INHERITANCE OF GREAT DANE COAT COLOR

By Jane Chopson

Revised September, 1992

This paper provides the reader with some fundamentals of coat color genetics and clarifies the meanings of some often used genetic terms. It will be presented in three parts: Part I deals with the basics of genetics; Part II, the basics of coat color inheritance; and Part III, coat color genetics and its application to a breeding program.

PART I

A single animal cell is the structure in which life begins. All living things are composed of cells which start the life cycle as a single cell, and through a process of cell division and specialization, the organism is formed. The dog reproduces by producing germ cells (gametes), sperm (male) and eggs (female), which fuse to give rise to a single fertilized egg cell (zygote). The zygote then grows and divides to form the embryo. The information to reproduce a complex many-celled animal is thus transmitted through two cells, sperm and egg.

The cell is made up of a cell wall inside of which is found cytoplasm and a nucleus. Within the nucleus are contained the thread-like structures known as the chromosomes. It is the chromosomes which carry the genetic material. They are arranged in pairs, one member of each pair being derived from each parent. The dog has 78 chromosomes (diploid number) in each cell with the exception of the sperm and egg cells which have only 39 chromosomes (haploid number), one from each pair. Hence when the sperm and egg fuse, the zygote will again have the diploid number of chromosomes, 78. The chromosomes contain DNA (Deoxyribonucleic Acid), RNA (Ribonucleic Acid) and protein. A gene is a particle on the chromosome which provides hereditary information and is made up of DNA. A gene can be thought of as a set of coded instructions or blueprints. In order to understand how DNA carries the genetic code, it is necessary to look at the structure of the nucleic acid molecule. Watson and Crick theorized the structure was that of a double helix. A double helix might be compared to two bedspring-like structures linked together. They stuggested sugar phosphate units forming two long chains. The chains are linked to each other by a series of nitrogen bases: Adenine, Thymine, Cytosine and Guanine; these chains forming the double helix. Adenine is always linked to Thymine and Cytosine is always linked to Guanine. Information is coded in each gene in a specific form or sequence of the four kinds of nitrogen bases within DNA. The sequence differs in each gene and thereby distinguishes one gene from another. Genes produce chemical compounds called enzymes. These enzymes govern the embryological development and growth of all structures in the body.

There are two types of cell division. The first is MITOSIS, a process in which the somatic cells (those other than gamete) of an organism are formed. In mitosis there is a daughter cell produced which received an exact copy of the diploid number of chromosomes in the mother cell. It is this method by which the somatic cells of the body reproduce.

The second kind of cell division is MEIOSIS, a process in which there is a reduction in the number of chromosomes from the diploid number to the haploid number. It is this type of cell division by which sperm and eggs are produced. The processes of producing sperm and eggs by meiosis are known as SPERMATOGENESIS and OOGENESIS, respectively.

Among the 39 pairs of chromosomes, there is one pair which determines the sex of the animal. This pair is known as the SEX CHROMOSOMES. They differ significantly from each other in size and shape, one being called the X chromosome and the other being called the Y chromosome. Males carry and X and a Y; females carry two Xs. Hence it is the sire who determines the sex of the puppies because his sperm can contain and X or a Y chromosome. The female’s egg cells can only carry an X. Therefore, if an X carrying sperm fertilizes the egg, the puppy is a female.

The remaining 38 pairs of chromosomes are known as autosomes. Each one is similar to its partner is size, shape, and function. In a pair of chromosomes, each one is known as being HOMOLOGOUS to its partner. Homologous chromosomes have the same gene sites or locations on them. These gene sites are known as the gene loci and only # gene may be found each loci on a chromosome. Alternate forms of a gene which occupy the same loci on homologous chromosomes and affect the same trait are known as alleles. If these two genes are the same, then the animal is said to be HOMOZYGOUS for that pair. If they are different, then the animal is said to be HETEROZYGOUS for that pair. If he is heterozygous, one gene may be DOMINANT to the other. Dominant genes mask RECESSIVE genes. An example of this would occur with the inheritance of tail shape. A dog who is homozygous for the dominant straight tail gene (SS) bed to a homozygous recessive screw tail (SS) would produce all heterozygous puppies (Ss). GENOTYPE of the puppies, the genetic make up, would be Ss while their PHENOTYPE, outward physical appearance, would be that of a straight tail dog. Since the straight tail gene is dominant to the recessive screw tail gene, the straight tail gene will mask the screw tail and the puppies will have straight tails. It should be noted that one gene was derived from each parent. The possible phenotypes and genotypes of puppies may be figured by the following method:

	EXAMPLE I	EXAMPLE II	EXAMPLE III
PARENTS	SS x ss	Ss x ss	Ss x ss
POSSIBLE GAMETES	S s	S or s s	S s
	s s S Ss Ss S Ss Ss	s s S Ss Ss s ss ss	s s s ss ss s ss ss
OFFSPRINGS: Phenotype Genotype	4/4 Straight tailed 4/4 Ss	½ Straight, ½ Screw Tail ½ Ss ss	4/4 Screw tail 4/4 ss

The expected outcomes are based on an infinite number of breedings. Although SS x ss and ss x ss breedings would always result in the expectd ration, the Ss x ss could deviate from the expected 50:50 ratio if too small a sampling were made. It is the same as tossing a coin for heads or tails. If we toss it a million times, we would probably get close to the expected 50:50 ratio, and if we toss it an infinite number of times, we would get the expected ratio. But if we just tossed a coin twice, we might not get what we should have theoretically based on the small sample. It should also be noted that certain cases of deviance may be accounted for by mutation. MUTATION is a spontaneous genetic change or may be induced by such things and radiation, temperature change, chemicals or other causes.

PART II

Dr. Clarence C. Little, in his book THE INHERITANCE OF COAT COLOR IN DOGS, mentions the following allelic series concerned with the inheritance of coat color. Remember, a dog will carry two genes from each of the series, one on each chromosome, with each one being derived from each parent. They may be the same gene (homozygous) or they may be two different genes (heterozygous). The G, P, and R gene loci have been omitted from discussion since almost all Great Danes are homozygous for one gene in each series and so are not important in determining different colors in Great Danes. In the G series, they are homozygous for the recessive non-greying gene (gg) and in the P series homozygous for the recessive non-pink eyed dilution gene (pp). In the R series they are probably homozygous for the recessive non-roan gene (rr). It should be noted, however, that there is some question involving the R genes and there is a slight possibility that Merle or Harlequin Danes may carry the R gene.

A SERIES; INFLUENCES THE DISTRIBUTION OF DARK AND LIGHT

PIGMENT IN ORDER OF DOMINANCE

A^s	Allows distribution of dark Brown or Black all over the body.
a^w	Produces the wile Agouti type of color.
a^y	Restricts dark or Black and produces Tan, Fawn or Sable
a^t	Produces Bi-color (Black & Tan, etc.)

A further gene in this series is postulated by some writers, but not by Little. It would restrict the dark to a Saddle –pattern.

B SERIES

In order of dominance:

B	Black
b	Liver or Chocolate

PRESENT IN GREAT DANES: B and b (Little does not include "b" in Great Danes but evidence is clear as to its presence).

C SERIES

In order of dominance:

C	Full depth of pigment
c^ch	Chinchilla, reduces the Red or Yellow pigment more than the Black.
c^a	Complete Albinism

PRESENT IN GREAT DANES: C and possibly c^ch. Little does not include c^ch in Great Danes, but it may account for extremely pale Fawns who are not dilutes.

D SERIES

In order of dominance:

D	Allows dense pigmentation
d	Dilution of pigment

PRESENT IN GREAT DANES: D and d

E SERIES

In order of dominance:

E^m	Black Mask
E	Allows formation of dark pigment (Black or Brown) over the entire body
e^br	Brindle
e	Restricts dark pigmentation to Red, Orange, Yellow

PRESENT IN GREAT DANES: E^m and E and e^br. There is a slight possibility that ee exists in a few Great Danes. This could possibly account for the production of Blacks from "Fawn to Fawn." If the two dogs involved were a^ya^yE^mE^m x A^sA^see, all A^sa^yE^me would result and they would be Black.

M SERIES

In order of dominance:

M	Produces irregular patches of dark pigment on a lighter background of the same pigment. Homozygous MM is semi-lethal and animals are often blind and/or deaf and/or sterile. They are usually mostly White. The M genes also tend to add White
m	Non-Merle

PRESENT IN GREAT DANES: M and m.

S SERIES

In order of dominance:

s	Solid coat; no White except possibly a small amount on the chest and toes.
s^i	Irish spotting; White on one or more of the following: muzzle, forehead, feet, neck, tail tip, chest, belly and/or throat.

There is some confusion is distinguishing animals in this series phenotypically due to the action of plus and minus modifying factors (various genetic factors independent of the main gene). There are also frequently situations of incomplete dominance occurring in this series. This is a quantitative, not a qualitative series.

PRESENT IN GREAT DANES: S and sⁱ and possible S^p.

T SERIES

In order of dominance:

T	Ticking; produces small flecks of color, ticking is not visible at birth
t	Non-ticking

PRESENT IN GREAT DANES: T and t.

There must also be at least one other gene, and possibly more, to account for the difference between Merles and Harlequins. Little attempts to explain the difference by an interaction of M and s^p and s^w genes, but this seems to be an inadequate explanation. Burns and Fraser suggest that Harlequins are EEMm and that Merles are EeMm or Ee^brMM, but this would also seem unlikely, since the occurrence of eemm or e^bre^brmm dogs is infrequent in Merle x Merle breedings and even more so because it doesn not explain how two Harlequins can produce a Merle, which they frequently do. Dr. Bagala, in 1966, hypothesized another gene series which he titled the W series (not to be confused with the W series of Iljin. This seems to possibly provide an explanation for the genetic difference between Harlequin and Merle. This is only a possibility, not yet a proven theory.

W SERIES (PROPOSED)

In order of dominance:

W	Clears ground Grey color of Merle to White, Harlequin. Lethal in the homozygous state, and death occurs in the embryo.
w	Non-clearing gene.

Since homozygous W is lethal, all Harlequins must be Ww and all Merles must be ww. This explains why we get no true breeding Harlequins. There is a question as to what effect W would have on Fawns, Brindles, Blues and Blacks, but for the time being we will assume they could be ww or Ww and be phenotypically the same.

Even if Bagala’s hypothesis is correct and we assume that the S series is quantitative in nature and that a number of situations of incomplete dominance exist, there are still some unexplained problems. The reports of Harlequin puppies from Merle to Merle breedings cannot be accounted for within Bagala’s theory.

Schaible and Brumbaugh in 1976 indicated their data suggested Harlequins were accounted for by an additional allele in the M series and gave it a provisional symbol of M^h. They further suggest that the gene responsible for Harlequins is subject to a high degree of germinal and somatic mutation. Spontaneous germinal mutation to different alleles within the M series could account for the colors produced by Harlequin breedings and Merle breedings as well as the absence of "true breeding" Harlequins.

Clearly, there is a need for further data collection and analysis of breeding within the Harlequin family to clarify the genetic difference between the Harlequin and the Merle.

SOME SAMPLE GENOTYPES	PHENOTYPE
A^sa^yBBCCDdEEmmSSww A^sA^sBbCCDDE^mEmmSSww A^sA^sbbCCDDEEmmSSww a^ya^yBBCCDDE^mE^mmmSSww A^sA^sBBCCDDEEMms^is^iWw a^ya^yBBCCddE^mE^brmmSSww	Black Black Liver or Chocolate Masked Fawn Harlequin Masked Dilute Brindle

Remember, for one phenotype, there can be many genotypes.

Below is a list of what a dog must carry to be a certain color. It is not what it would necessarily be best for it to carry as far as a breeding program is concerned. For example: a dog may carry one Liver recessive and still phenotypically be a good color, but he might not be a good dog to use in a breeding program. All dogs carry two genes in each series.

When the terms "Fawn, Brindle, Blue, Black and Harlequin" are used, they refer to the colors as described in the Great Dane Standard, unless otherwise noted.

A^s	All Blacks, Blues, Harlequins, and Merles must have one or two.
a^y	All Fawns and Brindles must have two. Blacks, Blues, Harlequins and Merles may carry one recessively.
B	All Blacks, Blues, Harlequins, Merles, Fawns and Brindles must have one or two.
b	All Chocolates or Livers (dogs with liver nose) must have two. Blacks, Blues, Harlequins, Merles, Fawns, and Brindles may carry one recessively.
C	All Fawns and Brindles must have one or two. Blues, Blacks, Harlequins and Merles may have none, one or two.
c^ch	Possibly some of the very light background Fawns and Brindles have two. Blues, Blacks, Harlequins, and merles may carry none, one or two.
D	All Blacks, Harlequins, Fawns, Brindles and Merles with Black pigment may have one or two.
d	All Blues, dilute Fawns, dilute Brindles, dilute Harlequins and dilute Merles must have two. All non-dilute dogs may carry one recessively.
E^m	All masked Fawns must have at least one. Most masked Brindles have one, but there is some confusion since the striping can tend to give the appearance of a mask when genotypically there is no masking gene. Blacks, Blues, Harlequins and Merles may have none, one or two.
E	All completely maskless Fawns have two. Brindles can have none or one. Blacks, Blues, Harlequins and Merles could carry none, one or two. Masked Fawns may carry one recessively.
e^br	All Brindles must have at least one. Blues, Blacks, Harlequins and Merles can carry none, one or two. Fawns can’t carry any. It must be remembered that E^m is dominant to e^br but E^m only affects a small area on the dog, and e^br expresses itself on the rest of the dog. Hence, a E^me^br animal will be a masked Brindle, providing, of course, that the genes in the other series also allow this.
M	Whites generally carry two; all Harlequins and Merles have one. Many who carry two are defective.
m	Blacks, Blues, Fawns and Brindles all carry two. Harlequins and Merles carry one.

In the S series there is considerable confusion, since plus and minus modifying factors (various genetic factors independent of the main gene) are operative. So there is some overlap between the genes. There are also frequent fluctuations due to incomplete dominance.

S	All Fawns, Brindles, Blues and Blacks must have two. Merles may have none, one or two. Harlequins can have none or one.
s^i	Harlequins and Merles can carry none, one or two. Fawns, Brindles, Blacks and blues generally carry none.
s^p	Same as s^i, if present in Danes.
T	All ticked Harlequins and Merles have one or two.
t	All non-ticked dogs have two.
W	(Proposed) All Harlequins have one. Assume that Fawns, Brindles, Blacks and Blues could have one undetected.
w	All Danes must carry one. Merles must carry two. Fawns, Brindles, Blacks and Blues could have two.

PART III

Since all ethical dog breeders strive to produce as few mis-marked puppies as possible and to produce as many correctly marked as possible, there is a need to understand and apply the knowledge of color inheritance to a breeding program. The breeder should have clearly in mind which colors he wishes to produce and strive to upgrade their quality. All breeders should be concerned with the upgrading of all five colors. In order to do this, he must not only know the phenotype of his animal, but also the genotype. To have a truly effective breeding program, one must take into account genotype rather than just phenotype. A Black is a Black is a Black is not the case. There are many genetically different Blacks and they must be treated as such.

If a breeder is to produce correctly marked offspring, he must know which genes to avoid when breeding. Below are some basic guidelines to produce correctly marked dogs. Remember, only color is being considered here. It should be noted, however, that color is only worth 8 points out of a hundred in our Standard, and it would be wrong to look only at color and forget the rest of the dog.

Sometimes, for the sake of upgrading conformation in a particular color, otherwise undesirable crossing might be done. However, if attempted they should only be attempted by the experienced breeder who has a thorough knowledge of color genetics and is willing to cull, control and test-breed the dogs he is working with until the undesirable color gene or genes have been eliminated from his new, hopefully upgraded, stock. The following is an outline for an ideal situation and would only be completely practical if all colors were of equal quality and popularity.

FAWNS:

Since it is undesirable to have dilute Fawns or Fawns with considerable white, or Merle Fawns, it would be best not to breed Fawns to Blues, Harlequins or Merles, or to any other color which carries a dilution, "d", White Spotting, "sⁱ or s^p", or a Merling, "M", allele. They could safely be bred to Fawns, Brindles or Blacks who do not carry those undesirable genes. These Fawns resulting from: Fawn x Brindle, Brindle x Brindle, Fawn x Black, Brindle x Black and Black x Black, are just the same as those from Fawn x Fawn breeding since they could not carry the dominant Brindle or Black gene and still be Fawn. It might be noted that the old idea of Fawn x Fawn over several generations washing out color is not true, provided that well-pigmented Fawns are always selected.

BRINDLES:

Like Fawns, dilution, white spotting and merling genes are to be avoided. Brindles can be bred to Fawns, Brindles, and to Blacks which do not carry those undesirable genes.

BLUES:

Fawn, brindle, merle, and white spotting genes should be avoided. Hence, Blues should be bred to Blues or Blacks not carrying those genes. It should be noted, however, that Blue bred to Black from Black breeding only, and hence not carrying a dilution gene, will yield only Blacks.

HARLEQUINS:

It is undesirable to have fawn, brindle and dilution genes in Harlequins. Hence, Fawns, Brindles, Blues and Blacks carrying fawn, brindle and blue dilution recessive must be avoided in breeding Harlequins, if we are to attain the highest degree of correctly marked offspring. If the proposed "W" is lethal in the homozygous state, and MM is semi-lethal, we will never have true breeding Harlequins since they would be MMWW. If they didn’t die in embryo, they could be defective. Hence, we must realize there will probably always be a certain number of Merles and mis-marked Blacks resulting from breeding within the Harlequin family.

Since it is presumed "W" is lethal in the homozygous state, it might be wisest to avoid producing WW dogs. One must also keep in mind that MM is semi-lethal and hence Harlequin to Harlequin breeding or Harlequin to Merle breeding may yield some defectives also.

Let us look at some results from harlequin, Merle and Black breeding. We will assume that WW is lethal, MM is semi-lethal. These are theoretically what should result:

Phenotype	Harle x Harle
Genotype	MmWw x MmWw
Possible Gametes	MW, Mw, mW, mw

GAMETES	MW	Mw	MW	Mw
MW	MMWW	MMWw	MmWW	MmWw
Mw	MMWw	MMww	MmWw	Mmww
MW	MmWW	MmWw	MmWW	MmWw
Mw	MmWw	Mmww	MmWw	mmww

MmWw	Harlequin
Mmww	Merle
Mmww	Black
MmWw	Black
MMww	White
MMWw	White
MmWW	Lethal in embryo
MMWW	Lethal in embryo
MmWW	Lethal in embryo

OFFSPRING:

Blacks	3/16
Merles	1/8
Dead in embryo	1/4
White	3/16
Harlequins	1/4
Total Acceptable Colors	7/16

(Note: No consideration is given to the amount of White on these Blacks and Harlequins, hence some may not be correctly marked.)

Harlequin (MmWw) to Merle (Mmww) by the same method yields:

OFFSPRING:

Blacks	1/4
Merles	1/4
White	1/4
Harlequins	1/4
Total Acceptable Colors	1/2 (i.e. without regard to white)

Harlequin (MmWw) to Black (mmWw) by the same method yields:

OFFSPRING:

Blacks	3/8
Merles	1/8
Dead in embryo	1/4
Harlequins	1/4
Total Acceptable Colors	10/16 (i.e. without regard to white)

Harlequin (MmWw) to Black (mmww) without W gene:

OFFSPRING:

Blacks	1/2
Merles	1/4
Harlequins	1/4
Total Acceptable Colors	12/16 (i.e. without regard to white)

White (MMWw) to Black (mmww) without W gene:

OFFSPRING:

Harlequin	1/2
Merles	1/2
Total Acceptable Colors	1/2 (i.e. without regard to white)

It can be seen from the above that 1/4 Harlequins is yielded from all these breedings, except the White to Black breeding. There are obvious differences in the % of Merles, Blacks and whites also produced and the number lost in embryo. It should be noted that the fractions are what are conceived, not actually whelped. In the breedings in which there is a loss of puppies in embryo, there would be more than ¼ Harlequins actually whelped, but only 1/4 Harlequins actually conceived.

BLACKS:

The Black Great Dane cannot be treated without regard to its genotype and have only its phenotype considered. Blacks should probably be divided into five groups as far as breeding is concerned:

1. Fawn or Brindle or Black-bred Blacks
2. Harlequin-bred Blacks
3. Blue and Black-bred Blacks
4. Black-bred Blacks
5. Combinations of the above

1. Blacks with fawn or brindle genes should not be used on Blues, Blue-bred Blacks, Harlequins or Harlequin-bred Blacks. If fawn and brindle genes are combined with dilution genes, there is a risk of producing dilute Fawns and Brindles. Harlequin white spotting factors can also produce undesirable effects on Fawns and Brindles. These Blacks may safely be bred to Fawns, Brindles, Blacks like themselves, and Blacks who are only Black-bred.
2. Blacks from Harlequin may carry recessive genes for white spotting and hence, it could be unsafe to use them on Fawns, Brindles, Blues or Blacks carrying Fawn Brindle or Blue. They can be bred to Harlequin, although there will probably be some Merles and Blacks with too much white produced. They can be bred to whites to produce Harlequins and Merles. They could also be bred to Blacks like themselves or Black from Black breeding, although again this could produce blacks with too much white.
3. Blue-bred Blacks should only be bred to Blues and Blacks like themselves to produce Blues and Blacks – or to Black-bred Blacks to produce Blacks. If would be undesirable to breed them to Fawns, Brindles, Harlequins or Blacks carrying these genes, because of the chance of producing dilution of these colors in future generations.
4. Black-bred Blacks (A^sA^sBBCCDDEEmmSSww): This is the one Great Dane which can be bred to all five colors with relative safety. The only exception likes in breeding to Harlequins. Since they may produce Merles and Blacks with too much white, some of the resulting Harlequins may have too little white. If used on Harlequin-bred Blacks, the later may contribute too much white to the offspring. They may safely be bred to Fawns, Brindles, Blues and Blacks carrying these genes or Blacks like themselves, but only Blacks will be obtained.
5. Blacks which carry genes from two of the following categories: (a) Fanw and/or Brindle; (b) Blue dilution; (c) Harlequin White Spotting. These could be termed problem dogs as far as a breeding program is concerned. Since dilute or white spotted Fawns and Brindles are undesirable, as are dilute Harlequins or white spotted Blues. Before such a dog was bred, a breeder should seriously ask themselves if such a dog has so much to offer that it would be worth the risk. If in spite of the sizable chance of producing mis-marked pups they still decide to breed the dog, they should decide what color to produce then breed to a dog pure for that color or carrying recessive for it, and who does not also carry the other undesirable gene that this dog carries. They then must carefully control and cull test breed resulting puppies until the undesirable gene has been eliminated.

Since the possibility of test breeding being done to determine whether or not a hidden recessive gene is present has been mentioned, it might be of use to explain how such breedings should be done. If, for example, we had a Black from a black (DD) x Blue (dd) breeding, there is no need to do a test breeding for the presence of the recessive dilution gene, since one member of each gene pair is derived form each parents. Hence, we know that the dog is Dd. But if the Black dog is Black from parents who are "D?", we would do a test breeding to determine the genotype of that Black offspring. Since he is Black, we know that he is either DD or Dd. To determine which, we breed him to a Blue. If any Blue (dd) puppies result, we can conclude he was Dd. But if no dilute offspring result among a sufficient number of puppies, we can probably conclude that he was DD. The same method can be used to test for any recessive gene, i.e. be breeding a dominant back to a double recessive.

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Copyright 1992 Jane Chopson. All rights reserved. Our thanks to the willingness to share this article for educational purposes.