A Gerbil Owner's Guide to Genetics

Glossery of terms/symbols used on this site:

locus - a "gene place". A Loci controls one aspect of gerbil colors. There are 6 known loci in gerbils. They are as follows:

A - controls whether or not an animal has ticking (dark hairs) or is all one color, and whether they have a white belly or not. The recessive at this locus is a which causes the white belly to disappear and the ticking as well. Agouti turns to black.

P - controls the intensity of the coat and the eye color. The recessive at this locus is p which turns the eyes to ruby and dilutes the coat. Agouti goes to argente or black goes to lilac, etc.

C - controls the coat pattern & the amount of pigment. Two recessives - c(b) which causes the burmese/siamese pattern. c(h) causes the white gerbils we know and love. Combining those 2 genes c(b)c(h) gives you siamese patterns.

E - controls the amount of yellow in the coat. Two reccessives. e causes lots of yellow to the coat creating colors like Dark Eyed Honey and nutmeg. e(f) causes the animal to fade at 8 weeks old. e(f)e is like a lighter version of ee.

G - controls the amount of golden in the coat. the recessive is g which takes all the golden out leaving behind grays. Agouti turns to grey agouti, etc.

Sp - controls white spotting. Sp is a dominant gene and gives a gerbil spots. sp is nonspotted. All spotted gerbils are Spsp because it is homozygous lethal so any spotted paired to any gerbil will always give a mix of spotted and nonspotted babies.

heterozygous - having 2 different genes at a locus. Aa is heterozygous for nonagouti; Cc(h) would be heterozygous for white; c(b)c(h) is heterozygous for burmese and white.

Nonagouti - having the genes aa making the body all one color. Also referred to as "self"

Split - similar to heterozygous. Being "split" for pink eyes, for example, is to be Pp.

phenotype - the way an animal looks, determined by the interaction of its genes.

genotype - an animal's gene code, ex. AaPPCCEEgg.

homozygous lethal - meaning that when a gerbil is homozygous for one trait, the embryo is inviable and dies before birth. This is the case with Sp in gerbils. All SpSp gerbils die in the womb, meaning that 2 spotted paired together would have approximately 25% less babies than they would have because 25% of their pups are SpSp ie inviable.

* or - are symbols that stand for "any gene". That means whether the gene there is dominant or recessive makes no difference, the animal would look just the same. Agouti for example is A*P*C*E*G*. A gerbil that is AAPPCCEEGG is agouti as is a gerbil that is AaPpCc(h)EeGg. It can also be written as A-P-C-E-G-.

Say you have an agouti gerbil. You want to introduce a new trait into the family so you decide to breed it to a black, and then try to introduce black into your gerbils.

To view chart without my comments click here

What happened here?

The agouti was paired to a black. As your family was AA, contained no nonagouti (black) gene, all the resulting pups were agouti. *But* these agouti pups were split for nonagouti (Aa). As you can see the best option is to pair one of these to another black (aa). That way you end up with 50% black and 50% agouti, and all the agoutis are split for black. Whenever you want to maximize the amount of a recessive trait you get from a carrier, you have to pair them to a gerbil that's homozygous for that trait.

Another option is to pair 2 of these agouti siblings together (or to another agouti you know carries a). But this yields only 25% black, and 75% agouti--and only 2 out of every 3 agouti will carry black. Also on this chart I showed 2 blacks being paired together. Since they were both homozygous for a, all of their babies had to be aa -- therefore they were all black.

If you were to pair one of the original pair's Aa offspring to an agouti that doesn't carry black, as you can see you get nothing but agoutis. But 50% of them carry black--but you won't know which ones.

Bear in mind this is all assuming that there are no other recessive genes involved.

If you want 2 traits in your line (for example if you want to breed lilac) things get a bit more complicated.

To see this chart without my comments, click here

To begin with you'd need to start with either A. two gerbils, one which expresses one trait, another which expresses another or B. a gerbil that expresses none of the traits and another that expresses both. In the above example we took a black (which is aa) and an argente (which is pp) and bred them together. In this example, again, we're assuming neither carried any other genes. All their babies were agouti, but they all carried a and p (they were AaPp). The best option here is to pair an agouti to a lilac. Lilac is homozygous for a and p, so you can get AaPp babies (agouti), Aapp (argente), aaPp (black) and aapp (lilac), all in equal amounts.

What I also did as an example was showed one of these heterozygous agoutis to a black. If the black carried no other genes you would only get 50% agouti and 50% black, but you got lucky here and the black was Pp. You got the same colors as from agouti x lilac but different *percentages* (I will explain that below). You got 37.5% agouti, 37.5% black, 12.5% lilac and 12.5% argente. Additionally, the agoutis and blacks aren't all Pp. about 2/3 will be and 1/3 won't. So this isn't the best option--it's always best to maximize your odds by pairing a heterozygous to a homozygous.

In another example I paired an agouti to an argente. Here you *didn't* get lucky and the argente was AA. So they had 50% agouti and 50% argente, and 2/3 would carry a and 1/3 wouldn't.

For 3, 4, and more genes things get more and more complicated, but folow the same basic formula. First step is combining parents that have those traits, and pairing their children to each other or two other gerbils that express some or all of those traits. You can't go wrong doing that.

Spotting is very common and so far is the only dominant mutation in gerbils (that is obvious). But it's tricky. Some people think that since it's dominant they will get ALL spotteds when a spotted is paired to a nonspot. This isn't the case. Approx. half will be spotted and half won't be, because ALL spotteds are Spsp meaning they are heterozygous. Paired to any gerbil, spotted or otherwise, they will always have spotted AND nonspotted babies. This is because homozygous (SpSp) spotteds are reabsorbed in the womb because it is "homozygous lethal". Here is another chart.

To view this chart without my comments click here

Well that seems simple enough.

Some people don't care for percentages but it's kind of a silly idea to plan a pair for a color that has only a 2% chance of that color if you don't want all the others. So here is how it works. You make a large square and divide that into four squares. On the outside of the top part you write the father's genes for just ONE locus. Then down the left side you put the 2 genes the mother has on that same locus. Then you cross multiply. You make one for every locus. Here:

Then, as you can see, you see how many squares out of four each trait is in. 1 out of 4 is 25%, 2 out of 4 is 50%, etc. If, in a box you have possibilities that are Xx and some that are XX then you lump them all together as X*, because they would all have the same phenotype. Write down all the percentages for the the possibilities in the box like I did. Then, once you're down, come up with as many combinations as you can (there are usually not as many as you might think. The possibilities for this pair are:

A*PpC*E*Gg - agouti
aaPpC*E*Gg - black
A*PpC*E*gg - grey agouti
aaPpC*E*gg - slate
A*Ppc(h)c(h)E*Gg - dark tailed white
aaPpc(h)c(h)E*Gg - dark tailed white
A*Ppc(h)c(h)E*gg - dark tailed white
aaPpc(h)c(h)E*gg - dark tailed white

You would lump all the c(h)c(h) (white) into one category since they'd all be essentially the same. Then you multiply the percentages at each loci to find out the percentage for each color. See here:

Agouti (A*PpC*E*Gg):

A* (75%) x Pp (100%) x C* (75%) x E* (100%) x Gg (50%) = 28.125%

aa (25%) x Pp (100%) x C* (75%) x E* (100%) x Gg (50%) = 9.375%

A* (75%) x Pp (100%) x C* (75%) x E* (100%) x gg (50%) = 28.125%

aa (25%) x Pp (100%) x C* (75%) x E* (100%) x gg (50%) = 9.375%

** (100%) x Pp (100%) x c(h)c(h) (25%) x E* (100%) x *g (100%) = 25%

Maybe that white is throwing you off. ** = 100%??? You say. Well, yes. * stands for any gene, right? So since c(h)c(h) makes it white no matter what the rest of the genes are, then a white could be AA or Aa or aa and it'd make no difference, so it's **. Any gene can be there.

With the punnet squares we can also draw other conclusions. For example take note of the E locus. The argente cream was Ee but the grey agouti wasn't, so we didn't get any ee pups. But as you can see 50% of the pups will inherit the e gene. However it's unlikely you'd be able to tell which ones. Similarly if you look at the A locus you can see that 66% of the gerbils that are A* (grey agouti or agouti) will be Aa and 33% will be AA. Again, you can't tell.

I Still Don't Get It!!

No I will NOT explain it again. But don't fret, for Joerg has created a Gerbil Gene Calculator! (yeah, I know, why didn't I tell you that to begin with ;)