino's normal allele. The letter b is an acceptable choice for blizzard, and + is for blizzard's normal allele. And "//" stands for a pair of chromosomes. So "+//a" is a pair of chromosomes with the Tremper albino mutant gene on one chromosome and the normal version of the gene in the same place (the a locus) in the other chromosome. "+//b" is a pair of chromosomes with the blizzard mutant gene on one chromosome and the normal version of the gene in the same place (the b locus) in the other chromosome. The a locus and the b locus are in different chromosome pairs.

2. Identify the relevant genes in each parent. Let's say that both parents are heterozygous Tremper albino and heterozygous blazing blizzard. The male is +//a +//b, and the female is +//b. (Edit. Sorry, I made a mistake. Here's the correction: The female is +//a +//b.)

3. Determine the possible gene combinations in the sperm and eggs.

If a +//a +//b parent produces an egg with a + gene at the a locus, it has either a + gene or a b gene at the b locus. If a +//a +//b parent produces an egg with an a gene at the a locus, it has either a + gene or a b gene at the b locus.

That makes four possible combinations of genes in the eggs:
+ +
+ b
a +
a b

The male has the same four possible gene combinations in his sperm because he has the same set of genes as the female.

4. Use those four gene combinations on both the top and the side of the Punnett square.

5. Make the combinations of sperm and egg. Add the identical combinations together, and assign phenotypes.

+//a +//b mated to +//a +//b produces
( or +//a +//b X +//a +//b -->)
1/16 +//+ +//+ = normal
2/16 +//+ +//b = normal
1/16 +//+ b//b = blizzard
2/16 +//a +//+ = normal
4/16 +//a +//b = normal
2/16 +//a b//b = blizzard
1/16 a//a +//+ = Tremper albino
2/16 a//a +//b = Tremper albino
1/16 a//a b//b = Tremper albino blizzard (blazing blizzard)

There are two possible genes for the a locus -- + and a. These can be combined into 3 pairs -- +//+, +//a, and a//a. There are two possible genes for the b locus -- + and b. These can be combined into 3 pairs -- +//+, +//b, and b//b. 

These possible gene pairs can be arranged into 36 possible matings. The result of one mating has been calculated for you. 
+//+ +//+ X +//+ +//+ 
+//+ +//+ X +//+ +//b 
+//+ +//+ X +//+ b//b 
+//+ +//b X +//+ +//b 
+//+ +//b X +//+ b//b 
+//+ b//b X +//+ b//b 
+//+ +//+ X +//a +//+ 
+//+ +//+ X +//a +//b 
+//+ +//+ X +//a b//b 
+//+ +//b X +//a +//b 
+//+ +//b X +//a b//b 
+//+ b//b X +//a b//b 
+//+ +//+ X a//a +//+ 
+//+ +//+ X a//a +//b 
+//+ +//+ X a//a b//b 
+//+ +//b X a//a +//b 
+//+ +//b X a//a b//b 
+//+ b//b X a//a b//b 
+//a +//+ X +//a +//+ 
+//a +//+ X +//a +//b 
+//a +//+ X +//a b//b 
+//a +//b X +//a +//b --> done; results above
+//a +//b X +//a b//b 
+//a b//b X +//a b//b 
+//a +//+ X a//a +//+ 
+//a +//+ X a//a +//b 
+//a +//+ X a//a b//b 
+//a +//b X a//a +//b 
+//a +//b X a//a b//b 
+//a b//b X a//a b//b 
a//a +//+ X a//a +//+ 
a//a +//+ X a//a +//b 
a//a +//+ X a//a b//b 
a//a +//b X a//a +//b 
a//a +//b X a//a b//b 
a//a b//b X a//a b//b 

<Mission Impossible theme> The result of one of the 36 matings has been calculated. Your assigment, should you choose to accept it, is to work out the result of each of the other 35 matings. </Mission Impossible theme>

Have fun!


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## atum (Jun 1, 2009)

whoa, all looks a bit different to what i'm use to :gasp:

tonight i will give that a proper read through and start attempting to work all them out


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## atum (Jun 1, 2009)

okay think i have the possible results for the first 5. not mega sure i've done it correctly though

+//+ +//+ = all normal

12/16 +//+ +//+= normal
4/16 +//b +//b normal

9/16 +//+ +//+ = normal
6/16 +//+ +//b = normal
1/16 +//+ b//b = blizzard

4/16 +//+ +//+ = normal
8/16 +//+ +//b = normal
4/16 +//+ b//b = blizzard

1/16 +//+ +//+ = normal
6/16 +//+ +//b = normal
9/16 +//+ b//b = blizzard


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## paulh (Sep 19, 2007)

Matings 1 and 4 are correct. The others are not.

The answer to mating 2 has "4/16 +//b +//b". I think the first +//b is a typo where you meant to put +//+. But the fractions would still be wrong.

Simplifying is good in this sort of thing because it reduces work and work time. 

Mating 2 is +//+ +//+ X +//+ +//b. The genotype for the right parent's a locus is +//+. Half the babies get the left chromosome with a + gene and half the babies get the right chromosome with a + gene. But the two + genes are identical, so it doesn't matter which baby gets which gene. So all the babies get a + gene at the a locus from the right parent. Half the babies also get the b locus + gene, and half the babies also get the b locus b gene from the right parent. So the right parent makes only two types of sex cells -- + + and + b.

We look at the left parent. Both genes in the a locus gene pair are +, and both genes in the b locus are +. So that parent has only one type of sex cell -- + +.

So put + + and + b on the top of your Punnett square and + + on the left side. Now you have a Punnett square with two boxes, which is much easier to do than a Punnett square with 16 boxes.

Try matings 2, 3, and 5 again. Simplify the sex cells as I did here and post the new results.


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