Egyptian Mau Coat Color Inheritance
(((taken from an article from the EMBFC)))
Egyptian Mau Coat Color Inheritance
This article summarizes coat color inheritance in the Egyptian Mau to aid breeders in breeding for certain colors. It begins with an overview of core principles of genetics then reviews the specific genes controlling Egyptian Mau coat color. It also provides guidance in determining the possible genotypes of your cat.
Basic Principles of Genetics
A cat’s appearance (“phenotype”) is determined by the genes that govern the expression of that trait (the “genotype”). For every trait, from eye color to coat length, a cat has two genes, one inherited from each parent. In some instances, a cat receives the same form of a gene from both parents in which case the cat is said to be “homozygous”. If the cat receives different forms of a gene, the cat is “heterozygous”. When a cat is homozygous, there is no question as to the cat’s phenotype. Since the cat only has one form of the gene, it expresses the form called for by that gene. Thus, a cat that is homozygous for green eye color has green eyes. However, when a cat is heterozygous, the different genes, generally, do not create a “blended” effect.1Instead, some genes are always expressed in the cat’s appearance while other genes are hidden unless the cat has two copies of that gene (i.e., the cat is homozygous). Genes that are always expressed are said to be “dominant” while genes that can be hidden are “recessive”. A cat that is heterozygous for a trait has one dominant and one recessive form of the gene for that trait. Since a dominant form is always expressed, the cat’s phenotype shows the type called for by the dominant form and the recessive form is hidden. For example, a shorthaired cat that is heterozygous for longhair carries a gene for longhair. However, because the shorthaired form of the gene is dominant to the long-haired form, the cat is shorthaired. A cat’s appearance can never express the recessive form of a trait and be heterozygous. However, certain phenotypes are controlled by the interaction of multiple gene pairs. Therefore, a cat may be homozygous recessive for one of the gene pairs and heterozygous or homozygous for the other. A cat that is homozygous for a trait only has one form of a gene; thus, it passes only that form to its offspring. However, a cat that is heterozygous has two different forms and can pass either form to its offspring. Thus, a shorthaired cat that is heterozygous for longhair will pass on either the shorthair or longhair gene to any one offspring. As a result, depending upon the genotype of the other parent (which contributes the offspring’s other gene for that trait), the shorthaired cat might produce longhair offspring. When writing a cat’s genotype, geneticists use a capital letter to indicate the dominant form of a gene and that letter’s corresponding lowercase form for its recessive counterpart. For example, shorthair (the dominant form) could be indicated by a capital “L” and longhair (the recessive form) could be indicated by lowercase “l”. When a dominant form of the gene is present but it is not known whether the other gene is the dominant form or the recessive form (i.e., whether the cat is homozygous or heterozygous), a “-” is used to indicate the “unknown” gene. Geneticists frequently use a tool called a “Punnett Square”2 or “checkerboard” to determine the genotypes that can be produced by the mating of two cats and the frequency with which those genotypes occur. The Punnett Square lists the genotype of one parent in columns across the top of the square and the other parent’s genotype in rows along the left side. The genotype of the offspring produced by the intersection of each column and row is written in each square. For example, to determine the possible genotypes produced by two shorthaired cats heterozygous for longhair (“Ll”), the Punnett Square is the following:
|L||LL (Homozygous for shorthair)||Ll (Shorthair heterozygous for longhair)|
|l||Ll (Shorthair heterozygous for longhair)||ll (Longhair)|
Thus, crossing two shorthair cats that are heterozygous for longhair results in 75% shorthaired kittens (of genotypes “LL” and “Ll”) and 25% longhaired kittens (genotype “ll”). It is also referred to as shorthaired kittens to longhaired kittens in a ratio of 3:1.
The Agouti and Inhibitor Genes
The four Egyptian Mau colors3 of silver, bronze, smoke and black are determined by the interaction of the agouti gene and the inhibitor gene. The agouti gene controls expression of the cat’s tabby pattern.4 In its dominant form, usually indicated by a capital “A”, the tabby pattern is expressed resulting in the silver and bronze Mau. In its recessive form (also called “nonagouti”), the pattern is not expressed and the cat is self colored as in the smoke and black Mau.5 The inhibitor gene restricts (“inhibits”) the expression of pigment – especially yellow pigment – in the hair shaft. In an agouti cat such as the bronze Mau, this changes the yellow base in each hair to white, creating a “silver” color.6 In a non-agouti or self-colored cat like the black Mau, it creates a white undercoat resulting in “smoke”. The dominant form of the inhibitor gene (indicated by capital “I”) calls for pigment restriction or a white undercoat as in the silver and smoke respectively. Its recessive form calls for normal pigment as in the bronze and black and is indicated by lowercase “i”. From the above, we can determine the genotypes for each color are as follows:
|Silver||AAII (homozygous silver)|
|AaII (silver heterozygous agouti & homozygous for the inhibitor gene)|
|AAIi (silver homozygous agouti & heterozygous for the inhibitor gene)|
|AaIi (silver heterozygous for agouti and the inhibitor gene)|
|Bronze||AAii (homozygous bronze)|
|Aaii (heterozygous bronze)7|
|Smoke||aaII (homozygous smoke)|
|aaIi (heterozygous smoke)|
The agouti and inhibitor genes work independently of one another to affect coat color. Therefore, a cat may be of recessive genotype for agouti (“Aa”) and of homozygous genotype for the inhibitor gene (“II”).8 What is important to note is that a cat of a genotype that combines dominant and recessive forms (bronze and smoke) does not “hide” the genotype produced by the dominant form for which it has a recessive form. Therefore, bronze does not hide smoke (which requires the dominant form of the inhibitor gene) and smoke does not hide bronze or silver (which requires the dominant form of the agouti gene.) We can use a Punnett Square to determine the results of multiple gene pairs such as those that govern Mau coat color. For example, to determine the genotypes produced by breeding two silvers each heterozygous for both the agouti and the inhibitor genes, one would go through the following analysis. First, each parent must have the genotype “AaIi.” Because the agouti and inhibitor genes are passed to the offspring independently of one another, each gene for agouti must be paired with a form of the inhibitor gene as illustrated below.
|AI||AAII (silver)||AaII (silver)||AAii (silver)||AaIi (silver)|
|aI||AaII (silver)||aaII (smoke)||AaIi (silver)||aaIi (silver)|
|Ai||AAIi (silver)||AaIi (silver)||AAii (bronze)||Aaii (bronze)|
|ai||AaIi (silver)||aaIi (smoke)||Aaii (bronze)||aaii (black)|
Therefore, the offspring produced by this pairing will be 62.5% silvers, 18.75% bronzes, 12.5% smokes, and 6.25% black, or silver to bronze to smoke to black in the frequency of 10:3:2:1.
Use the genotypes above to fill in the Punnett Squares and determine the genotypes that result from the following pairings: 1. Heterozygous bronze (Aaii)-top x homozygous smoke (aaII)-side
Color(s) and Percentage(s): ______________________________________________________ 2. Black (aaii) x silver heterozygous for agouti & homozygous for the inhibitor gene (AaII). [Fill in genotypes for the silver parent.]
Color(s) and Percentage(s): ______________________________________________________ 3. Silver heterozygous for agouti and the inhibitor gene (AaIi) x heterozygous smoke (aaIi). [Fill in the genotypes for both parents. Hint: fill in the genotypes for the smoke parent on the side and for the silver parent on the top.)
Color(s) and Percentages: ______________________________________________________ [The answers will be in the next issue of the newsletter.]
With the exception of black (which is always homozygous “aaii”), a Mau’s coat color alone is not enough information to determine its genotype or the possible genotypes and phenotypes of its offspring. Rather, we can use the information we have about the cat’s relatives to make an educated guess at the cat’s genotype – if not determine it with 100% certainty.
By the Parents
We can use the color of a cat’s parents to determine the cat’s possible genotype(s). For example, if we have a silver cat, we know it must have a “base” genotype of “A-I-”. If we know that our silver cat has a silver parent, we know that parent must also have a base genotype of “AI-”. If the other parent is bronze, we know the bronze parent must be, at a minimum, “A-ii.” Since both silver and bronze are caused by the dominant form of the agouti gene and therefore can “hide” that gene’s recessive form, non-agouti, we do not know whether either or both of the parents is homozygous (“AA”) or heterozygous (“Aa”) for agouti. However, we do know that the bronze parent is homozygous recessive for the inhibitor gene because bronze is always homozygous recessive for that gene. Thus, we know our silver cat must have received a recessive form of the inhibitor gene from its bronze parent and must be, at a minimum, “A-Ii.” Of course, if we had more information about any of our cat’s grandparents or its siblings, we might know more about the genotypes of its parents and as a result, the genotype of our cat. For example, let’s say our silver cat had a smoke sibling. We know that smoke is caused by the recessive form of the agouti gene (“a”). In order for a recessive form to be expressed, a cat must have two copies of the recessive form, i.e., it must receive one copy from each parent. Thus, in order for the parents identified above to produce a smoke, they must have the genotypes “AaI-” (silver) and “Aaii” (bronze). How does this information help us determine the genotype of our hypothetical silver cat? We can use a Punnett Square to plug in the parents’ genotypes and the possible genotypes of their offspring. (The silver parent’s genotypes are along the top, and the bronze parent’s genotypes are along the side.)
|Ai||AAIi (Silver)||AaIi (Silver)||AAi-(Bronze if “i” or silver if “I”)||Aai- (Bronze if “i” or silver if “I”)|
|ai||AaIi (Silver)||aaii (Smoke)||Aai- (Bronze if “i” or silver if “I”)||aai-(Smoke if “I” or black if “i”)|
Since our cat is a silver, we know it must have one of the genotypes indicated for silver: “AAIi” or “AaIi” produced in the ratio of 2 AaIi:1 AAIi.10 Therefore, our cat has a 67%11 of having a genotype that can produce a smoke.12 We can also use knowledge of the phenotypes of our cat’s grandparents to determine the possible genotypes of its parents. For example, let’s say that not only do we know that our cat has a smoke sibling, we also know that our cat’s silver parent has a black parent.13 We know that black is always homozygous recessive for the inhibitor gene (with genotype “aaii”); thus, it can only pass the recessive form of the inhibitor gene “i” to its offspring. Therefore, we know that our cat’s silver parent must have the genotype “AaIi”, and we can use that information to fill in the last two columns of the chart above.
By the Offspring
A cat’s offspring can also reveal much about its genotype.14 For example, if we have a silver cat which has produced bronze and smoke kittens, we know the cat must have the recessive forms of the genes that create those colors. Since silver has the dominant form of both the agouti (“A”) and inhibitor (“I”) genes, our silver would have the genotype “AaIi.”
Breeders can never have too much information when it comes to planning their matings. Hopefully, this article will help you more accurately determine your cat’s genotypes thereby more reliably planning the color of their offspring.
Ready to test your knowledge? Use the Punnett Square below to determine the genotypes and phenotypes that result from crossing a silver heterozygous for non-agouti, the inhibitor gene and dilution (AaIiDd) x bronze homozygous for agouti and heterozygous for dilution (AAiiDd). [Fill in the chart for the bronze parent.]
Color(s) and Percentages: ______________________________________________________
BIBLIOGRAPHY AND RESOURCES
Kerry Conway, Domestic Cat Genes Affecting Egyptian Mau Color Inheritance, EGYPTIAN MAU BREEDERS’ AND FANCIERS’ CLUB NEWSLETTER (Mar. 1989). Genetics 300-level undergraduate course, University of Michigan College of Literature, Science and the Arts (Winter Semester 2001). Dr. Susan Little, DVM, Colour Genetics of the Egyptian Mau, Part Two, EGYPTIAN MAU BREEDERS’ AND FANCIERS’ CLUB NEWSLETTER (Sept. 1986). Kathy Sinclair, Private Research. Carolyn M. Vella, Lorraine M. Shelton, John J. McGonagle et al., ROBINSON’S GENETICS FOR CAT BREEDERS AND VETERINARIANS, 4 ed., (Butterworth-Heinemann 1999). Bonnie Wydro and Melanie Morgan, Egyptian Mau, CAT FANCIERS’ ASSOCIATION YEARBOOK (1999), p. 130-138. 1 This article does not discuss incomplete dominance which results in such a blending as in the aquamarine eye color of the Tonkinese, which is a blending of the blue Siamese eye color and the yellow Burmese eye color. 2 So named because it was developed by 20th century British geneticist Reginald Punnett. 3 Blue Maus, caused by the recessive form of the “Density” gene also exist. The Density gene controls the spacing of pigment granules in the hair shaft. “D” is the dominant form and results in evenly spaced pigment granules, creating the black tabby coloration of all Maus. The recessive form, “d”, causes the pigment granules to clump together and the hair color to appear “blue” (actually gray). Since all Maus are black tabbies, the recessive form of the Density gene creates a dilute of blue for each: blue silver (“A_I_dd”), blue bronze (“A_iidd”), blue smoke (“aaI_dd”) and solid blue (“aaiidd”). 4 This article presumes the tabby pattern is “spotted” rather than classic tabby which also exists Maus. 5 The agouti ticked coat is itself caused by color restriction by the agouti protein, creating alternating bands of yellow (phaeomelanin) or black (eumelanin) pigment. In self-colored cats, the color restriction is ineffective and allows pigment along the length of the hair shaft rather than alternating color bands. 6 When this restriction is incomplete, the yellow pigment “breaks through” creating the effect called “tarnish”. 7 Technically, this should be referred to as “bronze heterozygous agouti & homozygous for the recessive form of the inhibitor gene”; however, due to the fact that bronze is always homozygous for the inhibitor gene, its heterozygous form can only be heterozygous for agouti. Likewise, with “heterozygous smoke”, which is more aptly referred to as “smoke homozygous non-agouti and heterozygous for the inhibitor gene”. However, because smoke is always homozygous non-agouti, its heterozygous form can only be heterozygous for the inhibitor gene. 8 In which case, the cat is a “smoke.” 9 A “_” is used because we do not know whether the silver parent is homozygous or heterozygous for the inhibitor gene. 10 Note: This ratio does not change if we confirm that the silver parent is of genotype “AaII”. In that case, the number of kittens produced with each genotype would be doubled 4 of AaIi and 2 of AAIi for a ratio of 4:2 or 2:1. 11 For clarification, this is not the percentage of silver kittens from all kittens but the percentage of silver kittens with genotype “AaIi” from silver kittens. 12 This genotype can also produce a black since the non-agouti gene can also produce a black and this genotype also the recessive form of the inhibitor gene which is necessary to produce a black. 13 Could our silver cat have a smoke sibling if both of its silver parent’s parents were either silver or bronze? Absolutely! The silver and bronze phenotypes are caused by the dominant form of the agouti gene which hides the presence of the recessive form, non-agouti. A cat can pass either of its genes to its offspring; therefore, one of the parents would have to be heterozygous for non-agouti (of genotype “Aa”). Could the other grandparent be homozygous for agouti (genotype “AA”) and our cat’s silver parent still be able to produce a smoke offspring? The answer is “yes.” Use a Punnett Square to figure out how this could occur. 14 Note: Recessive genes can lie hidden for generations; therefore, just because a cat never produces a kitten that is caused by a recessive gene nor does it have any relatives of a color that is caused by a recessive gene does not mean that the cat is – nevertheless – not carrying that recessive gene. However, the more offspring that are produced which do not have that recessive gene, the greater the likelihood that the cat does not carry it.