What Color is Your Horse? Coat Color Genetics: Part Two
what color is your horse?: coat color genetics: part II
by anita enander (ARABIAN HORSE WORLD – Aug 12)
This is the second in a series of articles on coat color. (See Part I in the July 2012 issue, page 216.) Next month we’ll look at the complexities found in Half-Arabians, where genes for colors that are found in other breeds come into play.
What Color Will the Foal Be?
The foal’s coat color is determined by the genotype of its parents. Colors don’t really “jump” generations. However, genes that are not expressed in the parents can be passed on to the foal and result in a color different from either parent. How can that happen? Each parent has two copies of each gene, but passes on only one copy to the foal. It is the combination that determines the foal’s color.
In last month’s article we identified the three genes that influence coat color: Extension (E or e) for Black/Chestnut; Agouti (A or a) to limit Black to the points and create Bay; and Grey (G or g) that causes gradual loss of hair color.
Plant a Radish, Get a Radish, Never Any Doubt
(apologies to The Fantastiks)
There are a few guaranteed results. Here are two:
1. Chestnut to Chestnut produces a Chestnut foal. That is because the coat color of both parents is ee (homozygous for the recessive base coat color) AND both parents are gg (homozygous for the recessive form of the Grey modifier). Therefore, the foal can also only be ee and gg.
Regarding gg there is one more predictable outcome:
2. If you breed two non-Greys together, you will get another non-Grey. It doesn’t matter if the parents are Black, Chestnut, or Bay. Because they are NOT Grey, they have to be gg and can only pass on the recessive (non-Greying) form of the gene.
To Grey, or Not To Grey, That is the Question
If there is one Grey parent (so at least one copy of the dominant G that causes hair pigment loss), there’s a possibility that the foal will be Grey; but the foal must inherit one copy of the G to turn Grey.
Chestnut x Bay =
Chestnut (5) – The Chestnut parent gives e. The Bay parent must be either Ee or EE. If it is Ee and gives an e, the foal will be Chestnut. The agouti of either parent doesn’t matter.
Black (6) – The Chestnut parent gives e. The Bay parent must be either Ee or EE. If it is Ee and gives an E, or if it is EE and therefore has to give an E, the base coat will be Black. Agouti matters. To remain Black, the foal must be aa. That means the Chestnut parent must be either Aa and give the a or aa and will always give an a. The Bay parent must be Aa and give the a.
Bay (7) – The Chestnut parent gives e. The Bay parent must be either Ee or EE. If it is Ee and gives an E, or if it is EE and therefore has to give an E, the base coat will be Black. Agouti matters. To be Bay, the foal has to inherit one A from either parent. That means the Chestnut parent must be either AA or Aa and give the A, or the Bay parent must be AA or Aa and give the A.
Black x Black =
Chestnut (8) – To get Chestnut from two Blacks, both parents must be Ee and both must give the e to the foal. That makes the foal ee. Agouti must be aa for the Black in both parents, so the Chestnut foal will also be aa, but it doesn’t matter for the color of this foal.
Black (9) – To be Black, the foal only needs one E from either parent. It doesn’t matter if the parents are EE or Ee. If both parents are EE, the foal will get an E from each parent, making the foal EE and Black (homozygous to homozygous gives homozygous). If one parent is EE and one is Ee, it will certainly get an E from the EE parent and be Black. If both parents are Ee, the foal only needs an E from either parent to be Black. Agouti can’t influence here because both parents must be aa to be black. So the foal will also be aa.
NOTE: Bay is not possible because neither parent can have an A.
Black x Bay =
Chestnut (10) – Both parents could be EE or Ee. But to get Chestnut, both have to be Ee and both have to contribute the e. Agouti can’t influence here, even though the Bay parent has to have at least one A.
Black (11) – Both parents could be EE or Ee. If the foal gets one E from either parent, the base coat will be Black. Agouti matters. To remain Black, the foal has to get A from both parents. The Black parent must be aa, so can only give a. The Bay parent could be AA or Aa. If the Bay parent is Aa and gives a, the foal will remain Black.
Bay (12) – Both parents could be EE or Ee. If the foal gets one E from either parent, base coat will be Black. Agouti matters. The Black parent must be aa, so will give an a. The Bay parent can be AA or Aa. If the Bay parent is AA it must give A and the foal will be Bay. If the Bay parent is Aa and gives A, the foal will be Bay.
Bay x Bay =
Chestnut (13) – Each parent could be EE or Ee. If both parents are Ee, and if both give the e, then the foal will be Chestnut. Agouti doesn’t matter.
Black (14) – Each parent could be EE or Ee. If either parent gives an E, the base coat will be Black. Agouti could matter. To remain Black, both parents have to be Aa and both must give the a.
Bay (15) – Each parent could be EE or Ee. If either parent gives an E, the base coat will be Black. Agouti matters. Each parent could be either AA or Aa. If at least one parent is AA, the foal will get an A from that parent and be Bay. If both parents are Aa and the foal gets an A from either parent, it will be Bay.
Let’s try two examples:
Example 1: We’ll start with two parents, both Grey, but with different genotypes. One is homozygous for Grey (GG) and the other is heterozygous (Gg). There are four ways for these genes to combine in the foal:
In all cases, the foal will be Grey because it will always inherit one dominant Grey gene from the homozygous (GG) parent. But two of the four possible outcomes are Gg, so there is a 50/50 chance that the foal will get a g from the heterozygous parent. That g will remain in the genotype, and can influence the next generation, even though it won’t change the color of this foal.
Example 2: The outcome can be different if both parents are Gg
In this mating, there is a one in four chance of a homozygous GG; two in four chance of Gg (gG is the same; the order is just reversed here to show which parent is contributing which gene); and one in four chance of homozygous gg. The gg would NOT go Grey. This is how you get a color from two Greys. For those who want the math, the probabilities are: a 25 percent chance of homozygous Grey, 50 percent chance of heterozygous Grey, and 25 percent chance of homozygous non-Grey (color). In terms of phenotype (what you see), there is a 75 percent chance of Grey (add together the probabilities for the homozygous GG and heterozygous Gg results) and a 25 percent chance of non-Grey (gg).
The Biologists’ Tool: Punnett Squares
The following chart is what biologists call a Punnett square. This is a tool that biologists use to figure out the possible genotypes in the offspring of two parents. The three possible genotypes of Parent 1 are in the column on the left. The three possible genotypes of Parent 2 are in the row at the top. The nine squares in the middle show the possible combinations from which you can calculate probabilities of the genotype of the resulting foal.
As you can see, when you breed two Greys together (assuming you don’t know the actual genotype), you will have either 75 percent or 100 percent chance of getting a Grey foal. The only way to make your chances 50/50 is to breed a Grey to a color. BUT, if at least one parent is homozygous for Grey (GG), you will ALWAYS get a grey foal.
GG (grey) Gg (grey) gg (color)
PARENT 1 GG (grey) GG GG Gg
GG Gg Gg
GG GG gG
GG gG gG
100% grey 100% grey (Example 1 from the text) 100% grey
GG (grey) GG GG Gg
gG Gg Gg
GG gG gg
Gg gg gg
100% grey 75% grey 50% grey
25% color (gg) (Example 2 from the text) 50% color (gg)
Gg (grey) gG gG gg
gG gg gg
Gg Gg gg
Gg gg gg
100% grey 50% grey 100% color
If Not Grey, Then What?
Now let’s look at the Chestnut, Black, and Bay colors. This is more complicated because there are two sets of genes involved: the base coat (Chestnut/Black) and the agouti modifier that causes the Black to be Bay by limiting the black hair to the points (legs, mane, tail).
Like a lot of logic puzzles, you have to go one step at a time and understand each step before going on to the next.
Study the gene combinations that are possible for each parent. If you don’t understand these, refer to last month’s article.
The next chart is an example of a modified Punnett square. Down the left side column are the three possible colors and associated genotypes for Parent 1. Across the top row are the three for Parent 2. The nine central boxes show the possible colors of foals, based on the various combinations of parents. We’ve assigned numerals in parentheses, which are explained below. You’ll note that, for example, a Black Parent 2 with Chestnut Parent 1 is the middle box in the top row, and shows three possible colors for the foal: Chestnut (2), Black (3), and Bay (4). If you reverse the colors of the parent — so Parent 1 is Black and Parent 2 is Chestnut, you’ll get the box in the middle row at the far left. You’ll also notice that the colors have the same numerals in parentheses, Chestnut (2), Black (3), Bay (4). That is because it doesn’t matter which parent is which color. Black crossed on Chestnut has the same possibilities regardless of which parent is the sire and which is the dam. You’ll see that the same thing applies for Bay x Bay and Bay x Black.
Chestnut-ee (AA or Aa or aa) Black-EE or Ee (aa) Bay-EE or Ee (AA or Aa)
PARENT 1 Chestnut-ee (AA or Aa or aa) Chestnut (1) Chestnut (2) Chestnut (5)
Black (3) Black (6)
Bay (4) Bay (7)
Black EE or Ee (aa) Chestnut (2) Chestnut (8) Chestnut (10)
Black (3) Black (9) Black (11)
Bay (4) Bay (12)
Bay-EE or Ee (AA or Aa) Chestnut (5) Chestnut (10) Chestnut (13)
Black (6) Black (11) Black (14)
Bay (7) Bay (12) Bay (15)
Geneticists would expand this diagram to show all the possible combinations of base coat and agouti. We’ve compressed them here for simplicity, and explained in the text below about the possible influence of the agouti gene, depending on the base coat color. As we explained last month, consider the base coat color first.
As we said above, Chestnut x Chestnut always gives Chestnut.
Now look at the middle box: Black x Black can give Chestnut or Black BUT NOT BAY. That’s because Black must be aa and so, because both parents are aa, neither can pass on the dominant form of the agouti modifier. You can get Chestnut, but you can’t get Bay. All other combinations can give you any color, depending on the genotype.
So here’s a third absolute to add to the two stated at the beginning of this article: 3. Black x Black cannot produce Bay
The numerals in parentheses are explained in text that follows the Punnett square.
Remember, these are possibilities, depending on the genotype of the parents. We’ll get to probabilities near the end of this article.
Chestnut x Chestnut =
Chestnut (1) – Both parents have to be ee, so the foal can only get an e from each parent and will therefore be ee. The agouti doesn’t matter; it can’t have any known effect on this foal.
Chestnut x Black =
Chestnut (2) – The Chestnut parent gives e. If the Black parent is Ee and gives the e, the foal will be Chestnut. The agouti in the Chestnut parent doesn’t matter. The Black parent has to be aa, and it wouldn’t matter either.
Black (3) – The Chestnut parent gives e. If the Black parent is EE or Ee and gives an E, the foal base coat will be Black (Ee). Agouti can matter. To be Black, the foal has to get an a from both parents. No problem with the Black parent; it has to be aa and can only give a. But the Chestnut parent has to be either Aa or aa to give an a.
Bay (4) – The Chestnut parent gives e. If the Black parent is EE or Ee and gives an E, the foal base coat will be Black (Ee). Agouti can matter. If the Chestnut parent is AA or Aa and the foal gets an A from the Chestnut parent, it will be Bay instead of Black.
Chances Are . . .
You can use your knowledge of the coat color for the parents of the stallion and dam (grandparents of your planned foal) to figure out some of the genotype for the parents in the planned mating. From that information, you can do a pretty good job of estimating the chances of a particular coat color of the foal.
Write out the genotypes as best you can, but put in “?” for anything that you don’t know.
— If your Black horse has one Chestnut parent, the Black must be Ee (and aa).
— Your Bay horse has to have at least one E and at least one A, so it’s E? A?, but if one parent was Chestnut, it has to be EeA?
— Your Chestnut can be set as ee?? if you don’t know its agouti combination; but if one parent was Black (therefore aa), then your Chestnut has to be eea?.
— If your Grey had one colored parent, it must be Gg.
Here’s an example based on one Black and one Bay parent. The Black had a Chestnut sire, so we know it is Ee. The Bay parent had two Bay parents, so we can only be sure it has at least one E. The other form of the gene remains unknown so we show it as ‘?’
First, do the base coat.
PARENT 2 (Black)
PARENT 1 (Bay) E EE Ee
? E? e?
No matter what the ? is for Parent 1, we can be certain that three of the four chances (75 percent) will give an E and the base coat will be Black. The only uncertainty is whether the fourth case will be ee (Chestnut) or Ee (Black). Therefore, there is NOT MORE THAN 25 percent chance the foal will be Chestnut. Now put that on hold, and look at the agouti.
The Black parent has to be aa. The Bay has to have at least one A, but we don’t know what the second gene is, so we designate it with?
PARENT 2 (Black)
PARENT 1 (Bay) A Aa aA
? a? a?
For two of the four, there will be at least one A. There are two that we can’t know (because of the ?), but we can conclude that there is at least a 50 percent chance that the foal will have at least one dominant agouti. If the two ? are actually a, the two that are a? in the square would be aa, and there would be 50 percent chance of no agouti modifier.
To get the probabilities for the foal, you multiply the known or maximum probabilities together (if you don’t remember this basic from your early math class, please take it on faith); 75 percent chance of Black base coat times 50 percent chance of getting an agouti A = 37.5 percent. So there’s AT LEAST a 37.5 percent chance the foal will be Bay. And because there’s not more than a 50 percent chance the foal will be aa, there’s also NOT MORE THAN A 37.5 percent chance the foal will be Black. There’s NO MORE THAN A 25 percent chance the foal will be Chestnut (no influence of the agouti matters).
For those of you who love math, you can create Punnett squares for the genotypes of your proposed mating and have a blast. Stick with the Black/Chestnut gene first. That will tell you the probability of Chestnut. Then do the agouti. Multiply the chance of agouti times the chance of Black and you’ll get your Bay percentage. Then figure the Black probability.
There are several calculators online that will predict the probability of foal color based on the genotype information you provide. However, I have found errors in them. Doing your own Punnett square will help solidify your understanding of coat color genetics. Have fun.
Next month: What about dun and palomino and pearl and silver and tobiano and sabino and overo and all the rest? We’ll look at the coat color genes found in other breeds that can give you these variations for Half-Arabians.